Quantum Sleeping Beauty and the Multiverse

Hidden in my papers with Chip Sebens on Everettian quantum mechanics is a simple solution to a fun philosophical problem with potential implications for cosmology: the quantum version of the Sleeping Beauty Problem. It’s a classic example of self-locating uncertainty: knowing everything there is to know about the universe except where you are in it. (Skeptic’s Play beat me to the punch here, but here’s my own take.)

The setup for the traditional (non-quantum) problem is the following. Some experimental philosophers enlist the help of a subject, Sleeping Beauty. She will be put to sleep, and a coin is flipped. If it comes up heads, Beauty will be awoken on Monday and interviewed; then she will (voluntarily) have all her memories of being awakened wiped out, and be put to sleep again. Then she will be awakened again on Tuesday, and interviewed once again. If the coin came up tails, on the other hand, Beauty will only be awakened on Monday. Beauty herself is fully aware ahead of time of what the experimental protocol will be.

So in one possible world (heads) Beauty is awakened twice, in identical circumstances; in the other possible world (tails) she is only awakened once. Each time she is asked a question: “What is the probability you would assign that the coin came up tails?”

Modified from a figure by Stuart Armstrong.

Modified from a figure by Stuart Armstrong.

(Some other discussions switch the roles of heads and tails from my example.)

The Sleeping Beauty puzzle is still quite controversial. There are two answers one could imagine reasonably defending.

  • Halfer” — Before going to sleep, Beauty would have said that the probability of the coin coming up heads or tails would be one-half each. Beauty learns nothing upon waking up. She should assign a probability one-half to it having been tails.
  • Thirder” — If Beauty were told upon waking that the coin had come up heads, she would assign equal credence to it being Monday or Tuesday. But if she were told it was Monday, she would assign equal credence to the coin being heads or tails. The only consistent apportionment of credences is to assign 1/3 to each possibility, treating each possible waking-up event on an equal footing.

The Sleeping Beauty puzzle has generated considerable interest. It’s exactly the kind of wacky thought experiment that philosophers just eat up. But it has also attracted attention from cosmologists of late, because of the measure problem in cosmology. In a multiverse, there are many classical spacetimes (analogous to the coin toss) and many observers in each spacetime (analogous to being awakened on multiple occasions). Really the SB puzzle is a test-bed for cases of “mixed” uncertainties from different sources.

Chip and I argue that if we adopt Everettian quantum mechanics (EQM) and our Epistemic Separability Principle (ESP), everything becomes crystal clear. A rare case where the quantum-mechanical version of a problem is actually easier than the classical version.

In the quantum version, we naturally replace the coin toss by the observation of a spin. If the spin is initially oriented along the x-axis, we have a 50/50 chance of observing it to be up or down along the z-axis. In EQM that’s because we split into two different branches of the wave function, with equal amplitudes.

Our derivation of the Born Rule is actually based on the idea of self-locating uncertainty, so adding a bit more to it is no problem at all. We show that, if you accept the ESP, you are immediately led to the “thirder” position, as originally advocated by Elga. Roughly speaking, in the quantum wave function Beauty is awakened three times, and all of them are on a completely equal footing, and should be assigned equal credences. The same logic that says that probabilities are proportional to the amplitudes squared also says you should be a thirder.

But! We can put a minor twist on the experiment. What if, instead of waking up Beauty twice when the spin is up, we instead observe another spin. If that second spin is also up, she is awakened on Monday, while if it is down, she is awakened on Tuesday. Again we ask what probability she would assign that the first spin was down.

beauties

This new version has three branches of the wave function instead of two, as illustrated in the figure. And now the three branches don’t have equal amplitudes; the bottom one is (1/√2), while the top two are each (1/√2)2 = 1/2. In this case the ESP simply recovers the Born Rule: the bottom branch has probability 1/2, while each of the top two have probability 1/4. And Beauty wakes up precisely once on each branch, so she should assign probability 1/2 to the initial spin being down. This gives some justification for the “halfer” position, at least in this slightly modified setup.

All very cute, but it does have direct implications for the measure problem in cosmology. Consider a multiverse with many branches of the cosmological wave function, and potentially many identical observers on each branch. Given that you are one of those observers, how do you assign probabilities to the different alternatives?

Simple. Each observer Oi appears on a branch with amplitude ψi, and every appearance gets assigned a Born-rule weight wi = |ψi|2. The ESP instructs us to assign a probability to each observer given by

P(O_i) = w_i/(\sum_j w_j).

It looks easy, but note that the formula is not trivial: the weights wi will not in general add up to one, since they might describe multiple observers on a single branch and perhaps even at different times. This analysis, we claim, defuses the “Born Rule crisis” pointed out by Don Page in the context of these cosmological spacetimes.

Sleeping Beauty, in other words, might turn out to be very useful in helping us understand the origin of the universe. Then again, plenty of people already think that the multiverse is just a fairy tale, so perhaps we shouldn’t be handing them ammunition.

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245 Responses to Quantum Sleeping Beauty and the Multiverse

  1. John Barrett says:

    Maybe you shouldn’t look at it as a division of the number of possible outcomes, but look at it as the odds of winning each event in succession. Sleeping Beauty would have a 1/2 chance of getting heads the first time. Then it would take getting another 1/2 in order for it to be Tuesday. That would mean that she would only get heads twice in a row 1/4 times. Then it would seem like she only had a 1/4 chance of it being Tuesday.

    The number of possible outcomes may not be the total number of worlds, but it could be the total number of possible worlds that may or may not even occur.

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  2. JollyJoker says:

    If Beauty were told upon waking that the coin had come up heads, she would assign equal credence to it being Monday or Tuesday. But if she were told it was Monday, she would assign equal credence to the coin being heads or tails. The only consistent apportionment of credences is to assign 1/3 to each possibility, treating each possible waking-up event on an equal footing.

    How is it not consistent to assign 1/2 each to heads or tails, then split heads into 1/4 Monday and 1/4 Tuesday? I feel about this pretty much like about the earlier Born rule post; if you ignore the probabilities and just count distinct options, you’ll get the wrong answer.

    Disclaimer: I read the Wikipedia page and still don’t understand the “thirder” position at all.

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  3. trivialknot says:

    @JollyJoker
    If Beauty assigns 1/4 probability to (heads & Monday), and 1/2 probability to (tails and Monday), and then if you told Beauty that it was Monday and not Tuesday, then she would be forced to assign 2/3 probability to tails. This is inconsistent.

    @Sean
    I’m not sure you can say I beat you to the punch, when it was right there in your paper!

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  4. Ted Sanders says:

    Hi Sean,

    As someone sympathetic to Many Worlds, I am trying hard to understand your paper as well as your last two blog posts, but to be honest, I find them opaque. I don’t have a specific question to bother you with, but I simply want to relay how how one member of your audience (me) has been reacting to your recent work (with confusion). The tone of your posts suggests this stuff is straightforward, but I suspect much of your audience is nonetheless confused (my friends and I have been, at least). I encourage you to keep refining and simplifying your message until it reaches a more convincing form. I still think the Born rule is a big problem for Many Worlds and I’m glad that you (and others) are tackling it. Good work so far, and good luck in the future.

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  5. Joe Polchinski says:

    I thought that your 1/3 sounded correct, but when I heard that certain other blogs were vigorously denouncing your answer I became certain that you must be correct.

    Most questions of probability can be settled by the question, which side would you bet on? When SB is awakened she is given the opportunity to make a bet that the coin is heads: if it is heads, she is +$1. If it is tails, she is -$x. For what x is this a fair bet? There is a 50% probability of heads, in which case she is +$2 for her two wakings. There is a 50% probability of tails, in which case she is -$x. So x=2 for a fair bet, meaning that the probability is 1/3.

    Presumably this argument is well-known. It is described on Wikipedia as the Phenomenalist position, which gives the result 1/3 unless the additional constraint is imposed that she can only bet on Monday, which is changing the problem: if we condition her probability on its being Monday, then of course it is 1/2.

    So what’s the problem?

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  6. Sean Carroll says:

    Joe, I think that’s a perfectly good argument. You can look at our argument as saying that an “epistemic” view of the problem gives an answer that is compatible with that more “operational” view (exactly as in the case of the Born Rule derivation). Having more than one way to get the right answer is a good thing, especially when people are skeptical of the right answer.

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  7. Daniel Kerr says:

    Before responding to the quantum version, I want to make sure I understand how you get the thirder position. The thirder position is based on the assumption that waking on each day has equal probability. However, this isn’t the case as waking up on Monday and Wednesday each have probability weights of 1 and Tuesday has a probability weight of 1/2. Each day itself has 1/3 probability from a perspective that isn’t Sleeping Beauty’s. So stated mathematically:

    P(Wake|H and Mon) = P(Wake|T and Mon) = P(Wake|Mon) = 1
    P(Wake|H and Wed) = P(Wake|T and Wed) = P(Wake|Wed) = 1
    P(Wake|H and Tue) = 1
    P(Wake|T and Tue) = 0
    P(H) = P(T) = 1/2
    P(Mon) = P(Tue) = P(Wed) = 1/3

    So we want to find:

    P(H|(Wake and Mon) or (Wake and Tue) or (Wake and Wed))

    This is the statement, “What is the probability of the coin flip being heads given she wakes up on one and only one of the days, Monday, Tuesday, or Wednesday?”

    P(H|(Wake and Mon) or (Wake and Tue) or (Wake and Wed)) =
    P((Wake and Mon) or (Wake and Tue) or (Wake and Wed)|H)*P(H)/P((Wake and Mon) or (Wake and Tue) or (Wake and Wed)) by definition of conditional probability.

    Let’s focus on calculating the denominator:
    P((Wake and Mon) or (Wake and Tue) or (Wake and Wed)) =
    P(Wake and Mon) + P(Wake and Tue) + P(Wake and Wed) =
    P(Wake|Mon)*P(Mon) + P(Wake|Tue)*P(Tue) + P(Wake|Wed)*P(Wed)
    1*1/3 + (1/2)*(1/3) + 1*1/3 = 5/6

    The intersection cross terms are omitted because P(Mon and Wed) = 0, since they are mutually exclusive days.

    Now let’s focus on the first term of the numerator:

    P((Wake and Mon) or (Wake and Tue) or (Wake and Wed)|H) =
    P(Wake and Mon|H) + P(Wake and Tue|H) + P(Wake and Wed|H) =
    P(Wake|Mon and H)*P(Mon) + P(Wake|Tue and H)*P(Tue) + P(Wake|Wed and H)*P(Wed) = 1*1/3 + 1*1/3 + 1*1/3 = 1

    So the whole expression is equal to:

    P(H|(Wake and Mon) or (Wake and Tue) or (Wake and Wed)) = 1*(1/2)/(5/6) = 3/5

    I don’t believe I made a mistake, but one of my probability assignments must be incorrect, meaning I misunderstand the statement of the problem.

    3/5 makes a certain sense for heads though. While each day has equal chance (1/3), waking up on each day does not. You have 2 sets of cases, being woken up on Mon, Tue, and Wed; or being woken up just on Mon and Wed. Each day has 1/3 probability with each set of days having 1/2 probability, so 1/6 for waking up on each day. As a result, you have 5/6 cases Sleeping Beauty is awake and 1/6 cases of her not being awake. 3/6 cases are heads, the other 3/6 are tails, but she can only answer the question in 5 of those cases.

    Regardless of the legitimacy of my attempted breakdown, I just don’t see how the odds of the coin being heads could ever drop below 1/2 if heads results in more wakeful days for sleeping beauty to sample versus tails.

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  8. Ignacio says:

    This situation reminds me a little bit of the Monty Hall problem where upon first inspection you are also certain the probability must be 1/2 but it is not 1/2. Here is the wikipedia entry:

    http://en.wikipedia.org/wiki/Monty_Hall_problem

    Nevertheless this has nothing to do with QM really.

    Everyone can make mistakes. Hopefully the other blog will admit it this time but I wouldn’t hold my breath.

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  9. Ted Sanders says:

    Would someone mind explaining where expression 3.13 (in the arXiv paper) comes from?

    To me, it looks like it assumes the Born rule to prove the Born rule. I’ll describe the situation:

    Expression 3.11 tells us to assume a state that’s Sqrt(2/3)*UP + Sqrt(1/3)*DOWN.

    Then expression 3.13 factors this expression into three terms, with a Sqrt(1/3) prefactor. It looks something like Sqrt(1/3)*(a*UP+b*UP+c*DOWN).

    To me, it seems that this step is the magic that leads to the derivation ‘proving’ that UP is twice as common as DOWN. But as far as I can tell, this step is not justified in the text. Was it a mistake? Am I totally misreading something?

    Thanks for any insight that you can offer.

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  10. Bill Bedford says:

    Interesting……

    There are four pieces of information, Monday, Tuesday, heads and tails and how they are presented matters to what Beauty can tell about the world. There are four possible scenarios:

    1/
    Beauty wakes up for the first time and is told it is Monday, so she knows that there is a 50% chance the coin has come up heads and 50% tails.
    Beauty wakes up for the second time and is told it is Tuesday, so she knows that there is a 100% chance the coin has come up tails.
    2/
    Beauty wakes up for the first time and is told the coin has come up heads, so she knows that there is a 50% chance that it is Monday. If she is told the coin came up tails, the chance it is Monday is then 100%.
    Beauty wakes up for the second time and is told the coin has come up heads, so she knows that there is a 50% chance that it is Tuesday.
    3/
    Beauty wakes up for the first time and is told the coin has come up heads, so she knows that there is a 50% chance that it is Monday. If she is told the coin came up tails, the chance it is Monday is then 100%.
    Beauty wakes up for the second time and is told it is Tuesday, so she knows that there is a 100% chance the coin has come up heads.
    4/
    Beauty wakes up for the first time and is told it is Monday, so she knows that there is a 50% chance the coin has come up heads and 50% tails.
    Beauty wakes up for the second time and is told the coin has come up heads, so she knows that there is a 50% chance that it is Tuesday.

    In all these scenarios there are two pairs of variables Monday and Tuesday, and heads and tails, being given any one variable allows us to determine whether the two variables in the other pair have either a 50% or 100% probability.

    What Elga seems to have done is to confuse the state of the coin given the day, with the day given the state of the coin when these two probabilities are entirely separate.

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  11. Sean Carroll says:

    Ted– That equation just relies on Pythagoras’s theorem, which is true no matter what you believe about quantum mechanics. The Born Rule is a statement about the probabilities of measurement outcomes.

    As mentioned in the previous post, the fact that the amplitude gets squared to give you probabilities is pretty trivial. What’s non-trivial is explaining why there are probabilities at all.

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  12. Daniel Kerr says:

    Ah, I caught my mistake in my analysis above, Sleeping Beauty is not interviewed Wednesday, this changes the probability of heads to 2/3, and tails to 1/3. I switched heads and tails following wikipedia’s description and not Sean’s, so this agrees with Sean’s answer. All is well. Sorry for spamming the comments with a lengthy calculation.

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  13. Eric Winsberg says:

    Joe, here’s the problem with that simple argument. (warning: I am a halfer).

    Imagine the following situation: I flip a coin. I give you an opportunity to bet on whether it is heads or tails. And I ask you for what X is it a fair bet. But now I add the following wrinkle. If its comes up tails, I am going to make you bet twice. Now, I would say, your estimation of X no longer reflects your credence. I have messed with the rules too much. I think this is happening in the SB case.

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  14. Eric Winsberg says:

    (and that’s why the halfer argues that you should condition on it being monday, so that SB is prevented from betting twice on the same coin flip, if and only if it comes up tails. But maybe the wikipedia doesnt explain that well.)

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  15. Daniel Kerr says:

    Eric, the issue with that simplification is that the fact you are even being asked to bet at all is contingent on the outcome of the flip.

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  16. Ignacio says:

    It is interesting that the famous mathematician Paul Erdos did not believe the solution of the Monty Hall problem until he was shown by a computer. This entry might elicit similar reactions. But that is mostly because once you’re “sure” you then proceed to become angry after you’re shown to be wrong. Then comes bargaining where you seek to alter the terms.

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  17. The coin is flipped only once. It could either be heads or tails, a 50-50 chance. When Sleeping Beauty is awoken either it would be Monday or Tuesday, no other choice, simply a 50-50 chance.

    When she is awoken Beauty will not know whether she was awoken before or not, again simply a 50-50 chance.

    When beauty is awoken what odds should she attribute to the probability that the coin toss was heads?

    When awoken Beauty could be having her first interview or she also could have had one the day before. She would not know which, again just two choices. She should consider the odds as being 50-50 concerning whether the coin toss was heads or tails since (if) she gained no new information during the experiment.

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  18. Mikkelsen says:

    The only consistent appointment of credences is to assign 1/3 to each possibility, treating each possible waking-up event on an equal footing.

    Problem is that the “probability” of a waking-up event has nothing to do with the “tails” probability: If SB always answers 1/3 for “tails” and you repeat the experiment e.g. 90 000 times, do you believe that “tails” will occur approximate 30 000 times?
    (Remember we are told the coin is fair)

    SB will get more correct interview-answers by saying “heads”, but she will still be correct in only about 50% of the experiments, i.e. coin-flips.

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  19. Mario Voelker says:

    After first being with the halfers I slowly reasoned myself into the thirders position.
     
    The halfers are only looking at the probability of the coin toss, disregarding the context in which the question was asked.
     
    They answer the general question “what is the probability of tails when a coin is tossed?”
    But the actual question is: “what is the probability of tails when the beauty wakes up?”.
     
    I think the thirder argument is slightly off though.
    Mixing up the information about if its monday or if heads came up seems wrong to me.
     
    The only information that seems to make sense is how many possible waking up events there are in an outcome in relation to the sum of all waking up events in all outcomes. (So for tails 1 event / 3 total events)
     
     
    There must be more to it though. This reasoning works only if heads and tails are equally likely.
    If we have something else than a coin with say 1/3 chance of heads and 2/3 chance of tails it fails.
     
    Guess I skip the quantum stuff – my head already exploded.

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  20. Daniel Kerr says:

    The trick is to realize there are 4 cases here, heads and Monday, tails and Monday, heads and Tuesday, tails and Tuesday. They each have 1/4 probability, but sleeping beauty only “samples” 3 of these cases, as she is not awake for tails and Tuesday. So of the 3 cases she sees, two of them are heads and if she wants to make money from betting, she’d better bet on heads.

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  21. Ignacio says:

    Daniel your explanation is clear. I like it.

    But it makes you wonder why people are publishing papers on this stuff, which is not uncertain and has nothing to do with QM.

    So conditional probabilities can be tricky so what’s the relevance to QM?

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  22. James Gallagher says:

    Sean says “As mentioned in the previous post, the fact that the amplitude gets squared to give you probabilities is pretty trivial. What’s non-trivial is explaining why there are probabilities at all.”

    erm, are you sure it’s trivial?

    Why is this experiment considered worthwhile then?

    You can notice that the probability function must be a positive definite form globally conservered by the unitary evolution (since probability is positive and globally equal to 1), but beyond that you are appealing to the consistency of the definition of probability for multiple outcomes – hence Gleason’s theorem. But that’s not a deduction, that’s us humans saying Nature wouldn’t be so stupid as to make the outcomes of measurements proportional to something other than the absolute square because it wouldn’t make sense compatible with us dumb humans understanding probability.

    We can’t deduce the power 2, we can only marvel that the universe seems to obey that law.

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  23. Daniel Kerr says:

    Ignacio, I don’t know why it’s debated. If you correct my first post to not include Wednesday, it’s as explicit of a probability calculation you can get. The only way for it to be 1/2 is if Sleeping Beauty is awake for an equal number of coin flips. The solution to the Monty Hall problem also took a while to gain acceptance, though its paradox is a more subtle mistake in my opinion.

    I think Sean is putting forth that the power of his branch probability assignments unequivocally have an answer to the problem. But I think probability does as well, as counting each state directly results in only one answer. It would be more interesting to apply ESP to uniquely conditional probability, quantum cases, like Born/Kochen-Specker type situations. An application to Bertrand’s paradox would actually be fascinating as a classical probability example.

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  24. Charlie says:

    Let me set up a different thought experiment. Again, the subject (me) knows the rules of the game. The rule is that my friend will knock me out, then role a pair of fair dice. If the dice come up snake eyes (both one’s), then he will put me in room A. Anything else and he will put me in room B. The two rooms are indistinguishable from the inside. So when I wake up, what odds should I take that I am in room A?

    Does anyone here want to argue that I should consider it 50/50 that I am in room A? If not, can you explain why this answer would be different than the “thirder” position above?

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  25. Ignacio says:

    Daniel it is not all that different. The bottom line is that you have to take the rules seriously. In the Monty Hall problem you need to understand that there is someone in the background who behaves predictably and has the answers. In this case the sampling is biased, I guess it is one way to characterize it.

    But it has NOTHING to do with QM. It is just trickier than first meets the eye. That’s all.

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  26. Daniel Kerr says:

    Charlie, in your example you will be in room B 11/12 of the time and thus should bet accordingly. The example is not analogous as you waking up is not contingent on the outcome of the dice roll. You sample every possible outcome. The trick of this problem is that not every outcome is sampled.

    Ignacio, the lesson I got from the Monty Hall problem was to be careful with how you define a state. The problem fools you because the “state” of having picked the wrong door is really 2 different states, despite them having the same qualitative description. In a problem of n-coin flips, it’s like specifying the state of k (k<n) heads. There's no single, unique realization of this description, the k heads can be distributed among the n (unique) coins in many possible ways. The doors are unique, that's the take-away, at least my take-away.

    As for QM, like I said, I think Bertrand's paradox would be an interesting test for how robust the probability emergence is.

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  27. Ignacio says:

    Daniel my takeaway of the monty hall problem is that for the analysis to be correct you have to assume that the game show host behaves predictably always in the same fashion, which is why he changes the probabilities. The game show host could not, say, choose to open the door to the car one time, or decline to open a door at all.

    All of this enters into the analysis. But the point remains the same. You just have to be very careful about the rules of the game for any game of probability.

    QM is based on probabilities and therein ends the relevance of this post.

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  28. Daniel Kerr says:

    Ignacio, I think it’s a valid interpretation, but I usually solve the problem by omitting the game show host as an agent and instead wrapping his actions up into the conditional probabilities. This comes down to state counting then and realizing that there are two different ways where you can switch to the door with the prize behind it. I’m sure modeling it as a two player game works too, but I think it’s really a one player game.

    I would agree with you, except Sean’s construction of probabilities in MWI is not grounded in a normal probabilistic formulation of QM. It is possible that his interpretation is incorrect and will diverge from some probabilistic phenomena. I don’t believe his paper shows how his interpretation results in statements in QM that satisfy probability axioms, so there’s always a possibility.

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  29. Charlie says:

    OK, I see my mistake above. And clearly I jumped at the problem without reading Sean’s second scenario. It’s not trivial though counting possible outcomes and what’s sampled (at least not to me).

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  30. Ignacio says:

    Daniel saying it is a one player game is by definition equivalent to saying that the game show host is a predictable automaton, not a free-will agent. An automaton with perfect knowledge of the answers.

    There is a level of deception in thinking such puzzles can serve to explain fundamental rules of probability as applied in QM.

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  31. Daniel Kerr says:

    Ignacio, sure, but the point is you don’t have to include him in the analysis at all. You can just map his choices as outcomes and nothing is lost.

    Obviously verifying that the interpretation produces the axioms of probability on its own would be enough. Don’t know why this isn’t standard protocol for interpretations to verify they satisfy the axioms of probability.

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  32. Avattoir says:

    Doesn’t SB being told that “it’s Tuesday” mean she simply gets to rely on her memory of whether or not she was woken on Monday, since her memory was NOT washed of having been woken up on Monday?

