No they don’t. Your “A or B” isn’t two separate things but one thing with prior probability 1. C and D each have a prior probability of 1/2; C happens if the coin lands heads and D happens if the coin lands tails, and the prior probability that a coin will land heads is 1/2. — Michael

1. Your question "what is your credence the coin will/did land on Heads" is asking SB to distinguish between the cases where your coin will/did land on Heads, and will/did land on Tails. So cases A and B, which depend on the same distinction, must be distinct outcomes to SB. Even if the coin isn't flipped yet, this is a part of your experiment that you insist others must recognize. So please don't be disingenuous and try to deny it again.
2. (A or B) is indeed certain to happen in the experiment. But since it is possible that SB is in the part labeled D, and part D is not in (A or B), the probability that she is in (A or B) has to be less than one. Do I need to provide a lecture on what probability means? Or are you ready to at least pretend to have an open mind?

But this can all be clarified, by considering my questions. I have little "credence" that you will, because you can't accept those answers.

What it means for one's credence to be 1/2 rather than 1/3 is a secondary matter. — Michael

Here's another stab at it.

Ramsey has that ingenious example when he's originally arguing for the very idea of subjective probability and both the possibility and the necessity of putting numbers on credences. You're walking from one town to another, but come to a point where you're not sure you're going the right way; there's a farmer working in a field alongside the road. You can count the steps you would take -- going out of your way -- to reach him and ask directions: the more steps you'd be willing to take, the less certain you are that you know the way, the fewer you'd take the more certain you must be that you know the way.

That's brilliant, but note there are no percentages here to start with, but there is specifically the possibility of comparing one level of confidence to another, and that leads directly to percentages, because you can say how much more confident one answer is than another.

But how does this analysis actually work? Is there a possible world in which Frank walks seventeen paces and another in which he walks thirty? It's all hypothetical, counterfactual even, and experiments that are not performed (I am told on good authority) have no results.

Your question "what is your credence the coin will/did land on Heads" is asking SB to distinguish between the cases where your coin will/did land on Heads, and will/did land on Tails. So cases A and B, which depend on the same distinction, must be distinct outcomes to SB. — JeffJo

She’s being asked her credence that step 3 happens.

Step 1 just isn’t two events with a prior probability of 1/4 each. It’s one event with a prior probability of 1. Step 2 has a prior probability of 1 and steps 3 and 4 each have a prior probability of 1/2.

what "subjective probability" could possibly be is also kinda what the whole puzzle is about. I just thought we could pause and consider the foundations. — Srap Tasmaner

I roll two six sided dice. I tell you the resulting sum is an odd number. What is your "credence"="subjective probability" that the dice landed on a 3 and a 4? Is it 2/36, or 2/18 (there are two ways they can land on 3 and 4)?

This is pertinent to your question if one thinks the answer to the SB problem is 1/2, solely because a fair coin toss has that probability. In that interpretation, the answer here is 2/36.

The puzzle is "about" whether SB receives "new information," similar to "the sum is odd," when she is awake. Halfers say she doesn't, and Thirders say she does. What you choose to call the probability has no relevance. In my opinion, arguments based on what you call it are used because they can't defend their usage of "new information or not," and they know that their counterpart cannot defend such a definition.

She receives new information. The reason some think otherwise is because they think "you sleep through the outcome" means "the outcome doesn't occur." There are four equally-likely experiment states that can arise that are distinct from each other. One of them means the subject is asleep (or not asked for credence, same thing), but it is a state nonetheless. If she is awake, she knows that she is in one of three.

I they don’t. She’s being asked here credence in the outcome of step 3. — Michael

You they do. In step 1, she is asked for her credence in the outcome of step 2, not step 3 (which also asks about it). So she must be able to distinguish between its two possible results when she is in step 1. You can't have it both ways; either the credence question is ambiguous in step 1 becasue the coin hasn't been flipped, or there are two recognizable future results that also distinguish A and B.

We’re talking about prior probabilities, i.e the probabilities as established before the experiment starts.

The prior probability that step 1 will happen is 1.
The prior probability that step 2 will happen is 1.
The prior probability that step 3 will happen is 1/2.
The prior probability that step 4 will happen is 1/2.

There is no prior probability equal to 1/4.

When she is asked her credence she cannot rule out any of these prior probabilities, and she is being asked her credence that step 3 occurs.

We’re talking about prior probabilities, i.e the probabilities as established when the experiment starts. — Michael

Which doesn't change the fact that A can be distinguished from B.

