So ChatGPT is saying that P(Heads | today is Monday or Tuesday) = 1/2 is trivially true. Doesn't that just prove my point?

And I think it's even better to not consider days and just consider number of times wakened. So first she is woken up, then put to sleep, then a coin is tossed, and if tails she's woken again. Then we don't get distracted by arguing that her being asleep on Tuesday if Heads is part of the consideration. It doesn't make sense to say that she's asleep during her second waking if Heads.

With this reasoning I think Bayes' theorem is simple. The probability of being woken up is 1 and the probability of being woken up if Heads is 1. That she's woken up a second time if Tails is irrelevant. — Michael

In fact there's an even simpler way to phrase Bayes' theorem, even using days (where "Mon or Tue" means "today is Monday or Tuesday").

P(Heads | Mon or Tue) = P(Mon or Tue | Heads) * P(Heads) / P(Mon or Tue)
P(Heads | Mon or Tue) = 1 * 1/2 / 1
P(Heads | Mon or Tue) = 1/2

Since the setup of the experiment doesn't even require that anyone look at the result of the toss before Monday night, nothing changes if the toss is actually performed after Sleeping Beauty's awakening. In that case the credences expressed on Monday are about a future coin toss outcome rather than an already actualized one. — Pierre-Normand

I think this is a better way to consider the issue. Then we don't talk about Heads & Monday or Tails & Monday. There is just a Monday interview and then possibly a Tuesday interview. It's not the case that two thirds of all interviews are Tails interviews; it's just the case that half of all experiments have Tuesday interviews. Which is why it's more rational to reason as if one is randomly selected from the set of possible participants rather than to reason as if one's interview is randomly selected from the set of possible interviews (where we distinguish between Heads & Monday and Tails & Monday).

And I think it's even better to not consider days and just consider number of times wakened. So first she is woken up, then put to sleep, then a coin is tossed, and if tails she's woken again. Then we don't get distracted by arguing that her being asleep on Tuesday if Heads is part of the consideration. It doesn't make sense to say that she's asleep during her second waking if Heads.

With this reasoning I think Bayes' theorem is simple. The probability of being woken up is 1 and the probability of being woken up if Heads is 1. That she's woken up a second time if Tails is irrelevant. As such:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

This, incidentally, would be my answer to Milano's "Bayesian Beauty".

I don't have access to Stalnaker's paper to comment on that.

If so, this would suggest a highly unusual implication - that one could acquire knowledge about future events based solely on the fact that someone else would be asleep at the time of those events. — Pierre-Normand

Elga's reasoning has its own unusual implication. In his own words:

Before being put to sleep, your credence in H was 1/2. I’ve just argued that when you are awakened on Monday, that credence ought to change to 1/3. This belief change is unusual. It is not the result of your receiving new information — you were already certain that you would be awakened on Monday.

...

Thus the Sleeping Beauty example provides a new variety of counterexample to Bas Van Fraassen’s ‘Reflection Principle’ (1984:244, 1995:19), even an extremely qualified version of which entails the following:

"Any agent who is certain that she will tomorrow have credence x in proposition R (though she will neither receive new information nor suffer any cognitive mishaps in the intervening time) ought now to have credence x in R."

I'm inclined towards double-halfer reasoning. P(Heads) = P(Heads | Monday) = 1/2, much like P(Red) = P(Red|Red or Blue 1) = 1/2. Even if the experiments are not exactly the same, I suspect something much like it is going on, again given the Venn diagram here. I just think the way the Sleeping Beauty problem is written makes this harder to see.

It looks like you may have misinterpreted Elga's paper. He doesn't define P as an unconditional probability. In fact, he expressly defines P as "the credence function you ought to have upon first awakening." Consequently, P(H1) and P(T1) are conditional on Sleeping Beauty being in a centered possible world where she is first awakened. The same applies to P(R) and P(B1), which are conditional on you being in a centered possible world where you are presented with a ball still wrapped in aluminum foil before being given a tequila shot. — Pierre-Normand

I don't see how this entails that P(A|A or B) = P(B|A or B) entails P(A) = P(B).

