It is a fair coin flip, so the chance of Monday and Heads is 50% and the chance for Monday and Tails is 50%. Since on one tails flip she is awakened on two days then the chance of Tuesday and Tails is also 50%. Therefore, Beauty has a 33% chance of being awake on a heads

It is a fair coin flip, so the chance of Monday and Heads is 50% and the chance for Monday and Tails is 50%. Since on one tails flip she is awakened on two days then the chance of Tuesday and Tails is also 50%. Therefore, Beauty has a 33% chance of being awake on a heads — Jeremiah

The chance that on Monday the coin flip was tails is 50%, but that chance that any given waking day is Monday and that the coin flip was tails is 25%.

Consider instead that if it was heads then she's woken on Monday and if it was tails then she's woken on Tuesday and Wednesday. What's the chance that she's woken on Monday? 50%. Therefore what's the chance that it was heads? 50%.

The chance that it was heads isn't reduced simply by changing which two days she's woken if it's tails. It's still going to be 50%. Or are you really suggesting that it's more likely to be tails if she's woken on the Monday rather than the Wednesday?

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

P(Heads) = 1/2
P(Heads|Awake) = 1/3

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). — Andrew M

Maybe you could turn that around a bit:

p(Heads) = 1/2
p(Tails) = 1/2

p(Awake|Heads) = 1/3
p(Awake|Tails) = 2/3

Which strikes me as just a restatement of the conundrum.

I prefer the Monty Hall problem.

Here's a variant that is structurally similar:

No sleeping theatrics.

Examiner tosses a fair coin, and then tosses another. If the first toss was tails, she asks Beauty her credence that the first toss came up heads; if the first toss was heads she only asks for Beauty's credence if the second toss was heads as well, otherwise the round is over.

Done this way, Beauty will know that when she was not asked the first toss was heads, but she can do nothing with that knowledge. She's not asked and the round is over. What matters is that she's always asked when it was tails and asked half the time when it was heads. So her credence that it was heads should be 1/3.

and in the case that it's tails it's only her bet on the last day that's accepted — Michael

I don't understand this provision. Beauty knows that she will be woken more often on tails than heads and should be allowed to show that in her betting behavior, even if she can't at the time know how many times she's been woken.

((Will try to put together a betting argument when I have time, because I haven't understood the ones presented so far. If the odds of heads really are 2-1 against, a fair book with no vig should give an expected payout of 0, and Beauty should be able to make a dutch book against anyone who thinks the odds are even. I don't understand how you and andrew are setting the odds.)

Here you might find these papers interesting.

https://www.princeton.edu/~adame/papers/sleeping/sleeping.html

http://fitelson.org/probability/lewis_sb.pdf

Consider instead that if it was heads then she's woken on Monday and if it was tails then she's woken on Tuesday and Wednesday. What's the chance that she's woken on Monday? 50%. Therefore what's the chance that it was heads? 50%. — Michael

And a 50% chance of that she is awakened on Tuesday and Wednesday.

Therefore a 33% of Monday and Heads. What you are not considering is that from the chance mechanism 50% is allocated to two of the possible awakening as the same outcome. Not split between them, but instead, in terms of the coin flip, they are not separate events, they are the same outcome.

I don't understand this provision. — Srap Tasmaner

Just pointing out that there's only one payout, not 99 in the case of tails.

So if you were in her shoes, what would you bet? £1 on heads or £99 on tails?

And a 50% chance of that she is awakened on Tuesday and Wednesday. — Jeremiah

So a 50% chance that it's tails and a 50% chance that it's heads.

I never disagreed with that, I don't really feel you are understanding what I am saying.

What I'm saying is that there is no reason for her to have a greater belief that it was tails than heads. When she's asked what her belief is that it was heads the rational answer is 50:50.

And a 50% chance of that she is awakened on Tuesday and Wednesday.

Therefore a 33% of Monday and Heads. — Jeremiah

No, if there's a 50% chance that she is awakened on Tuesday and Wednesday then there's a 50% chance of Monday and heads.

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? — Jeremiah

I hate to say there is no conundrum about this one. I've not read most of the posts past this point in the thread.
Jeremiah's answer here seems correct to me. The 50/50 answer is from one who has no additional data and knows only that there is a coin flip. Sleeper here has additional data: I am awake, which reduces the case to which of the three awakenings above is this one.

Sleeper here has additional data: I am awake, which reduces the case to which of the three awakenings above is this one. — noAxioms

But one of those awakenings is twice as likely as each of the other two, which is why the halfer answer is correct.

Right I agree with that, and have from the start.

Sorry, I thought you were arguing for the thirder answer.

Here's a variant that is structurally similar:

No sleeping theatrics.

Examiner tosses a fair coin, and then tosses another. If the first toss was tails, she asks Beauty her credence that the first toss came up heads; if the first toss was heads she only asks for Beauty's credence if the second toss was heads as well, otherwise the round is over.

Done this way, Beauty will know that when she was not asked the first toss was heads, but she can do nothing with that knowledge. She's not asked and the round is over. What matters is that she's always asked when it was tails and asked half the time when it was heads. So her credence that it was heads should be 1/3. — Srap Tasmaner

Here's my variation:

If it's heads then she's given a red ball. If it's tails then she's given a blue and black ball.