    If SB is at all relevant times fully informed about the rules that are in force, and then she’s woken on Tuesday and is TOLD, accurately and truthfully, “It’s Tuesday”, she simply needs rely to her own memory: if she does NOT remember on that Tuesday having been woken on the prior Monday, then necessarily, with 100% accuracy, she knows that the coin toss that occurred on Monday came up “Heads”. If she DOES remember on that Tuesday having been woken up the day before, then she knows that the coin toss that occurred on Monday came up “Tails”.

    Similarly, if she’s told “It’s Monday”, what she does NOT know at that point is whether there’s still a plan in place to erase her memory; IOW she can’t know from just that information whether the coin of decision came up “Heads” or “Tails”. She’s thus compelled to rely on the honesty of the coin, so her accuracy, over enough experiments, is going to fall at or around 50:50.

    If she’s not told what day it is, the condition of ‘knowing everything else’ is broken, because there are TWO pieces of information being kept from her: the outcome of the coin toss AND the day. And that’s where the ‘extra dimension of uncertainty’, the squaring of the uncertainty, kicks in.

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  33. Mark F. says:

    “Before going to sleep, Beauty would have said that the probability of the coin coming up heads or tails would be one-half each.” — Well, that depends on whether she’s a Halfer or a Thirder.

    Although I don’t mean that quite literally. Before going to sleep, Beauty will say that the probability of the coin coming up tails is 1/2, but the probability of her waking up to a world where it comes up tails will be 1/3. The fact that she “gets no new information” is a red herring; it would take new information to lead her to an opinion of 1/2 or 1. It also doesn’t matter that the coin needn’t be tossed till Monday night. When asked, she doesn’t know if it’s been tossed or not.

    It’s interesting to think about what happens after the final interview. Suppose they don’t tell her what day it is, put her back to sleep, and awaken her on Wednesday (keeping her asleep through Tuesday if it came up Tails). In that case, Wednesday morning I claim she says the probability was 1/2.

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  34. Charlie says:

    Why not run the top course (heads) out to 99 awakenings. Now Sleeping Beauty can say with 99% certainty upon awakening that the coin came out heads. Heck, let’s dispense with this whole charade entirely and quietly kill her if it came out tails (with her prior consent and understanding, of course). Now Sleeping Beauty can say with 100% certainty that she cannot experience death from this experiment and is, in fact, immortal. But is it a quantum immortality?

    This is all meant in good fun of course.

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  35. In the SB scenario the coin is only flipped once. There will be a 50-50 chance of it turning up either heads or tails. There can only be two outcomes regardless of what’s happening with Beauty.

    If you are a “thirder” reconsider the possibilities again. If the coin flip is tales she will be interviewed on Monday only. There is a 50% chance that this will happen. If the coin flip lands on heads there would be a 50% chance that this will happen also; then she will be interviewed on both Monday and Tuesday, a requirement of the scenario. Beauty is awoken and interviewed no more than twice to make an assessment, in Sean’s scenario.

    SB should realize that she cannot gain any knowledge at all during the experiment by its design and that a single coin flip, as in this scenario, can only have two outcomes of equal probability.

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  36. Ted Sanders says:

    “As mentioned in the previous post, the fact that the amplitude gets squared to give you probabilities is pretty trivial. What’s non-trivial is explaining why there are probabilities at all.”

    Oh, then I guess I’ve been misinterpreting this whole series. I thought the point of the paper and these posts was to derive the Born rule from a Many Worlds framework, not to explain why branching leads to apparent probabilities. Oops! (Many Worlds leading to probabilities already seems pretty straightforward to me. Though perhaps it’s not so straightforward to halfers.)

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  37. Ted Sanders says:

    (To briefly follow up, I did see that the last post gave Gleason’s theorem as a justification for the Born rule. But I had thought that the rest of the post and paper were going to give another way to derive the Born rule that was not Gleason’s theorem. Perhaps this is why I’ve been so confused!)

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  38. The “thirder” argument may have its basis in the many-worlds interpretation, granted, but outside this interpretation I think it makes about as much sense as Schrödinger’s cat being alive, dead, or neither until it is observed, 3 possibilities :)

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  39. John Pollard says:

    Just take an extreme case where SB is told she will be woken 100 times for Heads, but will not recall previous wakings. Then she can confidently say it is almost certainly Heads as I am sure I keep telling you! Forget 50/50

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  40. Joe Polchinski says:

    Amusing that I got 9 down-votes. Perhaps you all would like to play a little poker? A hint: if you can make money taking one side of a 50-50 bet, then it’s not a 50-50 bet. Anything more is overthinking the problem.

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  41. Joe Polchinski says:

    But if you do want to overthink it: the question is a conditional probability, what is the odds of heads GIVEN that SB has been wakened and asked the question. Now, does the latter constitute useful information? It obviously would if the conditions were that SB is not wakened at all on Tuesday, but here it is more subtle and I think that intuition can go either way. So we must look closely at the definition of probability: “that body of knowledge that is used to win money from the mathematically impaired.”

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  42. Daniel Kerr says:

    Joe, I have no idea why your answer has so many dislikes and mine has none when they essentially say the same thing, haha. I guess my initial calculation dissuaded users from reading past that point.

    Like or Dislike: Thumb up 4 Thumb down 5

  43. Joe Polchinski says:

    Thanks Daniel, also Ignacio. At the risk of belaboring the point: You are in a room. You are asked a question. You know that if you answer heads you will win the bet 2/3 of the time. If you answer tails you will win the bet 1/3 of the time. What are the odds that the answer is heads? (The fact that you might be asked the same question later, or before, is not relevant: your answer in either case has no effect on the subsequent offer).

    p.s. I just noticed a typo in my previous comment: it should say “SB is not wakened at all on _tails_”

    Like or Dislike: Thumb up 5 Thumb down 4

  44. Joe Polchinski says:

    I think that one thing that is causing confusion with the bet is the idea that there is a secret double-payout in one case. But this is not true: each decision of SB controls only one payout. If the experiment is repeated 200 times, and SB always guesses heads, in the end she will have made 300 guesses and will have 300 betting slips, of which 200 say `heads, you win’ and 100 say `tails, you lose.’

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  45. James Gallagher says:

    I can only assume Joe got down-voted because he gave a disappontingly elementary derivation rather than applying a brane-world scenario. And using a monetary argument comes across as bragging about his his recently acquired wealth. ;-)

    edit: and if he’s not the real Joe Polchinski then certainly deserves the down votes!

    Like or Dislike: Thumb up 2 Thumb down 0

  46. Phil Koop says:

    As with many surveys, Sleeping Beauty’s answer will depend on how the question is phrased. If you say “I flipped a coin on Sunday; what was the probability that it came up heads?”, SB will answer “1/2.” That is because everyone, including SB, agrees that P(H) = 1/2.

    On the other hand, if you say “I flipped a coin on Sunday; what is the probability that it came up heads?”, SB will answer “2/3″. That is because poor SB cannot observe the act of flipping the coin, even accounting for the fact that the outcome is masked. She can only observe the fact of having been woken. And everyone, including SB, agrees that P(W and H) = 2/3.

    Sleeping Beauty, then, belongs to a fairly extensive class of popular philosophical “problems”, such as Monty Hall, “why does a mirror reverse left and right but not up and down”, and the Two Envelopes problem which are not problems at all because when specified with sufficient precision, they have correct answers.

    It is therefore tempting to roll ones’ eyes in these cases, and certainly that is a reasonable reaction to some particular arguments. But I have to say that I found working through the permutations of the Two Envelopes problem required to eliminate contradictions to be quite stimulating; and no doubt many people feel the same about their own pet favorites.

    Like or Dislike: Thumb up 2 Thumb down 2

  47. Phil Koop says:

    I think it is interesting that by coincidence, Sabine Hossenfelder has just posted a physical and mathematical perspective on another of these philosophical “problems”: can you touch your nose?

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  48. James Gallagher says:

    Phil Koop

    “Why does a mirror reverse left and right but not up and down?” I’m surprised that is considered a philosophical problem – the (quite well-known) explanation is that the mirror just relects what’s facing it, so if you turn (yourself or an object) around a vertical axis to face it you reverse left and right, and if you turn around a horizontal axis to face it you reverse up and down. (Or write on a thin piece of paper, and hold it with the writing facing a light, then what you see through the back of the paper is what you see in a mirror reflection)

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  49. Ignacio says:

    Joe I think the real source of the confusion in this sort of situations is believing that your intuition is an automatically good description of the probabilities.

    For the Monty Hall problem the intuition is that you stand nothing to gain by switching doors and in a real game show that is probably correct because the host would just as often try to trick you. In this instance the problem is the thinking that the coin was tossed and it is either heads or tails so 50/50 has to be the answer.

    It is nothing complicated just a tendency in the human mind.

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  50. John Barrett says:

    I didn’t really care for the explanation of the Monty Hall problem either, that you would increase your odds to 50/50 for switching doors after a loser one was revealed. It didn’t seem like it would be truly 50/50 if you decided to simply take the same action every time. Your intuition would tell you that it would still just be 1/3, but it will just be one out of the three that you didn’t pick to begin with.

    I think the Sleeping Beauty problem is a lot easier than that. If you had a 1/2 chance of it her not being woken up on Monday from the original coin flip and a 1/4 chance of being woke on Monday or Tuesday, then that would add up to 100%.

    1/4 + 1/4 +1/2 = 4/4 = 100%

    It would be more like the question, “What are the odds of getting heads twice in a row?” The only thing is that, if you fail to get heads, you stop flipping the coin. If you flip the coin twice there are only 4 possible outcomes, it is just that some of the outcomes are not realized, because there would be no point of trying again if you already failed. Then you would have a 50% chance of failing on the first try, so half of the time Sleeping Beauty would just sleep through the whole ordeal. Then there would only be 1/4 chance that she was woken on Monday, since there was also a 1/2 chance that the coin was flipped again making it Tuesday due to her amnesia.

    I think it would go along more with what happens in everyday experience. It could mean the difference of rather everyone that goes to Vegas in the multiverse comes out a winner, or if mostly everyone that goes to Vegas is a loser. The odds are always stacked in the houses favor. Then mostly everyone always losses everything. Although, if there was an equal chance to win as there was to lose, then in the multiverse someone would always come out a winner. Then people always come back as a loser every time no matter how many times they vacation to Vegas. If every possibility in the multiverse was realized, then it would seem like if you gambled enough that anyone could eventually become a winner!

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  51. Daniel Kerr says:

    I don’t get why there’s debate, this problem can be realized experimentally or in simulation, never mind simply calculating it in full probability theory formalism. I certainly don’t get the strong language being used on either side of the debate, there’s no reason to make this personal.

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  52. Ignacio says:

    Daniel this has happened before as I mentioned concerning the Monty Hall. There should be no debate. You are quite right. These problems are not hypothetical. It is easy enough to program them.

    But people get caught saying the wrong thing and then there is denial, anger, bargaining, disillusionment, and acceptance.

    Now and again someone will get stuck at denial until he/she is forced to look at output from a code. It says a lot about humans, a little bit about probabilities and nothing about QM

    Like or Dislike: Thumb up 3 Thumb down 0

  53. Sean Carroll says:

    I’m sympathetic to the claim that the problem is not that hard. Run the experiment many weeks in a row. One-third of the times that Beauty woke up the coin will have been tails, and two-thirds of the time it will have been heads. But it obviously is “hard” in the sense that many people disagree!

    Also, the “quantum Sleeping Beauty” puzzle is not precisely the same as the classical one, at least in EQM where both branches actually exist. What Chip and I have argued is that there is a unique way of apportioning the credences that matches up with all the right expectations.

    Like or Dislike: Thumb up 5 Thumb down 2

  54. Moshe says:

    Any particularly convincing argument for the one-half position? the Bayesian sounding argument in the main post is incorrect I think: knowing the experimental protocol, SB certainly learns something from the fact that she is awaken, an event that is correlated with the outcome of the coin flip, and therefore should update her probability assignment accordingly.

    Like or Dislike: Thumb up 3 Thumb down 1

  55. Moshe says:

    Oh, sorry, Sean, missed that last comment. Still, since there is presumably a large body of work on this, some potentially confused but semi-convincing reasoning must exist…

    Like or Dislike: Thumb up 0 Thumb down 1

  56. Haelfix says:

    Like the Monte Hall problem, I find the intuition significantly easier if instead of having only two days, to instead make the problem last say 100 days. (eg sb wakes up on monday and is put back to sleep, then wakes up on tuesday and is put back to sleep etc).

    I do however agree that it depends a little on how you word the problem. I like the word ‘belief’ as opposed to probability here.

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  57. Stephen says:

    Suppose instead of “What is the probability you would assign that the coin came up tails?”, the question asked were “What are fair odds you would assign that the coin came up tails, such that you will have zero expectation value if you bet on these odds?”

    Note that the problem stipulates that she will be interviewed on every waking. It is an unambiguous result in mathematics that the answer is 1/3, and even many halfwits see this. What is puzzling is that they interpret these two questions differently.

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  58. Moshe says:

    Maybe the issue is the distinction between probability and conditional probability. A clearer version can be that SB is woken only if the coin falls on heads (perhaps with a daily coin toss, for humanitarian reasons). When SB is then woken and is asked for her opinion of what transpired with that coin toss, a rational SB with any definition of probability should have no uncertainty about the matter, given that she is now awake. This does not contradict the fact that the probability for her to be woken up is only 50%.

    Like or Dislike: Thumb up 4 Thumb down 1

  59. JollyJoker says:

    @Joe Polchinski

    You’re adding a random game on top of the original question and assuming SB’s answer to “What is the probability you would assign that the coin came up tails?” is the one that minimizes her own profit in this invented game. Any other answer would let her turn a profit by picking her guesses on whether it was heads or tails. You’re assuming the one who answers the question is playing _against_ SB when the problem statement clearly says _she_ is he one answering the question.

    The key, of course, is the question “What is the probability you would assign that the coin came up tails?”. What is SB trying to achieve here? Any clear definition of that gives a clear answer, but my opinion is that any invention of games that doesn’t have SB answer honestly, as the problem states initially, that the probability is 50-50 _or_ that she can answer whatever she wants because nothing gives her an incentive to _lie_, is just plain stupid.

    You can invent additional casinos or whatever that define what would optimize her expected value in iterated SB problems, but saying that 1/3 is the correct answer is just plain wrong unless you define a different game for which the answer is 1/3.

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  60. JollyJoker says:

    @Joe Polchinski

    You can disprove my claims by specifying what SB is trying to achieve in a clear enough fashion and getting someone to bet against you despite this goal being clear to both parties.

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  61. CB says:

    SB’s goal is to assign probabilities to the possibilities that 1) she just woke up, and the coin flipped at the beginning of the trial was heads, and 2) she just woke up, and the coin flipped at the beginning of the trial was tails.

    Since the experimental procedure clearly states that every time the coin comes up heads she will be woken up twice, and every time she comes up tails she will be woken up once, then it is obvious that over time the ratio of “woken and heads” to “woken and tails” will be 2:1. This is the same as saying the probabilities are 2/3 and 1/3. Literally the definition of probability.

    So while she knows that the odds of a fair coin flip in complete isolation is 50:50, *that’s not the case here*, and the result of the flip has consequences, and those consequences mean that 2/3rds of the time she awakes in a universe where a fair coin landed heads. Therefore that is the correct probability to give. And repeated trials would bear it out.

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  62. Jim Z says:

    Silly.

    So an experiment with Sleeping Beauty is more important than an experiment by Alain Aspect?

    Like or Dislike: Thumb up 1 Thumb down 3

  63. Galois Was Randomly Walking Here says:

    I am assuming the coin flips are mutually exclusive and the semantics of the problem indicate that the event of “coin flip” and the event of “waking up” are independent events . Since sleeping beauty is guaranteed to wake up the probability of her waking up after a coin 50/50 coin flip is one.

    P(Flip) = 1/2
    P(Waking up) =1

    If she has knowledge of what day it is that changes the semantics of the problem meaning that the mutually exclusive event of “coin flip” and the following event of “waking up” are conditionally dependent allowing her to assign different probability values (especially if it is a Tuesday). This appears to be a poorly worded problem. The problem most likely leaves sleeping beauty’s knowledge ambiguous to get people to ask the following question: What does sleeping beauty know and will she go out on a date with me (please check the box yes or no)? I could be wrong and what I love about probability theory is how it is deceptively awesome . Do I have any volunteers for this experiment?

    P.S. Did you remove my earlier post? Because I could have sworn I posted here before.

    Like or Dislike: Thumb up 1 Thumb down 1

  64. Rww says:

    It’s sleight of hand, no? The trickster gets us to focus on the odds of the flip rather than the number of interrogations that follow. Two-thirds of all interrogations follow heads and that is all we need to know

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  65. JW Mason says:

    Ignacio’s comments here are convincing to me.

    This post has solidified my impression that MWI is not an interpretation of quantum mechanics, but is rather a way of thinking about probability in the context of a single event. It doesn’t seem to say anything about where the probabilities come from.

    The minimal interpretation seems to be the ensemble interpretation. Given that science is in the business of describing general rules, why would we ever care about the “probability” of a unique nonrepeated event? And for repeated events, all these puzzles go away, no?

    Once you already have probabilities, what does it add to postulate an amplitude equal to their square roots? “The same logic that says that probabilities are proportional to the amplitudes squared also says you should be a thirder” seems trivially true, in the sense that if you have established probabilities by any means whatsoever, you should be a thirder.

    I’m a layman whose opinion doesn’t matter. But Sean might want to know that, for one reader anyway, the effect these posts are producing is the opposite of their goal. (Though they are certainly serving their larger goal, of being interesting and informative!)

    Like or Dislike: Thumb up 4 Thumb down 1

  66. Below is the original question in Sean’s scenario.

    “What is the probability you would assign that the coin came up tails?”

    But for SB to come to the “thirders” conclusion something like this should have been the question IMO:

    “What likelihood should SB assign to tails if she is told her sole objective is to guess correctly the most number of times?”

    From this perspective SB might make a different choice than with the original question.

    It also seems to me that the “thirders” position is not explained well above. In the SB scenario the coin is flipped only once. If she would guess heads for each interview and the coin flip came up heads, she would make two correct guesses, one on Monday and another correct guess on Tuesday. But if it comes up tails then she will be guessing wrong only once on Monday. So she is twice as likely to guess right if she guesses heads. This is the basis of the “thirder” position, but the wording of the question should have been different (similar to the “likelihood” question shown above IMO) for her to more likely come to this conclusion.

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  67. Goodmongo says:

    The answer is dependent upon the time frame. SB is being asked what the probability was for something that already happened based on what she knows. She has two clues to point to heads and one clue for tails. Therefore, she gives heads a 66% chance.

    The 50% is what the coin flip could produce in the future. Now if I changed the story to say that when SB awoke she overhears the researcher say that the coin was tails then when asked what the chance was that it was heads she would reply zero. She was told what it was tails in this case.

    In one problem we are trying to figure out the probability of a past event based on limited information. In the other it is what is the probability of it going to be. It’s a stupid little game and mind trick. I can’t believe all these PHD’s can’t see this.

    Like or Dislike: Thumb up 2 Thumb down 1

  68. Daniel Kerr says:

    JW Mason, I agree with you too. This treatment convinces me to view the set of universes as possible worlds in a modal logic framework, not in a “multiverse” view where each world is ontologically equal. I think the Ensemble Interpretation blends nicely with this approach since it would relax the assumption that the state is ontological. A key difference is that in MWI, upon making a measurement, the state branches into another quantum state. The Ensemble Interpretation argues that the universe itself never occupies a state, that a state gives you the distribution of possible outcomes and you are simply sampling values from that distribution. This avoids the preferred basis problem but still allows the same treatment of observers as Sean has carried out. I prefer to make less ontological assumptions with science, so even though I value Sean’s contributions to thinking of probability in a quantum context, I will just not be able to accept the universe is ever actually, physically in a wavefunction. I can accept it is the best possible description however.

    Like or Dislike: Thumb up 4 Thumb down 0

  69. CB says:

    forrest noble:
    I don’t think those two phrasing are actually different. “What is the most likely outcome of an event?” and “How would you bet on that event?” must have the same answers if we’re talking about someone rationally trying to maximize their return, or number of correct guesses.

    2/3rds of the time SB wakes up in a universe where the coin landed heads. That’s the probability, that’s how she should guess, and it’s the same either way.

    The question that could have been asked that I think gets at your point is: “What probabilities would you assign to the chance *that the researchers, who are not subject to this sleeping/amnesia potion nonsense* observed tails?” In other words, breaking out of the experimental protocol into a different one where we’re just flipping fair coins over and over and that’s all that’s happening.

    There’s nothing wrong with different observers observing different probabilities, that’s perfectly natural. And each of those observers can reason about what the other one’s subjective probability should be, given sufficient information about their situations. But I think it’s implicit in the problem that SB is being asked about her perspective.

    But here’s a wrinkle: Imagine that SB *didn’t* know about the use of the amnesia potion in the protocol. She would say the odds were 50:50, and would have no choice but to guess at which came up. But let’s also say that she was allowed to record and reference previous results. It wouldn’t take too long for her to realize that the researchers must be up to some form of chicanery — a loaded coin? — because her rationalization of the probability is clearly wrong.

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  70. JP says:

    The answer is 1/2. It’s a straightforward application of Bayes’ rule once you accept that we aim to calculate P( T | interview ), and assume that P( interview | T ) = P( interview | H) = 1. Thirders should address which of these assumptions they disagree with.

    Those of you applying a long-run frequency calculation to get the result 1/3 are neglecting the fact that “Monday & H” and “Tuesday & H” are not independent events. Indeed, they are completely determined by each other. So you can’t just aggregate them to estimate a probability. Incidentally, this is the error in Adam Elga’s paper linked to by the Wikipedia article.

    The betting odds arguments are fine, but they calculate the value of a bet, not a probability. The problem asked for a probability. Given certain fixed conventions about the structure of the bet these two things are equivalent – as in, e.g. The Dutch book argument – but those conventions do not apply here (since the number of bets depends on the bet’s outcome).

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  71. Daniel Kerr says:

    JP, I calculated exactly what you claim to calculate in my first comment, except I added an extra interview on Wednesday. You can eliminate Wednesday from the calculation and you will get 2/3, all of my assumptions are clearly laid out, feel free to let me know what is incorrect. And I disagree, the problem is asking for P(T|Interview), which is not in general equal to P(Interview|T).

    Like or Dislike: Thumb up 1 Thumb down 1

  72. JP says:

    Daniel Kerr,
    I believe the issue with your calculation is where you write P(wake and Monday OR wake and Tuesday) = P(wake and Monday) + P(wake and Tuesday). There’s another similar line in which the probabilities are conditioned on H. Since “wake and Monday” and “wake and Tuesday” are not mutually exclusive events you need to subtract P(wake and Monday AND wake and Tuesday). This is the same issue that I alluded to in my second paragraph.

    Like or Dislike: Thumb up 2 Thumb down 2

  73. Daniel Kerr says:

    JP, “wake and Monday” and “wake and Tuesday” are mutually exclusive events because Monday and Tuesday are mutually exclusive events. Simply conjugating Monday and Tuesday each with “wake” does not change this. Technically it should read “‘Monday or Tuesday’ and wake” as there’s no logical distinction between that and “‘Monday and wake’ or ‘Tuesday and wake.'” The latter is simply a more intuitive expression.

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  74. JP says:

    Daniel,
    You’re effectively saying that P(wake and Monday AND wake and Tuesday) = 0. But this is clearly wrong, since when the coin comes up heads both events occur.

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  75. Daniel Kerr says:

    JP, no, that’s clearly right you have to concatenate it to see. “Wake and Monday AND wake and Tuesday” = “wake and ‘Monday and Tuesday.'” Monday and Tuesday are mutually exclusive, it can’t be both days at once. You’re confusing waking on Monday and waking on Tuesday with these notions. What you want is “Monday -> wake” and “Tuesday-> wake.” In probability we’d represent these as P(wake|Monday) and P(wake|Tuesday) respectively. These conditional probabilities are independent and equal 1 and 1/2. However, P(wake|Monday and Tuesday) has no evaluation as “Monday and Tuesday” has a 0 probability of occurring.

