Since SB doesn't remember Monday, she cannot feel the difference but the structure of the experiment KNOWS the difference.So if she is asked twice, Monday and Tuesday, that only happens with tails outcome. Even without memory, her credence may shift, but because the setup itself is informative. — Kizzy

You are one of four volunteers gathered on Sunday Night. You see the combinations "Monday and Heads," "Monday and Tails," "Tuesday and Heads," and "Tuesday and Tails" written on four different note cards. They are turned over, shuffled, and distributed between you, but you can't look. You are told that after you go to sleep, a single fair coin will be flipped. Then, on Monday and again on Tuesday, three of you will be wakened asked some questions. The one who is left out will be the one whose card says the actual coin flip result, and the current day. Afterwards, you will be put back to sleep with amnesia.

Some time later, you find yourself awake and sitting in a room where you can see two of the other three volunteers on TV monitors (you are instructed to not try to communicate through them). One is labeled "Monitor A," and the other "Monitor B." You, and these other two, have their card face-down card on the table in front of them.

• Not knowing what your card says, you are asked for your credence that the coin result written on you card is the actual coin result. AND, the same question about your credence for it matching A's card, and B's card.
• Once you all have provided an answer (unseen by the others, of course), you are told to look at your card, without revealing it, and answer the same questions.

I say the answers in #1 cannot be anything but 1/3. You have the exact same information about each, and they have to add up to 1. If you disagree, please explain how it is possible. Note that the "structure of the experiment KNOWS" that there is a day, an a coin face, that apply. The importangt part is that yolu don't know these.

I say the answers in #2 can't change. Knowing the specific names applied to you "sleep day" does not change their existence what the "structure of the experiment KNOWS," in any way. You seem to think it can; that what the "structure of the experiment KNOWS" changes for you.

But the same applies to A and B. If it changes the same way, your answer for them changes the same way and eveybody's is 1/2. This is a paradox.

And if it changes in a different way for A and B, allowing you to say 1/4 for them, how did it change differently?

SB knows that Monday waking is guaranteed, no matter what the outcome of the coin toss, if so how can she eliminate the sleeping day and update the probabilities or her credence to 1/3 — Kizzy

SB does not know if a waking day is a Monday. Only that it is a waking day. She can eliminate the sleeping day because she knows this is a waking day.

Compare two versions:

• Three days where she is wakened and interviewed, and a fourth where she sleeps.
• Three days where she is wakened and interviewed, and a fourth where she is wakened and taken to DisneyWorld.

On a waking day in the second version, she clearly can eliminate the DisneyWorld day and the probability of Heads. Why is that different? Whether on not she would be awake on that "fourth day" is irrelevant. The important fact is not being able to observe it when it happens, it is being able to observe that it is not happening when it does not.

SB "locates" herself in one of the four possible states in the experiment. These states exist whether or not she would be able to observe them, That was the point of the Camp Sleeping Beauty experiment: the 36 possible "days" each have a prior probability of 1/36. If she participates in activity "A", the probability that the die rolled N is the number of times A appears in row N, divided by the number of times A appears in the 6tx6 calendar. This is true if all activities are waking activities, or if one is a sleeping activity she would not be able to observe. Her observation is that B, C, D, E, and F did not occur, not just that A occurred.

This is what determines the "probability mass." You correctly described how it is used, but you refuse to identify it correctly. I think you agreed (its hard to recall with all your pedantry) that this is true when all activities are waking ones, but I don't recall you addressing how sleeping activities affect it.THEY DON'T.

The "halfers run-centered measure" is precluded because you can't define, in a consistent way, how or why they are removed from the prior. So you avoid addressing that.

Sure, but Sleeping Beauty isn’t being asked what her credence is that "this" (i.e. the current one) awakening is a T-awakening. — Pierre-Normand

She is asked for her credence. I'm not sure what you think that means, but to me it means belief based on the information she has. And she has "new information." Despite how some choose to use that term, it is not defined in probability. When it is used, it does not mean "something she didn't know before," it means "something that eliminates some possibilities. That usually does mean something about the outcome that was uncertain before the experiment, which is how "new" came to be applied. But in this situation, where a preordained state of knowledge eliminates some outcomes, it still applies.

