The betting matters, and be honest: you were happy enough to use wagering arguments when it suited you. (What's more the examples don't show what you say is obvious; they assume it and it's irrelevant here anyway.) — Srap Tasmaner

I’ll admit I was too quick to think it relevant when I thought it agreed with me. But after testing it I realised the problem.

In this case, it’s not that tails has greater likelihood odds but that tails has greater payout odds, and these seem to be equivocated.

Imagine I offered 2:1 for correctly guessing a coin toss or 10:1 for correctly guessing a dice roll. Heads is more likely than rolling a 1 but rolling a 1 is the better bet. So that tails is the better bet isn’t that it’s more likely.

Since there are seven awake states (out of twelve distinct states), I would be indifferent between them and so distribute probabilities as follows:

That’s not right. Consider instead a dice with 5 red faces and 1 blue face. The three waking states are red and Monday, blue and Monday, or blue and Tuesday.

Or a weighted coin that’s 5/6 chance of heads.

Or a weighted coin that’s 5/6 chance of heads. — Michael

We shouldn't be indifferent between the two possible states for a weighted coin. So probabilities can't be distributed on that basis. But those two states can be transformed into different states that we can be indifferent between. That is, five states that come up heads and one state that comes up tails. Then the Sleeping Beauty result is again P(Heads|Awake) = 5/7.

Is that what I've done above?

P(Tails and Monday) = P(Tails) * P(Monday|Tails) = 0.5 * 0.5 = 0.25
P(Heads and Monday) = P(Heads) * P(Monday|Heads) = 0.5 * 1 = 0.5

Therefore, P(Monday) = 0.75 — Michael

A simple, discrete probability space consists of two things - a sample space, which is the set of all possible outcomes, called Events, and probabilities of each Event.

It seems from what you are doing here that your sample space consists of four Events:

1. Coin landed Heads and today is Monday
2. Coin landed Heads and today is Tuesday
3. Coin landed Tails and today is Monday
4. Coin landed Tails and today is Tuesday

I think most people would agree that 2 is impossible, so it must have probability zero. The other probabilities are at first unknown and must be inferred by other relationships we have been given.

One such relationship on which I think most people would agree is that is that the probabilities of 3 and 4 must be the same.

That leaves one degree of freedom, which is the probability of 1 which, since the prob of 2 is zero, is also the probability of Heads in this probability space - which we note may not necessarily be the same as the probability space of an independent observer (who will of course say the probability of Heads is 1/2), since this is an epistemological probability space based on the knowledge state of Beauty just after being woken up.

Let the probability of 1 be p. Then the probabilities of 3 and 4 will each be (1-p)/2.

If we set p=1/2 then each of 3 and 4 have probability 1/4 and the probability of Heads is 1/2.
If we set p=1/3 then each of 3 and 4 have probability 1/3 and the probability of Heads is 1/3.
If we set p=1/4 then each of 3 and 4 have probability 3/8 and the probability of Heads is 1/4.

It seems to me that there is no indisputable way of removing the free parameter p. To do so, we need to make an assertion of probability, but it can't be derived without circularity.

We can say p=1/2, in which case we are asserting a principle that Beauty should have the same epistemological probabilities for Heads and Tails as a non-amnesified, independent observer. IN that case we conclude that the probability Beauty assigns to Heads is 1/2.

Alternatively, we can say p=1/3, in which case we are asserting a principle that all non-impossible events in the sample space should have the same probability. In that case we conclude that the probability Beauty assigns to Heads is 1/3.

Or we can say p=1/4, in which case we are basing Beauty's assigned probabilities on the expected values of winnings from a betting strategy.

I expect there are other arguments, with various degrees of convolutedness, for other values of p.

I am inclined to conclude that, if we try to use probabilities rather than betting to solve this, we are left with a degree of freedom - the value of p - that cannot be removed without making a controversial assertion about how Beauty should assign probabilities in her epistemological probability space.

In short, it seems that Beauty's degree of belief that the coin came up Heads can reasonably be whatever she wants it to be - subject to it being in the range [0,1].

That is, five states that come up heads and one state that comes up tails. — Andrew M

Then there's 6 states, not 7. You're counting the tails state twice, which you shouldn't do. The two tails days need to share the probability that it's the tails state (1/6) giving each 1/12 which is the correct figure you get when you apply the probability rule:

P(A and B) = P(A) * P(B|A)

P(Tails and Tuesday) = P(Tails) * P(Tuesday|Tails)

P(Tails and Tuesday) = 1/6 * 1/2 = 1/12

I liked the solution I read on the physics forum from PeroK. Let's change the story shall we?

*** 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. The interview consists of her being asked whether the coin toss was heads or tails, if she guesses wrong, she'll be executed. If she still lives, she will be awakened on Wednesday without interview and the experiment ends. ***


Given these facts, what answer will Sleeping Beauty give?