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  76. phayes says:

    I think it’s useful to drop the assumption of a fair coin from the argument for the correct answer. Replacing that rather fragile-looking Elga/Wikipedia argument with a couple of expansions in conditionals yields P(Heads|.) = P(Heads|Monday, .)/(2 – P(Heads|Monday, .)) and the frequentist and non-frequentist arguments agree whatever the coin bias.

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  77. CB says:

    Any time you find yourself saying that the probability of each outcome is different from the the ratio of the occurances of each outcome after many repeated trials, you should realize that you’re doing something wrong.

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  78. Galois Was Randomly Walking Here says:

    Sorry Doc if I have commented here before, I have this short term memory problem . Were we talking about the paintings or that every time the stock market tosses a coin then I put a dollar into my no load index mutual fund? Did you know that the government (I forget which one) has a new deal and gives me another dollar in welfare/tax refund when the coin lands heads? So I get two dollars for every coin flip that lands on heads! My stock broker just called me and said the stock market flipped a coin what is the probability that it is heads or tails? If the stock market did 1000 coin flips today then what is the expected value of my portfolio?

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  79. @CB,

    “I don’t think those two phrasing are actually different. “What is the most likely outcome of an event?” and “How would you bet on that event?””

    I agree they are quite similar but neither one of them was the question asked SB.
    The question to be asked beauty on each interview was: ‘What is the probability you would assign that the coin came up tails?”

    There are no other qualifications to the question above in Sean’s scenario. In such a case I believe there could only by one right answer 50:50 since that is the exact indisputable probability of a coin flip.

    However if instead she was told that her only objective was to make the most right guesses concerning the results of the coin flip, during the whole experiment, then was explained again how the experiment will go, she might then realize that if she guesses heads she can be right twice during the experiment if the coin flip is heads, but she could be right only once if she picks tales. In this case upon being explained this different objective, she might realize that her odds are 2 to 1 if she picks heads, and 1 to 2 if she picks tales for her guesses.

    This is the “thirder” position. The key IMO is asking precisely the proper question for her to consider the 1/3 assessment. Although the odds for tales would be 1/3 based upon the guessing odds and the possibilities based upon both sides of the coin, there can only be two maximum interviews, and only one or two waking times in one coin toss and experiment.

    Of course my argument is semantic, but the correct wording is important to come to the correct conclusion concerning the meaning of anything.

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  80. JP says:

    Daniel: Thanks, I understand your point better now. And I agree that I really mean ” wake on Monday”. Where I now disagree is your assignment of probability 1/3 to each day of the week. Days themselves do not occur randomly, and each day occurs exactly once following each independent random trial. So they all have probability 1, and are not mutually exclusive.

    CB: Probability is what you get by applying the axioms of probability theory. It turns out that it may be calculated as long-run frequency over *independent* trials. My argument is that waking a on different days of the week in the SB problem are not independent.

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  81. Daniel Kerr says:

    JP, the days divide the probability space, from sleeping beauty’s point of view, it’s a random variable. It actually doesn’t matter though, as you could replace my whole analysis with wake|Mon or wake|Tue and you will get the same result.

    Forrest noble, I disagree with the notion of probability you are arguing. You seem to be assuming there is an “absolute” probability at heart here, which in the example is the coin flip, but in practicality is not what we mean when we ask for a probability. A conditional probability is still a probability, and in full Bayesian formalism, every probability is a conditional probability. The problem is asking for the probability from the point of view of Sleeping Beauty. A lot of the times the question is asked, they don’t say “probability,” they ask what belief she should assign to the outcome of the flip.

    Like or Dislike: Thumb up 2 Thumb down 1

  82. JP says:

    Daniel: OK, I guess by “Monday” you mean the event “A particular day that SB was awoken is a Monday.” Fine, then we can create a sample space with the four exclusive elements “Monday, H”, “Tuesday, H”, “Monday, T”, “Tuesday, T,” as per your earlier comment.

    Now the question is how to assign a prior probability to these four events. You have assumed a uniform prior. But SB knows all along that P(Tuesday, T) = 0 and P(Monday, T) = 1/2. The problem gives no principled basis for choosing P(Tuesday, H) and P(Monday, H), but we do know that P(Tuesday, H) + P(Monday, H) = P(H) = 1/2. We then introduce an event “woken” which is true for “Monday, T”, “Monday, H”, and “Tuesday, H.” So, finally, P(T | woken) = P(Monday, T) / (P(Monday, T) + P(Monday,H) + P(Tuesday, H)) = 1/2.

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  83. CB says:

    JP:
    The probability of an event can be calculated by repeated trials of that event, full stop. The trials must only match the conditions of the probability you’re trying to calculate. If they don’t match, then they will differ, but that means either you’re doing the trials wrong, or calculating the probability wrong. In this case, the trials are correct by definition.

    Events not being independent just changes how you have to calculate the probability. It doesn’t mean any event which was dependent on a previous one is ignored.

    If a random number generator produces a 0 or a 1 with equal probability when it produces a number randomly, but every time it produces a 1 randomly it also deterministically produces a 1 the next time it is queried then reverts back to random behavior, you don’t say “well the second 1 is not independent so that doesn’t change the probability of getting a 1 away from 1/2 even though twice as many 1s are being produced as 0s”. Well, you do, but that’s incorrect — as demonstrated by you arriving at the conclusion that this RNG is fair when it clearly isn’t. Sorry.

    Like or Dislike: Thumb up 3 Thumb down 0

  84. JP says:

    CP: We both agree that events not being independent changes how you calculate the probability. In your RNG example, I certainly would not conclude that it’s a fair RNG, as it has serial correlations. In fact, in your example it’s basically meaningless to ask for P(1). You could ask for the probability of a 1 on trial t, P(1 | t), and then hope that it converges as t -> infinity. But this is provably not the case. You could also ask for the expected fraction of 1s, and this I agree is 2/3, but that’s not the same thing as a probability.

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  85. CB says:

    So your new answer to the SB problem is “undefined”? And yes, they’re the same. Every time you query the RNG, you’re asking it “At the time of being queried (awoken), what was the result of the last random selection (coin flip) — which could be either a fresh flip, or the same result as a previous flip as defined by the algorithm/experimental protocol.” And the question “What are the odds that the last coin flip prior to this query was heads?” is the same as the probability that the RNG answers 1.

    The expected fraction of 1s is precisely the same as the probability of a 1. That’s what expectation means. That’s what probability means. There is debate over what probability means (if anything) with regards to a single trial — and I’m certainly in the camp that says that it is meaningful — but what *everyone* agrees is that probability is the same as what you would expect to see over repeated trials as t->infinity.

    I guess except you, with your new “undefiner” stance?

    The expected ratio of heads to tails is 2:1. If you don’t get 2/3rds for your probability, then you did the math wrong.

    Like or Dislike: Thumb up 3 Thumb down 1

  86. Daniel Kerr says:

    JP, your probability assignments are inconsistent. The arguments of your probability functions should follow propositional logic or set theory for the claims you are making. Your arguments clearly are not held to such a logical standard. P(Tuesday, T)=0 implies that given a sample space divided into partitions Heads, Tails, Monday, and Tuesday, the set of elements in Tuesday and Tails is empty.

    Keep in mind you haven’t even defined the “woken” state yet on this probability space and you’ve already assumed that the Tails and Tuesday combination cannot happen. This is equivalent to stating that regardless of Sleeping Beauty’s experience/observations, Tuesday and Tails is not a possible event. P(Tuesday, T) is not conditioned on Sleeping Beauty’s observations, it’s an absolute claim for Tuesday and tails for all possible observers. According to you, not even the experimenters can observe this combination. You’re assuming the conclusion from the very beginning.

    The fact is that all four of those cases each have a probability of 1/4. 1/2 of all events lie in Monday, the other half lie in Tuesday. Likewise for the coin flip. Because these two set of partitions are mutually independent, Heads and Tails splits each of those partitions by 1/2 too. So now when you substitute the correct probability assignments into your equation, you get 1/3.

    Like or Dislike: Thumb up 3 Thumb down 1

  87. JP says:

    CB: You’re right, my comment that P(1) in your RNG example is undefined was hasty. But I certainly have not changed my mind that the answer to the SB problem is 1/2, I just don’t think your P(1) is analogous to the quantity we wish to calculate in the SB problem.

    Daniel: This comes back to semantics, and what exactly you mean by the events “Monday” and “Tuesday.” In the comment you’re replying to I am working under the assumption that “Tuesday” is shorthand for “A specific day on which SB woke up is a Tuesday.” This is how I understand your notation. With this definition it is certainly the case that the intersection of “Tuesday” and “Tails” is empty for all observers, including the experimenter.

    You could instead define “Tuesday” as the event that “A Tuesday occurred during the week following the coin flip.” Then of course the experimenter may see a world containing both “Tuesday” and “Tails.” But then you have to give up mutual exclusion: P(Monday) = P(Tuesday) = P(Monday & Tuesday) = 1 and I’ve already addressed this case in an earlier comment. But if you want to hold that P(Tuesday & T) > 0 *and* P(Monday & Tuesday) = 0 then I fear you’re being inconsistent.

    Or you can simplify the whole calculation, and not try to break the event “SB woke up on a Monday” into “SB woke up” and “Monday.” This straightforwardly gives the answer 1/2, assuming you agree that we’re calculating P(T | SB woke up on at least one day).

    Like or Dislike: Thumb up 0 Thumb down 0

  88. phayes says:

    JP: If I were SB I’d define the events “Monday” and “Tuesday” as “Today is Monday” and “Today is Tuesday”: https://www.writelatex.com/1311844gmhppg#/3221529/

    Like or Dislike: Thumb up 0 Thumb down 0

  89. JP says:

    phayes: OK, but SB’s “Today” = “A day that SB woke up,” so this is the same as my phrasing “A specific day that SB woke up is a Tuesday.”

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  90. CB says:

    JP: Great, progress! Well, they’re exactly the same. Not analogous, identical.

    The odds that the RNG returns 1 on a query is the same as the odds that the most recent bit flip it performed was a 1. It performs a bit flip prior to this determination if 1) the last coin bit was 0 or 2) the last bit flip was a 1, and the RNG has since been queried a second time. Otherwise (on that second query in the case of a bit flip 1), the bit is not flipped again and so the RNG returns the same value as before, which is 1.

    This is exactly what is going on with SB. She is being asked what are the odds that the last coin flip was heads. Not *a coin flipped in isolation*, but the last coin flip prior to her being woken up and queried. A coin flip was performed prior to this question if 1) the last coin flip was tails or 2) the last coin flip was heads, and she has since been queried a second time. Otherwise (on the Tuesday when the Monday flip was heads), the coin is not flipped again, so the value of the coin flip is the same.

    Literally the same except for the words used.

    Remember, the question is not “considered in isolation, what would the odds of a fair coin flip be?”

    The question is “Given these rules, what are the odds that the last coin that was flipped was heads?”

    These are not the same. Since the rules state that the queries are not evenly distributed among coin flip outcomes, the odds are not the same as the coin flip considered in isolation.

    Remember, it’s perfectly clear that the outcome of repeated trials will be heads 2/3rd, tails 1/3rd. Which means that’s the probability.

    It’s just an issue of overcoming what you think the answer “should” be, and realizing that the answer depends on the rules, and the rules change the probability from the “simple” case.

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  91. CB says:

    No, JP, it’s P(T | She woke up). She wakes up twice for heads, once for tails, therefore P(T | she woke up) = 1/3rd.

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  92. JW Mason says:

    Daniel Kerr- I should have mentioned you along with Ignacio as making convincing responses to the post.

    I expect Sean will engage with these arguments at some point.

    Like or Dislike: Thumb up 2 Thumb down 0

  93. CB says:

    forrest_noble:
    So, you feel that the question “What is the most likely outcome of the event?”, which we agree is “Heads, with probability 2/3rd” is different than “What is the probability the event came out heads?”, and that this answer is 1/2 (you specifically said tails but that’s 1-Heads in any case)? You realize that’s self-contradictory, right?

    Like or Dislike: Thumb up 2 Thumb down 2

  94. DN says:

    The halfers and the thirders differ in how they want to assign interview state probabilities (by state I mean, say, Tails and Monday, TM):

    halfers: 1/2, 1/4, 1/4

    thirders: 1/3, 1/3, 1/3

    I suggest the halfers are essentially answering this question:

    “Here’s an experiment. Now pick one interview from the experiment. Which state is it?

    Since, in case of heads, we end up picking one interview and disregarding the other, we get a 1/2, 1/4, 1/4 distribution (and this, trivially, leads to a tails state 50% of the time).

    While the thirders are essentially answering this question:

    “Here’s an experiment. Pick all interviews from the experiment. Which states are they?”

    The experiment produces 0.5 of each interview state, so this gives a 1/3, 1/3, 1/3 distribution (and leads to a tails state 33% of the time). Note that this equality with all being 1/3 is NOT due to them being “indistinguishable, thus equally likely”. It is just a coincidental consequence of the concrete values in the game (fair coin etc.). If the coin was, say, 70% tails/30% heads, then the distribution would be 7/13, 3/13, 3/13 (but the halfers would say 7/10, 1.5/10, 1.5/10).

    To figure out why the halfers are wrong, let’s say we record all the interviews on video. Now the halfers will pick (“buy”) one and only one video from an experiment and then regard this as representing the SB situation.

    But the SB doesn’t “shop” videos (or interviews) in this way. In case of heads, she also stars in the video NOT picked by the halfer, and when woken up, she can’t disregard that video from the set of possible states.

    Let’s consider 4 weeks as illustration:

    Week 1: Tails, halfer buys TM video.
    Week 2: Tails, halfer buys TM video.
    Week 3: Heads, halfer buys HM video. (And disregards the HT video.)
    Week 4: Heads, halfer buys HT video. (And disregards the HM video.)

    Now the halfer says that, see, 2 out of 4 videos are tail videos, so the correct answer is 50% tails. (And he will point out that 2/3 of the Monday videos are tails videos.)

    But the SB, when woken up, will know she actually “was” (or will eventually be) recorded 6 times, even though the halfer only wanted to buy 4 videos (one video pr. experiment). Of those 6 videos, 2 are tails videos, while the other 4 are heads videos. (And half of the Monday videos are tail videos.)

    Another way to put this is that the halfer desire to pick only one video pr. experiment and then wonder about its state is more or less arbitrary and only superficially similar to the actual SB situation (in that, of course, the SB can only star in one video/interview at a time). But the SB actually participates in 1.5 interviews pr. experiment, not just one. And of those interviews, 0.5 are tails, and 0.5 + 0.5 = 1 are heads (and this is what she is effectively being asked about).

    You could say that the halfers are making a sampling mistake in that they calculate as if the SB is only interviewed once, no matter whether the coin lands heads or tails. (This is a little ironic of course, since usually it is the thirders that are accused of sampling bias by the halfers.)

    An explanation somewhat similar to the above was given by Berry Groisman.

    http://arxiv.org/ftp/arxiv/papers/0806/0806.1316.pdf

    Liberally paraphrasing, he distinguishes between two different experimental setups:

    1) Flip a coin, pick one video from the experiment, and show it to your wife. Which state is it?

    2) Flip a coin, pick all videos produced and put them in a box. Keep doing this, and eventually have your wife pick a video from the box. Which state is it?

    Now, the SB situation is not quite identical to either of these setups, but it is pretty much equivalent to 2), since the number of times the SB is interviewed depends on the coin clip. That is, to the SB, in each and every interview situation she is effectively picking from a box of 1.5 interviews, with only 0.5 being a tails interview.

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  95. JP says:

    CB: “No, JP, it’s P(T | She woke up).”

    You’re not giving poor SB enough credit. She has two pieces of information: 1) She woke up today. 2) She may have woken up yesterday. She has to account for (2) in order to arrive at a rational answer.

    Like or Dislike: Thumb up 0 Thumb down 1

  96. Daniel Kerr says:

    JP, All I’m going to say is to be careful with your semantics. My calculation started with the most general use of days of the week and assumed no other structure. I absorbed this in my definition of “wake” and qualified it’s relationship to Monday and Tuesday via conditional probabilities. When I started the calculation, I was neither a halfer or a thirder, this was my first time seeing the paradox. Assuming the most basic probability assignments and nothing extra, I got 1/3. You’re playing a risky game with how you define terms in this problem, short-handing a lot of structure and conditional statements within single words. I was honestly surprised by the 1/3 answer until I analyzed what my calculation implied.

    Like or Dislike: Thumb up 2 Thumb down 1

  97. CM says:

    Let’s say you want to test the problem experimentally. Let’s assume that after running the experiment for 1000 weeks, you have 1000 coin flips – 500 heads, 500 tails – and 1500 interviews, 1000 of them after a heads coin flip, and 500 after tails. Every interview has the same answer, 1/3. What do you do with these numbers?

    Like or Dislike: Thumb up 3 Thumb down 0

  98. Ramanujan says:

    The flip-weighted probability is 1/2. The interview-weighted probability is 1/3. The question is asking for the interview-weighted probability; “Each time [that she is awakened] she is asked a question: …”

    Like or Dislike: Thumb up 3 Thumb down 0

  99. @CB,

    You keep writing the question down wrong. You can understand the question best by quoting it verbatim. Semantics can change the opinion of the listener greatly if just one word is changed. Arguments based upon semantics can be a total waste of time. The wording of Sean’s scenario can by seen below again. You must write it down correctly word for word and then consider it from SB’s perspective.

    “What is the probability you would assign that the coin came up tails.” This is the question that will be asked SB in her interview(s). This is the only question she will be asked. There is no other information given her excepting for how the experiment will be conducted. Remember the coin will only be tossed once for the entire experiment. She does not know that she will be graded based upon her answers. As far as she is concerned one guess is as good as another based upon a “fair” coin flip.

    But if she were asked instead what do you think is the best strategy for you to come up with two correct guesses instead of just one for the experiment, she might realize that heads is the best answer by 2:1 odds, with 1/3 odds for tails being a correct guess in each interview, and 2/3 odds for heads being a correct guess in each interview, considering the possibilities (probabilities) of both sides of the coin.

    Like or Dislike: Thumb up 2 Thumb down 0

  100. Joe Polchinski says:

    Here is perhaps another way to look at this, which may make it clearer [to those to whom it is not already clear]. Suppose we repeat the experiment 100 times, so that there will approximately 100 wakings following a heads and 50 wakings following a tails. Suppose that we ask the question, “What is the probability that the 97th waking followed a tails?” This should be a routine exercise in probability, without any philosophical subtlety. I propose that you give this comment an up vote if you think the answer is [very close to] 1/3, and a down vote if you think it is [very close to] 1/2.

    Thumb up 16 Thumb down 12

  101. Sean Carroll says:

    It’s nice to win the Dirac Medal, but what’s really cool is tricking people into upvoting your blog comment.

    Well-loved. Like or Dislike: Thumb up 9 Thumb down 3

  102. Joe Polchinski says:

    :-) Well, anyone who dislikes the previous comment can downvote this one instead.

    Now, given agreement on the mathematical fact of 1/3 for the previous question, a series of questions:

    1. Suppose we tell SB that it’s the 97th interview. She then has all the information we have, so her probability is 1/3. Of course, this is a different situation, but…

    2. Now suppose that we just tell her that it’s the 65th (or whatever) experiment. So she doesn’t know if it’s the 93rd, 94th, 95th, … interview. But it doesn’t matter: there was nothing special about 97, they are all the same. So her answer is again 1/3.

    3. Now we don’t even tell her what experiment it is. The answer is the same, since the experiments are identical.

    Like or Dislike: Thumb up 4 Thumb down 5

  103. Ignacio says:

    Joe why not make it even simpler and say that you only wake the sleeping beauty if it is heads and never if it is tails. According to the sleeping beauty the coin always comes heads so 100 % instead of 50 %. Your sampling is perfectly biased and the sleeping beauty thinks it is a coin with two heads.

    The Monty Hall case has a similarly simple limit such that you choose 1 door out of a zillion doors + 1. The host opens a zillion doors – 1. Now you have two doors the one you chose at the outset and the only one the host left closed. The chances of guessing right if you switch is functionally 100 % for zillion sufficiently large.

    The entire thing is an empty exercise that people get crossed in their heads.

    Like or Dislike: Thumb up 4 Thumb down 0

  104. Joe Polchinski says:

    Ignacio – The halfer logic is that SB learns nothing new upon waking: she knew that she was going to wake up, and she did. If on tails she never wakes then she does indeed learn something new and all would agree.* The halfer argument has an intuitive appeal, but with probabilities intuition often fails; one must do the math, which shows that this intuition is simply wrong.

    *Well, I suppose there are some halfers who take an even less informed point of view: “The probability that a coin was tails is always 50-50.” Your example should show them that this can be influenced by new information.

    Like or Dislike: Thumb up 3 Thumb down 3

  105. Ignacio says:

    Ok Joe I will leave you to it then but as Sean mentioned your Dirac medal will not help you here.

    Like or Dislike: Thumb up 2 Thumb down 0

  106. Tom Snyder says:

    “Trial”: Flipping the coin and carrying out the waking procedure corresponding to the outcome of the coin toss.

    There are 3 types of “events” to consider:

    E1: waking SB on Monday after heads was thrown
    E2: waking SB on Tuesday (after heads was thrown)
    E3: waking SB on Monday after tails was thrown

    In a large number N of trials, each type of event (E1, E2, or E3) will occur approximately N/2 times. Thus, the relative “trial-frequency “ for each type of event is 1/2.

    In a large number N of trials, there will be a total number M of events that occur and M will be about 1.5 N. These M events make up a “pool” of events. Each type of event (E1, E2, or E3) will occur approximately M/3 times. Thus, the relative “pool-of-events-frequency” for each type of event is 1/3.

    Based on the trial-frequency, the probability that event E3 occurs is 1/2.

    Based on the pool-of-events-frequency, the probability that event E3 occurs is 1/3.

    E3 is the only type of event for which a tails was thrown. When SB is awakened she can assign a probability of 1/2 for tails being thrown if she defines the probability based on relative trial-frequency.

    Or, she can assign a probability of 1/3 for tails being thrown if she defines the probability based on a relative pool-of-events-frequency.

    These are not contradictory assignments of probability. They just use different definitions of “probability that tails was thrown.”

    Suppose we ask SB to say “heads” or “tails” when she is awakened and she knows we will pay her a dollar if what she says corresponds to what was actually thrown. She should say “heads” every time if she wants maximum payout after many trials. This is true even if she adopts the trial-frequency probability of 1/2 for heads or tails to occur. In this case she would state that the probability of tails is 1/2 and the probability of heads is 1/2. But, she knows that if heads was thrown then she will get paid twice in the same trial if she always says “heads”.

    [Suppose you are asked to predict heads or tails for a fair coin toss. I tell you that I will give you 2 dollars if heads occurs and you predicted heads and I will give you 1 dollar if tails occurs and you predicted tails. Clearly, you should always predict heads even though you would assign equal probabilities of 1/2 for heads or tails to occur.]

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  107. Richard says:

    Tom,

    Yes, the coin-flip trial frequency is 1/2 for each outcome. But SB is asked for her credence in the outcome of the flip. That is, given everything she knows (the background information about the experimental procedure, and that she is awake during the experiment), what is her credence that the coin came up heads vs. tails? And that has to be different from the prior assessment of 1/2, because it is modified by the knowledge that she is in one of the three possible events described by “in between being awoken during the experiment and learning the outcome of the experiment”. In those moments (but only in those moments), she must assign probabilities of 2/3 and 1/3, respectively.

    Like or Dislike: Thumb up 1 Thumb down 1

  108. Richard says:

    Modify the experiment as follows: At the start of the experiment, there are four boxes. Two boxes contain 5 pounds of iron. One box contains 5 pounds of gold. One box is empty.

