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Featured I The Sleeping Beauty Problem: Any halfers here?

  1. 1/3

    12 vote(s)
    33.3%
  2. 1/2

    11 vote(s)
    30.6%
  3. It depends on the precise formulation of the problem

    13 vote(s)
    36.1%
  1. Jun 2, 2017 #1

    Demystifier

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    The sleeping beauty problem is a well known problem in probability theory, see e.g.
    https://en.wikipedia.org/wiki/Sleeping_Beauty_problem
    http://allendowney.blogspot.hr/2015/06/the-sleeping-beauty-problem.html
    https://www.quantamagazine.org/solution-sleeping-beautys-dilemma-20160129
    or just google.

    Allegedly, there are many "thirders" who think that the correct answer is 1/3, but also many "halfers" who think that the correct answer is 1/2. For me, it is quite obvious that the correct answer is 1/3. Is there anybody here who is convinced that the correct answer is 1/2? If you are one of them, what is your argument for 1/2?
     
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  3. Jun 2, 2017 #2

    fresh_42

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    What is your argument of a third? She will not be interviewed on Wednesday, which leaves two possible days.
    As the coin decides on which day(s) she will be awakened, her a priori probability for heads (= Monday only) is .5.
    Now the question is, whether her a posteriori probability changed during the experiment. However, due to her amnesia, there is no way for her to gain additional information, except it is neither Sunday nor Wednesday which she knew already before. Consequently her guess still cannot be better (or worse) than .5.
     
  4. Jun 2, 2017 #3

    Demystifier

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    Let me present an argument for 1/3, by analysing a similar problem from my actual real experience. Let us divide a day into a two equal halves, namely the range 24:00-12:00 and the range 12:00-24:00. When I awake during a sleep, at first I am not fully conscious, so at first I don't know what part of the day it is. Nevertheless, almost always I automatically assume that I am somewhere in the 24:00-12:00 range. And almost always I turn out to be right, despite the fact that, a priori, the two ranges should be equally likely. That's justified because I rarely go to sleep in the afternoon. (But sometimes I do go to sleep at afternoon hours, and in such cases, when I awake, at first I get confused when I see what time it is.)
     
  5. Jun 2, 2017 #4

    Demystifier

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    Are you a genuine halfer, or just a devil's advocate? :biggrin:

    Anyway, perhaps the best way to see why 1/3 is the right answer is to go to the extreme. Instead of awaking the Beauty n=2 times, let us awake her n=1.000.000 times in the case of tails (and only k=1 times in the case of heads). So when she is awake, she can be almost certain that it was tails, so the probability for heads must be very small. And if n=1.000.000 is not convincing enough, consider the limit ##n\rightarrow\infty##. If this is still not enough, vary also k and consider the limit ##k\rightarrow 0##.
     
  6. Jun 2, 2017 #5

    fresh_42

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    But this changes her a priori chances: she knows beforehand that heads is basically impossible. If we agree on the fact, that the two probabilities (before and after) don't differ, then the question is only what she can say on Sunday, even without the entire experiment. And these chances are even. I simply don't see how additional information should enter the system, like it does in the three doors riddle.
     
  7. Jun 2, 2017 #6

    PeroK

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    It depends how you define "credence". You could extend the problem thus:

    If she guesses wrong, then another coin is tossed and if it's tails she is executed. Now, if she guesses heads every time, then she has a 50-50 chance of being wrong but only once. However, if she guesses tails every time she has a 50-50 chance of being wrong, but she will be wrong twice, so has an increased chance of execution.

    This leads to a bias for heads over tails.

    I would compare it to a similar scenario: toss a coin and ask one or two different people at random. If the people know this rule then the fact that they have been picked favours heads. If they don't know this rule, they must go 50-50.

    The princess is effectively two different people on account of the amnesia.
     
  8. Jun 2, 2017 #7

    PeroK

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    PS change the rule so that she gets woken every day for a year (heads) or only once a year (tails). I think that blows the 50-50 argument.
     
  9. Jun 2, 2017 #8

    fresh_42

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    Wouldn't this contradict
    So if you have a different person every day of the year, it is still 50:50 for each single person assuming they cannot communicate.
     