My example proves that this doesn't follow where P is the credence function I ought to have after being explained the rules of my game. Elga doesn't explain why it follows where P is the credence function I ought to have upon first awakening.

It certainly doesn't follow a priori, and so without any further explanation his argument fails.

To understand what P(R) entails, let's look at the situation from the perspective of the game master. At the start of the game, there is one red ball in one bag and two blue balls in the other. The game master randomly selects a bag and takes out one ball (without feeling around to see if there is another one). They hand this ball to you. What's the probability that this ball is red? — Pierre-Normand

1/2.

Here, P(R|R or B1) is the probability that the ball you've just received is red, conditioned on the information (revealed to you) that this is the first ball you've received in this run of the experiment. In other words, you now know you haven't taken a shot of tequila. Under these circumstances, P(R) = P(B1) = 1/2. — Pierre-Normand

There is a difference between these two assertions:

1. P(R|R or B₁) = P(B₁|R or B₁)
2. P(R) = P(B₁)

The first refers to conditional probabilities, the second to unconditional probabilities, and in my example the first is true but the second is false.

This scenario doesn't accurately reflect the Sleeping Beauty experiment. — Pierre-Normand

Even if it doesn't, it does show that Elga's assertion that if P(A|A or B) = P(B|A or B) then P(A) = P(B) is not true a priori, and as he offers no defence of this assertion with respect to the Sleeping Beauty experiment his argument doesn't prove that P(H₁) = 1/3.

I think the above in fact shows the error in Elga's paper:

But your credence that you are in T₁, after learning that the toss outcome is Tails, ought to be the same as the conditional credence P(T₁|T₁ or T₂), and likewise for T₂. So P(T₁|T₁ or T₂) = P(T₂|T₁ or T₂), and hence P(T₁) = P(T₂).

...

But your credence that the coin will land Heads (after learning that it is Monday) ought to be the same as the conditional credence P(H₁|H₁ or T₁). So P(H₁|H₁ or T₁)=1/2, and hence P(H₁) = P(T₁).

Combining results, we have that P(H₁) = P(T₁) = P(T₂). Since these credences sum to 1, P(H₁)=1/3.

There is a red ball in one bag and two numbered blue balls in a second bag. You will be given a ball at random. According to Elga's reasoning:

1. P(B₁|B₁ or B₂) = P(B₂|B₁ or B₂), therefore P(B₁) = P(B₂)

2. P(R|R or B₁) = 1/2, therefore P(R) = P(B₁)

3. Therefore, P(R) = P(B₁) = P(B₂) = 1/3

The second inference and so conclusion are evidently wrong, given that P(R) = 1/2 and P(B₁) = P(B₂) = 1/4.

So his reasoning is a non sequitur.

I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all. — Pierre-Normand

I've been thinking about this and I think there's a simple analogy to explain it.

I have one red ball in one bag and two blue balls in a second bag. I am to give you a ball at random. Your credence that the ball will be red should be $1\over2$.

Being told that it's Monday is just like being told that the second bag only contains one blue ball. It does nothing to affect your credence that the ball you will be given is red.

My current credence P(H) is 1/2, but if I were placed in this exact same situation repeatedly, I would expect the outcome H to occur one third of the time. — Pierre-Normand

I wouldn't say that the outcome H occurs one third of the time. I would say that one third of interviews happen after H occurs, because two interviews happen after every tails.

I think thirders commit a non sequitur when they claim that tails is twice as likely. Amnesia between interviews doesn't make it any less fallacious.

Sleeping Beauty's calculation that P(H) = 1/3 doesn't hinge on her participation in the experiment being repeated. She's aware that if the coin lands heads, she will be awakened once, but if it lands tails, she will be awakened twice. If we run this experiment once with three participants, and all three of them bet on T every time they are awakened, they will be correct 2/3 of the time on average, which aligns with their credences. — Pierre-Normand

This goes back to what I said before. There are two ways to reason:

1. I should reason as if I am randomly selected from the set of all participants
2. I should reason as if my interview is randomly selected from the set of all interviews

Why would Sleeping Beauty reason as if the experiment was conducted multiple times and that her current interview was randomly selected from that set of all possible interviews, given that that's not how the experiment is conducted?