I was referring to your second case, not the first. In the first case, one of three cards is picked at random. Those three outcomes are mutually exclusive by construction. In your second case, the three cards are given to SB on her corresponding awakening occasions. Then, if the coin lands Tails, SB is given the two T-cards on two different days (Mon & Tue). So "Mon & Tails" and "Tue & Tails" are distinct events that both occur in the same timeline; they are not mutually exclusive across the run, even though each awakening is a separate moment. — Pierre-Normand

And.... you continue to ignore the obvious point I am making. You keep looking at an "outcome" as what occurs over two days. The only "outcome" SB sees occurs on one day.

But if you really want to use two days, do it right. On Tails, there are two waking days. On Heads, there is a waking day and a sleeping day. The sleeping day still exists, and carries just as much weight in the probability space as any of the waking days. What SB knows is that she is in one of the three waking days.

Each day carries a 1/4 prior probability. Since SB knows she is in a waking day, the sleeping day is "eliminated" and she can use conditional probability to update the probabilities of the three waking days to 1/3 each. And it is no different than if you always wake SB, but have three interview days and one DisneyWorld day. The three days that have a common description, when that common description is what SB sees, each have a probability of 1/3. This is true regardless of what the other one description is.

And you have offered no counter arguments except "but what if SB wants to look across all of the days."

In the second case, which mirrors the Sleeping Beauty protocol more closely, two of the possible outcomes, namely "Monday & Tails" and "Tuesday & Tails," are not mutually exclusive. — Pierre-Normand

Oh? You mean that a single car can say both "Monday & Tails" and "Tuesday & Tails?" Please, explain how.

"What is your credence in the fact that this card says "Heads" on the other side? This is unquestionably 1/3.

"What is your credence in the fact that the coin is currently showing Heads?" This is unquestionably an equivalent question. As is ""What is your credence in the fact that the coin landed on Heads/i]?"

I realize that you want to make the question about the entire experiment. IT IS NOT. I have shown you over and over again how it leads to contradictions. Changing the answer between these is one of them.

Now, the fact that the coin shows Tails on Tuesday, if a question can be asked, certainly is the same fact as it was on Monday. But SB's knowledge set does not allow a connection between these.

In modal logical terms, one is "actual" if and only if the other is

And how is this relevant to SB?

even though they do not occur at the same time.

No. BECAUSE ONE EXISTS IN HER "WORLD," AND THE OTHER DOES NOT.

Picking "Monday & Tails" guarantees that "Tuesday & Tails" will be picked the next day, and vice versa. They are distinct events but belong to the same timeline. One therefore entails the other.

And how does this affect what SB's credence should be, when she does not have access to any information about "timelines?"

AGAIN: You are constructing invalid and inconsistent logic to support the conclusion you want to reach.

Your argument in favor of the Thirder credence that the coin landed Tails (2/3) relies on labeling the awakening episodes "the outcomes". — Pierre-Normand

Uh, yeah?

Write "Heads and Monday" on one notecard. Write "Tails and Monday" on another, and "Tails and Tuesday" on a third. Turn them over, and shuffle them. Then write "A," B," and "C" on the other sides.

Pick one. What is the probability that it says "Heads" on the other side? What is the probability that it says "Tails" on the other side? Call me silly, but I'd say 1/3 and 2/3, respectively.

Each morning of the experiment when SB is to be awakened, put the appropriate card on a table in her room, with the letter side up. Hold the interview at that table.

What is the probability that the card, regardless of what letter she sees, says "Heads" on the other side? Or "Tails?" This "outcome" can be defined by the letter she sees. But that does not define what an outcome is, being the description of the experiment's result, in SB's knowledge, is. If she wakes on a different day, that is a different result. Being determined by the same coin flip does not determine that.

Now, did these probabilities change somehow? For which letter(s) do they change? Or are they still 1/3 and 2/3?

Within SB's knowledge, is not the outcome where it says "Heads" the exact same outcome where the coin is showing Heads? And the same with "Tails?" If it says "Tails and Monday," is there not another interview along this same path where it says "Tails and Tuesday?" Does that change the probability that this card says "Tails?" How does that carry over to the one time it would say "Heads?"

Again, halfers are constructing inconsistent logic to support the answer they desire. Not using valid logic to answer the question.

But what is it that prevents Halfers from labelling the experimental runs "the outcomes" instead?

Because it is not both Monday, and Tuesday, when she is asked the question? What else may or may not happen is irrelevant.

