• JeffJo
    73
    Probability is about expectations.
    No, it isn't.

    Statistics is about expectations. Statistics uses Probability Theory to calculate expectations. In conventional experiments, we find that the value of the expectation is the value of the probability. The concepts of "probability" and "expectation" are very different.

    The Sleeping Beauty Problem bends conventionality, and makes statistics inapplicable, by making the number of trials depend on the result for which you are trying to assess a probability. So arguments about frequency and expectation are meaningless until the "bent" issues are resolved.

    Specifically: people disagree about whether there are two, three, or four disjoint events that comprise Beauty's sample space. On Sunday Night, the answer is clearly "two" since how the concept "today" applies at that time is different than when she is awakened during the experiment.

    Halfers want the number to be "three" inside the experiment. They treat it like Tuesday doesn't happen if Heads is rolled. The problem with the rationale they use, it that it only supports "two," which nobody thinks applies within the experiment. The inconsistency in their argument is that Monday is different than Tuesday if TAILS was flipped, but not if HEADS was flipped. In that case, Tuesday does not exist.

    You reject the existence of Tuesday yourself, when you say things like:
    it sure does look like the design of the experiment involves conditioning heads on ~Tuesday,
    That is not what the condition is. It is {AWAKE}, which is can also be written as ~{HEADS&Tuesday} or {HEADS&Monday,TAILS&Monday,TAILS&Tuesday}.

    I've presented an alternative problem that unequivocally demonstrates the solution. You give no reason for "not getting .. its equivalence to SB." It is my opinion that you "do not get it" because doing so would make you change your answer. To "get it," all you need do is let Beauty be wakened on HEADS+Tuesday, but do SOTAI.

    One thing I'm generally uncertain about is how strongly to lean on "what day today is" being random.
    "Random" is not a property of what you are looking at in an experiment. It is a property of what you know about it, but can't see. Either because the experiment hasn't happened yet, or it has but you can't see what happened.

    Here's an example that I've used in this thread, that directly addresses your uncertainty: Forget the coin, and the two possible awakenings. Put Beauty to sleep, and roll a six-sided die. Call the result of the roll R. Leave her asleep on five days during the following week, waking her only after R nights (so R=1 means Monday, R=2 means Tuesday, etc.). When she is awake, ask her for her confidence that today is Wednesday.

    Is the answer 1, meaning "what day today is, is not random," or is the answer 1/6, meaning "what day today is, is random" ?

    The reason it is random, is because Beauty has no evidence of what day it could be. Does this sound anything like the Sleeping Beauty Problem to you?
  • Srap Tasmaner
    2k

    Thanks. We've moved on here, but I appreciate your thoughts. You've addressed a lot of the gaps in my understanding of this stuff, and I especially appreciate you taking the time to do that.

    Your scheme for removing the time element is pretty cool, and I'm going to spend some more time looking at it.
  • Andrew M
    459
    Still some things to puzzle through, but I'm convinced. My sojourn in the land of halferism is at its end.Srap Tasmaner

    Welcome back!

    I do still disagree about how to interpret this thing though. The failure rate of my tails-guessing Beauties is still 1/2, no matter how much they pat themselves on the back. The argument you give here totally justifies conditioning on being interviewed, so the epistemic issue isn't there; it's in this conflict between the two ways of measuring success.Srap Tasmaner

    I regard them as two distinct epistemic perspectives. One is Beauty's on Sunday (or Wednesday) who doesn't condition on being interviewed and the other is Beauty's on Monday or Tuesday who does.

    If you take a step back, SB looks a bit like a fucked up way of doing two trials of a single experiment. (No worries about the single coin flip -- the trial is asking different subjects for their credence.) But whichever way you split, by toss outcome or by day, it's not two trials: it's one trial each for two different experiments and which experiment is being run is determined by the coin toss, and is thus the source of Beauty's uncertainty.Srap Tasmaner

    Yes. You could say the setup is that there is one of two experiments which is randomly selected. In one experiment, only one random person gets interviewed. In the other, everyone does. You find yourself getting interviewed. So what is the probability that you're in the single interview experiment? 1/(1+N) where N is the number of people.

    As a result, it is structurally unlike a biased coin or regular betting scenario. There is no one-to-one mapping when the coin toss outcome itself has a different number of interviews (or agents) associated with it. Instead the sample space must properly account for those degrees of freedom with probability being the measure of which state an agent is currently in.
  • Srap Tasmaner
    2k
    Welcome back!Andrew M

    It was worth the wandering just for the return. I'm still kind of stunned by the elegance of the argument that convinced me. The symmetry of it. In standard SB, on (H & Tue) our Beauty receives no information at all, is not even conscious; in Informative-SB, she receives nearly all the possible information. We gather all of it together into one box -- and then close the lid. Just like that, transforming one into the other.

    It's quite beautiful. And a fine reminder to look for the general problem of which the one at hand is only a special case.
  • JeffJo
    73
    In standard SB, on (H & Tue) our Beauty receives no information at all.Srap Tasmaner
    But she isn't asked for her confidence in that situation, so this argument is a red herring. A very appealing red herring, as you go on to describe, but irrelevant nonetheless.

    The issue is, does she receive information on ~(H & Tue)? And the answer can't depend on how, or even if, she would receive information on (H & Tue).

    So, instead of the (H & Tue) protocol being "let her sleep," make it "take her to DisneyWorld." Just before she finds out where she is going, the Law of Total Probability says:

      Pr(H) = Pr(H|Interview)*Pr(Interview) + Pr(H|DisneyWorld)*Pr(DisneyWorld)

    Here,

      Pr(H) = 1/2 is the probability of Heads.
      Pr(Interview) = X is the probability she will be taken to an interview.
      Pr(DisneyWorld) = 1-X is the probability she will be taken to DisneyWorld.
      Pr(H|Interview) = Y is the probability of Heads in an interview (what the original problem asks for).
      Pr(H|DisneyWorld) = 1 is the probability of Heads at DisneyWorld.

    So we can now say that:
      1/2 = Y*X + (1-X)
      (1-Y)*X = 1/2

    If Y=1/2, as you believe, that means X=1. This is a contradiction, since it means she cannot be taken to DisneyWorld. Whether or not you accept that Y=1/3, the fact that there is a chance of going to DisneyWorld proves that Y<1/2.
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