The prior probability that step 1 will happen is 1. — Michael

And the prior probability that the current waking, is a step-1 waking, is 1/2. The prior probability that this is an "A" waking is 1/4. Same for B, C (yes, she is wakened in C in your system), and D. EACH IS A DISTINCT OCCURRENCE IN YOUR SYSTEM and so has a 1/4 prior [probability.

And guess what? If she is asked for a credence, that 1/4 prior probability for C is "ruled out."

But you don't agree with this because you refuse to recognize the difference between B and D, as well as A and C (which you deny).

And guess what else? My experiment bypasses all these issues. Which is why you won't discuss it.

And the prior probability that the current waking, is a step-1 waking, is 1/2. — JeffJo

Prior probabilities are established before the experiment starts, so there is no “current waking” prior because there is no “current waking” before the experiment starts.

Prior probabilities are established before the experiment starts, so there is no “current waking”. — Michael

There is no theory of when prior probabilities are established. But if there were, it would be fom the start, not before the start. This is the same principle that allows you to ask for the credence of an occurrence that hasn't happened yet. Any time parameter you need has to extend over the entire experiment, so it sees the future possibilities and can distinguish them.

But this is the entire controversy behind the Sleeping Beauty Problem. One that I have shown can be trivially removed. And that is why you ignore it.

Now, I have addressed everything you have tried. Answer my questions.

There is no theory of when prior probabilities are established. But if there were, it would be fom the start, not before the start. — JeffJo

So when is this alleged P(X) = 1/4 prior established if not before the experiment starts?

It cannot be when she is asked her credence as you’ve said that in being asked her credence this prior is reduced to 0.

But it also cannot be before because there is no “current interview” before she’s asked her credence.

So this alleged prior just makes no sense.

But this is the entire controversy behind the Sleeping Beauty Problem. One that I have shown can be trivially removed. And that is why you ignore it. — JeffJo

I'm not ignoring it. I'm showing you that your version with the P(HH) = 1/4 prior that is ruled out when asked is not the same as the normal problem because the normal problem doesn't have an equivalent P(X) = 1/4 prior that is ruled out when asked.

So when is this alleged P(X) = 1/4 established if not before the experiment starts? It cannot be when she is asked her credence as you’ve said that in being asked her credence this prior is reduced to 0. — Michael

If you have so little understanding of probability theory, you should not be trying to explain it to those who do. The (somewhat simplified) basics are:

1. A probability experiment is any set of actions that has multiple possible, but unpredictable, results.
2. An outcome is a measurable result.
3. A sample space is any set of outcomes that:
   
   • Are all distinct.
   • Include all possibilities.
4. Each outcome in the sample space is assigned a probability value such that:
   
   • Each probability is greater than or equal to zero.
   • The sum of the probabilities for the entire sample space is 1.
5. Sets of outcomes are called events.
   
   • The probability of an event is the sum of the probabilities in it.
6. These can also be called "prior" probabilities.
   
   • This has absolutely nothing to do with when the probabilities are "established."
   • It refers to before any information about an outcome is revealed.
7. The companion to a prior probability is a posterior probability. It is sometimes called a conditional probability.
   
   • It does require timing, but not directly.
   • Specifically, it refers to a time after the result has been determined, and some information about that result has been revealed.
   • I said it was not directly related to timing because the result can be hypothetical. As in "If I roll a die, and the result is odd, the conditional probability that it is a 3 is 1/3."

When looking at your SB experiment as a whole, there are only two distinct outcomes. The sample space is {Heads,Tails}. We know that both (A or B) and (C or D) will occur, so this distinction does not qualify as different results.

But when looking at it from SB's awakened state, four are necessary:

• A and B belong to distinct outcomes, because "Heads" and "Tails" are distinct results. SB knows that only one is (or could be) true. Please note that it does not matter that SB does not see this result, since she knows of the distinction and that only one can apply to her at the moment.
• B and D are distinct outcomes by the same logic. The measure that distinguishes them is "first possible interview" and "second possible interview." Or "Monday" and "Tuesday" in Elga's solution (not the experiment he described). However you name this quality, SB knows for a fact that one value applies, and the other doesn't.
  
  • It does not matter if SB is awakened in step 3, or not. The same quality that distinguishes B from D also distinguishes A from C.
  So the sample space is {A,B,C,D}. The prior probabilities, for an awakened SB, are 1/4 for each. The "new information" she receives is that C is ruled out. The posterior probability for A, the only outcome where the coin is Heads, is 1/3.

But all of this is unnecessary in my scenario. The sample space for either waking is {HH,HT,TH,TT}. Being awake rules out HH.