That's right, and this is a good argument favoring the Thirder position but it relies on explicitly introducing a scoring procedure that scores each occasion that she has to express her credence: once for each awakening episode. — Pierre-Normand

A "scoring procedure" based on imagined repeats is a way of testing your probabilities, not of defining it. It does not work in the SB problem, as should be painfully obvious, because each side will define the number of trials differently since repeated runs require looking at more than one outcome, and the number changes based on the subject event.

The reason this reference is made (to the future verification conditions) is to disambiguate the sense of the question, ... — Pierre-Normand

Perhaps you didn't parse correctly. There is no ambiguity. If she is asked to project her state of knowledge on Wednesday, or to recall it from Sunday, of course the answer is 1/2.

Where, exactly, do you think projection or recollection is implied or stated? You are forcing this issue into a place where it does not belong, in order to justify saying that 1/2 is a possible answer.

Remember: SB isn't betting on the card (neither is she betting on the current awakening episode). She's betting on the current coin toss outcome.

I keep looking at the problem, and I can't find a reference to betting anywhere. The reason I don't like using betting is because anybody can re-define how and when the bet is made and/or credited, in order to justify the answer they like. One is correct, and one is wrong.

They ask her one question after each time she awakens, however: What is the probability that the coin shows heads.

Do you see the words "landed when it was flipped" here? No? How about "will be showing after the experiment ends"? Still no? Then stop inserting them. What is the probability that the coin is showing "Heads"? This is in the present tense.

So, if a bet were to exist, and assuming she uses the same reasoning each time? She risks her $1 during the interview, and is credited her winnings then also. If she bets $1 on Heads with 2:1 odds, she gains $2 if the coin landed Heads, and loses 2*$1 if it landed on Tails. If she bets on Tails with 1:2 odds, she loses $1 if the coin landed Heads, and gains 2*$0.50=$1 if it landed Tails.

But if she bets $1 on Heads with 1:1 odds, she gains $1 if the coin landed Heads, and loses 2*$1=$2 if it landed on Tails. If she bets on Tails with 1:1 odds, she loses $1 if the coin landed Heads, and gains 2*$1=$2 if it landed Tails.

The answer, to the question that was asked and not what you want it to be, is 1/3.

On the occasion of an awakening, what is Sleeping Beauty's expectation that when the experiment is over ... — Pierre-Normand

This is what invalidates your variation. She is asked during the experiment, not before or after. Nobody contests what her answer should be before or after. And you have not justified why her answer inside the experiment should be the same as outside.

The issue with her remembering or not is that if, as part of the protocol, she could remember her Monday awakening when the coin landed tails and she is being awakened again on Tuesday, she would be able to deduce that the coin landed Tails with certainty and, when she couldn't remember it, she could deduce with certainty that "today" is Monday (and that the probability of Tails is 1/2). That would be a different problem, and no problem at all. — Pierre-Normand

Gee, if she is given a different set of information (knowing that it is not Tuesday, after Heads is information, as is all of these) produces different conditional probabilities. Possibly certainties. And?

Here's one more attempt. It's really the same thing that you keep dodging by changing the timing of the question, and claiming that I have "vallid thirder logic" while ignoring that it proves the halfer logic to be inconsistent.

• Get three opaque note cards.
• On one side of different cards, write "Monday and Heads," "Monday and Tails," and "Tuesday and Tails.
• Turn the cards over, shuffle them around, and write "A," "B," and "C" on the opposite sides.
• Before waking SB on the day(s) she is to be woken, put the appropriate card in the table in her room, with the letter side face up.

Let's say she sees the letter "B." She knows, as a Mathematical fact, that there was a 1/3 probability that "B" was assigned to the card with "Heads" written on the other side. And a 2/3 chance for "Tails."

By halfer logic, while her credence that "Heads" is written on the "B" card must be 1/3, her credence that the coin landed on Heads is 1/2. This is a contradiction - these two statements represent the same path to her current state of knowledge, regardless of what day it is.

Your justification for considering the halfer logic is that there may be a different path that _includes_ her her current state of knowledge and another that has a different day written on the card. My reason for rejecting that justification outright, besides the contradiction that it produces, is that this other day is not a path of the the path she sees.

