## Mathematical Conundrum or Not? Number Five

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The Sleeping Beauty 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 awakened, 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 awakened and interviewed on Monday only. If the coin comes up tails, she will be awakened and interviewed on Monday and Tuesday. In either case, she will be awakened on Wednesday without interview and the experiment ends.

Any time Sleeping Beauty is awakened and interviewed, she is asked, 'What is your belief now for the proposition that the coin landed heads?

So what is Beauty's credence that the coin landed on heads?

There is currently no agreed upon resolution to this problem. You could consider this an exercise in philosophy or probability, it should overlap both.
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If the amnesia-inducing drug is effective, Beauty cannot know how many times she was woken, or even if she was woken. Beauty is reduced to guessing, using the information we all know about tossed coins: they come up heads about 50% of the time, on average. But 'on average' does not apply to individual coin tosses. The statistics are useful only for large numbers of coin-tosses, or for large numbers of subjects like Beauty. I think Beauty can only guess, and her only 'evidence' is the statistical 50% probability.
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Well let's break it down.

If the coin flip is heads then Beauty is awakened on Monday but sleeps though Tuesday.

If the coin flip is tails Beauty is awakened on Monday and awakened on Tuesday.

So we have:

H: A, S
T: A, A

So since there are three possible awakenings and only one is when the coin comes up heads, then won't that mean she has a 33% chance of it being heads?
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She should flip a coin. If the original flip was heads then she’ll have a 50% chance of being right, but if it was tails then she’ll have a 75% chance of being right at least once.
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So there's a conflict between guessing which interview this is -- and you weight by the number of possible interviews -- and guessing which day this is, right? It's not Wednesday, because you're being interviewed; if it's Monday, you're interviewed either way; so what are the odds that it's Tuesday?
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Consider it this way:

P(Tails|Monday) = P(Tails|Tuesday) Right? Because they are on the same flip they have the same probability.

Now if the coin is fair then P(Heads and Monday) = P(Tails and Tuesday)

Therefore P(Tails and Tuesday) = P(Tails and Monday) = P(Heads and Monday).

It must add up to one, thus the 1/3.

Btw, there are other ways to look at this, Pattern-Chaser was not necessarily wrong. I only presented this side because they led with the 1/2 argument; however, they are correct in pointing out Beauty has gained no additional information. Really all she knows is what she was told before the experiment.
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Yeah that's pretty clean.

I realized right after I posted that the question is almost literally what is the probability we got heads given that you have been awakened and are being interviewed? So we want the conditional P(H|A).

That's what I'm working at. Not sure what I've got so far. Nice puzzle for me because my probability skills be weak, so thanks.
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Now if the coin is fair then P(Heads and Monday) = P(Tails and Tuesday)

Therefore P(Tails and Tuesday) = P(Tails and Monday) = P(Heads and Monday).

It must add up to one, thus the 1/3.

P (Tails and Monday) and P (Heads and Monday) are mutually exclusive. Also, P (Tails and Monday) and P (Tails and Tuesday) are not independent. So you don't add their probabilities. Also, you left out P (Heads and Not Tuesday).

Anyway. Different experiment - much more ethical. SB comes in on Sunday and agrees to the experiment. She leaves. They flip a coin. On Monday, they place a red jellybean under an opaque cover. SB comes back and is asked whether the flip was heads or tails. On Tuesday, if the coin came up heads on Sunday, they place a green jellybean under the cover along with the red one. If the flip was heads, no jellybean is added. SB comes in and is asked again. Then on Wednesday she comes back in and they give her the jelly bean(s).
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The coin is only flipped once.

P (Tails and Monday) and P (Heads and Monday) are mutually exclusive

Clearly . . . .

Also, P (Tails and Monday) and P (Tails and Tuesday) are not independent.

They are the same flip. It is pointless to argue independence. That is like saying I rolled a die and got 4 and it is not independent because I got 4. If the coin is tails she will be awakened on Monday and awakened on Tuesday, therefore they have the same probability.

Also, you left out P (Heads and Not Tuesday).

She is not awakened on Tuesday if the coin is heads.