The prior probabilities, for an awakened SB, are 1/4 for each. — JeffJo

It makes no sense to say that when she wakes there is then a prior probability that she’s “asleep” of $1\over4$ that is immediately ruled out. If some prior probability is ruled out when she wakes then it must be that the prior probability is established before she wakes. But the prior probability that she will wake a second time if the coin lands heads is 0, not $1\over4$.

If she’s awake then it is just the case that either the coin hasn’t been tossed or it landed tails.

I can set out an even simpler version of the experiment with this in mind:

1. Sleeping Beauty is given amnesia
2. She is asked her credence that a coin has been tossed
3. A coin is tossed
4. If the coin lands tails then:
4A. She is given amnesia
4B. She is asked her credence that a coin has been tossed

The prior probability that step 2 will happen is 1 and the prior probability that step 4B will happen is $1\over2$.

When Sleeping Beauty is given amnesia she knows that she is in either step 1 or step 4A. When asked her credence she knows that she is in either step 2 or step 4B.

No prior probability is ruled out when she is given amnesia or asked her credence.

You have to say that in step 2 her credence is $1\over3$. I have to say that in step 2 her credence is $1\over4$. Of note is that neither of us can just apply the principle of indifference and say that in step 2 her credence is $1\over2$.

Thirder reasoning is that because step 2 is twice as likely to occur as step 4B then I am twice as likely to be in step 2 as step 4B.

Halfer reasoning is that because step 2 is twice as likely to occur as step 4B and that because if 4B will occur then I am equally likely to be in step 2 as step 4B then I am three times as likely to be in step 2 as step 4B.

It makes no sense to say that when she wakes there is then a prior probability that she’s “asleep” of 14 that is immediately ruled out. — Michael

That is the only thing that does make sense. You are completely ignoring that fact that the experiment that SB sees when she is awake, is not the experiment that she volunteered for. In fact, this is the mistake all halfers make, and they do it because they do not know how to model the difference so they claim the difference doesn't exist.

1. The experiment she volunteered for involves the possibility of two wakings.
2. The experiment she see involves exactly one waking.

The reason it is hard to model, is that experiment #2 still has to account for the other possible waking. But it has to do so while accounting for the fact that it is not the current waking. I have proposed a way to implement the actual SB problem (not the "most frequent" one that Elga used to account for the other waking) that removes this difference, And that is why you ignore it. Your argument about "no prior=1/4 to rule out" is explicitly saying you do not recognize the difference.

Now, I can see why some would say we should agree to disagree on this issue. But to make that agreement, you have to at least make an effort to say what you disagree with, and "it can't use the solution I choose, and doesn't get the answer I want" is not such an effort. I have told you why I think all of your arguments are wrong. You display intellectual dishonesty by refusing to address what you think could be wrong with mine. It pretty much implies you can't find anything. I don't mean to be so blunt, but you can easily disprove this by making such an effort.

I can set out an even simpler version of the experiment with this in mind:

1. Sleeping Beauty is given amnesia
2. She is asked her credence that a coin has been tossed — Michael

Um, no. You are asking if it has occurred when you know it hasn't, while correct implementation has to refer to the flip whenever it might occur.

The prior probability that step 2 will happen is 1 and the prior probability that step 4B will happen is 12 — Michael

Again, no."Prior" refers to before information revealed, not to before that information is "established." You do not help your argument by ignoring how probability theory works.

You are asking if it has occurred when you know it hasn't — JeffJo

Sleeping Beauty doesn't know that it hasn't occurred. She has amnesia.

"Prior" refers to before information revealed — JeffJo

So after waking, and before new information is revealed, the prior probability that the coin landed heads and that she is being woken for a second time is 1/4? Then what new information is revealed after waking that allows her to rule out that prior probability? It can’t be “being awake” because that isn’t new information. And it can’t be “being asked her credence” because if she has just woken then she knows with certainty that she is about to be asked her credence.

The simplest answer is the correct one. The prior probability that the coin will or did land heads and that she is being or will be woken for a second time is and always was 0. That’s just a rule of the experiment.

Nothing is ruled out when woken or asked her credence that wasn’t already ruled out before the experiment started.

Even Elga understood this:

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

Nothing is ruled out when woken or asked her credence that wasn’t already ruled out before the experiment started.

Even Elga understood this: — Michael

I believe Elga was mistaken about this. There actually is some information that becomes available to Sleeping Beauty when she awakens, though the nature of this information is rather peculiar. I discussed the nature of this information with GPT-4 in this earlier post.