The experiment is conducted by tossing a coin, and so it is only rational to reason as if she was randomly selected from the set of all possible participants.

If we run this experiment once with three participants, and all three of them bet on T every time they are awakened, they will be correct 2/3 of the time on average, which aligns with their credences. — Pierre-Normand

2/3 bets are right, but that’s because you get to bet twice if it’s tails. That doesn’t prove that tails is more likely. With 4 participants, 1/2 of participants are right whether betting heads or tails. You can frame bets to seemingly support either conclusion.

Although you literally said in your previous post that betting is irrelevant, so why go back to it?

Most of her awakenings occur on the rare occasion when 100 tosses yield heads, which forms the basis for her credence P(100H) being greater than 1/2. — Pierre-Normand

Except the experiment is only conducted once. Either all her interviews follow one hundred heads or all her interviews (one) follow not one hundred heads.

The second is more likely. That’s really all there is to it. I would say it’s irrational for her to reason any other way.

However, the Sleeping Beauty problem specifically inquires about her credence, not about the rationality of her attempt to maximize her expected value, or her preference for some other strategy (like maximizing the number of wins per experimental run rather than average gain per individual bet).

Even if she were to endorse your perspective on the most rational course of action (which doesn't seem unreasonable to me either), this wouldn't influence her credence. It would simply justify her acting in a manner that doesn't prioritize maximizing expected value on the basis of this credence. — Pierre-Normand

And this is precisely why the betting examples that you and others use don’t prove your conclusion.

The only significant divergence lies in the frequency of opportunities: the hostage can't be provided with frequent chances to escape without invalidating the analogy, whereas Sleeping Beauty can be given the chance to guess (or place a bet) every single day she awakens without undermining the experiment.

However, we can further refine the analogy by allowing the hostage to escape unharmed in all instances, but with the caveat that he will be recaptured unknowingly and re-administered the amnesia-inducing drug. This would align the scenarios more closely. — Pierre-Normand

This is heading towards a betting example, which as I've explained before is misleading. There are three different ways to approach it:

1. The same participant plays the game 2¹⁰⁰ times. If they bet on 100 heads then eventually they will win more than they lose, and so it is rational to bet on 100 heads.

2. 2¹⁰⁰ participants play the game once. If they bet on 100 heads then almost everybody will lose, and so it is rational to not bet on 100 heads (even though the one winner's winnings exceed every losers' losses).

3. One participant plays the game once. If they bet on 100 heads then they are almost certain to lose, and so it is rational to not bet on 100 heads.

Given that the very premise of the experiment is that it is to only be run once, a rational person would only consider 3.

And I really don’t see any counterexamples refuting 3. I will never reason or bet that 100 heads in a row is more likely.

But even if the same participant were to repeat the experiment 2¹⁰⁰ times, they don't bet on 100 heads because they think it's more likely, they bet on 100 heads because they know that eventually they will win, and that the amount they will win is greater than the amount they will lose.

And with this variation, do you not agree that the probability of it being heads is 3/8? Would you not also agree that the probability of it being heads in this scenario must be less than the probability of it being heads in the traditional scenario, where being woken up on your assigned day(s) is guaranteed? If so then it must be that the probability of it being heads in the traditional scenario is greater than 3/8, i.e. 1/2.

However, consider a different scenario where the hostage has a small, constant probability ε of discovering the means of escape each day (case-2). In this scenario, stumbling upon this means of escape would provide the hostage with actionable evidence that he could use to update his credence. Now, he would believe with a probability of 6/11 that he's in safehouse #2, thereby justifying his decision to pick up the torch. Consequently, given that 6 out of 11 kidnapped victims who find the means to escape are surrounded by lions, 6 out of 11 would survive. — Pierre-Normand

Then this is a different scenario entirely. If we consider the traditional problem, it would be that after the initial coin toss to determine which days she could be woken, another coin toss each day determines if she will be woken (heads she will, tails she won't).