The possible outcomes:

- - M T
H: A, S
T: A, A

If she is awakened before Wednesday, there is one awake on heads and two awake on tails. Of the possible awakens there is a 1/3 chance it is Monday and heads.
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It is a fair coin, fair means 50% chance the coin come up heads and and 50% chance it comes up tails. When Beauty is awakened she has to provide the probability it is heads. I am arguing at this time 33%, as there are three possible outcomes where she is awakened and only one of those outcomes is heads.
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Is she told what day it currently is when she is awakened?
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They are the same flip. It is pointless to argue independence. That is like saying I rolled a die and got 4 and it is not independent because I got 4. If the coin is tails she will be awakened on Monday and awakened on Tuesday, therefore they have the same probability.

Here's the additive law of probability:

If events A and B are mutually exclusive (disjoint), then P(A or B) = P(A) + P(B)

In your case, P (Tails and Monday) and P (Tails and Tuesday) are not mutually exclusive, so this rule does not apply.
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Unlike the others, this one is not a misunderstanding of a well-understood and resolved problem. There is an interesting discussion to be had about it.

What is the correct answer depends on what meaning one attaches to the word 'credence'. The use of that unusual word rather than 'probability' in posing the problem is deliberate.

There's a long discussion about it here on physicsforums.

I can't remember, off-hand, whether I was a thirder or a halfer.
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So we want the conditional P(H|A).
The trouble is that we cannot use conditional probabilities.

A conditional probability P(H|A) is the probability of event H given the probability of event A, and A will be an event that is known to be true. But the only event that Beauty knows to be true is that she has been woken at least once - because she has just been woken, so the known event is not a proper (ie smaller) subset of the entire sample space, call it S. So if H is heads and A is 'I am woken at least once' then P(H|A) = P(H|S) = P(H) = 1/2. The conditional probability is the same as the unconditional one.

To use conditional probabilities à la Bayes' Rule, we must have some information that narrows down the set of possibilities - that tells us we are in a proper subset of the sample space. But Beauty has no such information. All she knows is something she already knew before the experiment started.

I think it's to avoid that straightforward solution that the word 'credence', or sometimes 'degree of belief' is used instead of 'probability'.

I'll add that statements like P(Tails and Monday) or P(Tails | Monday) are ambiguous, because the 'Monday' could be the event 'I get woken on a Monday', for which the probability is 1, or it could mean 'today is Monday', and it's not immediately obvious how to set up the probability space so that 'today is Monday' is a well-defined event, without losing the connection to the coin toss.
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P (Tails and Monday) and P (Tails and Tuesday) are the same flip. Are you really suggesting they have a different chance of occurring when they both occur on the same chance event?
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P (Tails and Monday) and P (Tails and Tuesday) are the same flip, Are you really suggesting they have a different chance of occurring when they both occur on the same chance event?

P (Tails and Monday) and P (Tails and Tuesday) are the same event. You can't (legitimately) count it's probability twice.
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Presented in a way that suggests we should look for a conditional, but that won't work. I could see that, but I'm still mulling it over.

Everything else seems to point to 1/3. Twice as many possible awakenings are on tails, leaving just 1/3 for heads. 2/3 of the awakenings are on Monday and half of those are heads, so 1/3 again.

But now I don't know if that's misdirection too.
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The possible outcomes:

- - M T
H: A, S
T: A, A

If she is awakened before Wednesday, there is one awake on heads and two awake on tails. Of the possible awakens there is a 1/3 chance it is Monday and heads.

As she's awake we have to dismiss Heads-Tuesday as an outcome. The only outcomes are Heads-Monday, Tails-Monday, and Tails-Tuesday. But is it right to treat these outcomes as equally likely? Perhaps not. If it it's heads then there's 100% chance that it's Monday, but if it's tails then there's a 50% chance that it's Monday. So Heads-Monday is twice as likely as Tails-Monday (and twice as likely as Tails-Tuesday).