    SB will be awakened both days. Amnesia drug is still given when she goes back to sleep.

    If heads was flipped, then on the first day she will be handed a box of iron, and on the second day she will be handed the other box of iron. If tails was flipped, then on the first day she will be handed the box of gold, and on the second day she will be handed the empty box. The empty box weighs less than a pound.

    SB is awakened and given a box. She can tell that it weighs several pounds. What is the probability that the box contains the gold?

    Like or Dislike: Thumb up 1 Thumb down 1

  109. Sleeping Beauty Problem is Classical says:

    The answer is 1/3, I think its pretty obvious.

    But this has absolutely nothing to do with Quantum Mechanics.

    Like or Dislike: Thumb up 3 Thumb down 5

  110. Logicophilosophicus says:

    Instead of waking/asking, suppose we tell SB that we will put a token in a jar if the result is x; if it is y we will put a token in the jar, later taking it out but immediately putting it back again. Clearly examining the contents of the jar at any moment (ignoring the out-in singularity) tells her only that the trial has been completed, and gives no extra information about the result.

    There are not three separate states offering evidence. I’m a halfer.

    Like or Dislike: Thumb up 1 Thumb down 1

  111. Tom Snyder says:

    Richard,

    I agree that the “thirder” specification of probabilities is valid. It’s just that I have some doubt whether it is the only valid way for SB to specify the probabilities.

    SB knows that in a large series of repeated trials:

    1/3 of all awakenings will be “tied” to tails and 2/3 of all awakenings will be tied to heads.

    1/2 of all trials will be tied to tails and 1/2 of all trials will be tied to heads.

    A “thirder” is drawn to the first statement and takes the “probability of tails” to be 1/3. As I see it, by this she means nothing more and nothing less than that 1/3 of all awakenings are tied to tails.

    But why can’t SB be a halfer and take the “probability of tails” to be 1/2? By this, she means nothing more and nothing less than that half of all trials are tied to tails.

    It seems to me that both assignments of probability are valid. They correspond to SB adopting different definitions of “the probability of a tails”. You can start with the thirder probability assignments and deduce the halfer probability assignments. And vice versa.

    I believe both a thirder and a halfer would agree on the answer to any “objective” question that can be asked regarding the scenario. For example, take this question:

    Every time SB is awakened we ask her to predict whether heads or tails was thrown. What should she say in order to maximize the number of correct predictions?

    You can easily show that both the thirder and the halfer predict that, in a large number of trials, there will be twice as many awakenings tied to heads as awakenings tied to tails. So, they both agree she should always say “heads”.

    I don’t see how one could prove that the thirder interpretation of “probability of tails” is the one and only “correct” interpretation. If it is, in fact, the only reasonable interpretation, then it would not change the content of the problem to explicitly spell out this interpretation in the statement of the problem. Then, I bet almost everyone would agree on what SB should give as an answer to the question.

    Anyway, as I see it there is an ambiguity in the meaning of “probability of tails” in the original wording of the problem. And that allows one to be either a thirder or a halfer. But everyone should agree on the answer to any “objective” question!

    Like or Dislike: Thumb up 4 Thumb down 0

  112. germo says:

    Joint distribution (probabilities x,y,z are unknown at first and x+y+z=1):
    P(wakeup, monday, heads) = x
    P(wakeup, monday, tails) = y
    P(wakeup, tuesday, heads) = z
    P(wakeup, tuesday, tails) = 0

    By the problem statement we know that:
    P(heads)=1/2, P(tails)=1/2
    which implies (by definition of marginal probability) :
    x + z = 1/2 => x=z=1/4 and y=1/2

    So we can simply calculate the required probability:
    P(tails|wakeup) = P(tails, wakeup)/P(wakeup) = P(wakeup, monday, tails) = 1/2

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  113. Daniel Kerr says:

    Germo, be careful with your probability assignments, I would define P(day, flip) first then define wake conditioned on them. When skipping steps it’s easy to miss insight.

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  114. Richard says:

    Tom,

    Since the question to SB is what probability *she* would assign, then we can’t really count trials as counted by the experimenter. We have to count what SB is cognizant of. For her, in her quest to hypothesize what she can about the current state of affairs, she sees a “trial” as an instance of being awake. The semantics of the experimenter, who sees a trial as a coin toss followed by a protocol dictated by the toss’s outcome, does not apply.

    Like or Dislike: Thumb up 1 Thumb down 3

  115. Tom Snyder says:

    Richard,

    When you say “she sees a ‘trial’ as an instance of being awake”, then it looks to me that you are redefining the meaning of “trial” from how I was using the word so that it would correspond to what I was calling an “event”. This could get confusing. I believe SB would understand and be able to use the word “trial” as I was using it.

    May I ask you to give your definition of the phrase “probability *she* would assign for a tails” so that there is little or no leeway for interpretation of the phrase? If the words used in the statement of the problem do not have the same meaning to everyone, then it is not surprising that there will be differences in opinion on the “correct” answer to the problem. I’m not trying to be difficult, I’m just struggling to get to the heart of the problem. At this point, I believe that much of the disagreement between the halfers and the thirders lies in an ambiguity of “probability” as used in the problem statement.

    Like or Dislike: Thumb up 3 Thumb down 0

  116. Logicophilosophicus says:

    Tom Snyder et al,

    You’re right that there is a minefield for misunderstanding probability. (What IS an a posteriori probability?) But that has nothing to do with this puzzle.

    Before the experiment, SB is told the protocol. She knows at that moment that she will be woken and asked the question. She knows no more when she is actually woken. The thirders are therefore claiming that the protocol predetermines the best estimate of the probability of the toss. (Knowing the protocol, if SB accepts the thirder “logic” she can say: “Save yourself the trouble of doing the trial – I know ALREADY that the answer will be 1/3…”)

    In “The Hunting of the Snark” the Bellman nonsensically states: “What I tell you three times is true.” Are the thirders claiming that “what I tell you twice is twice as likely to be true”?

    Well-loved. Like or Dislike: Thumb up 6 Thumb down 1

  117. phayes says:

    “She knows no more when she is actually woken.”

    Of course she does. She knows that now it’s either Monday or Tuesday.

    Like or Dislike: Thumb up 2 Thumb down 3

  118. Logicophilosophicus says:

    Phayes,

    She did know it was going to be “either Monday or Tuesday” – it was in the protocol. She has nothing new requiring or enabling her to update her information about the trial result.

    Like or Dislike: Thumb up 2 Thumb down 1

  119. phayes says:

    Logicophilosophicus,

    More precisely: she knew she was going to wake up knowing that it’s either Monday or Tuesday. Her being told the protocol beforehand gave her enough information to model the trial and deduce [what] her inference [should be] during it.

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  120. DN says:

    Richard,

    I am a thirder and all, but I think that your example with the gold and iron boxes doesn’t quite capture the intricacies of the SB experiment. Something “weird” or deceptive is going on in the SB experiment due to the drugging + repeated interviews and due to the 100% chance of interview when tails as well as when heads.

    Slight change of setup:

    1)
    If tails, we produce gold bags. (according to some protocol)
    If heads, we produce stone bags. (according to some other protocol)

    2)
    After this, we reveal our production (the bags, not their contents) to you according to one of three protocols:

    A) (“You asked for a bag.”) OK, here is one of the bags, in case we made any bags.

    B) Here are ALL the bags we produced.

    C) We show you ALL the bags we produced, but one by one by using the drugging procedure.

    What gold odds do you give during the reveal?

    Under a lot of different production protocols, the choice of revealing protocol would not matter and nobody would dispute the heads and tails odds for a revealed bag(s). But in the SB case, the revealing protocol matters a lot, and I think this is what is hard to get an intuitive grasp on (when you are more used to operating in your mind with certain revealing protocols).

    EXAMPLE 1.

    If tails, we produce one gold bag 10% of the time.
    If heads, we produce one stone bag 20% of the time.

    Here the revealing protocol doesn’t matter, and I think that nobody disputes that a reveal event (in case anything is revealed) is going to be one bag with 1/3 gold odds. But notice how protocol B and C was the same as A only due to the fact that no more than one bag would be produced.

    EXAMPLE 2:

    If tails, one gold bag 10% of the time.
    If heads, 2 stone bags 10% of the time.

    Here the reveal protocol matters.

    A gives 50/50 odds, in case we are shown anything.
    B reveals one bag or two bags, in case we are shown anything. Of course, the last reveal looks entirely different than the first.
    C reveals one bag at a time (with the 10% proviso), 1 bag in case of tails, 2 bags in case of heads.

    I think some halfers may intuitively agree here that each reveal under C is going to be 1/3 gold. The point is that (as in example 1) the mind is led to notice that we may not always see a reveal, so we begin to think in frequencies.

    Example 3:

    If tails, one gold bag 10% of the time.
    If tails, one stone bag 10% of the time, 1 gazillion stone bags 10% of the time.

    A: 1/3 gold, in case we are shown anything (this is like in example 1 since we only ask for one bag to be shown pr. experiment).
    B: 1/3 of the time we are shown one gold bag, 1/3 of the time we are shown one stone bag, 1/3 of the time we are shown a gazillion stone bags.
    C: Might a halfer agree that any one reveal under C is going to be ~0% gold?

    Example 4:
    If tails, 1 gold bag 100% of the time.
    If heads, 2 stone bags 100% of the time.

    This is the SB situation.

    Here it is my contention that halfers are somehow seduced by reveal protocol A due to the fact that a bag is now going to be revealed. Protocol A is now similar to saying “show me a bag” (while in the other examples, this is meaningless to say without the “in case any was made” restriction which leads us to think in frequencies).

    So when the SB problem talks about an unspecified interview (at least one of which we know is bound to happen), halfers feel justified to think in terms of A, since this (coincidentally) always produces one reveal. “Since isn’t this what the SB problem is all about, one reveal (at a time)?”

    Only it is not ONLY one bag revealed as pr. experiment description. We are INDISPUTEDLY asked to use reveal protocol C, which is some kind of hybrid between B and A. First we can say we give a middleman all bags produced (as in B), and THEN he reveals all of those to you by the drugging procedure.

    I don’t really know what else to add (but all the different ways to explain this are surely helpful). If you look in discussion threads about the SB problem, practically all the halfer math presented (rather than just the “no new info” argument”) is based on using reveal protocol A, only one reveal pr. experiment. They justify this sometimes by saying stuff like “we need to split the pie in case of heads” which is really a tacit admission that they are now using protocol A.

    They also sometimes give the “repeated bet” argument, that we somehow have to adjust for the repeated reveal in case of heads, which is, again, just another way to say that we should convert to protocol A rather than stick with protocol C.

    I suggest trying to consider EXAMPLE 2 above. Hopefully a lot of halfers would agree that reveals here are going to be 1/3 gold, since we “have” to think in frequencies. Then one may note that the SB experiment is more or less the same as example 2, except we now choose to produce bags more of the time (but still with the same ratio).

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  121. Daniel Kerr says:

    Logicophilosophicus, nobody’s arguing that the number of times the question is asked affects the probability of the flip. The thirder position is arguing that if the number of times you are asked is dependent on the outcome, then you’ve been given more information about the outcome. Half of the time it’s tails, they are not bothering to ask you while every time it’s heads, you are being asked.

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  122. DN says:

    The doctor has a patient. The patient is sick with 50% probability, well with 50% probability.

    If sick, the patient will visit the doctor 99 times in a year.
    If well, only one time.

    The doctor tells his assistant all this by year end.

    The doctor then says: “Here are the notes from one of the visits.”

    What are the odds that the notes are from a visit where the patient was well rather than sick?

    *ALERT* The assistant can’t tell yet, since he doesn’t know the condition for being shown the notes…

    “Here are the notes from one of the visits, as you requested.”

    Then it’s 50/50.

    “Something peculiar happened at one of the visits. Here are the notes from that visit.”

    Then it’s 1/100 (with obvious nitpicking provisos).

    “My dear assistant, I will show you the notes from all the visits, using the infamous SB drugging procedure.”

    Then it’s 1/100, but halfers will say 1/2 since they conflate this scenario with the “request” scenario.

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  123. phayes says:

    Here’s another puzzle in which the distinction between an inference and an inference about an inference is important. Without it, it appears the islanders learn nothing they didn’t already know when the Guru speaks, and the puzzle can remain somewhat puzzling even when the solution is known.

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  124. Scot Cannon says:

    Well if SB is rational she will realize that there was only one coin flip and the odds are the same for it being heads or tails, so a 50% of it being tails. What are the chances of SB being that rational and not buying into that screwy thirder logic? You start getting into nonlinear processes now.

    A real life example of non-rational or nonlinear thinking is the whole dark energy question. That logic goes like this we do not understand the expansion of the universe from what we are seeing so let’s invent dark energy, which we really do not understand, to explain it, just to not be arguing from ignorance.

    Now nonlinear solutions can be rational also and if there is more than one possible conclusion from a set of facts, the most you can say is about the probability of any given solution.

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  125. DN says:

    phayes,

    yes, that blue eyes puzzle is worth keeping in mind to remind us how easily one can overlook that new information is available.

    I find it particularly easy to see that new information is available when you consider the case with N=2 blue-eyed islanders. Then one of them now knows that the other one knows that there is at least one blue-eyed person. Which he (the first one) didn’t know before the Guru spoke.

    You can then explain the step from 2 to 3 with some expectation of acceptance to get things rolling, and from 3 to 4.

    What I still can’t quite figure out about this puzzle is whether it is enough to have the islanders think “I know that he knows that” and then the induction logic will be enough on its own for the islanders (and you), of if you actually need to deal with a cascade of “I know that he knows that he knows that…” unfolding for each day passing in order to solve the puzzle.

    http://en.wikipedia.org/wiki/Common_knowledge_(logic)

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  126. DN says:

    Another slightly tricky “did we obtain new info” example:

    You know one ball out of 12 weighs less or more than the other 11 that are all identical.

    You split them up 6 and 6 on a lever weight.

    Will you learn something new?

    (The challenge of the full puzzle is to not only identify the “fake” ball in only 3 weightings, but also to tell if it is lighter or heavier than the others.)

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  127. phayes says:

    DN,

    Well I would say that the induction steps occur in the modelling – the inference about the inference – which each blue eyed islander makes and so shouldn’t be considered the knowledge learned in the ‘real’ setting. So what the ‘real’ islanders learn from the guru is just that they actually are in the setting of the puzzle, and each blue eyed islander deduces that if the 99 others they can see are still there on the 100th day, their own eyes must be blue too.

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  128. Scot Cannon says:

    Sean Carroll’s modified SB story where he inserts atomic spin for it’s quantum implications, but it is actually not a quantum problem as there is no imaginary dimensional quanity, so no need to square the wave function to get the probability. Could not follow his logic about the observers, where exactly are they and what they are observing? Not clearly defined and there is another problem of why he put an up arrow on the top lower Tuesday rung.

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  129. DN says:

    phayes,

    thanks, that’s a great way to put it, I think. I usually have liked to point out that we somehow “need something to get the induction rolling”, but it might be better to paraphrase the situation after the Guru speaks as:

    “Oooops, so now it certainly seems like I am in the blue-eyed puzzle scenario with all the well-known consequences. I damn better wait and see whether those 99 blue-eyed islanders around me are here on day 100″.

    The concise proof given here:

    http://terrytao.wordpress.com/2011/04/07/the-blue-eyed-islanders-puzzle-repost/

    also doesn’t make any use of any higher order “common knowledge”.

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  130. Brian Beverly says:

    I could have sworn I just submitted a comment. Does this blog have a glitch? What is the probability that a coin flip is tails?

    Answer: 1/2

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  131. Logicophilosophicus says:

    Phayes, Daniel Kerr,

    Certainly the result of the trial is encoded in the number of times the wake-question routine is run; unfortunately that information is explicitly withheld from SB – else the puzzle evaporates. The thirders thinking is that since SB will be asked twice after a heads result, being asked on any one occasion is twice as likely to occur after a heads result as opposed to a tails result, which is true; but this is then bizarrely supposed to imply “possible universes” where the heads-result universes outnumber the tails-result universes 2:1. The experimenters have created a multiverse of sorts by a whimsical Groundhog-Day choice of protocol.

    Meanwhile, in any one universe – and, to those of us who are multiverse-averse, there is only one universe – the trial was always a 50-50 proposition. There is no elimination of an outcome à la Monty Hall, no extra information.

    Here is an easy non-Groundhog-Day parallel. The Unsleeping Beauty is told that a coin will be tossed. If the result is heads, the words “Trial Completed” will be written down twice, i.e. on two separate pieces of paper. If tails, only once. UB will subsequently be given a piece of paper on which the magic words appear, and asked to assign an a posteriori probability to a tails result. If the result was tails, the two pieces of paper for the heads result just don’t exist, they don’t get to vote if you like, except in the multiverse. In fact it is just such paradoxical results that cast doubt on the multiverse concept.

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  132. phayes says:

    Logicophilosophicus,

    Well I’m averse to multiverse mind-projections too and I’d be mortified if I thought the “thirder” position, if correct, somehow gave support to the ‘reality’ of some of the concepts used to arrive at it. But there is no need to worry about that, and resist accepting the “thirder” S.B. solution itself, because a single universe, ‘pure’ probability calculation (like the one I posted upthread) also arrives at the “thirder” position.

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  133. Tom Snyder says:

    The SB “problem” is not really a problem. It’s a very interesting opinion poll.

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  134. DN says:

    The experimenter tells SB:

    “Next week, I am going to wake you up once, and kill you 5 mins later, the day of the week determined by the roll of a dice.”

    SB credence in being killed on any one of the 6 days: 1/6

    SB is put to sleep, then woken up.

    “SB already knew she was going to be woken up for sure, so she has no new information. Hence, SB credence in being killed on any one day of the week is still 1/6. Specifically, she has credence 1/6 that she is going to be killed today.”

    Surely this is sound reasoning!?!? ^^

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  135. DN says:

    Logicophilosophicus,

    In your example with the pieces of paper, there is a subtle change compared to the SB problem. Your problem is an analogy to a change in the SB problem such that the SB is awakened on either Monday or Tuesday, not both. In that case, the odds are 1/2 (trivially).

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  136. Logicophilosophicus says:

    Phayes,

    I think you are missing something: the three waking-askings, and the three slips of paper, and (in my earlier version) the three tokens simply do not coexist. There is therefore no meaning to the idea of drawing from that non-existent population.

    Of course, in a multiverse they coexist; but then we are excluded from drawing from them. And, if we want to be really silly, we can imagine poor SB trying to factor in all the alternative realities where she misremembered the protocol, or the experimenters misexplained it, or the universe just popped into existence on Monday night. Everything happens in an infinite multiverse: here be dragons – don’t go there.

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  137. Logicophilosophicus says:

    DN,

    You’re right, of course, that there is a difference, but it is not useful to SB.

    a) The proposition “The result was heads” is encoded as (or has the same “truth value” as) paired waking-asking episodes.

    b) If it’s Tuesday, the evidence has been totally destroyed, of course. If SB knew it was Tuesday, she would reason: “There is a 50% chance I was woken yesterday, dependent on a 50-50 coin toss.”

    c) If it’s Monday, the evidence will be destroyed before it becomes evidence, since there are not (yet) paired or unpaired waking-asking episodes. If she knew it was Monday, she would reason: “The evidence isn’t in yet – the coded message isn’t complete – so all I can say is that a coin toss is a priori 50-50.”

    d) If she doesn’t know whether it’s Monday or Tuesday, she can reason: “I suppose I should take some kind of weighted average of the Monday and Tuesday situations – but whatever weighting I choose, it’s going to come out at 50%.” (Just to be clear: the weighted average the thirders want is that 2/3 of all possible waking-asking episodes are tied to heads; fine, 2/3 of 50% + 1/3 of 50% = 50%…)

    Destroyed evidence is here entirely equivalent to withheld evidence i.e. no evidence.

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  138. DN says:

    The SB experiment is tweaked to consist of 1000 weeks, 1000 coin flips, etc. The SB is never told which weak it is when interviewed.

    Everybody agrees that a random awakening is now 1/3 tails?

    The SB now applies halfer logic, though.

    1. “If this awakening is in week 1, then tails credence is 50%. If week 2, also 50%, and so on and so forth.”
    2. “Since the credence is 50% in every week, and every week is equally likely, the credence of my current awakening is 50%.”

    Of course, this reasoning is wrong. The credence is not 50% in week 1, or in any other week. Indeed, why should it matter if she is suddenly told “by the way, this is week 267″?

    This example illustrates what can happen if you, for some reason, try to insist on other odds than the “long-term betting odds”. Like saying “this is only one week, where heads, or tails, but not both can happen, hence the odds are different than the long term odds.”

    If you do that, it’s trivial to expand your “one time only” experiment to a “many times” experiment that has the same odds as your “one time only” experiment, leading to a contradiction with already established “long-term betting odds”.

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  139. DN says:

    Logicophilosophicus,

    You write: “If SB knew it was Tuesday, she would reason: “There is a 50% chance I was woken yesterday, dependent on a 50-50 coin toss.””

    Knowing it’s Tuesday, she surely gives a 100% chance to being woken yesterday?

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  140. Logicophilosophicus says:

    DN,

    Sorry, my last was misstated. I was conflating two versions of the puzzle – my (b) belongs in the interesting SB-not-woken-until-Tuesday-if-tails version: worth discussing, but we weren’t.

    In the original version, b) should read “… So it was heads – probability therefore not at issue.” (Hence my “What is an a posteriori probability?” remark.) SB can lay odds against an undetermined result, but that does not apply here.

    I don’t feel I’ve expressed that very well. The real issue (for thirders) is that the population of three waking-askings does not exist.

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  141. Daniel Kerr says:

    Logicophilosophicus, are you really surprised by the result that probabilities might vary depending on the observer? In this case we have an absolute universe where the probability is 1/2, but every “measurement” sleeping beauty takes, she’s biased simply by the methodology of that measurement. That’s all this puzzle is getting at. Sleeping Beauty’s universe is different than the experimenters, the good thing is that it’s strictly a subset of absolute universe here because her universe is constructed via biased sampling on the original.

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  142. DN says:

    Logicophilosophicus,

    I agree with your revision:

    If Tuesday, then it’s 100% heads.
    If Monday, it’s 50% heads.

    Now, as you suggest, we can do a weighted average. And without giving precise odds for the day, we can at least agree that the chance for Tuesday is more than 0%. And then the weighted average trivially sums to more than 50%.

    So if one wants to save the 50% credence, one ends up having to claim another credence on Monday, before the coin is tossed.

    Some people do this (and claim 2/3 Monday chance), but this can only work if you change the protocol of the experiment to be either Monday or Tuesday, but not both (in case of heads). And this would require a coin flip before the wakening up decision on Monday, unlike in the real SB experiment.

    The real issue (for thirders) is that the population of three waking-askings does not exist.

    I don’t see this as a problem at all. Consider this provocative revision:

    If heads, 1 awakening.
    If tails, 1 awakening.

    Here, as you say, the population of two waking-askings does not exist. Still, we have no problem giving 50/50 odds.

    I agree that it’s a confusing issue that the SB is waked-asked more than one in case of heads. But once you do the math then, it’s going to be 1/3 for each waking-asking (but only because it is a fair coin etc.).

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  143. Logicophilosophicus says:

    DN,

    First, your “provocative revision” – there is “no problem” because we assign the 50-50 odds without any consideration of the awakenings. We assign it because the coin has two sides, and from that population of two we are going to select one at random.

    Back to the awakenings, there is no population of three.

    Even if there had been such a population, the two awakenings for heads are not even partially independent, so there is no reaching into the jar twice: two of the candies are stuck together. But there is no jar. The only choice was made by the toss of a coin.