What informs Sleeping Beauty about the likelihood that the coin landed (or will land) tails, allowing her to update her credence from 1/2 to 2/3, is the fact that she awakens and that, whenever she awakens, the coin landed (or will land) tails two times out of three. After the experiment is over, and she is awoken on Wednesday (assuming she always receives the amnesia-inducing drug after each interview), this information is lost to her, and her credence reverts back to 1/2. The reason why she can't retain the information available to her during each awakening is that this information pertains specifically to the state of the coin in relation to her current episode of awakening. Upon awakening on Wednesday, she loses this information because she loses the ability to refer deictically to her past episodes of awakening (not even knowing how many of them there were).

This loss of information can be emphasized further by modifying the experiment in such a way that the information is not lost by her on Wednesday. Suppose we update the protocol so that on rare occasions, which present themselves with equal probability on each awakening episode, Sleeping Beauty is able to write down a note saying "I have now been awakened and interviewed." She can retain this note and read it again on Wednesday. Upon rereading the note on Wednesday, she can reason that it is twice as likely that such a note was produced if the coin landed tails since she would have been twice as likely to write it during such an experimental run. Therefore, she can retain her updated credence P(T) = 2/3 through ordinary Bayesian reasoning.

The key realization is that the same information that allows Sleeping Beauty to update her credence P(H) from 1/2 to 2/3 upon receiving the note, is inherent in every awakening she experiences due to the causal structure of the experiment. Each awakening serves as an implicit notification of her being in one of the two potential kinds of awakening episodes, which are twice as likely to occur if the coin landed tails. This causal relationship between coin toss results and awakenings, established by the experimental setup, provides information that is available to her in every awakening, even when she doesn't have the opportunity to physically write it down. Essentially, the note merely serves to extend this causal relationship to her Wednesday state, providing her with twice as many opportunities to receive the note if the coin landed tails.

Suppose we update the protocol so that on rare occasions, which present themselves with equal probability on each awakening episode, Sleeping Beauty is able to write down a note saying "I have now been awakened and interviewed." She can retain this note and read it again on Wednesday. Upon rereading the note on Wednesday, she can reason that it is twice as likely that such a note was produced if the coin landed tails since she would have been twice as likely to write it during such an experimental run. — Pierre-Normand

Not necessarily.

Assume a probability of 1/2 each time. The probability of writing it if the coin landed heads is 1/2. The probability of writing it (at least once) if the coin landed tails is 3/4. It is 3/2 times as likely to have been tails.

Assume a probability of 2/3 each time. The probability of writing it if the coin landed heads is 2/3. The probability of writing it (at least once) if the coin landed tails is 8/9. It is 4/3 times as likely to have been tails.

But notice that as the probability of writing a note each time approaches 1 the "greater likelihood" of it having been tails gets smaller, approaching 1.

Also, apply the principle of this reasoning to this.

A fair coin toss is equally likely to be heads as tails, she is less likely (not guaranteed) to wake if tails, therefore if she does wake then she reasons that it's less likely to be tails. In waking she rules out TTT.

If the answer to this problem is 4/7 then the answer to the normal problem is 1/2. If the answer to the normal problem is 1/3 then the answer to this problem is 1/2.

@Pierre-Normand

These cannot all be true:

1. Credence "is a statistical term that expresses how much a person believes that a proposition is true"
2. My current interview is a heads interview iff I have been assigned one heads interview
3. The fraction of interviews which are heads interviews is $1\over3$
4. The fraction of experiments which have one heads interview is $1\over2$
5. My credence that my current interview is a heads interview is equal to the fraction of interviews which are heads interviews
6. My credence that I have been assigned one heads interview is equal to the fraction of experiments which have one heads interview

Propositions like 5 and 6 might usually be true, but they are not true by definition.

Given 1 and 2, my credence that my current interview is a heads interview is equal to my credence that I have been assigned one heads interview.

Therefore given 1, 2, 3, and 4, one or both of 5 and 6 is false.

So how will your reasoning let you choose between 5 and 6 without begging the question?

I accept 6 and reject 5. My credence that my current interview is a heads interview isn't equal to the fraction of interviews which are heads interviews.

My argument is:

P1. If I am certain that A is true if and only if B is true then the degree to which I believe that A is true is equal to the degree to which I believe that B is true
P2. I am certain that my current interview is my only interview if and only if I have been assigned only one interview
C1. Therefore the degree to which I believe that my current interview is my only interview is equal to the degree to which I believe that I have been assigned only one interview (from P1 and P2)
P3. I am certain that if I have been assigned at random by a fair coin toss either one or two interviews then the probability that I have been assigned only one interview is $1\over2$
P4. I am certain that I have been assigned at random by a fair coin toss either one or two interviews
C2. Therefore I am certain that the probability that I have been assigned only one interview is $1\over2$ (from P3 and P4)
P5. The degree to which I believe that I have been assigned only one interview is equal to what I am certain is the probability that I have been assigned only one interview
C3. Therefore the degree to which I believe that I have been assigned only one interview is $1\over2$ (from C2 and P5)
C4. Therefore the degree to which I believe that my current interview is my only interview is $1\over2$ (from C1 and C3)
P6. I am certain that my current interview is my only interview if and only if the coin landed heads
C5. Therefore the degree to which I believe that the coin landed heads is $1\over2$ (from P1, C4, and P6)