If the toss is heads, she will awaken Monday. If she'd then say heads, she lives. If she'd say tails she'd be dead.

If the toss is tails, she will awaken first on Monday. If she then says heads, she dies. If she says tails, she'll live.

If she said tails, she will awaken again on Tuesday. If she then says heads, she dies. If she says tails, she'll live.

In the above we see there's only one event where saying tails gets you killed and two events where saying heads gets you killed. Sleeping Beauty would be smartest to state tails. — Benkei

I don't feel that this changes the situation, because although there are two events at which saying Heads can get Beauty killed, if she decides to say Heads, she will never make it to the second event, so the event of being killed at the second waking has a zero probability of occurring.

This assumes that Beauty doesn't randomly choose what to say, by tossing a coin herself after being woken. But so far that possibility has not been canvassed (and I don't think it leads anywhere productive). If the choice is not random then whatever reasoning she uses on Monday will be used again on Tuesday, since the days are indistinguishable to her, so she will give the same answer on both days.

I am inclined to conclude that, if we try to use probabilities rather than betting to solve this, we are left with a degree of freedom - the value of p - that cannot be removed without making a controversial assertion about how Beauty should assign probabilities in her epistemological probability space. — andrewk

I don't think there's any freedom. We just apply the Kolmogorov definition:

$P(A|B) = \frac{P(A ∩ B)}{P(B)}$

$P(A ∩ B) = P(A|B) * P(B)$

$P(Monday ∩ Heads) = P(Monday|Heads) * P(Heads)$

$P(Heads) = 0.5$

$P(Monday|Heads) = 1$

$P(Monday ∩ Heads) = 0.5$

Have I made a mistake somewhere? If not then this must be the answer.

I don't feel that this changes the situation, because although there are two events at which saying Heads can get Beauty killed, if she decides to say Heads, she will never make it to the second event, so the event of being killed at the second waking has a zero probability of occurring. — andrewk

She knows it's either Monday or Tuesday but not which day it is. I understand from previous posts you cannot apply a probability to which day it is. Nevertheless, if it's Tuesday only tails will save her. If it's Monday there's a 50% chance saying tails will save her.

I don't see how answering heads would be an equally safe bet... Also, I sucked at probabilities in high school so don't expect me to understand your answer. :razz: :rofl:

I don't see how answering heads would be an equally safe bet. — Benkei

It wouldn't be an equally safe bet, but that's irrelevant.

If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely.

The mistake is in going from "more likely to win if tails" to "more likely that tails".

It wouldn't be a safe bet, but that's irrelevant.

If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely. — Michael

Ok, I can see that but that seems a meta-position. I thought this was about what credence Beauty gives to the question. That makes the difference to me. Even if she knows there's a 50% chance of either, the fact that it could already be Tuesday would change my assessment if I were in her shoes.

Even if she knows there's a 50% chance of either, the fact that it could already be Tuesday would change my assessment if I were in her shoes. — Benkei

But applying the probability rule she knows that there's only a 25% chance that it's Tuesday, whereas a 75% chance that it's Monday. It stills works out as 50% heads (as it's 25% tails and Monday).

That doesn't mean it's more likely to be tails. It just means that tails is the better bet. — Michael

You say it is 50% odds, but tails is the better bet. This seems contradictory to me.
The whole point of my answer was which was the better bet. If the odds were 2:1 instead of even (33% heads), then Beauty would make or lose no money on average by betting.

The odds of the flip are 50% from nobody's point of view. They seems to be 33% (the one point where neither is the better bet) from Beauty's POV, and they are 100% from everybody else's POV since they know the outcome of the flip during any of the wakings. It is 50% only from the external POV (not Beauty) only before the toss, which is not during any of the wakings.

You say it is 50% odds, but tails is the better bet. This seems contradictory to me. — noAxioms

If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely.

Tails isn't the better bet because it's more likely but because it has a better payout.

The whole point of my answer was which was the better bet. — noAxioms

The topic is about credence. In my example of the free lottery tickets, my credence is that heads and tails are equally likely, despite knowing that tails is the better bet.

The odds of the flip are 50% from nobody's point of view. They seems to be 33% (the one point where neither is the better bet) from Beauty's POV, and they are 100% from everybody else's POV since they know the outcome of the flip during any of the wakings. It is 50% only from the external POV (not Beauty) only before the toss, which is not during any of the wakings. — noAxioms

If you tell me that you flipped a coin ten minutes ago I'm going to say that there's a 50% chance that it landed heads.

And if you tell me that I'll get £1 for guessing correctly, and that you'll let me guess twice if it's tails, I'm going to guess tails because I'll win £2 if I'm right rather than £1 if it's heads. That doesn't mean that tails is more likely. It's just the better bet.