  10. Jun 2, 2017 #9

    PeroK

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    Not if they know the rule about two or more people getting woken if it's heads. I see I got the role of heads and tails mixed up, but if I persevere with the way I wrote it.

    If you toss a coin and if it's heads you ask everyone in Belgium, but if it's tails you pick one person at random and ask only them. If everyone knows this rule, even if they cannot communicate, they know that by virtue of being asked it is almost certainly heads.

    If there were a cash prize for being right, it would be foolish to guess tails.
     
  11. Jun 2, 2017 #10

    PeroK

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    PS having read a bit more. One of the halfer arguments is that the princess "learns nothing new" about the coin when she is woken, so should stick with 1/2. Therefore, being woken involves learning nothing new.

    Now, change the problem so that she is only woken if it is heads and not tails. She has still learned nothing new if she is woken, so sticks with 1/2, even though it's now 100% heads.
     
  12. Jun 2, 2017 #11

    TeethWhitener

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    I don't think this is obvious at all. The probability that she will be woken up is 1 (regardless of whether the coin is heads or tails). She has no way of knowing whether she has previously been woken up, so I don't see how having n=a zillion, k=1 changes anything at all. I think of it this way: let's say we have a two-day period. We flip a coin, and if it's heads, sleeping beauty is awake for the entire 2-day period. If it's tails, she's awake for a random subset of that period whose measure equals 1 day. At some point while sleeping beauty is awake, we receive a note in that period saying "Sleeping beauty is awake right now." At least as far as I'm interpreting the problem, we receive the note regardless of what the outcome of the coin toss is. I can't see how we've gained any knowledge about the situation by receiving the note, and yet to me, the problem is essentially equivalent to the initial problem.
     
  13. Jun 2, 2017 #12

    fresh_42

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    All these counter arguments against the .5 fraction involve a change on the set up and the knowledge of it. Therefore it doesn't affect the original question with a .5 chance plus the argument, that there is no additional information available during the experiment. It's exactly this lack of information that makes the situation different from the standard three doors example for conditional probabilities.
     
  14. Jun 2, 2017 #13

    PeroK

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    The flaw in your argument is this: The question you imagine asking her at the start is:
    If I toss a coin now, what is the probability it is heads? To which the answer is 1/2.

    But, later, this is not precisely the question. The question now is: We have woken you up ...

    If you asked her the same question at the beginning it would be:

    If we toss a coin and wake you up twice if it's heads and once if it's tails, then at the time we wake you,what is the probability it is heads?

    To which the answer is 2/3.

    So, the answer to the precise question she is asked under the precise circumstances is 1/3 - 2/3. It is never 1/2 to begin with.

    That's the fundamental flaw. The answer to the question is not 1/2 in the first place, so has no need to change. The information is all there at the outset about the precise circumstances under which the question will be asked.

    The question is always asked under conditional circumstances. It is never asked under the unconditional circumstances of a single coin toss.
     
  15. Jun 2, 2017 #14

    PeroK

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    PS and, moreover, it's never 1/2 for the experimenters. It's always known to them by the time they wake her and ask the question.

    When the question is asked the coin toss is known. But, the princess has the knowledge she always had that when she is wakened the coin is more likely to already have been a head than a tail.

    The experimenters know this because they know that if they are waking her it is more likely to have been a head.
     
  16. Jun 2, 2017 #15

    fresh_42

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    Let's take the Wikipedia setting: heads = Monday only ; tails = Monday and Tuesday
    The chances to get awoken twice equals the chances to get awoken once. That doubles the chances for a Monday.
    However, in case of Monday she still don't know anything about the coin flip. But she will be asked: "What is your credence now for the proposition that the coin landed heads?" So although the chances for Monday are biased, the chances for the coin flip are not.
     
  17. Jun 2, 2017 #16

    PeroK

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    PPS here's my solution:

    At the outset, the experimenters and the princess have the same information about the experiment.

    You ask the experimenters first: If you are waking the princess, what is the probability the coin is a head. This is clearly 1/3.

    But, as the princess has the same information at the start, she must answer the same:

    If you are being woken, what is the probability the coin is heads. She must answer 1/3.