[...]But notice that as the probability of writing a note each time approaches 1 the "greater likelihood" of it having been tails gets smaller, approaching 1.[...] — Michael

Yes, indeed, which is why I edged my specification by stipulating that the occasions to write a note were rare.

If Sleeping Beauty would receive two notes on Wednesday, she'd be able to infer that there were two awakenings and hence that the coin didn't land heads. On the earlier occasions when she was writing those notes, by contrast, she wasn't able to know this. When the probability that she would be able to write a note on each awakening occasion is exactly 1/2, the overlapping cases are just numerous enough to enable her to infer on Wednesday, when she receives one single note, that P(H) = 1/2.

As the occasions to write a note become rarer (e.g. 1/n with n >> 1), the frequency of those overlapping notes become negligible (n times as many single notes are received as double notes) and Sleeping Beauty's epistemic state (i.e. the value of her credence) approaches asymptotically her epistemic state as she was writing the note. And, as I had suggested in my previous post, this is because when she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note.

I'll address the other cases and analyses you have presented separately.

As the occasions to write a note become rarer (e.g. 1/n with n >> 1), the frequency of those overlapping notes become negligible (n times as many single notes are received as double notes) and Sleeping Beauty's epistemic state (i.e. the value of her credence) approaches asymptotically her epistemic state as she was writing the note. And, as I had suggested in my previous post, this is because when she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note. — Pierre-Normand

If heads and n = 100 then the probability of writing a note is 1/100

If tails and n = 100 then the probability of writing exactly one note is 1/100.

So if she finds exactly one note on Wednesday then her credence in heads is 1/2.

If heads and n = 100 then the probability of writing a note is 1/100

If tails and n = 100 then the probability of writing exactly one note is 1/100.

So if she finds exactly one note her credence in heads is 1/2. — Michael

Her probability of writing a note is 1/100 on each occasion she awakens. Since she awakens twice when the coin lands tails, her probability of writing a note is 2/100 when a T-experimental run occurs (discounting the 1/10000 cases where she writes two notes).

If tails then:

The probability of her writing on Monday is 1/100
The probability of her writing on Tuesday is 1/100 — Michael

Yes.

The probability of her writing on both Monday and Tuesday is 1/100 * 1/100 = 1/10,000
The probability of her writing on neither Monday or Tuesday is 1 - (1/100 * 1/100) = 9,999/10,000

The latter is actually 99/100*99/100 = 0.9801 ("both" and "neither" aren't complementary cases.)

The probability of her writing on Monday or Tuesday but not both is (1/100 + 1/100) / 2 = 1/100

It's actually 1 - "both" - "neither" = 1 - 0.0001 - 0.9801 = 0.0198 ≈ 2%, which is roughly twice the probability of writing a note in the H case.[/quote]

Yes, I got the maths wrong there (and deleted my post before you replied; apologies).

Though I don't see why I should accept your claim that if "she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note."

As you said yourself, if the probability of her writing a note is 1/2 then if she finds exactly one note on Wednesday then her credence in Heads is 1/2.

Though I don't see why I should accept your claim that if "she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note." — Michael

This is because, when the experimental protocol is expanded to enable Sleeping Beauty to hand notes to her future self in such a manner, the episodes of her receiving a note on Wednesday are produced twice as often in the long run when the coin has landed tails. On the occasion where she awakens and is offered the opportunity to write a note, Sleeping Beauty therefore is enabled to reason thus:

"When I will read again the note that I am currently writing, on Wednesday, I will be able to rationally infer that it is twice as likely that this note was written by me on the occasion of a T-awakening. But it can only be the case that it will have been written on the occasion of a T-awakening if I am now experiencing a T-awakening. Therefore, it is now twice as likely that I am experiencing a T-awakening."

Notice also that, since the probability that Sleeping Beauty would be offered an opportunity at any given awakening to write a note is the same regardless of whether it is an H-awakening or a T-awakening, being offered such an opportunity gives her no independent ground to update her credence.

Note that, as you said yourself, if the probability of her writing a note is 1/2 then if she finds exactly one note then her credence in Heads is 1/2.