I did and I agreed with you that it was a fine explanation of the rationale behind the Thirder interpretation — Pierre-Normand

You may have read it. You did comment on it from that aspect. But you did not address it. The points it illustrates are:

• That each "day" (where that means the coin toss and the activity that occurred during that awakening), in Mathematical fact, represents a random selection of one possible "day" from the NxN grid. If that activity appears S times in the schedule, and R times in the row, then the Mathematically correct credence for the random result corresponding to that row is R/S. This is true regardless of what the other N^2-S "days" are, even if some are "don't awaken."
• There is no connection between the "days" in a row. You call this "T-awakenings" or "the H-wakening." in the 2x2 version. They are independent.

SB does know the setup of the experiment in advance however. — Pierre-Normand

Yep. What makes it an independent outcome, is not knowing how the actual progress of the experiment is related to her current situation. This is really basic probability. If you want to see it for yourself, simply address the Camp Sleeping Beauty version.

Yes, that makes the answer 1/2 BECAUSE IT IS A DIFFERENT PROBLEM. — JeffJo


It isn’t a different problem; — Pierre-Normand

It's s different probability problem based on the same coin toss. SB has no knowledge of the other possible days, while this answer requires it.

so when those events aren't occurring in a way that is causally (and probabilistically) independent of the coin flip result. — Pierre-Normand

When she is awake, what knowledge does she have, related to any other day or coin result?

This is what seems difficult to accept. SB's "world" consists of one day, and one coin result, and due to the amnesia both are independent of any other "world" represented in another awakening. Illustrating that was the point of my "Camp Sleeping Beauty" variation.

It's not wrong — ProtagoranSocratist

His explanation for "double halfers" used two coin flips. There is only one coin flip. So it is both incorrect mathematics, and incorrect about the double-halfer's claim.

I think the problem was created more or less just to see what answers people would come up with, how they would project their logic onto what they read.

It was created to justify epistemic reasoning, where it does not apply.

1/2 makes since, since theoretical coinflips

1/2 does not make sense because it treats the problem unconditionally. It makes the "outside the experiment" interpretation that single outcome can be represented by two different awakenings.

Another exit rule could be that SB gets to go the Atelier Crenn at the end of the experiment — Pierre-Normand

Yes, that makes the answer 1/2 BECAUSE IT IS A DIFFERENT PROBLEM. SB is asked once on each waking day, not once at the end. To even try to make it similar, you'd need to take her to two restaurants if the coin landed Tails. In the original, that is equivalent to saying a single day is Tails&Mon and Tails&Tue AT THE SAME TIME.

This is what I mean by inconsistent reasoning designed to get a specific answer, not to be correct.

I think the double halfer reasoning is faulty because it wrongly subsumes the Sleeping Beauty problem under (or assimilates it with) a different problem in which there would be two separate coin tosses. — Pierre-Normand

The point is that, like you, they construct the reasons in order to get the result they want. Not because the reasons are consistent in mathematics. But your explanation is wrong. They argue that there is a single, but somehow the Law of Total Probability does not apply. That Tails&Monday is both the same, and a different, outcome than Tails&Tuesday depending one which way you try to use that Law.

Well, firstly, the Halfer solution isn't the answer that I want since my own pragmatist interpretation grants the validity of both the Halfer and the Thirder interpretations, but denies either one being the exclusively correct one. — Pierre-Normand

Like I said, you want the halfer solution to have validity, so you manufacture reasons for it to be. There can't be two valid answers. Your logic fails to provide ANY solution to my last (repeated) variation.

Viewed from outside the experiment - i.e., not SB's viewpoint - there are two paths with two distinct days each (For this logic, I'm calling Heads&Mon and Tails&Mon different days). And each path has a 50% probability, and the days are not distinguished.

SB's viewpoint inside the experiment sees that only one day is happening. But recognizes that there are three other days, including one she would sleep through, that exist as possibilities but are not this day. I don't recall if I've used it here, but of course I have an equivalent version that clarifies this.

There are three Michelin three-star restaurants in San Francisco, where I'll assume the experiment takes place. They are Atelier Crenn, Benu, and Quince. Before the coin is tossed, a different restaurant is randomly assigned to each of Heads&Mon, Tails&Mon, and Tails&Tue. When she is awoken, SB is taken to the assigned restaurant for her interview. Since she has no idea which restaurant was assigned to which day, as she gets in the car to go there each has a 1/3 probability. (Note that this is Elga's solution.) Once she gets to, say, Benu, she can reason that it had a 1/3 chance to be assigned to Heads&Mon.