And this is consistent with the fact that we know that, given a fair coin toss, there's a 50% chance that it landed heads. We shouldn't change our view of that just because we might be woken up twice rather than once.
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I didn't count it twice. I am saying P (Tails and Monday) and P (Tails and Tuesday) have the same likelihood of occurring because they are determined by the same chance event, but for Beauty the coin flip generates three possible answers. You have to distinguish between the coin flip and Beauty determining the probability it landed on heads based on the possible times she could be awakened.
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One way to attack the question of what 'credence' or 'degree of belief' means is to interpret it in terms of 'what would you bet, if you were Beauty'? The answer to that depends on the rules of the betting game that is offered.

Consider two different betting games, and we assume Beauty wants to maximise her expected profit.

Game 1: Beauty places her bet before the experiment starts, paying $1, and at the end she is paid$2 if in all her interviews she guessed the coin outcome correctly, otherwise she is paid nothing.

Under this game, her expected winnings are maximised at zero, whichever she chooses. But she must decide before going to sleep the first time which side she is going to guess, because her expected profit becomes negative if it is tails and she makes one guess of heads and another of tails.

Under this game, interpreting the betting strategy as 'degree of belief', we could say her 'degree of belief', at the time of being interviewed, that the coin has landed on Tails, is 1/2.

Game 2: At each interview, Beauty bets $1 to guess what coin came up, and loses that dollar if wrong or wins$2 if right.

Under this game, Beauty's expected winnings are maximised at 50 cents if she guesses Tails.

Under this game, interpreting the betting strategy as 'degree of belief', we could say her 'degree of belief', at the time of being interviewed, that the coin has landed on Tails, is 3/4.

I find it interesting that Game 2, which seems perfectly natural interpretation to me, is consistent with a 'degree of belief' of 3/4 rather than 1/2 or 1/3.

I wonder what betting game would be consistent with a probability of 1/3. I expect there must be a pretty natural one, since most people answer either 1/2 or 1/3 to this question.

I'm off for a jog on the beach now. Maybe it will come to me.
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But now I don't know if that's misdirection too.

None of it is misdirection, this problem has several possible rational answers. We are not talking about just different points of view, or different readings of the semantics, it can rationally be answered in multiple ways from the same point of view and the same interpretation of the semantics. This is because the problem is probing the relationship between knowledge and probability, in this case Beauty's knowledge.

And actually, Pattern-chaser's first answer is still a valid take. Beauty can't be sure of what day she is awakened, as she is given no new information when she is awakened, which means a 1/2 chance for Monday is a rational response. This would follow Bayesian philosophy on probability which suggest we should update our probability models when we get new information.
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And this is consistent with the fact that we know that, given a fair coin toss, there's a 50% chance that it landed heads. We shouldn't change our view of that just because we might be woken up twice rather than once.

Yes, Beauty is aware the coin is fair, but she has also been told the details of the experiment and knows there are three possible events in which she is awakened. If awaken on Tuesday she would not know it is Tuesday; she only knows there are three possible outcomes in which she will be awakened. To Beauty, who does not know if it is Tuesday or Monday when she is awaken, Tails and Tuesdays and Tails and Monday are both valid outcomes. Beauty has to consider three possibilities and only one of them is the desired outcome.
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And yet the other one was generating more discussion.
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The trouble is that we cannot use conditional probabilities.

Finally got some time to come back to this and I think you can use conditional probabilities.

I have Beauty treating Monday-Tuesday as another 50-50 coin toss. Not conditional on her being awakened, mind you -- will come back to that -- just something she has no way of knowing by some other means so she forms no opinion either way.

Here's the simple table @Jeremiah posted:
       Mon   Tue
Tails   A     A


$\small P(H \mid A) = \cfrac{P(A \mid H)P(H)}{P(A)}$
$\small \phantom{P(H \mid A)} = \cfrac{\cfrac{1}{2} \cdot \cfrac{1}{2}}{\cfrac{3}{4}}$
$\small \phantom{P(H \mid A)} = \cfrac{1}{3}$

That makes sense to me. Am I missing something?

We can also ask, what is the chance that it's Monday, given that she's been awakened?