The probability that it's Monday given the fact that it's heads/tails — Michael

I know what the notation means. I don't understand what these probabilities represent. Are you sure they're even defined?

If I offer you one free lottery ticket if you correctly guess heads and two free lottery tickets if you correctly guess tails then tails is the better bet even though equally likely.
Tails isn't the better bet because it's more likely but because it has a better payout. — Michael

This is not the situation faced by Beauty. She is offered one bet that wins or loses one ticket. There is no double payout during any of here wakings.

It's not. It's about her credence. — Michael

Yes, it is about credence. Beauty has information about the coin toss, and that alters the credence from the 50/50 credence that exists to nobody in the scenario.

If you tell me that you flipped a coin ten minutes ago I'm going to say that there's a 50% chance that it landed heads. — Michael

Yes, because I've been given no more information, so the odds remain 50%.

Yes, because I've been given no more information, so the odds remain 50%. — noAxioms

See my edit.

She gets one bet

And if you tell me that I'll get £1 for guessing correctly, and that you'll let me guess twice if it's tails, — Michael

This is a different scenario. When do I find out that I get a second guess? You would have to either tell me before the first guess, or after it. If after, odds are 50% on the first bet and 100% on the 2nd. If before, then 100% on both.
Beauty only gets one guess. If she didn't have the memory-wipe, then the situation would be as you describe it here. She would only have the information she needs on Tuesday, when the odds are 100% tails.

There are three awakenings (A, B, D), each with equal 50% probability of eventual occurrence. In that sense, each awakening has an objective 50% probability of happening, depending on the coin toss.
But Beauty is apparently being asked not if this occurrence will happen (that is obviously certain during any particular waking), but rather which of A, B, D it is. Each has a 1/3 chance if their odds are equal, and they are equal since each equally has a 50% chance of eventual occurrence. B and D do not have 25% (less than A) chance of occurrence, as you seem posit. All three have equal probability, but only A is heads.

Are you sure they're even defined? — Srap Tasmaner

Yes, it's the original Kolmogorov definition:

$P(A|B) = \frac{P(A ∩ B)}{P(B)}$

$P(Monday|Heads) = \frac{P(Heads ∩ Monday)}{P(Heads)}$

$P(Monday|Heads) = \frac{0.5 * 1}{0.5}$

$P(Monday|Heads) = 1$

In the Monty Hall problem, the host gives you information that changes the probabilities that you assign to each door. That information is new to you.

Similarly, in the Sleeping Beauty problem, awakening provides information that enables you to rule out one of the four states. However since you have no information distinguishing the remaining states, you should be indifferent about which state you are currently in. — Andrew M

Beauty doesn't gain relevant new information when awakening, she knew all this before hand. If we do the experiment on you, then you have a prior belief that it is 1/3, what new information would then update that? Priors need relevant new information which would allow us to update it, not just any old information that you think happened.

B and D do not have 25% (less than A) chance of occurrence, as you seem posit. — noAxioms

There's a 50% chance that both B and D will happen, but there's a 25% chance that any specific day is B.

The fact that we can come up with the 1/3 and justify it without the experiment even being ran just proves that it is not a posterior, it is a prior.

That is were both Lewis and Elga error, in assuming that is the linchpin between which one is right.

Suppose we change the game slightly. Beauty wakes up each day, regardless of the coin toss. But if it Tuesday and Heads, we tell her it is Tuesday and Heads. Otherwise we tell here nothing. Each time we do the brain wipe and go back to sleep. Beauty knows ahead of time we are going to do this (the whole 33% odds business depends on Beauty knowing the scenario).

I say nothing has changed. If it is Tuesday and Heads, she is informed of that situation, and is therefore not allowed to place a bet on the known coin toss. All of A B C D occur with equal probability, and any given waking has 25% odds of being any of the 4 (A,B,C,D) or 50% odds of eventual occurrence. But on C, she is told it is C. On the other three, she is not informed which waking it is. All she knows is that it is not C. 33% odds of being any of the other ones.

I ask now what the odds of heads is when she's informed that it is not scenario C this time? Because this is exactly the information she has been given.

There's a 50% chance that both B and D will happen, but there's a 25% chance that today is B. — Michael

'will happen' implies the coin has not yet been tossed. It has, and today has happened, which yields information that changes the credence. That changes the odds of B and D to 66%, 33% each.

That changes the odds of B and D to 66%, 33% each. — noAxioms

Why? If the flip had a 50% chance of being heads and if heads guarantees Monday then there's a 50% chance that today is heads and Monday. You seem to just be asserting these probabilities without adequately explaining how you got there. I get to my probabilities by applying an axiom of probability.

I think the mistake is you're going from

P(A will happen) = P(B and C will happen)

to

P(A is now) = P(B is now) = P(C is now)

But I don't see how that follows.