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  144. Logicophilosophicus says:

    I could have made this a lot easier by pointing out that this was (in a sense) resolved decades ago after heated correspondence relating to one of Martin Gardner’s “Mathematical Games” columns in Scientific American. The puzzle was (as near as I recall): “A woman has two children; one is a boy; what is the probability that the other is also a boy?” The end result of a discussion somewhat parallel to the above was that “the” answer depends on whether we select the women and look at their families, or select the boys and look at their families. It is assumed that both children are available for classification. In the SB problem the “children” are wake-ask and (tails + Tuesday) don’t-wake-ask. The amnesia requirement guarantees that the ambiguity is hidden and therefore unresolvable. We don’t get to pick on the basis of the kids, just the mothers – heads or tails.

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  145. DN says:

    Sure, the boys puzzle depends on whether you specify the boy:

    1. “One of the children is John, a boy.” 1/2

    2. “At least one of the two children are a boy.” 1/3

    Anyway, the SB puzzle. The point of asking about the provocative

    If heads, 1 awakening.
    If tails, 1 awakening.

    case is to make it obvious that we are talking about possible worlds, not realized worlds. I claim that the reasoning to reach 50% here is:

    1. How many awakenings are produced by the protocol, on average? 0.5*1+0.5*1 = 1
    2. How many tails awakening are produced, on average? 0.5*1 = 0.5
    3. Ratio of tails awakenings is 0.5/1 = 1/2.

    Consider instead a roll of a dice:

    1: 1 awakening.
    2: 1 awakening.
    3: 0
    4: 0
    5: 0
    6: 0

    What are the odds for an awakening to be “even”? You can’t just say 1/6 now, of course, we HAVE to do something apart from just looking at the default 1/6 odds from the dice roll.

    1. How many awakenings are produced by the protocol, on average?
    —— 1/6*1 + 1/6*1 = 2/6
    2. How many “even” awakenings are produced by the protocol, on average?
    ——- 1/6*1 = 1/6
    3. What is the ratio?
    ——- (1/6)/(2/6) = 1/2

    And the SB case:

    If tails, 1 awakening.
    If heads, 2 awakenings.

    1. How many awakenings are produced by the protocol, on average?
    ——– 0.5*1 + 0.5*2 = 1.5
    2. How many tails awakenings are produced by the protocol, on average?
    ——– 0.5*1 = 0.5
    3. Ratio of tails awakenings to all awakenings:
    ——– 0.5 / 1.5 = 1/3

    When SB is awoken, she can’t just ignore the protocol because “durdle, durdle, coins are 50/50″. She has to consider carefully what happens in the full experiment.

    When you add prediction shares (betting) to the puzzle, the math becomes increasingly clear. Upon awakening, how much could she pay for the share “This share pays 100$ if the coin was tails”? If she pays more than 33$, she is going to lose money. Simply because only 1/3 of the awakenings are tails, and not 1/2.

    I have given a concise proof why the odds could not be 1/2. (http://www.preposterousuniverse.com/blog/2014/07/28/quantum-sleeping-beauty-and-the-multiverse/#comment-7295910552604296032)

    The “only” way to circumvent this proof is by changing the experimental protocol so that the SB was awoken either/or, not both. But the protocol is specifically designed so that she is awoken more often in case of heads than in case of tails – hence why there are more heads awakenings on average.

    Consider a dice again:

    1: 0.5 awakenings.
    2: 0.8 awakenings.
    3: 2.3 awakenings.
    4: 0.2 awakenings.
    5: 7.2 awakenings.
    6: 1.7 awakenings.

    Now halfer logic is tantamount to saying that we should “normalize” like this:

    1: 0.5 awakenings.
    2: 0.8 awakenings.
    3: 1.0 awakenings.
    4: 0.2 awakenings.
    5: 1.0 awakenings.
    6: 1.0 awakenings.

    It’s just not warranted, and from a betting perspective it becomes completely absurd.

    Final example. Compare

    if tails, 0.5 awakenings
    if heads, 5237 awakenings

    to

    if tails, 0.5 awakenings
    if heads, 1 awakening

    Do you really want to give equal tails credence in these two cases? Also when you consider betting? If yes, what is your tails credence?

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  146. Ziggy says:

    Apologies if I’m being repetitious here, it’s a long thread. With issues of probability it is important to look carefully at the definition. To define P(X|Y) we repeat the experiment many times, count the number of occurrences of Y and of X AND Y, and P(X|Y) = N_(X AND Y) / N_Y. Here, by a reading of the problem, X = “Aurora is interviewed”, and Y = “the most recent flip was tails”. There seems to be general agreement that with this definition P(X|Y) = 1/3.

    The halfer position seems to be that X = “the coin is flipped”, call this X’, that P(X’|Y) = 1/2 (true), and that P(X|Y) = P(X’|Y) because X and X’ are equivalent (no new information). But the halfer position is self-contradictory, since P(X|Y) is not = P(X’|Y). So X and X’ are not equivalent, “I will wake up” and “I have been waked” are not the same. The blue-eyed puzzle is indeed a nice illustration of new information that is not at all intuitive.

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  147. Logicophilosophicus says:

    DN,

    The boys thing actually goes like this:

    a) We can assume that the family is selected from the population of all families; the one-is-a-boy statement is then generated.

    b) We can assume that only those families with boys form the population from which we select – the one-is-a-boy statement is built into the protocol.

    The assumption is that we can go on to see the other child while keeping track of the already identified child – no problem with identical twins, for example… But in the SB problem that assumption does not hold good, and the twins are always impossibly identical (SB can have no cue that she has, for example, been involved in the trial for three days rather than two, no sense that her perfume has worn off, her hair needs washing, her nails actually do just need cutting – so identical that, in the multiverse, being indistinguishable they are actually one and the same I suppose). The first phase, waking-1, is not only indistinguishable from waking-2, SB is not allowed to know whether it took place already.

    Anyway, let’s not get bogged down. The thirder reasoning is that a waking chosen at random is more likely to be from a pair than a singleton. This statement could only be verified by looking at both: she can’t. The further inference to the likely result of the coin toss depends on that intermediate step. So, no, 1/3 does not work for me.

    If the result is heads, we will tell you twice that the coin has been tossed, but you won’t know we’ve told you twice, you’ll only know we’ve told you. You will know the a priori 50-50 probabilities. Go figure.

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  148. Daniel Kerr says:

    Logicophilosophicus,

    Let’s change the problem slightly. Let’s say if the result is heads, she wakes only on Monday and is asked what the result was. If the result is tails, she wakes every day for the rest of eternity and is asked the result. What belief should she assign to the result of the flip? Keep in mind, we’re not asking for the probability of the flip, we know she believes it’s 1/2, she’s not dumb. We’re asking for the result of that particular flip.

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  149. Logicophilosophicus says:

    Daniel Kerr,

    Memory-wiped then woken, we suppose. For the sake of avoiding a separate discussion about infinite-set paradoxes, let’s say she is woken every day for fifty years after a tails result. After even ten years she will be certain that it is not just next Monday, and that certainly highlights one of the sources of error in the puzzle – that the wakings cannot be identical.

    Anyway, back to 2 indistinguishable wakings for, let’s say, heads, since that’s the version I’ve been assuming. We combine the SB puzzle instructively with its very different version, Bertrand’s Box. After the coin-toss trial, the experimenter picks up one of two boxes, looks in, and according to the protocol selects a gold coin, which he gives her. The protocol specifies that one box contains two gold coins, the other a gold and a silver. If the result was heads, the GG box must be picked; if tails, the GS box. Then the twist: if heads, SB is memory-wiped, the first coin is hidden, and she is given an apparently identical repeat performance, where the coin is of course the second G, though she cannot know this. Her rational thoughts at any particular waking are: “A 50-50 coin toss determined which box is picked; the experimenter’s selection of a gold coin was MANDATORY and therefore devoid of information; I have no reason to update the 50-50 expectation yet.” The difference with the original Bertrand’s Box was that there was a random pick of box and a random selection of coin, and therefore (and for no other reason) we could draw a statistical conclusion from the fact that a silver coin was not selected. In the SB version, a silver coin cannot be selected – the protocol bars that source of information.

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  150. Peter McLoughlin says:

    Dr. Carroll
    I was just reading a Salon article where Noam Chomsky claims that we had yet another near run in with worldwide nuclear annihilation with the Bin Laden raid. Add to that the India/Pakistan crisis of 2003, a misidentified Norwegian rocket in 1996, NATO’s able archer war games in the fall of 1983, and of course the Cuban missile crisis in 1962 and we have a lot of near misses with extinction. Is our string of luck just that or is the many worlds interpretation right and we have been conducting a worldwide quantum suicide experiment since at least 1945. What are your thoughts on our string of luck.

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  151. DN says:

    Here’s another example that demonstrates that the “no new information” argument is flawed.

    1. If tails, 50% chance of an awakening.
    2. If heads, 100% chance of an awakening.
    3. If SB is not awakened, we let her sleep and start over the next week with a new coin flip. We carry on until SB is awakened. (SB knows this.)

    The first two rules alone give a 1/3 tails chance. And I submit that they also give a 1/3 tails chance when rule 3 is added. These rules have been specifically chosen for making this 1/3 claim “obvious” and hopefully easy to agree with.

    However, halfers can argue that when awakened under just the first two rules alone, she gets new info about the outcome of the 50% flip after tails, so it’s “ok” that she now has 1/3 credence rather than 1/2.

    However, with the addition of rule 3 in the protocol, SB knows that she will be awakened 100% of the time after experiment start. Hence, “she gets no new information” according to the standard halfer argument.

    So either:

    A) The halfer must insist on also 1/2 tails credence under this 3-point protocol, or
    B) The “no new information” argument must be flawed (or for some reason only apply in the original SB protocol and not in this 3-point protocol).

    I think the choice is pretty easy. And we have seen many others point out that “I will be awake (under such and such conditions)” is just not the same (as much) information as “I am awake right now (under such and such conditions)”.

    I gave another absurd example here:
    http://www.preposterousuniverse.com/blog/2014/07/28/quantum-sleeping-beauty-and-the-multiverse/#comment-7295910552604296023

    [Logicophilosophicus, it seems we are not really addressing each other’s arguments, so I agree to disagree.]

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  152. Daniel Kerr says:

    Logicophilosophicus, since you’re not using probability theory to frame your arguments, I am having a hard time following your reasoning here. For your analysis to hold, it seems to me that SB cannot know P(Wake|Day) for any day, for if she did that would the information in itself. So phrased another way, you’re assuming SB is daft and doesn’t know the setup of the experiment, yet according to the problem statement she is clearly aware of the setup. The calculation that results in 1/3 probability for heads is not a mistake, it’s derived given SB’s knowledge of P(Wake|Day), which you seem to be contesting that she should know that.

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  153. Logicophilosophicus says:

    D Kerr et al,

    Well I guess that makes me daft, since it is my reasoning I project onto the hypothetical SB.

    A coin is tossed. A result slip is printed reading “RESULT HEADS” or “RESULT TAILS”. This is explained to a statistico-psychological guinea pig, as is the fact that he will be hypnotized to forget his first update in the Heads scenario.

    The update consists of a statement by the experimenter: “There is at least one letter E on this results slip.” In the case of heads, he will be told twice, but cannot know whether this has happened yet, of course.

    If the experimenter had to stick a pin at random in the text, and chanced upon an E, the thirder conclusion would be correct. But the experimenter was mandated to make the E statement, therefore no new information.

    The remarkable conclusion of the SB-thirders is that the constraint (only an E statement allowed/mandated) is devoid of statistical information, since they still come up with 1/3. Actually it is an astounding conclusion. The Monty Hall problem hinges entirely on the fact that Hall is constrained – replacing the constraint with random selection by MH leads to a thirder solution; adding the constraint leads to a halfer solution.

    In any other probability problem a key issue is random vs non-random, and it is always a game changer. What is different about SB? (I’ll suggest an answer to my own question: SB-thirders seem to think that the two wakings in the heads timeline and the single waking in the tails timeline are not only psychologically indistinguishable, but are [therefore???] of equal probability 1/3 – rather than the more justifiable 1/4, 1/4 and 1/2 – and are all simultaneously available for random selection like some kind of Three Card Monte.)

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  154. DN says:

    Logicophilosophicus,

    You write:

    SB-thirders seem to think that the two wakings in the heads timeline and the single waking in the tails timeline are not only psychologically indistinguishable, but are [therefore???] of equal probability 1/3 – … – and are all simultaneously available for random selection like some kind of Three Card Monte.

    Well, not quite. The calculation is like this:

    Number of interviews pr. trial: 1.5
    Number of tails interviews pr. trial: 0.5
    Ratio: 0.5 / 1.5 = 1/3

    If the coin had been 70/30 tails-heads, the math would have been 1.3 in total, 0.7 tails, and 7/13 chance of it being a tails interview.

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  155. br says:

    Strangely enough, I find myself saying that the best answer is P=0. This is clearly the answer when a cumulative reward is offered, say $1 per correct answer, and the purpose is to maximise the payoff.

    Halfer:
    0.5(0.5+0.5) + 0.5(0.5)=3/4

    Thirder:
    o.5(2/3+2/3) + 0.5(1/3) = 5/6

    Zeroer:
    o.5(1+1) + 0.5(0) = 1

    These are in units of ‘long-term average dollars’, with P=0 giving the maximum return. When thinking about halfer and thirder positions, I can’t help but take the phenomenalist’s stance and wonder what the point is, from Sleeping Beauty’s point of view? What is she supposed to do with P=1/3? It won’t maximise her number of correct responses, for example.

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  156. DN says:

    Halfers and thirders disagree on “how to assign” an interview to the SB, or how to account for all the interviews.

    Halfers want to split along the coin flip. Thirders counter that you then effectively change the rules of the game to be only one interview on each outcome, rather than 2 interviews after heads.

    Thirders want to lump all interviews together. Halfers counter that either tails or heads happen, but never both simultaneously, so we have to split up along the coin flip.

    Below, I will present a new way to argue for the correctness of the thirder lump together frequency approach. It is written to help convince any halfer doubter who might be willing to be convinced.

    We first look at a game with obvious odds 1/3, and we then tweak back to the SB game in such a way that the “lump together” frequency approach should still be seen to be ok.

    —- Zero or one interview pr. day, interviews Monday, Tuesday and Wednesday.
    —- The Monday interview is red, the other two are blue. (Imagine the interviewer holding a red or blue ball behind his back.)
    —- SB is memory drugged after each day, she knows the rules.

    *** 1 ***
    A coin is tossed.
    If tails, none of the interviews will take place.
    If heads, all the interviews will take place.
    Upon being awakened, what is SB’s credence that it is a red interview?

    *** 2 ***
    A coin is tossed each day (so 3 flips in total) to decide if that day’s interview will take place.
    Upon being awakened, what is SB’s credence that it is a red interview?

    *** 3 ***
    A coin is tossed to decide for the red interview, another coin is tossed to decide for the blue interviews (both or none take place).
    Upon being awakened, what is SB’s credence that it is a red interview?

    *** 4 ***
    A coin is tossed.
    If tails, the red interview takes place.
    If heads, the blue interviews take place.
    Upon being awakened, what is SB’s credence that it is a red interview?

    As you may notice, case 4 is just like case 1, except the red interview is “switched” so that it takes place after tails rather than after heads.

    Also, you may notice that case 4 and case 1 are just like case 3, except case 3 may, or may not, align the red interview with the blue interviews.

    In case 2, we really don’t know the number of interviews to take place at trial start. It could be 0, 1, 2, or 3.

    Anyway, the math is still simple:

    Number of blue interviews pr. trial: 0.5*1 + 0.5*1 = 1
    Number of red interviews pr. trial: 0.5*1 = 0.5
    Interviews pr. trial: 1 + 0.5 = 1.5
    Red interview chance: 0.5 / 1.5 = 1/3

    The math is exactly the same in all 4 cases (nitpicking; you may want to put some brackets when a toss aligns several interviews, but the results are the same).

    To make the final conclusion, note that in case 4, we have:

    credence it is a red interview = credence it is a tails interview

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  157. DN says:

    br,

    Suppose SB is offered this share when interviewed:

    This share will pay 100$ if the coin flip was tails

    Then the fair price for that share is 33$. If she pays more, she will lose money, if she were to offer less, she would be outbid by another guy who would still make a profit. But 33$ breaks even.

    The fair price for a share on a given outcome is a pretty accepted way to define the probability of that outcome.

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  158. br says:

    DN says:
    “Below, I will present a new way to argue for the correctness of the thirder lump together frequency approach.”

    Thank you for that, but I would like to ask what a credence of 1/3 really means, from her point of view. Does it mean, for example, that if you also asked her the question “guess which side is up of the coin I’m holding?”, she would answer, on average, tails once and heads twice, for every three times asked. Is that correct? If so, it would lose to the position where she always answers heads, if one is counting the number of correct replies (or dollars) over a large number of trials. Hence I don’t see P=1/3 as useful for anything, and if it has no consequence then she may as well say P=1/2!

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  159. br says:

    DN:
    “This share will pay 100$ if the coin flip was tails
    Then the fair price for that share is 33$.”

    Seems we cross-posted. OK, I accept this is the fair price. Hence I accept that one gets different answers depending on how one rephrases the question!

    Still, I feel that my approach has significance, because if she has to guess ‘heads’ or ‘tails’, then the winning strategy is to guess ‘heads’ at 100%.

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  160. Logicophilosophicus says:

    DN,

    Me: “SB-thirders seem to think that the two wakings in the heads timeline and the single waking in the tails timeline are not only psychologically indistinguishable, but are [therefore???] of equal probability 1/3 – … – and are all simultaneously available for random selection like some kind of Three Card Monte.”

    You: “Well, not quite. The calculation is like this:
    Number of interviews pr. trial: 1.5
    Number of tails interviews pr. trial: 0.5
    Ratio: 0.5 / 1.5 = 1/3″

    To quote Catch-22, “I may be stupid, because the distinction escapes me…”

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  161. DUHOC says:

    Well, if she knew in advance about the plan, than the probability of her waking up on Monday would be 1/2. That is always the case. The question is what happens when she wakes up Tuesday and told it was Tuesday. Since there were two possible outcomes for Monday we could say that the wave amplitude was two. So when one of those events is realized the possibility for the next event would be four satisfying the Born rule 1/2 x 1/2 equals 1/4. However, if she is told it is Tuesday, she knows that one of those possibilities did not realize, in other words she has a clear view of the past which has allowed her to eliminate one possibility. So by knowing that a specific event is required in order to satisfy a condition for her observation the Born rule is broken, and by my reckoning that leads to a finite number of possibilities that is reduced on successive observations.

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  162. DN says:

    Logicophilosophicus,

    OK then, here we agree on what the reasoning is (even if you may not agree with it). I just like to point out that part of why is it 1/3 is that it is a fair coin. Otherwise the different awakenings would not be equally likely.

    In other words: They are not equally likely (only) because they are indistinguishable, but because they are “produced” equally often (+ are indistinguishable).

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  163. Daniel Kerr says:

    Logicophilosophicus, I urge you to write down your calculations like I did in my first comment, things are a lot more apparent when you actually compute the problem. You are describing things only in an informal language and the meaning is getting lost because the syntax of English does not preserve logical correspondence. You’ve replaced the waking event with this statement regarding the letter “E,” but not just that, you’ve also replaced the day structure. You’ve changed the underlying probability space, this is not an analogy.

    In the previous problem there was a distinction between the probability state of “wake” and “day.” You’ve obliterated this distinction and combined them under the letter “E.” As a result, tails and being told the statement twice has de facto a probability of 0. This analysis is already different from the Sleeping Beauty problem. Your probability space has only 2 outcomes, heads and E twice; tails and E once. The guinea pig still has a sample space of 3 due to having no memory, but two of the samples are coupled and are really only one sample together. The probabilities are different for GP as compared with SB since there are only 2 outcomes and not 4 as in the original problem.

    You take issue with the original problem as you think the probability assignments should be 1/2, 1/4, 1/4, but you have no justification for this except for this invalid analogy. I’ll tell you my probability assignments and I’ll justify them. Heads and Monday, 1/4; Tails and Monday, 1/4; Heads and Tuesday, 1/4; Tails and Tuesday, 1/4. There are 2 random variables here rather than 1 as in your analogy. The day is a variable in addition to the coin flip. Each are mutually exclusive and the two are independent, so each outcome MUST have an equal probability, leading to the 1/4 assignments above. SB samples only 3 of them, much like your analogy, but she uniquely samples 3 of the 4, hence getting the thirder result.

    SB’s knowledge of the days of the week completely inform her probability assignments. She knows that the interviews are conditioned on the days as well, which gives her more information about how to assign her conditional probabilities. GP has no such knowledge and thus cannot incorporate such information in its calculation. For GP, the experimenters are not conditioning their actions on a second random variable, everything is determined by the coin flip uniquely, hence GP sees an equal probability for the flip no matter how many times GP is asked on heads.

    In pseudo-code, you have the following structures:

    GP:
    If coin = H: Ask result, Ask result
    If coin = T: Ask result

    SB:
    Coin flip = random_uniform [H,T]
    For Day = [Monday, Tuesday]:
    -If Day=Monday: Ask result
    -If Day=Tuesday & Coin flip = H: Ask result

    If we then wrote a program to optimize the value to assign result to be right most of the time, the answers for these two problems would be different.

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  164. Daniel Kerr says:

    I should add that in both cases (GP and SB), if we were to catalog a table of the correct answers for each time GP or SB is asked, we will get a table where 2/3rds of the entries will be heads. Thus when we are asked to guess an entry in the table, naively we would expect to guess heads.

    However, by your argument you say that not every entry is equal, the probability of being asked for each entry in the table is not equal. You would say that any two heads combined have an equal probability being asked for as any single tails. My pseudocode for GP does not reflect this, probably because it’s fallacious reasoning. Ultimately we’re asking to guess an entry in the table described above, since we do not know if we’re double guessing or not, we should assume we are to maximize our result. If we guess heads 100% of the time, we will be right 2/3rds of the time.

    So you can ignore my argument about the days of the week/second random variable. Those help to stratify the probability space and clarify the analysis, but the result holds regardless.

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  165. DUHOC says:

    But I think the point being made here is that if there is consciousness, i.e. if sleeping beauty is awake, one of the possibilities must be eliminated. If on Monday the coin is flipped and sleeping beauty never wakes up there isn’t another coin flip. We may even say that the definition of consciousness is that she realizes there was a possibility she would never wake up, namely the addition of the set containing zero elements. In the case where there were two possibilities heads or tails, the chance is one out of two. But if me or sleeping beauty is thinking about the possibilities one of the chances had to have been eliminated by adding the set of zero elements. Therefore in the “thirder view” introducing consciousness reduces one element of the probable outcomes. And this is a very crucial point because although the universe is composed of magnitudes which behave in accordance with the rules which govern magnitudes, namely mathematics, these magnitudes are really only approximations which constrains an observer who views outcomes as the result of strictly math wrong on certain occasions.

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  166. Logicophilosophicus says:

    D Kerr,

    I shall give my informal calculation based on a biased coin, which lands tails 51% of the time. No doubt if I wrote it in the form P(x|y) or used Born’s rule it would work out differently – you can put me right.

    Remember, the protocol is that if the coin comes up heads, SB will be woken-and-asked, memory-wiped, woken-and-asked; if tails, just woken and asked. Calendars and other clues are not permitted. SB is aware that tails are favoured 51:49 a priori.

    Not being as daft as I am, she will bet that the coin came up heads when she is woken and asked. In the new protocol she is asked to bet HER LIFE on the result she selects – I expect she would be well advised to go with the best estimate of the probability.

    And, unsurprisingly, 51% of the time the smart SB-thirders end up dead. Meanwhile the daft halfers get it right.

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  167. Richard says:

    DN,

    I don’t think the percentage matters (the difference between your cases 2C and 4C).