    When she is being woken, nothing has changed for her - no new info - so she sticks with 1/3.

    It's an illusion that the probability is ever 1/2 and must somehow change to 1/3.
     
  18. Jun 2, 2017 #17

    fresh_42

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    This is what I don't see. The chances are 50:50 on Tuesday and 50:50 on not Tuesday as rest. But I get the feeling this riddle is very similar to one I once read in a little book from Martin Gardner. A criminal has been sentenced to death within a week and the judge says, he won't know the day. His lawyer celebrated this as a victory, because it obviously can't be on Sunday, which would be the last day and in which case he knew on Saturday. But for the same reason it can't be Saturday for he would know on Friday and so on. Finally he was very surprised as it happened on Wednesday morning.
     
  19. Jun 2, 2017 #18

    PeroK

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    That's just elementary conditional probability theory, surely. No tricks. No embellishments.

    The conditional probability of a head given the instance of a waking.

    You can do it with a simple probability tree.
     
  20. Jun 2, 2017 #19

    fresh_42

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    I know. That's why the day of the week is 2:1, but that doesn't affect the coin flip, which remains 1:1. And she isn't asked for the day, only for the coin.
     
  21. Jun 2, 2017 #20

    PeroK

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    If you take away all the confusing complications you are left with - for the experiment at least - a simple conditional probability. You can't look at it as a simple coin flip. It's the conditional probability of heads given some event.

    Effectively you have a random event that occurs twice after a tail and once after a head. The conditional probability of a head, given the event is 1/3.

    This is basic stuff. The problem has muddled this - with its talk of sleeping beauty - and obscured this.

    The coin flip isn't affected as such by the event, but the probability the event resulted from a head is not necessarily 1/2.

    For the experimenters, how is this problem different from any conditional probability?

    If we hadn't been introduced to the sleeping beauty, we would never have thought twice about this.

    The problem with the 1/2 answer - for the experimenters at least - is that it essentially denies that the conditional probability of a coin toss can ever be different from 1/2, because the subsequent event selection process cannot affect the original coin toss.
     
  22. Jun 2, 2017 #21

    PeroK

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    Here's another argument. Suppose the answer is 1/2. She gets woken. It's probability 1/2 that the coin was a head and hence Monday. And, it's probability 1/2 that the coin was a tail, hence 1/4 that it's Monday and 1/4 that it's Tuesday.

    The princess, therefore, calculates that it's 3/4 Monday and 1/4 Tuesday.

    In other words, out of every 4 times she gets woken, 3 are Monday and one is Tuesday.

    But, she knows from the outset that if she gets woken it's 2/3 Monday and 1/3 Tuesday.

    It can't be 1/2, therefore, because the calculation of how likely it is to be Monday goes wrong.

    It should be clear that she is always woken on a Monday and only 1/2 of the time on a Tuesday. So, if she is woken and has no other information, it must be 2/3 that it's.Monday and 1/3 that it's Tuesday.
     
  23. Jun 2, 2017 #22

    fresh_42

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    I do follow your argument, that's not the point. But there is only one coin flip, and one experiment. And as you mentioned earlier, the coin flip already took place when she wakes up. So there is no way she can improve her chances, as they are the same as they were when she fell asleep. And due to missing information on former wake-ups and current day, she can't improve on it. It's the lack of information that makes me think, that it is not a case of conditional probabilities. Being awake cannot influence the outcome of a past event which had been 50:50.
    And as I mentioned earlier, and also in the discussion of it on the Wiki page, all arguments against a 50:50 call rely on different set ups like repetitions or similar. Her wake up call is a post selection event.

    I guess there's something to the summary on Wiki: "All this seems to be consensual among philosophers. Therefore, the Sleeping Beauty problem is not about mathematical probability theory. Rather, the question is whether subjective probability or credence are well-defined concepts, and how they must be operationalized."
     
  24. Jun 3, 2017 #23

    PeroK

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    This is missing the point. The experimenters looked at the coin and this resolved, with 100% probability, what the coin actually was. That has changed. What the coin landed is now known.