That was only in the specific case where n = 2. As n grows larger, P(H) tends towards 1/3.

When I will read again the note that I am currently writing, on Wednesday, I will be able to rationally infer that it is twice as likely that this note was written by me on the occasion of a T-awakening. — Pierre-Normand

That depends on the probability that you will be given the opportunity to write a note. If that probability is 1/2 then it won't be rational on Wednesday to infer that it is twice as likely that this note was written by me on the occasion of a T-awakening.

That was only in the specific case where n = 2. As n grows larger, P(H) tends towards 1/3. — Pierre-Normand

And as it grows smaller, P(H) tends to 1. I don't understand the relevance of any of these three answers.

Why is the correct answer given by any of these situations, let alone by the situation where n is arbitrarily large?

So after waking, and before new information is revealed, the prior probability that the coin landed heads and that she is being woken for a second time is 1/4? — Michael

You are trying really hard to not understand this, aren't you? Of course, all of this would become moot if you would openly discuss other people's ideas, instead of ignoring them while insisting that they discuss only yours. (See: intellectual dishonesty.)

AT ANY TIME in an experiment, the prior probability for any event is based set of all possibilities that could occur, and how they could occur. Not what (or when) information is revealed. And not on how they can be observed.

But a person's BELIEF in an event is based on what information is revealed to her after a result has been realized. And that can depend on how it is observed.

The problem here, is that many experiments can be described by different sets of results. The roll of two dice can be described by 36 different combinations or 11 different sums. Neither is invalid, but only the 36 combinations allow for prior probabilities to be easily assigned.

And that is important here, because you are insisting that a two-day collection of events (I'll call your two passes Monday and Tuesday since only the order matters to anything). You are calling Monday+Tails and Tuesday+Tails the same event. But to SB, who can only observe one at a time, they are distinct events that each have half the prior probability that you assign to the combination.

And you not only call Monday+Heads and Tuesday+Heads the same event in the same way, you deny that Tuesday+Heads happens. It does. Even if (as in "the most frequent" version) SB can't observe it, she knows (A) that it can happen, (B) that it has a non-zero prior probability, and (C) that it is not what is happening if she is awake.

But in this version:

1. Sleeping Beauty is given amnesia and asked her credence that the coin will or did land heads
2. The coin is tossed
3. If the coin lands heads then she is sent home
4. If the coin lands tails then she is given amnesia, asked her credence that the coin will or did land heads, and sent home — Michael

She does observe it.

the prior probability for any event is based set of all possibilities that could occur — JeffJo

And her being woken a second time if the coin lands heads can't occur, which is why its prior probability is 0, not $1\over4$.

And that is important here, because you are insisting that a two-day collection of events (I'll call your two passes Monday and Tuesday since only the order matters to anything). You are calling Monday+Tails and Tuesday+Tails the same event. But to SB, who can only observe one at a time, they are distinct events that each have half the prior probability that you assign to the combination. — JeffJo

There aren't two days in my example.

And as it grows smaller, P(H) tends to 1. I don't understand the relevance of any of these three answers.

Why is the correct answer given by any of these situations, let alone by the situation where n is arbitrarily large? — Michael

The issue with making n small is that it allows Sleeping Beauty on Wednesday to decrease her credence P(H) regarding the origin of the single note. This is because (1) she did not receive two notes and (2) in a significant proportion of cases where a T-run occurs, two such notes are generated instead of one. This makes her epistemic situation dissimilar to her situation when she experiences a particular awakening episode. During such episodes, she can never know that there are two of them due to her amnesia. Making n large makes Sleeping Beauty's epistemic situation on Wednesday, when she receives a note, nearly identical to her situation when she wrote the note, since the Bayesian updating she can perform on the basis of the note being unique is negligible.

When evaluating P(H) on Wednesday, assuming n >> 1, the question Sleeping Beauty is asking is:

"What are the odds that this single note that I received was written by me during an H-awakening?"

The answer is approximately 1/3. However, the note could only have been written during an H-awakening if the coin landed on H. Therefore, P(H) is 1/3.

The second step in the reasoning is to consider that when Sleeping Beauty awakens and finds an opportunity to write a note, she knows that when she reads it on Wednesday (except on the very rare occasion when she finds two notes) she will be able to rationally infer that the odds that the note was written during an H-awakening are 1/3. Since it is now certain that she will read the note on Wednesday and will possess no more information regarding the circumstances of production of the current note than she currently has, she can already infer that this note is being written by her during an H-awakening with 1/3 odds.

A streamlined version of Sleeping Beauty's inference is: "Since I now know that I will soon rationally infer that this note was written during an H-awakening with probability 1/3 (on the basis of no new information), I can already infer this right now." (Here, I am making use of van Fraassen's reflection principle.)