    They don’t matter to a thirder. They shouldn’t matter to a halfer. If the probability of producing bags is upped from 10% to 99.99%, why should that 0.01% chance of “no bags” affect the halfer’s conclusion?

    What I really want to know is: What does a halfer make of the experiment I proposed? And, if the halfer agrees that the probability of gold is 1/3, what does the halfer think is the key difference (from SB’s epistemic point of view) between my experiment and the original SB experiment?

    Here is a another variation: On each day of the experiment on which an interview is conducted, a person on the street is selected at random (without replacement, to ensure the same person is not selected twice in the case of heads) to witness the interview. This person is told all the details of the experiment, except the result of the toss and whether it is the first or second day that SB is awakened. What credence should the outside witness give to the proposition that “the toss was heads”?

    Assuming that halfers give a different answer for the outside witness than they give for SB, what information does the witness have that SB does not?

    * Neither knows the result of the toss.
    * Neither knows which day of the experiment it is.
    * Both know the experimental protocol.

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  168. Daniel Kerr says:

    Logicophilosophicus, I realized phrasing my argument in terms of the calendar is misleading, the second random variable manifests as a second question, the structure of the days do not matter. However you still have evaded to provide a justification or calculation. Since this does not seem forthcoming, I suggest you take the perspective of the table I outlined above as I think we can both follow it easily.

    We do agree that such a table for 50-50 odds would have 2/3rds of its entries as heads, do we not? For every time the coin is heads, two entries in the table are heads too. So your argument comes down to justifying that when SB is asked to guess a random entry of the table, why she should treat each entry has having different probabilities of being asked. Since she has no memory of each interview, she has no basis for assigning the entries of the table different weights. If she guesses heads every time, then in the cases where she’s interviewed twice, she will be right and in the cases where she’s interviewed once, she’ll be wrong. Thus 2/3rds of the table will be filled out correctly by SB, yet 1/2 of the time the result will be heads. The two are not in conflict. If her life was bet on it, 1/2 of the time she would live if guessing heads but be right 2/3rds of the time.

    The information SB has is that she has a 1/2 probability of being asked the same question twice, but only if the result was heads. If each question corresponded to its own coin flip even if a second coin flip was only done if the previous result was heads, the probability would then be 1/2. But anytime she’s being asked a question, there’s a chance it’s the same question yet again, so that’s information she has about the problem is independent of the calendar days. The calendar days in my above analysis substituted for this knowledge, but in general the days do not need to be referenced.

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  169. Logicophilosophicus says:

    Daniel Kerr,

    Right but DEAD? I suppose she can have the table engraved on her tombstone.

    P(alive) = 49%, P(dead) = 51% and the thirders lose. No argument.

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  170. Richard says:

    Logicophilosophicus,

    No matter what she decides the best betting strategy is, her credence is still 2/3 heads, 1/3 tails (slightly modified by your loaded coin). But you can’t expect her to bet according to her credence under the terms you gave. In particular, she places only one bet (repeating the same bet on Tuesday doesn’t count if she only has one life to lose).

    Note, if she did have two lives to lose, she should bet on heads. If she’s right, she still has both lives after Tuesday. If she is wrong, she still has one life left.

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  171. Richard says:

    H = heads
    T = tails
    M = Monday
    Tu = Tuesday
    I = the evidence of being awake and interviewed

    p(H|I) = p(H|MI) * p(M|I) + p(H|TuI) * p(Tu|I)
    p(T|I) = p(T|MI) * p(M|I) + p(T|TuI) * p(Tu|I)

    If SB is informed at the start of the interview that it is Monday, then she should assign 0.5 to the probabilities of heads and tails:

    p(H|MI) = 0.5
    p(T|MI) = 0.5

    If SB is informed at the start of the interview that it is Tuesday, then she should assign:

    p(H|TuI) = 1.0
    p(T|TuI) = 0.0

    So,

    p(H|I) = 0.5 * p(M|I) + 1.0 * p(Tu|I)
    p(T|I) = 0.5 * p(M|I) + 0.0 * p(Tu|I)

    Halfer position:

    p(H|I) = p(T|I), which means the solution to the simultaneous equations requires 0 = p(Tu|I). The halfer position is valid only if there is to be no interview conducted on Tuesday, regardless of the toss.

    That doesn’t match the problem statement. Let’s try the thirder position:

    p(H|I) – p(T|I) = 1/3 = p(Tu|I)
    p(M|I) = 2/3

    That’s better: If SB is being interviewed, it’s more likely to be a Monday than a Tuesday, but Tuesday can’t be ruled out.

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  172. Daniel Kerr says:

    Logicophilosophicus, with a name like that I thought you’d be interested in pursuing how her guess odds are 2/3rds but her life odds are 1/2. Both are true. However from her perspective, her life odds are 2/3rds guessing heads, provided she’s killed only after the whole process and not immediately after the interview.

    Also thanks Richard, for having a much more succinct calculation above unlike my first comment’s misreading of the problem!

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  173. hubdog33 says:

    You cover your eyes and flip a fair coin. The coin lands. Without looking, what is the probability that it is heads? 50%. Now assume you look and it is heads. What is the probability it is heads? 100%. Halfers, who refuse to accept that the new information changes the probability from the perspective of the person receiving the new information, would incorrectly say still 50%.

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  174. Richard says:

    hubdog, it isn’t that halfers refuse to believe that new information changes the probability. The problem is that they don’t recognize that the informational context of the interview is different from that at the beginning of the experiment, in the first place. In other words, they think p(H|I) = p(H) because they don’t see the interview “I” as being informative.

    But SB is giving p(H) = 0.5 at the start of the experiment (Sunday night) because she isn’t being asked in the context of an interview. When asked during the interview, she must give p(H|I) which is a different thing. If she wants to calculate it in long form she can do it according to my previous post above: p(H|I) = p(H|MI) * p(M|I) + p(H|TuI) * p(Tu|I). For that she needs:

    p(H|MI) = probability toss was heads if this is a Monday interview = 0.5
    p(M|I) = probability that this is Monday if SB is being interviewed = 2/3
    p(H|TuI) = probability toss was heads if this is a Tuesday interview = 1.0
    p(Tu|I) = probability that this is Tuesday if SB is being interviewed = 1/3

    This is clearly different from the context of Sunday night. On Sunday, she knows p(H) = p(HSu) = p(H|Su) * p(Su) = 0.5 * 1.0 = 0.5.

    The knowledge (on Monday or Tuesday) that she is experiencing an interview prescribed by the experimental protocol replaces the full knowledge of what day it is, which was lost when she was put to sleep on Sunday night. Losing information also changes the informational context, and so changes her credence. (She can restore her original 50/50 credence by learning that it is Monday, or increase her credence to 100% by learning that it is Tuesday — but neither of these bits of information is relevant on Sunday.)

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  175. Logicophilosophicus says:

    Richard, Daniel Kerr,

    “…her guess odds are 2/3rds but her life odds are 1/2. Both are true.” But she already KNEW those true “life odds” at the original briefing on the protocol. She cannot rationally know one set of odds to be true and crucial, and yet deny them.

    The idea “if she had two lives to bet” is crazy, but let’s follow it through. You claim that by killing SB I have unfairly truncated the experiment, depriving her of the opportunity to make a second bet. Not so. We’ll take a more “realistic” position, that she is going to bet $3.00 on the outcome. Let’s say the coin is unbiased. In the tails timelines an improbably generous bookmaker offers her the true odds, i.e. evens. Betting on tails, she would win $3.00, but betting on heads she would lose $3.00. You claim that in heads timelines, she would lose $6.00 betting on tails, but win $6.00 backing heads, therefore – across both timelines – her expectation from betting on heads is +$3.00.

    But what you have totally failed to take into account is that SB’s induced confusion over the day – Monday or Tuesday – is entirely confined to SB’s mind. If you can find me a bookmaker so improbably generous as to take a bet on the winner after the stewards have ruled on the photofinish, I’ll get rich, too. In the real world, AND IN ANY BRANCH OF A CREDIBLE EVERETTIAN MULTIVERSE, SB breaks even with a fair coin, and loses when the true probability of tails is 1/2 < P(t) < 2/3 – i.e. all those other fairy tale scenarios where "thirder logic" suggests she should have "credence" that she would win.

    The "credence" involved was defined as an assignment of probability out there, where the heads-tails uncertainty is resolved. That probability is known, and if you believe in it you give it credence.

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  176. br says:

    There’s been so much discussion, I’m not sure if this alternative procedure has been discussed:

    Rules are similar, but with only one possible awakening. If the coin comes up heads, then she is woken and interviewed once on Monday. If the coin is tails, she is not woken up on Monday. If she is interviewed she is asked “What is the probability you would assign that the coin came up tails?”. Either way, on Wednesday she is woken up and told what happened. End of rules.

    I hope we can agree that if interviewed she will be 100% sure it is heads, so the credence she gives to it being tails is 0?

    By implication, this supports the P=1/3 answer to the question, as phrased on this blog (other blogs may have different versions!).

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  177. Logicophilosophicus says:

    br,

    You miss the point – basically you have written Monday’s business out of the protocol. Tuesday was always going to be definitive: result revealed. Monday was always going to be free of information about the toss result. The SB problem asks us to see what probabilistic conclusions, if any, she can draw from situations where Monday and Tuesday, or Before and After, are indistinguishable. Without that confusion there is no problem.

    All these problems have evolved from the analysis of games of chance beginning in the 17th century. The point of assigning a probability is to determine the appropriate action – that’s as true of deciding which medicine to take as it is of deciding which card to bet on. Thirders think that SB’s decision to bet is influenced by her Monday-Tuesday confusion, and they are right, just not in the way they think. She should attempt to make a bet based on the known probabilities (assuming that the odds offered are favourable) whenever she is asked, so that she does not miss the opportunity to profit. She knows a Tuesday bet is impossible, but she just attempts to place a bet whenever asked, and thus always succeeds in making a (Monday) bet. She never gets to place two bets in the Heads timeline. In your scenario, she never gets to place one bet. In any related scenario, the probabilities (the ones she is asked to assign) are “in the coin”, not the protocol. Thirders are wrong.

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  178. br says:

    Yes, my previous post was not exactly helpful – so I just downvoted it! :)

    I came up with this statement though:
    “I know the coin is fair so will come up heads in 1/2 of the trials. If I receive a dollar for each correct guess to the question ‘Is the coin I’m holding heads or tails?’, then remembering the structure of the trial (=information!) I should always pick heads, 100:0, to maximise profit. However I won’t be correct in all the guesses – with this strategy I believe 1/3 of the guesses will be wrong, averaged over many trials”.

    Receiving money can be replaced by ‘maximising the number of truthful statements I make in my life’, a more noble aim. Then one needs to interpret the question “What is the probability you would assign that the coin came up tails?”, which is different than the one I just posed ‘Is the coin I’m holding heads or tails?’. The question ‘what is the probability…?’ can indeed be seen as a question about the coin, which has P=1/2 and knowledge of the structure of the trial won’t change this, whereas the betting question ‘Is the coin heads or tails?’ will be affected by the structure of the trial – as to maximise profit she will guess heads 100% but be wrong 1/3 of the time, which some might interpret to mean P=1/3. So this is a problem that can distinguish ‘pure’ probability from ‘betting’ probability?

    Which I reckon makes thirders wrong, as P=1/3 doesn’t answer the question! (ducks for cover!)

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  179. Richard says:

    Logicophilosophicus,

    Please don’t insult my intelligence by suggesting that your “guess right or die” scenario was a fair bet. In fact she doesn’t even receive a payoff for guessing right. But more to the point, clearly she is under extreme duress if she accepted such terms. Or maybe she doesn’t want to live in the first place. After all it is a dreary life if the only thing that ever happens in it is you sleep or are being interviewed.

    If you want to find a bookmaker who will offer a fair bet, find one who doesn’t have inside information (such as whether today is day 1 or day 2 of the experiment). For example, try the outside witness I talked about in an earlier post.

    I made three separate cases for the thirder position. None require any assumptions about fair or favorable bets (I really don’t like to apply the gambling argument, since it is too easy to make mistakes similar to yours; Bayesian thinking is not so error-prone). Feel free to respond to any or all of them if you think you can make the halfer case as to why each is wrong. If there is such a case, I want to know what it is.

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  180. Richard says:

    L-P, you wrote (emphasis mine):

    “You miss the point – basically you have written Monday’s business out of the protocol. Tuesday was always going to be definitive: result revealed. Monday was always going to be free of information about the toss result. The SB problem asks us to see what probabilistic conclusions, if any, she can draw from situations where Monday and Tuesday, or Before and After, are indistinguishable. Without that confusion there is no problem.”

    Revealing the result of the flip on Tuesday never was part of the experiment. It is not relevant to SB’s answer, unless the reveal comes before she has to answer. And if she knows in advance that it will be revealed before she is asked (after all, she is told the full protocol of the experiment, remember?), then not being told before an answer is demanded is a sure signal that it is Monday. So if there is a reveal, it must be after she has given the answer, in which case it does not influence the answer. You are either changing the rules of the game or to intentionally confuse readers by muddying the waters with a false statement.

    Then: “She knows a Tuesday bet is impossible, but she just attempts to place a bet whenever asked, and thus always succeeds in making a (Monday) bet. She never gets to place two bets in the Heads timeline.”

    Being offered a bet on Monday but not Tuesday is another tip-off as to what the result of the flip was. If we want to include betting in the protocol (though it’s superfluous), the bet needs to be offered on both days, otherwise she doesn’t have the necessary ignorance for the experiment to be meaningful. The “bookmaker” in that case is the experimenter, and I am sure the cost was written into the research grant application, so you don’t need to assume the experimenter is “generous”.

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  181. Daniel Kerr says:

    Logicophilosophicus, you’re stuck in informal language, seriously, write out a formal calculation, your error has been outlined at least 5 different ways so far and your response is only to shift the language. The logic of your argument is to restate the true odds are 50-50 thus the thirders will not make their bets. If you’re going to make an informal argument, stop asserting it’s 50-50 before making your point. Derive that it’s 50-50, you haven’t done that yet.

    Your analysis of her “true life odds” are from an unbiased point of view. SB is biased as we’ve already established, she has two chances to guess heads if it is heads. You seem to be under the impression that the coin determines whether she lives or dies….but that’s not quite true. In that sense her life odds are 50-50, but as long as she gets an interview as outlined in the problem, provided the flip is heads she has another chance to guess.

    Let’s assume she’s not guessing optimally, i.e. guessing randomly. Following the protocol, what happens when you wake her on Monday and she guesses Tails but on Tuesday guesses Heads? Do you kill her? If she was betting, she’d take a loss on Monday but recoup her loss on Tuesday, she’d break even. If she guessed Heads on both days, then she’d gain twice. If the flip was Tails and thus she was interviewed only on Monday, then on guessing heads she’d lose once. So 50% of the time she gains twice as much as she loses the other 50% of the time. That’s 2:1 odds last I checked.

    If you decide that you’re going to kill her if she guesses wrongly even once, then her life odds are 50-50, but if you allow any correct guess to save her, then her odds are 2:1. I was hoping you would derive this yourself by finally analyzing it, and I’m still giving you the chance as all of this has been informal (but logically valid). I hoped the apparent contradiction I laid out would have grabbed your interest but instead you took it literally. There are no true odds, depending on how her life is decided, it’s either halfer or thirder. Betting is always thirder though.

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  182. Richard says:

    Questions for halfers:

    1. See my earlier variation on the experiment, using boxes of iron or gold. Is your answer for this experiment 1/3? If not, then why not? If so, then what difference can you point to between this and the classic SB experiment to explain the difference in your answers?

    2. See my post in which I introduced the idea of an outside witness. Should the witness assign a probability of 2/3 to heads? If not, then why not? If so, then what difference can you point to between the witness’s state of mind and SB’s state of mind to explain the difference in their answers?

    3. See my post in which I calculate p(Tu|I) assuming, respectively, the halfer and thirder positions. How do you reconcile the halfer position with its implication that the probability of an interview happening on Tuesday is 0?

    I do hope to hear from a halfer who is willing to engage in answering these questions. I am eager to know how such problems are resolved in the halfer world.

    Thanks.

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  183. Louis Wilbur says:

    From http://en.wikipedia.org/wiki/Sleeping_Beauty_problem we have the problem:

    “–Sleeping Beauty volunteers to undergo the following experiment and is told all of the following details: On Sunday she will be put to sleep. Once or twice, during the experiment, Beauty will be wakened, interviewed, and put back to sleep with an amnesia-inducing drug that makes her forget that awakening. A fair coin will be tossed to determine which experimental procedure to undertake: if the coin comes up heads, Beauty will be wakened and interviewed on Monday only. If the coin comes up tails, she will be wakened and interviewed on Monday and Tuesday. In either case, she will be wakened on Wednesday without interview and the experiment ends. Any time Sleeping Beauty is wakened and interviewed, she is asked, “What is your belief now for the proposition that the coin landed heads?”–”

    If the experiment is repeated one thousand times then, on average, she will be awakened and asked about her belief fifteen hundred times during the experiments. On average, five hundred of those fifteen hundred times she is asked (i.e. one third of the times) the coin will have landed heads and so the proposition “the coin landed heads” will be true. On average, one thousand of those fifteen hundred times she is asked (i.e. two thirds of the times) the coin will have landed tails and so the proposition “the coin landed heads” will be false.

    So sleeping beauty knows several facts. She knows that in repeated trials of the experiment, on average, one out of every three times she is awakened the proposition “the coin landed heads” will be true. She knows she was just awakened. So it is reasonable for her to give one third as an answer. She also knows that in repeated trials of the experiment, on average, one out of every two times that the experiment is run the proposition “the coin landed heads” will be true. She knows the experiment is being run. So it is reasonable for her to give one half as an answer. Her answer to the question should be: My belief is that, on average, the proposition is true one out of every two times the experiment is run and, on average, one out of every three times I am awakened and asked my belief about it.

    L

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  184. BobC says:

    Sean,

    We could really use another post here, else everyone will keep beating this one to death. How about a list of the papers you’ve most enjoyed recently?

    I just finished “From Eternity to Here”. Every page (at least once, sometimes more, despite the tiny font). Every endnote (despite the microscopic font, favorite: 273). Even a selected sampling of the bibliography (great choices, nice touch including the Buffyverse).

    My arm-waving is now far more nuanced concerning entropy and black holes, thermodynamics, de Sitter / Minkowski / anti-de Sitter space, various multi-verses and baby-verses, the inflaton, and the critical importance of (and problems with) low-entropy initial conditions.

    What I think I enjoyed most was your friendly conversational tone, as though you were explaining this over dinner. Next is your ability to foresee where a reader could have problems, and patiently provide ways through them, all without ever sounding condescending or rushed. I’m amazed at how little math you included, yet you still managed to communicate the “feel” in a meaningful and constructive way. Lucid and inviting.

    Then I made certain my clocks were all on time. Whatever that is.

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  185. Sleeping Beauty Problem is Classical says:

    Its funny that my comment – that the sleeping beauty problem has nothing to do with quantum mechanics – got voted down. Its funny because of all the 184 comments here, not ONE of them is about quantum mechanics. Yet that was the whole point of Sean’s post; he claims he was able to solve the problem by appealing to quantum mechanics. So everyone kind of tacitly agrees with me that the problem has nothing to do with quantum, but won’t admit it.

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  186. Tom Snyder says:

    Sorry for the length of this post.

    Suppose we let SB perform the coin toss on Sunday. She catches the coin in the palm of one hand and then covers up the coin with the palm of the other hand. It doesn’t matter if she sees the result of the toss since we assume that being put to sleep erases her memory of the result. When SB is awakened according to the standard protocol, we ask her, “What’s the probability that if you uncover the coin in your hand you will see a tails?”. Now SB knows that in a large number of repetitions of the experiments, 1/3 of the time she is awakened she will be holding a tails in her hand. So, she naturally answers the question, “the probability that I will see tails if I uncover the coin is 1/3”.

    Even a halfer should agree with that. But, a halfer will argue that the question we are now asking SB is not the same question as stated in Sean’s statement of the problem. In the original statement, the question asked of SB is, “What is the probability you would assign that the coin came up tails?” The halfer interprets Sean’s question as not being equivalent to the question about uncovering the coin when awakened.

    A halfer might argue as follows. In one half of all experiments the result of the coin toss will be heads and in the other half of the experiments the coin will be tails. Since SB knows that she is in an experiment anytime she is awakened, she gives a 50-50 chance that the coin came up tails.

    The halfer can go on and assign probabilities to the individual awakening events. But the halfer assigns these probabilities based on considering a randomly selected experiment out of a large set of repeated experiments. Let E1 and E2 denote the two awakening events that would occur following a heads and let E3 be the one awakening event that would follow a tails. In one half of the set of all experiments, E1 and E2 (but not E3) will occur sometime during the experiment. And in the other half of the set, E3 (but not E1 or E2) will occur during the experiment. So, she can say that for an experiment picked at random, E3 will have a probability of 1/2 of occurring during the experiment. (E1 and E2 will also each have a probability of 1/2 of occurring sometime during the experiment.) E3 occurs if and only if the coin came up tails, so she claims that the probability that the coin toss came up tails is 1/2.

    If SB is a halfer, then when she is awakened she realizes she is in one of the events E1, E2, or E3. But she has no way to know which event she is in. Since she knows that she is in a particular experiment, she assigns a probability of 1/2 that she is in event E3. She also assigns a probability of 1/2 that she will experience both E1 and E2 in this experiment. If she is an experiment where both E1 and E2 (and not E3) occur, then she knows that when she is awakened she cannot tell if it is E1 or E2. So, she assumes a 50-50 chance of being in E1 or E2 if she is awakened in an experiment in which both E1 and E2 will occur. So, the overall probability that she would assign to being awakened in event E1, say, would be 1/2 * 1/2 = 1/4 . The first 1/2 factor is the probability of being in an experiment where events E1 and E2 will occur, and the second factor of 1/2 is the probability of being in E1 out of the set {E1, E2}. So, on any awakening, she would assign a probability of 1/4 of being in event E1. Likewise, 1/4 for E2.

    If upon any awakening she is asked for the probability that the coin came up heads, she would say the answer is the sum of the probability that the current awakening is an E1 event plus the probability that the current awakening is an E2 event. Thus, the probability that the coin was heads is 1/4 + 1/4 = 1/2.

    Moreover, the halfer-SB can use her probabilities to come to the same conclusion as the thirder-SB in regard to the scenario where SB holds the coin in her hand. If SB uncovers the coin in her hand every time she is awakened, the halfer will deduce that she will see tails 1/3 of the times she is awakened in a large number of repeated experiments.

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  187. DN says:

    Tom Snyder,

    The halfer in your post is making the usual mistake in this discussion: Regarding the experiment as either Monday or Tuesday, but not both (in case of Heads). This is the only justification she has for “picking” one of these possibilities and multiply their chance by 0.5 (to arrive at 1/4).

    The SB experiment has the specific property that it actually matters whether you go

    –> pick experiment –> then pick one interview

    or

    –> pick all interviews, one by one while memory drugged

    and the wording is unambiguous in that it specifies the second procedure.

    A couple of further proofs that the halfer is wrong:

    1) The math you give, dear halfer, fits perfectly if the experiment is EITHER Monday OR Tuesday. Do you really claim that it wouldn’t make a difference if we instead chance the experiment to BOTH Monday AND Tuesday? (All this assuming SB’s drugged point of view, as in the actual SB formulation.)

    2) The simple proof that 50% must be too low for Heads is that, given Monday, the chance for Heads is 50%, given Tuesday, it’s 100%, and there is a non-zero chance it’s Tuesday. (And notice that this proof works exactly because we have BOTH. If we had either/or, the chance for Heads given Monday would only be 1/3.)

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  188. Logicophilosophicus says:

    Daniel Kerr,

    Your first comment way back was: “The trick is to realize there are 4 cases here, heads and Monday, tails and Monday, heads and Tuesday, tails and Tuesday. They each have 1/4 probability, but sleeping beauty only “samples” 3 of these cases, as she is not awake for tails and Tuesday. So of the 3 cases she sees, two of them are heads and if she wants to make money from betting, she’d better bet on heads.”