    The experimenters then follow a course of action based on the actual outcome of the coin flip. This gives someone else the ability to deduce the coin flip from their actions. Or to adjust the likelihood that it was heads.

    One of the reasons you cannot allow the game to be changed at all is that there are factors that allow 1/2 to seem like a plausible answer. It's not that different numerically from 1/3. But if you change the game so that your answer, by the same logic, remains 1/2 while the alternative reduces to 1/365, say, then it becomes harder to justify 1/2 as it's become numerically absurd.

    Another change that exposes the issue is to extend the game by a day. If it's heads she gets woken on Monday and if it's tails on Tuesday and Wednesday. The game ends on Thursday.

    Now, if she answers 1/2 for heads, she must also answer 1/2 for Monday. But, if all she knows is that she has been wakened, it's equally likely to be Monday, Tuesday or Wednesday. How can she tell it's (more likely to be) Monday?

    How does she know it's twice as likely to be Monday as Tuesday?

    Again, extend this to 365 days and the bias towards her thinking it's the first day becomes numerically absurd.

    Finally, the argument that the experiment only happens once, so we can't use relative frequencies is very weak. That argument effectively scuppers the use of probability theory in any situation where there is only one or a small number of experiments.
     
    Last edited: Jun 3, 2017
  25. Jun 3, 2017 #24

    PeroK

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    PS the key is simply this: The experimenters look at the coin and do a certain thing more often if it is tails than jf it is heads. Someone who experiences this thing - and knows their rule - knows it's more likely they are experiencing that thing through a tail than a head.

    That's conditional probability and is the case here, once you see through the fog.

    The original probability of a head or a tail never changes and never needs to change.

    Relative frequencies and probability trees are applicable tools, as they always are.
     
  26. Jun 3, 2017 #25

    stevendaryl

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    But there hasn't been a satisfactory account of why the probability of heads changes. Let's restate the experiment slightly differently:
    1. The experimenter flips a coin, and asks Sleeping Beauty what is the probability that the result is "heads".
    2. Presumably, she says 1/2.
    3. He then tells her that he plans to ask her the question again, either once or twice depending on the result.
    4. Now, he asks her again what the probability of heads is, given all this information.
    5. Presumably, she again says 1/2.
    6. Now, he asks her what answer she will give when she is waken up.
    7. Now she says 2/3.
    So she knows with certainty that her answer in the future will be 2/3. So why isn't it 2/3 now? You may say that the 2/3 is a conditional probability, while the 1/2 is the unconditioned probability. But normally, conditioning on an event that is certain doesn't change the probability.

    There is a related thought experiment with the same numbers where the conditional probability works out sensibly. Suppose you do the following:
    1. Flip a coin.
    2. If it's tails, you pick a random name out of the phone book, and ask him what he thinks the probability of heads is.
    3. If it's heads, you pick two random names out of the phone book, and ask them both what they think the probability is.
    In this case, assuming there are [itex]N[/itex] names in the phone book, your test subject can reason as follows:
    1. The probability that I will be picked if heads is [itex]P(me | H) = \frac{2}{N}[/itex]
    2. The probability that I will be picked if tails is [itex]P(me | T) = \frac{1}{N}[/itex]
    3. So the cumulative probability of being picked is [itex]P(me) = P(me | H) \cdot P(H) + P(me | T) \cdot P(T) = \frac{2}{N} \cdot \frac{1}{2} + \frac{1}{N} \cdot \frac{1}{2} = \frac{3N}{2}[/itex]
    4. The probability of heads and my being picked is: [itex]P(me \wedge H) = P(me | H) \cdot P(H) = \frac{2}{N} \cdot \frac{1}{2} = \frac{1}{N}[/itex]
    5. The conditional probability of heads given that I was picked is: [itex]P(H | me) = \frac{P(H \wedge me)}{P(me)} = \frac{\frac{1}{N}}{\frac{3N}{2}} = \frac{2}{3}[/itex]
    The numbers work out the same as in Sleeping Beauty. But in this case, the fact that I am picked is additional information that changes the conditional probability of heads. In the Sleeping Beauty case, the fact that she is asked the probability upon waking is no additional information, since it was a certainty that that would happen.
     
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