So the probability that the coin landed heads and she wakes on Monday is 1/4, the probability that the coin lands tails and she wakes on Monday is 1/4, and the probability that the coin lands tails and she wakes on Tuesday is 1/4. A simple application of Bayes' theorem is:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{{1\over2}*{1\over2}}\over{2\over3}}\\&={3\over8}\end{aligned}$

As compared to the normal situation which would be:

$\begin{aligned}P(Heads | Questioned) &={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

Also, as an aside, if you correctly reason that it's tails then you escape on the first day, and so you can rule out today being the second day (assuming you understand that you would always reason as if it's tails).

Your credence in each possibility is based on the number of ways in which you could find yourself in your current situation given the possible outcomes of the specific coin toss. — Pierre-Normand

That's the very point I disagree with, and is most evident with the example of tossing a coin 100 heads in a row. The possible outcomes have no bearing on my credence that the coin landed heads 100 times in a row. The only thing I would consider is that the coin landing heads 100 times in a row is so unlikely that it almost certainly didn't happen, and I think any rational person would agree.

This is less clear to see with the Sleeping Beauty problem given that heads and tails are equally likely, and so prima facie it doesn't matter which you pick, but given that there are two opportunities to win with tails there's no reason not to pick tails.

If you don't like to consider my extreme example because the numbers are too high then let's consider a simpler version. Rather than a coin toss it's a dice roll. If 1 - 5 then safehouse 1 with crocodiles for one day, if 6 then safehouse 2 with lions for six days. Any rational person would take the wooden plank, and 5 out of every 6 kidnapped victims would survive.

It's worth noting that your provided sequence converges on 1/3. If the captive is not keeping track of the date, their credence should indeed be exactly 1/3. — Pierre-Normand

I don't think this is relevant to the Sleeping Beauty problem. It's a different experiment with different reasoning.

In this case you're in safehouse 1 if not tails yesterday and not tails today and you're in safehouse 2 if either tails yesterday or tails today. Obviously the latter is more likely. I think only talking about the preceding coin toss is a kind of deception.

Also it converges to 1/3 only as you repeat the coin tossing, whereas in the traditional problem the coin is only tossed once.

First day P = 1/2, second day 1/4, third day 3/8, fourth day 5/16, etc.

Not sure what this is supposed to show?

Actually that’s not right (starting third day). Need to think about this. First two days are right though.

Not sure how this is at all relevant though.

Then on the first day P = 1/2, the second day P = 1/4, the third day P = 1/8, etc.

I don’t quite understand this example. There are multiple coin flips and no amnesia?

I’ve explained the error with betting examples before. Getting to bet twice if it’s tails doesn’t mean that tails is more likely.

Consider my extreme example. There are two ways to reason:

1. $2\over3$ of all interviews are 100 heads in a row interviews, therefore this is most likely a 100 heads in a row interview
2. $1\over2^{100}$ of all participants are 100 heads in a row participants, therefore I am most likely not a 100 heads in a row participant

I would say that both are true, but also contradictory. Which reasoning it is proper to apply depends on the manner in which one is involved.

For the sitter, his involvement is determined by being randomly assigned an interview, and so I think the first reasoning is proper. For the participant, his involvement is determined by tossing a coin 100 times, and so I think the second reasoning is proper.

We might want to measure which reasoning is proper by appeals to bets or expected values or success rate or whatever, but then there are two ways to reason on that:

1. $2\over3$ of all guesses are correct if every guess is that the coin landed heads 100 times in a row
2. $1\over2^{100}$ of all participants are correct if they all guess that the coin landed heads 100 times in a row

How do we determine which of these it is proper to apply?

So maybe there is no right answer as such, just more or less proper (or more or less compelling). And all I can say is that if the experiment is being run just once, and I am put to sleep and then woken up, my credence that the coin landed heads 100 times in a row would be $1\over2^{100}$.