What are the odds that she's given a red ball?

I am, you are reallocating the 50% for a tails flip across two days; however, it is not reallocated across two days, it is the two days. When Beauty is awakened there are three possible events in which she is awakened. Two of them are determined by the same outcome of the chance event, which is the coin flip, they both have a 50% chance of happening, so they are equally likely. The other possible outcome is on heads, and has a 50% chance of happening so it is equally likely as well. Since all three are equally likely, Beauty's credence that it is Heads and Monday is 33%.

But one of those awakenings is twice as likely as each of the other two, which is why the halfer answer is correct. — Michael

Yes, the Monday awakening is twice as likely as the Tuesday one. It doubles the weight of that awakening. So it adds up to 33% since there is a 50/50 shot on the heavier awakening, and a 0% shot on the Tuesday one.

Since all three are equally likely, Beauty's credence that it is Heads and Monday is 33%. — Jeremiah

They're not equally likely. Heads and Monday is 50%, because heads is 50% and heads guarantees Monday.

P(H) = P(H & M).

Yes, the Monday awakening is twice as likely as the Tuesday one. It doubles the weight of that awakening. So it adds up to 33% since there is a 50/50 shot on the heavier awakening, and a 0% shot on the Tuesday one. — noAxioms

I don't know what you mean by adding up to 33%.

P(Heads) = 50%
P(Monday|Heads) = 100%
P(Heads & Monday) = 50%

P(Tails) = 50%
P(Monday|Tails) = 50%
P(Tails & Monday) = 25%
P(Tuesday|Tails) = 50%
P(Tails & Tuesday) = 25%

However you look at it, nothing can change the fact that the chance of heads on a fair coin toss is 50%, so any solution that says otherwise must be invalid. No other information available to Sleeping Beauty suggests that tails is more likely. The multiple (forgotten) awakenings in one case is a red herring.

I don't know what you mean by adding up to 33%.

P(Heads) = 50% — Michael

This one is begging a different answer. From Sleeper's perspective, this has not been established.

Variation 2

Suppose I toss a fair coin; if it comes up heads, I ask you if it was heads or tails; if it comes up tails, I don't. When asked, you should always guess "heads".

Now move the likelihood that I'll ask in each case.

Variation 3

If the coin comes up heads, I ask you once to guess; if it comes up tails, I ask you twice.

This is different from our case because for each round you know the first question is the first question. That one's 50-50, but once I ask again, you know to answer "tails".

Now suppose there were a way to fix it so you didn't know how many times you were being asked or whether a question was a first or a second. All you know is that on tails you'll be asked twice. What do you guess?

Tails grantees Monday as well. She is awakened on Monday regardless of if it is tails or heads.

I agree the 1/2 argument has its merits as a valid argument, but not for the reasons you are listing. Probability is not reallocated evenly between Tuesday and Wednesday since she is awaken both of those days from the same chance mechanism. From the coin flip she has a 50% chance of being awakened on Tails and Tuesday, and a 50% chance of being awakened on Tails and Monday. Not 25% each as you suggested. This means they are equally likely as Heads and Monday.

I say bet £1 on heads. — Michael

You're correct. But that's because of this:

(and in the case that it's tails it's only her bet on the last day that's accepted). — Michael

If a bet is instead placed every time Beauty awakes, then the £99 bet on tails is the best bet.

Thirders and halfers will agree on how to bet for any given scenario. So that's not really the issue.

↪Andrew M 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. — Jeremiah

It's self-locating information that she can update on. P(Heads|Awake) = 1/3 which Beauty already knew before the experiment began. When she awakes within the experiment, she knows she is awake so the probability of heads for her at that time updates to 1/3.

She also knew before the experiment started that P(Heads|Tails) = 0 and P(Heads|Heads) = 1. So after the experiment, when she learns the result, the probability of heads also updates, this time to either 0 or 1.

I prefer the Monty Hall problem. — tom

The same idea really. One should update one's probabilities when given new information.

This one is begging a different answer. From Sleeper's perspective, this has not been established. — noAxioms

She already knows that it was a fair coin toss and that a fair coin toss has a 50% chance of landing heads. Nothing can change that.

From the coin flip she has a 50% chance of being awakened on Tails and Tuesday, and a 50% chance of being awakened on Tails and Monday. Not 25% each as you suggested. — Jeremiah

If it's heads then I'm given a box with a red ball on Monday. If it's tails then I'm given a box with a blue ball on Monday and a box with a black ball on Tuesday.

I'm woken up and given a box. What is the chance that it's a red ball, a blue ball, or a black ball?

I say 50% chance of it being a red ball because there was a 50% chance that the coin flip was heads and heads guarantees a red ball. I say 25% chance of it being a blue ball because there was a 50% chance that the coin flip was tails and there's a 50% chance, if it was tails, that today is Monday. And then of course 25% chance of it being a black ball.

Tails grantees Monday as well. She is awakened on Monday regardless of if it is tails or heads. — Jeremiah

That doesn't change the fact that P(H) = P(H & M) = 50%.

And that does not change the fact that P(T) = P(T & M) = 50%

And that does not change the fact that P(T) = P(T & M) = 50% — Jeremiah

You're looking at it wrong; see the example with the balls in boxes.