    Let me EXACTLY paraphrase that to describe the version of Bertrand’s Box where the experimenter peeks and ALWAYS extracts a gold coin first to show the subject. (The 2 boxes we shall call H and T contain H:Gold+Gold and T:Gold+Silver.) This version is equivalent to the SB protocol.

    “The trick is to realize there are 4 cases here, H and Gold(1), T and Gold, H and Gold(2), T and Silver. So of the three cases she sees, two of them are H and if she wants to make money from betting, she’d better bet on H.”

    Wrong. Do you see what you’ve done? Pulling out the second gold coin IDENTIFIES the box as A – there is nothing to bet on any more; just as waking SB on Tuesday identifies the timeline as heads. This was already pointed out to you: only in a fairytale land of bookmakers who take bets after the result – i.e. not in ANY real world – does a Tuesday bet have any meaning. Reality is a zero sum game.

    Well, you conjured up such a world. Instead of having to find a bookmaker, SB is to be GIVEN $1 each time she “guesses” correctly. Now you are able to say she should “guess” heads, thus winning $2 the half of the time she is right, rather than $1 if she “guessed” tails.

    Wrong. Do you see what you’ve done this time? You have changed the problem so that the experimenters have a payoff of $2 set aside for CHOOSING (definitely NOT guessing or “giving credence to”) heads and only $1 for CHOOSING tails. When SB says “heads” she does not “assign probabilities” to heads and tails in the ratio 2:1; she CHOOSES the 50:50 chance with the higher payoff – the ratio 2:1 is in the payoff table, not the coint toss probabilities. The Tuesday dollar is an extra freebie.

    Back in the fifth comment in this discussion, Joe Polchinski endorsed “the result 1/3 unless the additional constraint is imposed that she can only bet on Monday, which is changing the problem…” I shall EXACTLY paraphrase that to reflect the real world: The correct assignment of probability to tails, given a true coin and the stated protocol, is “1/2 unless the additional constraint is imposed that a bookmaker must take a bet on Tuesday (BUT only on heads, and after the result is published), which is changing the problem…” If the non-naturalistic element of a post-result bet is not stated in the problem, it is ridiculous to assume that the correct answer depends on importing it. (In the real world, even the withdrawal of a horse AFTER the bets are laid reduces the payoff – “Tattersall’s Rule 4″ I believe. But if the withdrawal is known, there can be no bets to win on a one-horse race.)

    I haven’t mentioned the induced amnesia element at all in that analysis, because it is not relevant to the odds. Consider: without amnesia, SB could bet either way on Monday with equal chances of success, while on Tuesday she is only allowed to bet if heads came up, i.e. there is no “bet”; with amnesia, she can maximise her payoff by saying “heads 2/3″ because she will get access to the possible Tuesdsay freebie, plus the result of Monday’s fair bet (where heads was as good as tails anyway).

    These results are not artefacts of plain language which the correct symbolism would somehow explode – the correct symbolism only works with correct inputs (GIGO). The correct symbolism is, in any case, the payoff matrix.

    DN, etc,

    Unless you can come up with a real-world scenario where the 2:1 assignment of probability has useful CONSEQUENCES (i.e. outperforms 50:50 in any APPLICATION) your assignment is wrong. The proof of the pudding.

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  189. Logicophilosophicus says:

    Whoops. I omitted the paraphrase of this sentence:

    “They each have 1/4 probability, but sleeping beauty only ‘samples’ 3 of these cases, as she is not awake for tails and Tuesday.”

    becomes:

    “They each have 1/4 probability, but sleeping beauty only ‘samples’ 3 of these cases as the experimenter is committed to showing the Gold coin, i.e. disallowed from revealing the Silver, when box T has been selected.”

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  190. Daniel Kerr says:

    Logicophilosophicus, my patience grows thin for reading informal arguments. Many mathematical calculations have been provided in these comments and result in the thirder result. My informal explanations are interpreting these calculations, they are not arguments in themselves. So as you attack my attempts to translate mathematical truth into informal language to convey understanding, you’re not attacking the argument at all. Ultimately your argument lies in showing how the calculation is in error. I have yet to see that and as such my interest in and ability to be persuaded by your arguments is rapidly diminishing.

    I will just say, in your analysis above, you clearly reject the notion of a conditional probability. You don’t understand that the pay-off is a result of a probability weight and is completely compatible with the coin odds being 50-50. Of course she goes with the 50% chance that has a higher payoff and that’s because she gets to bet twice. Your analysis seems to omit that if she guesses randomly the extra payoff is not free. The extra payoff is received only if she guesses correctly both times for a coin flip of heads.

    And arguments like this are not satisfactory:
    “Wrong. Do you see what you’ve done? Pulling out the second gold coin IDENTIFIES the box as A – there is nothing to bet on any more; just as waking SB on Tuesday identifies the timeline as heads. This was already pointed out to you: only in a fairytale land of bookmakers who take bets after the result – i.e. not in ANY real world – does a Tuesday bet have any meaning. Reality is a zero sum game.”

    First of all, your analogy doesn’t hold as in the original SB she can’t remember previous interviews. The second gold is tantamount to her having the knowledge that it’s Tuesday. You said yourself this is irrelevant to the problem, I don’t understand why you would choose an analogy that makes it absolutely integral to her state of knowledge and one where the analogy doesn’t hold as she has to know that it’s indeed Tuesday/Gold(2).

    The correct analogy is to say that she has no memory of the coin she has already drew and she doesn’t know how many coins are still left in the box, then it’s described by the same mathematical calculation you have yet to address.

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  191. Logicophilosophicus says:

    Daniel Kerr,

    “…she goes with the 50% chance that has a higher payoff and that’s because she gets to bet twice. Your analysis seems to omit that if she guesses randomly the extra payoff is not free. The extra payoff is received only if she guesses correctly both times for a coin flip of heads.”

    If she gets to bet twice then “the additional constraint is imposed that a bookmaker must take a bet on Tuesday… after the result is published” which is a non-naturalistic condidition. She does not get to bet twice unless you impose that strange condition. Money may not be as important as life itself, but this is just as strange as the idea you suggested earlier, that if she had two lives to lose etc etc.

    Please clarify “not free” – where does the extra dollar come from? A bookmaker prepared to take a bet on a one-horse race? Or is it built into the known protocol? Let me put this differently. Suppose the protocol includes this sentence: “Your assignment of a probability is taken to mean that you would be as happy to place as not to place a bet on that result (H or T) at those odds against that result on the universal understanding that a bet is only acceptable before the result is published.”

    You say my box analogy is inappropriate because she would have to be unaware whether a coin had already been shown. No. The indistinguishability of Monday and Tuesday is analogous to the indistinguishability of the coins. G(1) does not mean “the first gold coin to be taken from the box”: it is the identity of the coin (e.g., if you like, a microscopic number scratched into its surface). The temporal distinction between M and T, concealed by induced amnesia, is fully analogous to the physical distinction between G(1) and G(2), concealed by perfect similarity of manufacture.

    GIGO refers to the fact that your calculation may be impeccable – I haven’t checked – but since there is some “garbage in” there is necessarily “garbage out”.

    [edit: I missed a point. You wrote that “the second gold is tantamount to her having knowledge that it is Tuesday” – but the point is that that “the second gold is fully analogous to the fact that on Tuesday EVERYONE ELSE knows that it is Tuesday, that the result is heads, that no bet can be placed.” SB knows that on ONE of her awakenings in a heads-timeline no bet is possible, and she factors that knowledge into her assignment.]

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  192. Richard says:

    L-P,

    You wrote:

    “Back in the fifth comment in this discussion, Joe Polchinski endorsed “the result 1/3 unless the additional constraint is imposed that she can only bet on Monday, which is changing the problem…” I shall EXACTLY paraphrase that to reflect the real world: The correct assignment of probability to tails, given a true coin and the stated protocol, is “1/2 unless the additional constraint is imposed that a bookmaker must take a bet on Tuesday (BUT only on heads, and after the result is published), which is changing the problem…””

    You are misapplying Joe’s statement. Joe correctly said that adding a constraint that bets are only accepted on Monday is “changing the problem”. The original problem does not have the constraint. So when you start with that constraint in place (to get a probability of 1/2), and remove it to allow SB to place a bet on Tuesday if she is awake, then the result of 1/3 isn’t from “changing the problem”, it is from returning to the problem as originally formulated.

    You are persistent in artificially requiring that the only bookmaker that SB is allowed to place a bet with must be one who is both (a) in possession of inside information that SB does not have and (b) motivated to maximize profit. Of course, SB would not accept a bet at 2:1 odds from such a bookmaker, if she knows the bookmaker fits this description. That is because knowing that the bookmaker is offering the bet at all tips SB off to the fact that it is Monday, with the result that she now should not accept the bet because it is not a fair bet. But that is not the state of her knowledge at the time the interview takes place in the original experiment.

    I already gave two examples of fair bookmakers – one who fails to meet condition (a) and one who fails to meet condition (b). Any representation of SB’s assignment of probability by way of fair betting odds needs to assume a fair bookmaker, not one using inside information to serve a profit motive.

    I went back to find your gold coin argument. It is indeed analogous to the SB problem, but it also fails to support the halfer position. Once you formalize the probabilities, it is easy to see why.

    H = heads
    T = tails
    Au = the evidence of seeing a gold coin
    Au1 = this instance is the first instance of seeing a gold coin
    Au2 = this instance is the second instance of seeing a gold coin

    The “1” and “2” indices are hidden variables from SB’s point of view, but she understands that

    p(H|Au) = p(H|Au1) * p(Au1|Au) + p(H|Au2) * p(Au2|Au)
    p(T|Au) = p(T|Au1) * p(Au1|Au) + p(T|Au2) * p(Au2|Au)

    Noting that:

    p(H|Au1) = 0.5
    p(T|Au1) = 0.5
    p(H|Au2) = 1.0
    p(T|Au2) = 0.0

    leads to this:

    p(H|Au) = 0.5 * p(Au1|Au) + 1.0 * p(Au2|Au)
    p(T|Au) = 0.5 * p(Au1|Au) + 0.0 * p(Au2|Au)

    The second equation shows that p(Au1|Au) = 2 * p(T|Au), so substitution into the first equation gives:

    p(H|Au) = p(T|Au) + p(Au2|Au)

    Conditional probabilities (like other probabilities) cannot be negative. This means that either the second gold coin can never be shown (p(Au2|Au) = 0), or p(H|Au) is greater than p(T|Au). But in the case of heads, the second gold coin is required to be shown, so p(Au2|Au) is greater than zero, and the halfer position that p(H|Au) = p(T|Au) is disproven.

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  193. Daniel Kerr says:

    Logicophilosophicus, as I said before, my patience ran thin. Your argument against the betting analogy is that a bookmaker wouldn’t take a bet for an event that has not yet transpired. This is a non sequitur and I really have no response to that. You say there’s garbage in, and I’ve asked you multiple times to correct my probability assignments, and you haven’t yet. Instead you resort to poorly formed analogies to make your points. Perhaps this works in some circles, but I like logic and frankly, analogies explicitly don’t necessarily preserve logical invariance, so you’re not making an argument.

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  194. Logicophilosophicus says:

    p(H|I) seems to be the probability of getting a heads result given that an interview has already taken place. How does the protocol allow that sequence? There’s more that a hint of time travel there. Using normal Games Theory practice, I would draw a tree diagram and base any strategy on probabilities based on total enumeration. In this really simple scenario we have just two branches,
    H->I->I->?
    and
    T->I->?
    To reason backwards in time is a little odd, but just because SB can’t tell the difference between these three actually quite distinct interviews cannot make the I in the second branch lead back to an H in the first branch (we can’t jump from one branch to another); nor especially can it make the two I’s in the first branch lead back to two H’s, creating a third branch out of nowhere. Reasoning back in time from, say, the second I in the Heads branch leads back through the first I to the single H.

    There was an old Bhutanese Sherpa who spent his summers escorting climbers from the tourist centre T to the hill camp H. The tourists arrived by plane on the Saturday, were driven as far as possible to T, and on Sunday morning began the arduous trek to H, which they reached on Tuesday evening. The old BS would accompany returning climbers from H to T, which was just as hard terrain, but took a day less. Often he would wake, still exhausted and confused, and wonder for a moment whether he was half way down, or a third of the way up, or two thirds of the way up. But he never once reasoned that in the long run he would end up at H twice as often as T…

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  195. Logicophilosophicus says:

    Richard,

    “I went back to find your gold coin argument. It is indeed analogous to the SB problem, but it also fails to support the halfer position. Once you formalize the probabilities, it is easy to see why…
    Au1 = this instance is the first instance of seeing a gold coin
    Au2 = this instance is the second instance of seeing a gold coin”

    Well I went back to my argument because I was sure I deliberately and explicitly said something entirely different…
    “G(1) does not mean ‘the first gold coin to be taken from the box': it is the identity of the coin (e.g., if you like, a microscopic number scratched into its surface). The temporal distinction between M and T, concealed by induced amnesia, is fully analogous to the physical distinction between G(1) and G(2), concealed by perfect similarity of manufacture.”

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  196. Logicophilosophicus says:

    Daniel Kerr,

    “Logicophilosophicus, as I said before, my patience ran thin. Your argument against the betting analogy is that a bookmaker wouldn’t take a bet for an event that has NOT YET transpired. This is a non sequitur and I really have no response to that.”

    I believe I actually asserted the very opposite:

    “But what you have totally failed to take into account is that SB’s induced confusion over the day – Monday or Tuesday – is entirely confined to SB’s mind. If you can find me a bookmaker so improbably generous as to take a bet on the winner AFTER the stewards have ruled on the photofinish, I’ll get rich, too.”

    As I indicated in my reply to Richard, I thought I did have an understanding of conditional probability: p(I|H) is the probability of an interview, given that the result was (i.e. in the past) heads. I think p(I|T)=1, and p(I|H)=1. What is your understanding? And what does p(H|I) imply?

    (BTW, just a quibble: The statement about when a bet can be accepted is just, as I stated, a “universal understanding”. I didn’t deduce it, illogically or otherwise, from other propositions, so whatever else it is, it cannot be a “non sequitur”. If you actually “like logic”, that is an odd mistake to make. Well, universal except for one or two people here. That was EXACTLY the thrust of my “What is an a posteriori probability?” remark.)

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  197. George Ellis says:

    As several commentators have said, this post and the discussion here essentially deals with intriguing classical probabilities, not quantum theory. If we return to quantum theory and the Everett proposal as in the last post, the key statement there is:
    “Like the textbook formulation, Everettian quantum mechanics also comes with a list of postulates. Here it is:
    1. Quantum states are represented by wave functions, which are vectors in a mathematical space called Hilbert space.
    2. Wave functions evolve in time according to the Schrödinger equation.”

    But that is *not* what happens in the steps set out in the inset calculation later in that post. There are four steps in that process. I leave out the common factor of (1/ \sqrt{2}), which makes no difference, and give both the written description of the process, and the short summary given in the diagram (square brackets)
    1. Everything is in its starting state, before the measurement:

    Observer: |O_o>, electron: (|up> + |dn>), Apparatus: |A_o>, environment:|e_o> .
    2. The apparatus interacts with the system to be observed and becomes entangled. (“Pre-measurement.”) [Apparatus measures]:
    |O_o> (|up> + |dn>) |A_o>|e_o> -> |O_o> (|up>|A_up> + |dn>|A_dn>) |e_o> ) (1)
    3. The apparatus becomes entangled with the environment, branching the wave function. (“Decoherence.”) [Decoheres]:

    |O_o> (|up>|A_up> + |dn>A_dn>) |e_o> -> |O_o> (|up>|A_up>|e_up>| + |dn>|A_dn>| e_dn>) (2)
    4. [Self -locating uncertainty]:
    |O_o>(|up>|A_up>|e_up>| + |dn>|A_dn>| e_dn> ) = |O_o>|up>|A_up>|e_up> + |O_o>|dn>|A_dn>| e_dn> (3)
    5. The observer reads off the result of the measurement from the apparatus. [Measurement complete]:
    |O_o>|up>|A_up>|e_up> + |O_o>|dn>|A_dn>| e_dn> -> |O_up>|up>|A_up>|e_up> + |O_dn>|dn>|A_dn>| e_dn> (4)
    The overall process is thus
    |O_o> (|up> + |dn>) |Ao>|eo> -> |O_up>|up>|A_up>|e_up> + |O_dn>|dn>|A_dn>| e_dn> (5)
    and so the wave function has split into the two possible options (with equal weights).

    Now the intriguing point is that, despite statement 2. above, at no point in this set of steps is the Schrodinger equation used. On the contrary, this process involves a series of three irreversible, hence non-unitary, projections.
    Take step (1). By the linearity of quantum theory (|up> + |dn>) |A_o> = (|up> |A_o>+ |dn>|A_o>), so the essential dynamics is
    (|up> |A_o>+ |dn>|A_o>) -> (|up>|A_up> + |dn>|A_dn>) (6)
    This is in fact double non-unitary projection, each projection occurring with numerical factor unity, so representing equal probabilities of these two projections:
    |A_o> -> |A_up>, |A_o> -> |A_dn>. (7)
    This is irreversible (and hence non-unitary) because you can’t determine the initial state |A_o> from the final states |A_up> and |A_dn>. The information needed to determine |A_o> simply isn’t there at the end. Essentially the same happens at stages (2), which is irreversible projection of |e_o>, and stage (4), which is irreversible projection of |O_o>. Stage (3) does not involve projection and does not have to be inferred by means of some philosophical assumption (“Self -locating uncertainty”): it just follows straightforwardly from the linearity of quantum theory.

    Now one might try to follow the decoherence route and say, things are in fact unitary, so while there are indeed other terms in these equations, they are not shown because they are so incredibly small that we can neglect them and effectively get the progression above by use of the Schrödinger equation alone, even though the dynamics (1)- (4) is non-unitary. There are extra very small terms in these equations that have been deleted because they are of no consequence.
    But it is precisely that step of deleting these very small terms that makes the whole dynamics irreversible and hence non unitary. This is essentially the same as in the case of the claim that if we could reverse precisely the velocities of every atom in a broken wine glass and its surroundings after it has fallen on the floor, it would jump back up and reassemble on the table to reform the pristine object. This reversal is theoretically possible, even if not remotely practicable, because the underlying dynamics is Hamiltonian. However if you neglect these incredibly small velocities of the remnants and their surroundings and set them to precisely zero, that theoretical option is no longer available. But that is what the above progression amounts to.

    In summary: the dynamic progression (5) does not follow from assumptions 1. and 2., as claimed. It has extra assumptions which in effect amount to wave function collapse (see (6) and (7)).

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  198. Johan says:

    Sorry, I did not have the courage to read all comments, so this has probably been said many times before. Nonetheless, why do so many again and again let themselves be fooled by playing the silly wordgames philosophers are so renowned for.

    The answer to the question “What is the probability you would assign that the coin came up tails?”” is of course 1/2.

    The answer 1/3 is the correct answer to three totally different questions:
    1)” What is the probability that you are now being interviewed on a monday and won’t be interviewed again on tuesday?”
    2) “What is the probability that you are now being interviewed on a monday and will be interviewed again on tuesday?”
    3° “What is the probability that you are now being interviewed on a tuesday?”

    And of course, the answer to the question “What is the probability that you are now being interviewed on a monday” would have to be 2/3.

    The fact of being interviewed on a particular day does not influence the outcome of tossing a coin one or two days before. Causality.

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  199. DN says:

    Johan,

    You write:

    The fact of being interviewed on a particular day does not influence the outcome of tossing a coin one or two days before. Causality.

    No, but the fact that you are being interviewed may certainly influence your credence in the outcome of the tossed coin, particularly when said fact depends on the outcome of said toss.

    Consider the extreme revision where you are not interviewed at all after heads. Then your credence in heads is 0% if/when interviewed.

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  200. Johan says:

    DN writes “No, but the fact that you are being interviewed may certainly influence your credence in the outcome of the tossed coin, particularly when said fact depends on the outcome of said toss.”

    OK, in that case I would call it a “mind game”.
    Suppose (when heads) you don’t wake up SB two times in identical circumstances, but instead three times, four times, etc. … ad infinitum?
    The “credence” as you call it of coming up tails would go from 1/3 to 1/4 to … in the limit zero.
    From a “psychological” point of view I can understand why SB would want to believe that (and fairy tale figures have the advantage of living forever), but in that case I would call it misleading to refer to probability (theory).

    As for the connection with the Many Worlds Interpretation and Multiverses, it’s always tempting to get human consciousness involved, but that’s a can of worms I’m not going to open.

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  201. Daniel Kerr says:

    Logicophilosophicus, you deduced betting conventions from a “universal understanding” that is itself an external proposition to the problem. It’s a non sequitur, again analogies are not helping your argument. Stick with the math, which is the formal translation of the problem. As for my statement, yes I meant to say “already transpired,” haha, don’t know why I said the opposite. Clearly an actual bet usually takes place before the chances have played out.

    My understanding of P(H|I) is exactly as calculated in my very first comment. It’s equal to P(I|H)*P(H)/P(I). P(I|H) is clearly the probability of the interview given the coin flip is heads, which is equal to 1 in this problem. P(I)=3/4 since P(I|Wake_1 & H) = 1, P(I|Wake_1 & T) = 1, P(I|Wake_2 & H)= 1, P(I|Wake_2 & T) = 0. So using arithmetic, I’m dividing 1/2 by 3/4, which is 2/3.

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  202. Richard says:

    Philosophicus, you wrote:

    “p(H|I) seems to be the probability of getting a heads result given that an interview has already taken place. How does the protocol allow that sequence? There’s more that a hint of time travel there.”

    Thanks, that insight makes everything perfectly clear. I can now state with full confidence that “The Sleeping Beauty problem is a probability puzzle that separates people into two groups: those who understand conditional probability, and halfers.”

    You also wrote this:

    “To reason backwards in time is a little odd, but just because SB can’t tell the difference between these three actually quite distinct interviews cannot make the I in the second branch lead back to an H in the first branch (we can’t jump from one branch to another); nor especially can it make the two I’s in the first branch lead back to two H’s, creating a third branch out of nowhere. Reasoning back in time from, say, the second I in the Heads branch leads back through the first I to the single H.”

    Thanks to that masterful deconstruction, we can now dismiss the whole silly “science” of forensics and empty the prisons of those it has wrongly been used to convict.

    As nobody with a grasp of conditionalization has answered any of my three challenges with a viable defense of halferism, the thirder position is unvanquished. Game, set, match. My work here is done.

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  203. Daniel Kerr says:

    George Ellis, provided the measurement guarantees that the pre-measurement state is unchanged by the measurement and as long as the Hamiltonian is not separable in the observables corresponding to the apparatus, environment, observer, and target, then you cannot break up the tensor products to analyze unitary evolution. It is certainly possible that:

    U(t-t’)(|up> + |dn>)⊗|A_o> = (|up>⊗|A_up> + |dn>⊗|A_dn>)

    This can very well be a result of unitary evolution. The state,
    (|up> + |dn>)⊗|A_o>, can be uniquely recovered by time reversal of
    (|up>⊗|A_up> + |dn>⊗|A_dn>) under the same unitary operator. In this case, clearly the unitary operator U cannot be decomposed onto U_spin and U_apparatus unitary operators.

    However when coupling to the environment and assuming a general initial state (mixed or coherent superposition), asserting that the nondiagonal entries vanish is not as defensible.

    Regarding Logicophilosophicus, I didn’t even notice his interpretation that P(H|I) is “heads following an interview” in time. Rest assured, I think LP has acknowledged his unfamiliarity with conditional probabilities.