$\small P(M\mid A) = \cfrac{P(A \mid M)P(M)}{P(A)}$
$\small \phantom{P(M\mid A)} = \cfrac{1 \cdot \cfrac{1}{2}}{\cfrac{3}{4}}$
$\small \phantom{P(M\mid A)} = \cfrac{2}{3}$

That looks right, and also makes sense. Of course it's twice as likely to be Monday given that she's been awakened, just as it's twice as likely that the coin toss came up tails, given that she's been awakened.

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I only presented this side because they led with the 1/2 argument; however, they are correct in pointing out Beauty has gained no additional information. Really all she knows is what she was told before the experiment.

That's not correct. Beauty knows that she is awake and that is relevant information.

Whether 1/2 or 1/3 is assigned depends on whether one interprets the experiment as being about a coin toss event (1/2) or an awakening event (1/3).
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No. Nice analysis!
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Yes, Beauty is aware the coin is fair

Then she knows that there's a 50% chance that it landed heads. It doesn't matter if she's only woken once if it's heads but twice if it's tails; a fair coin toss is always going to be 50%.

but she has also been told the details of the experiment and knows there are three possible events in which she is awakened. If awaken on Tuesday she would not know it is Tuesday; she only knows there are three possible outcomes in which she will be awakened. To Beauty, who does not know if it is Tuesday or Monday when she is awaken, Tails and Tuesdays and Tails and Monday are both valid outcomes. Beauty has to consider three possibilities and only one of them is the desired outcome.

But each outcome is not equally likely. There's a 50% chance that it's Heads-Monday, 25% chance that it's Tails-Monday, and 25% chance that it's Tails-Tuesday.
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That makes sense to me. Am I missing something?

Why is P(A) 3/4? Given that she's awake (and knows it), P(A) is 1. She can dismiss P(S) as a possible outcome. And for the same reason P(A|H) should be 1, which then gives the chance that it's heads 1/2, consistent with what we actually know about coin tosses.
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I think @andrewk's example of betting is a good thing to consider. Let's say that you're woken up once if it's heads but 99 times if it's tails. You're asked to place a bet of either £1 on heads or £99 on tails, with a prize of £200 in either case if you're right (and in the case that it's tails it's only her bet on the last day that's accepted).

Using the thirder reasoning there's a 1/100 chance that it's heads, and so the £99 bet on tails is by far the best best. Using the halfer reasoning there's a 1/2 chance that it's heads, and so the £1 bet on heads is by far the best bet.

So put your money where your mouth is. What should you bet? I say bet £1 on heads.
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The trouble is that 'today is Monday', for which you have used the label 'M', is not an Event in the Kolmogorov sense of being a well-defined subset of the sample space and, in the absence of its being such an Event, it cannot be used in probability statements, as a probability statement is a probability measure of an Event.

The reason it can't be an event is that, so far as I can see, the definition of an event in the Kolmogorov framework, which is used, to the best of my knowledge, in all modern probability theory, is timeless. It cannot be relative to a particular time. So 'Beauty is woken on a Monday' and 'Beauty is woken on a Tuesday' are events because the time references they make are absolute, but 'today is Tuesday' is not an event because the reference 'today' is relative.

The same applies to 'I am awake', which is what I think you mean by the label 'A'. Because the 'am' is a relative time reference, it cannot be an Event in a probability space, and so cannot have a probability in the usual sense, that allows use of things like conditional probability rules.

To use a philosophy of time analogy, the question 'is today Tuesday', in the context of this experiment, has as much meaning in probability theory as it does when asked to a pan-dimensional being that is looking at our 4D spacetime from the outside, if McTaggart's B theory of time is the case.

Perhaps a probability space can be constructed in which 'today is Monday' or 'I am awake' is an event but so far I have not managed to construct one. I thought I had a lead, but it turned out to be a dud.

Unless we can construct a probability space in which 'today is Monday' is an event, and which also contains all the other information about the experiment, it cannot be meaningful to talk of the probability - conditional or unconditional - of the event 'today is Monday' or 'I am awake', so we cannot use conditional probabilities to calculate with them.
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That is not new information, she knew she'd be awakened beforehand. New relevant and significance information to reallocating creditably would be if she was told what day it was on Monday.
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