Sorry, I meant to say that he can rule out it being the case that the coin landed heads and that this is Day2. — Pierre-Normand

They ruled that out before the experiment begun. You might as well say that they can rule out it being the case that the coin landed heads and that this is day 3.

Do you agree that from the sitter point of view, the probability that the coin landed tails is 2/3?

Yes.

Once John Doe awakens, he can rule out the possibility of it being Day 2 of his participation, and so can you. — Pierre-Normand

Why?

Once you have been assigned to John Doe, your credence (P(H)) in this proposition should be updated from 1/2 to 2/3. This is because you were randomly assigned a room, and there are twice as many rooms with participants who wake up twice as there are rooms with participants waking up once. — Pierre-Normand

Yes, if you are randomly assigned an interview from the set of all interviews then the probability of it being a tails interview is greater than the probability of it being a heads interview.

The question is whether or not it is rational for the participant to reason this way. Given that the experiment doesn't work by randomly assigning an interview to them from the set of all interviews, I don't think it is. The experiment works by randomly assigning an interview set from the set of all interview sets (which is either the head set or the tail set), and so I believe it is more rational to reason in this way.

In this modified scenario, imagine Monty picks one of the unchosen doors at random and reveals a goat by chance. — Pierre-Normand

It's not exactly comparable as in my example he can only put to sleep one of the two people who will be put to sleep; he cannot put to sleep someone who won't be put to sleep.

A more comparable example would be if there are four doors, two containing a goat and two containing a car. You pick a door (say 1) and then Monty opens one of the two doors that contain a goat (say 2). What is the probability that your chosen door contains a car? What is the probability that the car is behind door 3, or door 4?

So there are two ways for the participants to approach the problem:

1. I should reason as if I am randomly selected from the set of all participants
2. I should reason as if my interview is randomly selected from the set of all interviews — Michael

To apply this to the traditional problem: there are two Sleeping Beauties; one will be woken on Monday and one on both Monday and Tuesday, determined by a coin toss.

What is their credence that they have been or will be woken twice?

Do they reason as if they are randomly selected from the set of all participants or do they reason as if their interview is randomly selected from the set of all interviews?

Which is the most rational?

The math can prove that the former reasoning gives an answer of $1\over2$ and the latter an answer of $1\over3$ but I don’t think it can prove which reasoning is the correct to use. It might be that there is no correct answer, only a more compelling answer.

Given the way the experiment is conducted I find the former reasoning more compelling. This is especially so with the example of the coin landing heads 100 times in a row.

From the external point of view of the experimenter, it makes sense that the tree probabilities add up to more than one since the three outcomes are non exclusive. The events (Tails & Monday) and (Tails and Tuesday) can (and indeed must) be realized jointly. The intersection zone in the Venn diagram highlights the fact that those two events aren't mutually exclusive. The perspective (and epistemic position) of Sleeping Beauty is different. When she is awoken, those three possible states are mutually exclusive since it is not possible for her to believe that, in her current state of awakening, the current day is both Monday and Tuesday. Her credences in the three mutually exclusive states that she can be in therefore must add up to 1. — Pierre-Normand

I think this is best addressed by a variation I described here:

There are four people, each assigned a number (unknown to them) between 1 and 4. Two of them are to be put to sleep at random, determined by a single coin toss: if heads then 1 and 2 are put to sleep; if tails then 3 and 4 are put to sleep.

After the coin toss one of the two is put to sleep first. For each of the remaining three, what is the probability that they will be put to sleep? Is it 1/3, because there are three of them and only one is to be put to sleep, or is it 1/2, because that was the probability before the first was put to sleep?

Before anyone was put to sleep, for each of the four participants the probability of being put to sleep was 1/2, which of course doesn't add to 1.

And depending on your answer to this scenario, for each of the remaining three participants the probability of being put to sleep is 1/2.