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  204. Daniel Kerr says:

    Correction to/clarification of above, meant to say it’s not as defensible that the off-diagonal entries (when traced over the environment) vanish.

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  205. Logicophilosophicus says:

    p(H|I) – which “I”?

    You both needed to look at those independent branches. There is no die or coin or roulette wheel where the 3 I’s are presented for selection and have similar connections to H. The I in the T->I branch precludes H. The second I in the H… branch is 100% dependent on H->I…

    The SB and the BS both know that the fact of waking does not change the probability of Heads, because they experience the confusion but also know the protocol. You are both apparently incapable of realising that the conclusion that H is always more probable than tails is paradoxical, which as Euclid could tell you means that (assuming the symbolic manipulation obeys the rules) means that the premisses are at fault (GIGO).

    I pointed out that you both attribute assertions to me which are the opposite of what I stated, but get no explanation. I asked for some fairly straightforward explanations but you prefer to be abusive.

    Well, boys, enjoy. I’m out of here.

    [Ironically, the typical problem with forensic evidence is the conditional probability fallacy. Not that that is the problem here.]

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  206. Daniel Kerr says:

    Logicophilosophicus, there’s no abuse here, I’m not disliking your comments solely because I disagree with them, I’m not disliking them at all actually. Conditional probability has nothing to do with time in the same sense logical inference does not. I preceding H/T is not a problem. Again, the problem is not asking you to determine the probability of a coin flip given this experiment, that is clearly 1/2 no matter what. The problem is asking you to determine the most likely outcome given the sampling scheme SB undergoes in the experiment. There is no reason why this is necessarily 1/2 independent of whatever the experiment is. We’re not saying an absolute coin flip has higher odds of Heads because of how you observe it, we’re saying that how you selectively count the result of coin flips will affect the apparent odds you measure.

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  207. Richard says:

    Daniel, I think when he mentioned “conditional logic fallacy”, he was intending to mean the base rate fallacy. I don’t know why he thinks that is ironic. The base rate fallacy is a fallacy committed by those who don’t understand conditionals, not by those who do (unless they are careless). So it is ironic that he brought it up, but it is not ironic in the context of valid arguments about conditionality.

    I don’t know what statement I attributed to him that he did not actually make so I can’t respond to that accusation.

    On the whole, he is committing a petitio principii fallacy, as shown by his last post:
    “You are both apparently incapable of realising that the conclusion that H is always more probable than tails is paradoxical, which as Euclid could tell you means that (assuming the symbolic manipulation obeys the rules) means that the premisses are at fault (GIGO).” Translation: The logical conclusion is inconsistent with halferism; halferism is correct (therein lies the petitio ptincipii); this means that the logical conclusion that halferism is wrong must be based on erroneous premises; therefore, halferism must be right after all.

    Had he wished to engage in an honest argument, he could challenge any of the anti-halfer assumptions, independently of conclusions. This has been tried: I have seen an article advocating rejection of the assumption that one can conditionalize on temporally indexed discoveries (or as a corollary, losses of information such as what SB experiences). Certainly he could have gone that route, but ultimate acceptance requires changing the standard axioms of probability and so would require a stronger case than was made in that article.

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  208. Daniel Kerr says:

    Richard, I was forgetting the name of that fallacy, thanks! Like I said in a previous comment, he assumed the conclusion from the beginning. Still, it reaffirmed my understanding of it by tackling these fallacies, so I got something from the discussion whether it was honest or not.

    I’m interested in Ellis’s post above more right now actually, wondering how justified orthogonality of the environment really is for a measurement where the measured object’s state is not changed by the measurement. For a weak measurement, one where the measured state would not be influenced, I definitely would think e1 and e2 would be highly correlated with each other and e0 except in the quantum numbers that correspond to the apparatus and observer states.

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  209. Bob Rehbock says:

    Amazing how much this has stirred up. We have only four outcomes possible. new information is knowing that she is awake thus only three outcomes are possible. Asking her the probability among each of four choices before she sleeps and asking her the probability when she knows only three equally probable choices remain changes her subjective probability from 1/4 to 1/3. Why is that wrong or difficult?

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  211. Hal Swyers says:

    The correct probabilities

    P(Tails given Tuesday) = 1
    P(Tails given Monday) = 0.5
    P(Heads given Monday) = 0.5

    This is a pretty straightforward calculation.

    http://thefurloff.com/2014/08/14/sleeping-beauty-is-smarter-than-most-physicists/

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  213. Daniel Kerr says:

    Hal, that blog has to justify why P(M)=2*P(T). See my calculation in my first comment for a correct application of conditional probabilities to the problem. I assumed P(M)=P(Tu), but P(I|M)=/=P(I|T), as in, I assumed that interviewing was a different set in the venn diagram than the days themselves. That Venn Diagram in that blog is too hastily put together, P(M)=P(Tu) in the experiment if you remove SB’s biased sampling from the problem. SB’s sampling should be yet another circle on the diagram overlapping H, T, M, Tu circles. The setup of the problem is thus incorrect as laid out in that blog.

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  214. Richard says:

    Hal, Daniel’s points are correct.

    I showed (Aug. 7, 3:46 pm post above) that the halfer position leads to an inconsistency (the conclusion that no interview can take place on Tuesday), given only the following assumptions, which most people will agree are reasonable given the problem statement:

    p(H|MI) = 0.5
    p(T|MI) = 0.5

    (i.e., it should be equally probable that the toss was heads or tails, if SB is interviewed on a Monday, and these outcomes are mutually exclusive and exhaustive under that condition).

    p(H|TuI) = 1.0
    p(T|TuI) = 0.0

    (i.e., it should be certain that the toss was heads and not tails, if SB is being interviewed on Tuesday).

    I did not prove that the thirder position is correct. In fact, the above assumptions are not sufficient to show that it is. I did, however, show that the thirder assumption avoids the contradiction that the halfer assumption leads to. To prove the thirder position, you need the additional assumption that

    p(M|I) = 2 * p(Tu|I)

    although I think most people would also grant that this assumption is reasonable (twice as many interviews will take place on Monday as on Tuesday, given a large number of trials of the experiment).

    Conditioning on interviews rather than days of the week allows for proper normalization of the calculated probabilities; it is not correct to say that Monday is twice as likely as Tuesday.

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  215. phayes says:

    Richard

    You don’t actually need that additional assumption (re-link).

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  216. Richard says:

    Phayes,

    I do see an assumption in your second set of equations:

    Pr(Mon|.) = Pr(Mon|Heads, .) Pr(Heads|.) + Pr(Mon|Tails, .) Pr(Tails|.)
    = Pr(Heads|.) + (1/2)*(1 – Pr(Heads|.))

    (Heads/Tails roles are interchanged between your document and Sean’s statement of the problem.)

    This requires that Pr(M|T,.) = 1/2, which is not explicitly given nor can it be derived from other givens. (Sure, it’s intuitively obvious, but we should declare our obvious intuitions as assumptions.) I believe it is equivalent to my additional assumption, although I don’t have time for the derivation right now.

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  217. phayes says:

    Richard,

    Yes it’s an assumption – at least it is if you regard the application of the principle of indifference here as an assumption rather than as a consequence of a consistency requirement [á la Jaynes, chap. II].

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  218. Daniel Kerr says:

    Richard, I derived the thirder position using only the following probability assignments:

    P(Wake|Mon) = 1
    P(Wake|H and Tue) = 1
    P(Wake|T and Tue) = 0
    P(H) = P(T) = 1/2
    P(Mon) = P(Tue) =1/2

    If the last one is the assumption you speak of, I’m not sure I’d agree that it’s an assumption. The days are a random variable in this problem with equal likelihood from the perspective of the experimenters (though ignoring their memories).

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  220. Hal Swyers says:

    I have provided some more explanation of the correct calculation and what happens when Wednesday is excluded.
    One thing I want to comment on. There appears to be a lot of confusion about how to include Tuesday into the calculation. The problem statement is pretty clear, the probabilities conditioned on Tuesday depend on whether Sleeping Beauty has been told its Tuesday. This is entirely from Sleeping Beauty’s vantage point. The probabilities would be different from an external observer’s vantage point who knew exactly what day it was. Sleeping Beauty has no knowledge of whether it is Monday or Tuesday unless she is told.

    The other point of confusion, which I address in the article below is that Sleeping Beauties belief of the probability of heads is 50% unless she is told what day it is. The probability of a fair coin is not dependent on the day.

    http://thefurloff.com/2014/08/15/sleeping-beauty-and-the-importance-of-wednesday/

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  221. Daniel Kerr says:

    Hal, you haven’t included SB in your Venn diagram at all though, you’ve skipped to collapsing it to her point of view. You have to build the Venn diagram as outlined by the experiment first, then draw SB’s circle overlapping the correct regions. You’ve skipped this step.

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  222. Richard says:

    phayes,

    Yes, I regard indifference as an assumption, for the sake of completeness. The argument I gave renders the halfer position inconsistent with the assumptions I originally gave, and compels the thirder position with the additional assumption. If halfers want to refute the thirder position, they only must deny one of the assumptions (including this one). If they want to assert their own position, they must deny one of my original assumptions — denying the indifference principle as applied to M/Tu probabilities is not sufficient for that purpose.

    I don’t know of any argument that the principle of indifference is mandated for consistency, so my support for the principle of indifference is simply on the basis of common sense.

    Daniel:

    “P(Wake|Mon) = 1
    P(Wake|H and Tue) = 1
    P(Wake|T and Tue) = 0
    P(H) = P(T) = 1/2
    P(Mon) = P(Tue) =1/2″

    Well, I would prefer to see things conditioned on interviews (or at least, on SB being awake), but I will assume that underlies these implicitly. Your first equation follows
    directly from the problem statement. So do the second and third. The fourth is in the problem statement if the problem calls for a fair coin, but by convention I think we can agree that it is the intent if not stated otherwise.

    I find p(M) = p(Tu) = 1/2 to be a little more problematic, since it doesn’t condition on being awake or being interviewed. I went with “p(M|I) = 2 * p(Tu|I)” to avoid questions about measures and normalization, or of how sampling bias is handled in the calculation (it expresses the bias explicitly and gives the same result). Phayes’s assumption “Pr(M|T,.) = 1/2″ likewise avoids such issues. (By the way, I assume his “.” is the same as my “I”, and I know he is using H and T in the reverse sense of Sean’s version.) But yes, I would label it an assumption and not a given condition, lest someone comes along with a nonstandard notion of probability. Then they are free to deny it, but let them justify the denial.

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  223. Hal Swyers says:

    Daniel,
    I think its pretty clear in the problem statement that we are interviewing SB and not the interviewer.

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  224. Richard says:

    Hal, in your third article you wrote:

    “This means the probability of heads or tails given Monday changes dramatically.
    P(H|M) = P(H & M)/P(M) = 0.5/0.66 = 0.75
    P(T|M) = P(T & M)/P(M) = 0.167/0.66 = 0.25
    The reason for the shift is that we are no longer allowed to consider P(T & M) to be a simple product of P(T) x P(M) . The correlations between the two mutually exclusive sets are too strong.”

    This is the kind of error trap you can fall into when you don’t condition on the sampling criterion (what I called “I” for interview). You argued that P(H&M) = 0.5 because you did not allow for the equal overlaps (assuming indifference :-) between H and M and between H and Tu. That SB is not awake to witness the latter case, so that there is no “I” in “H and Tu”, is what brings the probabilities for the remaining combinations to 1/3 each when you conditionalize properly.

    A Wednesday awakening isn’t relevant to SB’s calculation at all, assuming she will be told it’s Wednesday and no interview will take place on that day. Then when she is interviewed on Monday or Tuesday, she knows it isn’t Wednesday and does the calculation just as if there will be no Wednesday for her.

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  225. Ramanujan says:

    A question for halfers [perhaps some variant of this already appears in the thread]: Suppose that we change the SB puzzle, but only in the protocol for Tuesday after tails (T/t). The new protocol is that she is awakened as in the other cases, but, after a pause of a minute, is put back to sleep (but only on T/t). Again, she knows the full protocol in advance. In the first minute after awakening, she assigns equal probabilities for M/h, T/h, M/t, T/t. After a minute, if she is still awake, she excludes T/t and is left with equal probabilities for M/h, M/t, T/h. So in this case the 1/3 is correct. Do you agree with this result for the modified problem? If so, how can the changed protocol for T/t affect her relative weightings of the other three cases?

    The main halfer argument is that when she awakens she gets no new information. But she does: by awakening she learns that the current situation is not T/t. In the modified puzzle, she gets the exact same information but in a different way.

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  226. Richard says:

    Ramanujan,

    “The main halfer argument is that when she awakens she gets no new information. But she does: by awakening she learns that the current situation is not T/t. In the modified puzzle, she gets the exact same information but in a different way.”

    That is indeed what they claim, but the error comes from failing to realize that information loss (not just gain) affects conditionalization. Consider that on Sunday, she knows that “It is before the effects of the toss on SB have been realized” (I will call this information “B”). But on each awakening, she does not know B. If the odds for tails vs. heads are 1:1 with B, then without B, the odds need to be multiplied by p(B|H)/p(B|T). And (assuming the principle of indifference), that multiplier is equal to 0.5/1.0. Those probabilities are derived from background information which she retains throughout the experiment, rather than information gained on waking. So the new odds are tails:heads = 1:2, or 1/3 probability for tails. Now, if she is told B after the interview, she can reverse that calculation and recover the original odds.

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  227. Daniel Kerr says:

    Ah, I see your point Richard. I’m treating the days as random variables for the experiment and I did not condition them on interviews. The way I see it, the experiment is carried out on 1 of 2 days, and at any given time, it could be one of those days. Perhaps it makes sense to condition them on the interview them just to clarify them.

    Hal, I’m not suggesting you interview the interviewers, I’m suggesting you build the full probability sample space before you collapse it down to a conditional probability.

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  228. Hal Swyers says:

    Nothing more to add

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  229. Ramanujan says:

    I have had an interesting exchange with another blogger, a proponent of 1/2, about my delayed-information experiment four comments above. He agrees that before we get the information that it is or is not Tuesday+tails, the conditional probability P(tails | Tuesday+tails) = 1/3. The definition of P(A | B) is that if we are given the additional information that B is true, then the prior P(A | B) becomes the new P(A). He argues that this is not true, that P(A | B) can change retroactively when we learn that B is true. I believe that this is simply not true, ever, that it runs directly counter to the definition of P(A | B).

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  230. Daniel Kerr says:

    The definition of conditional probability is P(A|B) = P(A&B)/P(B), I don’t think you can get any clearer than that. It never changes whether A or B is true or not, it doesn’t “collapse” any more than “if” statements “collapse” in modus ponens.

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  231. Ramanujan says:

    typo in previous post: P(tails | NOT Tuesday+tails) = 1/3

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  232. Mark Srednicki says:

    Very late to the party here, but last week I was discussing this with Sean at the Quantum Foundations conference, so I can’t resist throwing in my 2 cents.

    I agree with Joe that the “phenomenalist” position is the correct one. It’s worth noting that, according to wikipedia, the phenomenalist position was originally articulated in a post by “Alice Atlas (Zorti Zandapheri)” (that’s what it says on the author’s facebook page) on the rationalist website Less Wrong: http://lesswrong.com/lw/32o/if_a_tree_falls_on_sleeping_beauty

    However, I think Joe misstates Alice’s main point. She did not argue that the “thirders” are correct, but rather that the answer to the question “What probability should Sleeping Beauty assign to the coin having come up tails?” depends on what Sleeping Beauty is going to do with this probability. Sean and Joe believe that the betting protocol (outlined by Joe) is obviously what was meant. But other meanings are possible. For example, after many runs of the experiment (with no betting), Sleeping Beauty could decide to look at the experimenter’s logbook of coin tosses, see what fraction of those came up tails, and compare her probability assignment to that fraction. (The first time I heard about this problem, this is what I thought Sleeping Beauty should do.) This protocol leads to the “halfer” position. Alice points out that there would be no controversy if the appropriate meaning of “Sleeping Beauty’s probability” was clearly spelled out at the beginning.

    Way to go Alice! You have, in my opinion, clearly outthought a whole bunch of professional philosophers, none of whom (as far as I know) made this key point.

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  233. Daniel Kerr says:

    The problem statement, as far as I know, doesn’t ask SB for the probability, it asks her what her belief is for the coin that determined her being asked this question had turned up heads.

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  234. Mark Srednicki says:

    Daniel, read the original post. Sean writes:

    Each time she is asked a question: “What is the probability you would assign that the coin came up tails?”

    But as Alice Atlas pointed out, there are several possible (more precise) versions of this question. Here are two (quoting Alice):

    Each interview consists of Sleeping Beauty guessing whether the coin came up heads or tails, and being given a dollar if she was correct. After the experiment, she will keep all of her aggregate winnings.

    Each interview consists of one question, “What is the limit of the frequency of heads [in the experimenter’s logbook] as the number of repetitions of this experiment goes to infinity?”

    (Here I’ve added the text in brackets to further improve the clarity.)

    I think that almost everyone would agree that in the first more-precise version, guessing “heads” gets Beauty twice as much money (on average) as guessing “tails”. I also think that just about everyone would agree that the answer to the second more-precise version is “one half”.

    So I agree with Alice Atlas that the whole problem is simply one of imprecise language.

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  235. phayes says:

    Mark,

    They’re not “more precise” versions of the question: they’re different questions and thus different experiments. The first one would require S.B. to calculate a conditional probability representing her own state of knowledge, as in the original experiment, and the second one would require her to ‘calculate’ a probability representing the experimenter’s (or any other ‘external’ observer’s) state of knowledge.

    I believe the appropriate meaning of “Sleeping Beauty’s probability” is clearly spelled out at the beginning and I’m afraid I don’t see any outthinking of philosophers in that Less Wrong article. In all these baroque rationalisations of the “halfer” position (or worse!) all I see is errors and misconceptions – like that “no new information” one, the failure to (consistently) condition, the confusing of a probability of a coin flip outcome event with a coin ‘propensity’ probability, the failure to distinguish between an inference and a meta-inference,… Instead of [an attempt to justify why] a simple and correct conditional probability calculation [is wrong] we’re given modifications to the S.B. Problem accompanied by long-winded appeals to decision theoretic or phenomenalist requirements or whatever. In that Less Wrong article this sort of folly leads to the rejection of [the use of] probability theory in its full generality – as a theory of events and representations of (observer/information-dependent) states of knowledge.

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  236. JimV says:

    I agree with Mark that there is some ambiguity. For example, the answer to “What is the probability you would assign that the coin came up tails?” should, it seems to me, revert to 1/2 as soon as SB starts to get out of bed, since at that point she knows the experiment is over, and whether or not she was awakened previously, she has just been awakened as part of the experiment for the last time, which happens once (per experiment) for heads and once for tails.

    The model that I use (because it seems most natural to me) is:

    The probability that the coin landed heads and SB will be awoken for the first time is 1/2.
    The probability that the coin landed heads and SB will be awoken for the second time is 1/2. (Both events will always occur for heads.)
    The probability that the coin landed tails and SB will be awoken for the first time is 1/2.
    The probability that the coin landed tails and SB will be awoken for the second time is 0.

    Therefore when SB wakes it could be one of three different events which have equal probabilities, so the probability she should assign to any one of them should be (1/2)/(1/2+1/2+1/2) = 1/3. However once she knows the experiment is over (but before she discovers how many days the experiment has taken) then her last awakening is is one of two different events with equal probabilities.

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  237. Daniel Kerr says:

    Mark, Sean was paraphrasing the wikipedia article which asks about belief. I think it’s fair to say that the problem is trivial if it’s asking SB’s probability assignment of a coin flip in general and not her bias in sampling this particular coin flip. Supposing all variants are probably asking the same question, the thirder version of the question is consistent across all variants, while the halfer is not. In light of this, I would phrase your more precise question as follows:

    Each interview consists of one question, “What is the limit of the frequency of heads [in the experimenter’s logbook in which the coin flip results are indexed by the interviews of SB] as the number of repetitions of this experiment goes to infinity?”

    Clearly tails would be double counted in such a log book even though one experiment is responsible for both entries. If the question isn’t asking this, then it’s a trick question, but the halfer arguments in this comment section are answering the above precise version of the question and not your own.

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  238. Mark Srednicki says:

    Well it was philosopher David Lewis who first articulated the “halfer” position; Alice was simply summarizing it (and not endorsing it).

    If we specify that “Beauty’s probability” is defined by the betting protocol in which bets are made and paid on both days (and Beauty keeps all winnings), then the “thirder” position is validated by a simple calculation, and we’re done.

    At that point, the entire argument is about whether or not the original language was clear enough to imply this betting protocol unambiguously. You think it was; I think it wasn’t. But an argument over precision of language is not one I’m very interested in.

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  239. Mark Srednicki says:

    Note for clarity: my last comment was in reply to phayes; the comments by JimV and Daniel Kerr were posted while I typing it.

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  240. Mark Srednicki says:

    Daniel,

    I do not accept your rephrasing; the logbook I had in mind shows the result of the coin flip, as seen by the experimenter (who only flips the coin once for each run). If the coin is fair, on average half the entries are heads and half are tails.

    I repeat that, when I first read the SB problem, my instantaneous reaction was that Beauty’s probability should be 1/2, because half the entries in that logbook are tails, end of story. It just doesn’t matter how many times she gets woken up. So you would say that I misread the problem. I would say (following Alice) that greater precision in language would have cleared things up.

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  241. Daniel Kerr says:

    Mark, I read the problem the same way as you did initially as Sean posted it, I did not reach the thirder position until I read the wikipedia link and calculated it myself. Again, my “precise” wording of the problem is following wikipedia’s phrasing. Alice’s “precise” wording is of a different question. Both are equally valid. The original “probability” phrasing is vague as it does not specify if its SB’s conditional probability we are finding or the absolute probability. The latter is trivial, so I assume the problem asks for the former. And again, halfers assume the former as well by and large.

    And it’s important to recognize that SB’s conditional probability is 1/3 whether she’s betting or not. And that’s because the interview logbook double counts tails. This is the logbook SB has access to, not the experimenter’s logbook. Even still, if the experimenters organized the logbook so that the result of the coin flip was indexed by days, then both heads and tails would be double counted. If we then looked at the subset SB was actually interviewed over, 1/2 of the heads results would not be included. This is SB’s logbook of the experiment.

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  242. Mark Srednicki says:

    Mark, I read the problem the same way as you did initially as Sean posted it

    This is just more experimental evidence that the language of the problem is not unambiguous.

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  243. Daniel Kerr says:

    Researching the history of the problem, it’s evolved from the paradox of the absent minded driver, which asks about maximizing a payoff between choices, so I’m not going to fault Sean for restating a problem using vague language!

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  244. Bob says:

    Hidden due to low comment rating. Click here to see.

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  245. Peter says:

    Hi,

    I have some naive layperson questions regarding the multiverse interpretation of quantum mechanic. I’m writing a SF novel and I would like to use this interpretation, but a few points are still not clear…

    I don’t understand why we observe interferences. In the sense that if I’m in universe A, where the result of measuring M will give M_A (the value of M in universe A, that’s why we call it universe A in the first place), why would some experiments give me the impression that M is in a superposition of several states? I guess you will say that’s what we observe and this is quantum mechanic dude… But I thought I would ask anyway, in case I’m missing something…

    Also, what does this interpretation of QM have to say about Delayed Choice Quantum Eraser Experiment? If I understood it well, this experiment says that a photon (well, a pair truly) correlated with another one (another pair), knows what its buddy is going to experiment, let’s say, 10 seconds before it experiments it. If we assume that the multiverse interpretation of QM is true, what does it have to say about this? How would a parent explains to his children the delayed choice quantum eraser in the scope of the multiverse interpretation? “Listen daughter/son, there will come a time where delayed choice quantum erasers will keep you awake up to late at night, so hear my piece of advise on this topic…”

    PS: English is not my first language, so excuse the strange formulation from time to time…

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