The last step in the argument requires reflecting on the case where Sleeping Beauty doesn't find an opportunity to write a note. In that case, when she awakens, she can reason counterfactually:

"If I had had an opportunity to write a note to myself, I would then have inferred on Wednesday that P(H) (regarding the current awakening episode that is the source of the note) is 1/3, and hence known now that P(H) is 1/3. But the odds that I am currently experiencing an H-awakening are probabilistically independent of my finding an opportunity to write a note. Therefore, they are 1/3 and the only reason why I will be unable to know this when I awaken on Wednesday (and rather infer that P(H) = 1/2) is because I will have lost the special causal connection that I currently have to my present awakening episode.

Note that when Sleeping Beauty doesn't receive a note on Wednesday, her credence P(H) = 1/2 doesn't merely differ in value from her credence P(H) = 1/3 during awakenings; the predicates P() also have different meanings. During awakenings, P(H) refers to the odds that her current awakening episode is occurring during a coin toss that landed heads. On Wednesday, P(H) refers to the odds that the experimental run she is exiting from was an H-run. While in each case the biconditionals "I am now in an H-awakening iff I am now (and will be) in an H-run" or (on Wednesday) "I was in an H-awakening iff I am now in an H-run" hold, the probabilities don't necessarily match due to the two-to-one mapping between T-awakenings and T-runs.

To emphasize this last point, suppose Sleeping Beauty writes a note on each awakening occasion and the experiment is run many times. She ends up with a collection of identical notes, approximately two-thirds of which were written during T-awakenings. She now has lost track of the pairing between the notes. Two things can now be true at the same time:

(1) Since 1/3 of those notes are H-notes, Sleeping Beauty was right during the occasions where she wrote them to believe P(H-note) = 1/3 and hence that P(H) = 1/3.

(2) Since at the end of each experimental run, Sleeping Beauty received either one H-note or two T-notes with equal probabilities, the probability that the individual note(s) she received were T-notes (or H-notes) was 1/2 (or 1/2). In other words, in advance of counting how many notes there were on any given Wednesday, Sleeping Beauty could point at the note(s) and say that they were equally likely to be H-notes or T-notes.

This analysis again highlights how Halfers and Thirders can both be right at the same time but talk past each other when they fail to attend precisely to their respective definitions of P(H), and especially how H and T outcomes are to be individuated and counted.

Here, I've asked GPT-4 to summarise the argument and highlight the main points:

Since I now know that I will soon rationally infer that this note was written during an H-awakening with probability 1/3 (on the basis of no new information), I can already infer this right now. — Pierre-Normand

Then before the experiment starts the thirder will say "since I now know that I will soon rationally infer that the coin will have landed heads with probability 1/3 (on the basis of no new information), I can already infer this right now, before the coin is tossed."

But I think this is wrong.

Making n large makes Sleeping Beauty's epistemic situation on Wednesday, when she receives a note, nearly identical to her situation when she wrote the note, since the Bayesian updating she can perform on the basis of the note being unique is negligible. — Pierre-Normand

They're not nearly identical. On Wednesday she knows that she only had the opportunity once. When she wrote the note she didn't know that it was her only opportunity. So contrary to the above, there is new information on Wednesday.

Note that when Sleeping Beauty doesn't receive a note on Wednesday, her credence P(H) = 1/2 doesn't merely differ in value from her credence P(H) = 1/3 during awakenings; the predicates P() also have different meanings. During awakenings, P(H) refers to the odds that her current awakening episode is occurring during a coin toss that landed heads. On Wednesday, P(H) refers to the odds that the experimental run she is exiting from was an H-run. While in each case the biconditionals "I am now in an H-awakening iff I am now (and will be) in an H-run" or (on Wednesday) "I was in an H-awakening iff I am now in an H-run" hold, the probabilities don't necessarily match due to the two-to-one mapping between T-awakenings and T-runs. — Pierre-Normand

I address this here.

And her being woken a second time if the coin lands heads can't occur, which is why its prior probability is 0, not 14. — Michael

In your experiment as you listed it, she isn't put to sleep and isn't woken. Yet you keep describing it as being woken, probably because that language is used in the "most frequent version." I have not mentioned this, because I recognize the need to read what you write with an open mind.

There aren't two days in my example.

But the time period after the coin is flipped still exists, and the coin can be Heads during that time. AS I DESCRIBED, I wanted a name for that period. So I choose to call it "Tuesday" regardless of when it occurs.