I agree with the idea that Sleeping Beauty's credence in H is updated to 1/2 after she learns that her current awakening is occurring on a Monday. The question is what was this credence updated from, before she learned that the current day was Monday. At that point she could not rule out that the current day was Tuesday. Does that not imply that her credence in H was lower than 1/2 before she learned that the day was Monday? Before she learned that, all three mutually exclulsive areas of the Venn diagram represent possible states for her to be in. She therefore should have non-zero credences for each one of them. — Pierre-Normand

There are two approaches. The normal halfer approach is:

P(Heads) = 1/2
P(Tails) = 1/2
P(Monday) = 3/4
P(Tuesday) = 1/4

The double halfer approach is:

P(Heads) = 1/2
P(Tails) = 1/2
P(Monday) = 1/2
P(Tuesday) = 1/2

The double halfer approach does entail:

P(Heads & Monday) = P(Tails & Monday) = P(Tails & Tuesday) = 1/2

This reflects what I said before:

The probability that the coin will land heads and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Monday is 1/2.
The probability that the coin will land tails and she will be woken on Tuesday is 1/2.

As the Venn diagram shows, there are two (overlapping) probability spaces, hence why the sum of each outcome's probability is greater than 1.

I'm undecided on whether or not I'm a double halfer.

As one of the 2^100 participants who has just been awoken, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, given that half the participant interviews (plus one) are with participants assigned the winning ticket. — Pierre-Normand

I think this is a non sequitur. That most interviews are with a winner isn't that I am mostly likely a winner. Rather, if most participants are a winner then I am most likely a winner.

So there are two ways for the participants to approach the problem:

1. I should reason as if I am randomly selected from the set of all participants
2. I should reason as if my interview is randomly selected from the set of all interviews

I don't think the second is the rational approach. The experiment doesn't work by randomly selecting from the set of all interviews and then dropping each participant into that interview (in such a case the second might be the more rational approach, although how it would be determined which participant(s) are chosen to have more than one interview and the order in which the participants are placed is unclear, and it may be that there is more than one winner). The experiment works by giving each participant a ticket (and in such a case I think the first is the more rational approach).

I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all. — Pierre-Normand

Well, consider the Venn diagram here (which you said you agreed with).

There are two probability spaces. Monday or Tuesday is one consideration and Heads or Tails is a second consideration. Finding out that today is Monday just removes the blue circle.

We must rather imagine that in the unlikely event that the coin lands heads 100 times in a row, 2^101 persons get pulled from the waitlist (or else, only one person does). In that case, if you are one random person who has been pulled from the waitlist, it is twice as likely that your have been so pulled as a result of the coin having landed heads 100 times in a row than not. It is hence rational for you to make this bet and in the long run 2/3 of the people pulled from the waitlist who make that bet with be right. — Pierre-Normand

This isn't comparable to the Sleeping Beaty problem because being a participant isn't guaranteed. That makes all the difference.

Compare these two scenarios:

1. There is a probability of 1 that I will be a participant if the coin lands heads 100 times and a probability of 1 that I will be a participant if the coin doesn't land heads 100 times

2. There is a probability of 1 that I will be a participant if the coin lands heads 100 times and a probability of $1\over2^{101}$ that I will be a participant if the coin doesn't land heads 100 times

In the first case it is rational to believe that the probability that the coin landed heads 100 times is $1\over2^{100}$

In the second case it is rational to believe that the probability that the coin landed heads 100 times is $2\over3$.

It certainly isn't the case that the two have the same answer. Surely you at least accept that? If so then the question is which of 1 and 2 properly represents the traditional problem. I say 1. There is a probability of 1 that she will be a participant if the coin lands heads and a probability of 1 that she will be a participant if the coin lands tails.

Regarding betting, expected values, and probability:

Rather than one person repeat the experiment 2¹⁰⁰ times, the experiment is done on 2¹⁰⁰ people, with each person betting that the coin will land heads 100 times in a row. 2¹⁰⁰ - 1 people lose, and 1 person wins, with the winner's winnings exceeding the sum of the losers' losses. The expected value is greater than the cost, but the probability of being a winner is still $1\over2^{100}$.

Even though I could win big, it is more rational to believe that I will lose.

It is a non sequitur to claim that a greater EV means a greater probability.

I'm not sure why you think this is absurd. Compare again my lottery study example. Suppose there are one billion people on the waiting list. If a coin lands heads 20 times in a row, then, 100 million people get pulled from the list. Else, one single person gets pulled from the list. I am then informed that I got pulled from the list (but not whether I am alone or one from 100 million). Is it absurd for me to believe that the coin landed heads 20 times in a row? My credence in this proposition should be roughly 99% — Pierre-Normand

The difference is that the unconditional probability of being called up is very low, and so just being called up at all affects one's credence. In the Sleeping Beauty case (both the normal and my extreme version), she's guaranteed to be woken up either way.

In Sleeping Beauty's case, your intuition that her credence in the high probability of the sequence of heads ought to be absurd apparently stems from your unwillingness to contemplate the possibility of her being able to updating it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. — Pierre-Normand

She's certainly able to update it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. I'm saying that it's irrational of her to.

The only rational approach, upon waking, is to recognize that it landing heads 100 times in a row is so unlikely that it almost certainly didn't, and that this is her first and only interview.

There is a space of possible awakening/interview events A that are being characterised by the day in which they occur ((M)onday or (T)uesday) and by the state of a coin that has been flipped prior to them occurring ((H)eads or (T)ails). P(H) = P(T) = 0.5. — Pierre-Normand

There's actually two spaces. See here.

One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating. — Pierre-Normand

In my extreme example she wins in the long run (after 2¹⁰⁰ experiments) by betting on the coin landing heads 100 times in a row.

It doesn't then follow that when the experiment is run once that Sleeping Beauty's credence that the coin landed heads 100 times in a row should be greater than her credence that it didn't.

As I have said before, that she can bet more times if one outcome happens just isn't that that outcome is more probable.

I would rather say that the experience works by ensuring that Sleeping Beauty finds herself being awoken in circumstances that she knows to be twice as likely to occur (because twice a frequent) as a result of a coin having landed heads than as a result of this coin having landed tails. — Pierre-Normand

Then you have to say the same about my extreme example; that even when she knows that the experiment is only being run once, Sleeping Beauty's credence that the coin landed heads 100 times in a row should be greater than her credence that it didn't.

And I think that's an absurd conclusion, showing that your reasoning is faulty.

But in that case P(Awake|Heads) should, consistently with this interpretation, refer to your being awakened at all conditioned on the case where the coin landed heads. This is (1/3+2/3)/2 = 0.5 — Pierre-Normand

Me being awaked at all conditioned on the case where the coin lands heads is 1/3, given that if it lands heads then I am only woken up if I was assigned the number 1.

If you mean P(Awake) to refer to the probability of your being awakened at all (on at least one day) then P(Awake) is indeed 0.5. — Pierre-Normand

And it's the same in the normal case. The probability of being awakened at all (on at least one day) is 1. That's what should be used in Bayes' theorem.

I am then using Bayes' theorem to deduce the probability of a random awakening having occurred on a Tuesday. — Pierre-Normand

Then this goes back to what I said above. These are two different questions with, I believe, two different answers:

1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?

2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads?

Thirders answer the second question, which I believe is the wrong answer to the first question. The experiment doesn't work by repeating the experiment several times, randomly selecting an interview from the set of all interviews, and then dropping Sleeping Beauty into it.

This is most clear with my extreme example of 2¹⁰¹ interviews following a coin toss of 100 heads in a row. Any random interview selected from the set of all interviews is most likely to have followed 100 heads in a row. But when we just run the experiment once, it is most likely that the coin didn't land 100 heads in a row, and so Sleeping Beauty's credence should only reflect this fact.

So how would your reasoning work for this situation?

Let's say that there are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.

My reasoning is that P(Awake) = 1/2 given that there are 6 possible outcomes and I will be awake if one of these is true:

1. Heads and I am 1
2. Tails and I am 2
3. Tails and I am 3

My reasoning is that P(Awake | Heads) = 1/3 given that if it is heads I will only be awake if I am number 1.

This gives the correct Bayes' theroem:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over3}*{1\over2}}\over{1\over2}}\\&={1\over3}\end{aligned}$