So we can label four distinct periods of consciousness for SB during your experiment, by fake-name day and coin result. Any such moment has a 1/4 prior probability to be any one of these. Including Heads+Tuesday.

But this was the subject of some of my questions. Even with sleep, the time period I choose to call "Tuesday" still exists, and can still have "Heads", regardless of SB being awake for it. The important part is that she knows when it isn't happening.

If you have any integrity at all, you'd discuss my questions. You won't. because you have no counter argument for them. I have debunked everything you have said.

Your "version" with sleep.
1. Sleeping Beauty is given a 10 minute sleep drug that induces amnesia.
2. Twenty minutes after she is given it (so ten minutes after she wakens with amnesia) a bell sounds, and she is asked her credence that the coin will or did land heads.
3. After answering, she is given the drug again.
4. The coin is tossed.
5. If the coin lands heads she is given a booster that extends the effect of the drug for thirty additional minutes.
6. The same bell sounds twenty minutes after the dose in step 2. If she is awake when it rings, she is asked the same question.
7. After answering, she is given the drug a third time.
8. She wakens about the same time regardless of the coin, and is sent home.

There are four situations where the sound of the bell could fall on SB's ears: (Ring #1, Heads), (Ring #1, Tails), (Ring #2, Heads), (Ring #3, Tails). Each has a prior probability of 1/2 to an outside observer who can remember both. But only 1/4 to to be the one that just fell on SB's ears, even if she can't hear it. If the bell rings and SB is asked a question, she knows that (Ring #2, Heads) is ruled out, and her confidence in Heads is 1/3.

Your error, as it has always been, is considering (Ring #1, Tails) and (Ring #2, Tails) to be the same result. Because both happen if the coin lands on Tails. But the point is that don't happen at the same time, and this makes them different outcomes to SB since in here world there is only ever one ring at a time. Even if she is asleep.

Then before the experiment starts the thirder will say "since I now know that I will soon rationally infer that the coin will have landed heads with probability 1/3 (on the basis of no new information), I can already infer this right now, before the coin is tossed."

But I think this is wrong. — Michael

You are correct that this would be wrong. The entire aim of my variation (and the Leonard Shelby variation before it) was to highlight that there is indeed some new information available to Sleeping Beauty upon awakening, and that this information can be retained by her on Wednesday through a note. This information wasn't available to her before the experiment began, and isn't available to her on Wednesday when she doesn't receive a note.

The objective of my discussion was also to highlight the unique nature of this information. It's not a form of information indicating, consistent with this information, a higher proportion of possible worlds in which the coin landed tails. Indeed, the proportion of possible worlds remains exactly the same. It's rather information that is generated by placing Sleeping Beauty in a higher proportion of centered possible worlds (her distinct awakening episodes within a single experimental run) within the T-run timelines.

This type of information is the same as the information that is transmitted from her to her own future self (on Wednesday) when she awakens, by selecting twice as many future recipients of the note in the long run when it is a T-note. This is akin to the information you gained that someone you met was a Tunisian with a probability of 2/3, not because there are twice as many Tunisians as there are Italians (there were actually as many of each in the city), but because Tunisians venture outside of their homes twice as often, doubling your opportunities of meeting them. Likewise, the setup of the Sleeping Beauty experiment makes the coins that land tails twice as likely to "meet" her on the occasion of her awakenings.

They're not nearly identical. On Wednesday she knows that she only had the opportunity once. When she wrote the note she didn't know that it was her only opportunity. So contrary to the above, there is new information on Wednesday. — Michael

I would argue that the situations are nearly identical since this new knowledge is almost inconsequential. Suppose n = 100. On Wednesday, Sleeping Beauty knows that she only had a single opportunity and can thus rule out the miniscule 0.01% chance that she would have had two opportunities. The probability that the note she obtained was an H-note (produced on the occasion of an H-awakening) therefore is P(H-note|single)/(P(H-note|single)+P(T-note|single)) where

P(H-note|single) = P(single|H-note)P(H-note)/P(single) = (1)(0.5%)/(0.0149) ≈ 0.3356
(since P(single) = 0.5%+(1/2)*0.0198 = 0.0149, and we had calculated the 0.0198 before)

and

P(T-note|single) = P(single|T-note)P(T-note)/P(single) = (99%)(1%)/(0.0149) ≈ 0.6644

As expected, P(H-note|single) and P(T-note|single) sum up to 1, and the probability that the note Sleeping Beauty obtained on Wednesday was an H-note rises to 0.3356. This figure is slightly larger than 1/3 only because the rare cases of H-runs where two opportunities to write a note were present are discounted when she only receives one.

I address this here. — Michael

Thank you! I'll respond to this separately.