The point is that each restaurant represents one of three possible "waking" days, not the path that it is a part of. Outside the experiment there is a pair that represent the Tuesday path, but that is irrelevant to SB.

How about this schedule:

. M T W H F S
1 A E E E E E
2 A B E E E E
3 A B A E E E
4 A B A B E E
5 A B A B A E
6 A B A B A B

When A happens, if E is treated as only an unseen portion of a different "experimental run," should not B be also considered that? What happened to "straightforward Bayesian updating procedure"?

OR, what if E is "If a coin flipped in Sunday landed Heads, leave SB asleep. But if it landed Tails, wake SB and take her to Disneyworld without an interview." Now you have to use a two models - not different probabilities, two different probability models, for each E day.

Halfers don't condition on the propostion "I am experiencing an awakening". — Pierre-Normand

Right. And this is they get the wrong answer, and have to come up with contradictory explanations for the probabilities of the days. See "double halfers."

They contend that for SB to be awakened several times, rather than once, in the same experimental run

Right. Which, after "Tails," requires SB's observation to be happening on both days, simultaneously. See "on a single day."

Halfers, however, interpret SB's credence, as expressed by the phrase "the probability that the coin landed Tails" to be the expression of her expectation that the current experimental run,

Why? How does something that is not happening, on not doing so on a different day, change her state of credence now? How does non-sleeping activity not happening, and not doing so on a different day, change her experience on this single day, from an observation of this single day, to an "experimental run?"

You are giving indefensible excuses to re-interpret the experiment in the only way it produces the answer you want.

How about this schedule:

. M T W H F S
1 A E E E E E
2 A B E E E E
3 A B A E E E
4 A B A B E E
5 A B A B A E
6 A B A B A B

When A happens, if E is treated as only an unseen portion of a different "experimental run," should not B be also considered that? What happened to "straightforward Bayesian updating procedure"?

OR, what if E is "If a coin flipped in Sunday landed Heads, leave SB asleep. But if it landed Tails, wake SB and take her to Disneyworld without an interview." Now you have to use a two models - not different probabilities, two different probability models, for each E day.

I don't see how it bears on the original problem — Pierre-Normand

Then try this schedule:
. M T W H F S
1 A E E E E E
2 A A E E E E
3 A A A E E E
4 A A A A E E
5 A A A A A E
6 A A A A A A

Here, A is "awake and interview."

If E is "Extended Sleep," the Halfer logic says Pr(d|A)=1/6 for every possible roll, but I'm not sure what Pr(Y|A) is. Halfers aren't very clear on that.

But if E is anything where SB is awoken but not interviewed, then the straightforward Bayesian updating procedure you agreed to says Pr(d|A)=d/21, and if Y is an index for the day, Pr(Y|A)=Y/21.

My issue is that, if A is what SB sees, these two cannot be different.

the variation that you actually propose, when only one activity is being experienced on any given day, yields a very straightforward Bayesian updating procedure that both Halfers and Thirders will agree on. — Pierre-Normand

Thank you for that. But you ignored the third question:

• Does it matter if E is "Extended sleep"? That is, the same as Tuesday&Heads. in the popular version?

I don't see how it bears on the original problem where the new evidence being appealed to for purposes of Bayesian updating isn't straightforwardly given — Pierre-Normand

Then you don't want to see it as straightforward. Tuesday still exists if the coin lands Heads. It is still a single day, with a distinct activity, in the experiment. Just like the others in what you just called straightforward.

And what makes it "a very straightforward Bayesian updating procedure" is observing that it does not match what is happens on the day SB is experiencing. That is the straightforward Bayesian methodology.

I didn't provide a detailed response to your post because you didn't address it to me or mention me. — Pierre-Normand

It's "addressed" to what I thought was a discussion forum. You know, to discuss this problem and the approach to its solution. And more specifically, to the unnamed in the discussion who try to obfuscate what is a simple conditional probability problem.

Using this table, in the Camp Sleeping Beauty setup:

. M T W H F S
1 A D C F E B
2 F B B C C F
3 A B C D F D
4 F E B D B C
5 C D F C E E
6 E E F C C F

• On each single day, after activity X, what is the probability/SB's credence that the die roll was d, for d=1 to 6?
• On each single day, after activity X, what is the probability/SB's credence that the day is Y, for Y in {M,T,W,H,F,S}?
• Does it matter if E is "Extended sleep"? That is, the same as Tuesday&Heads. in the popular version?

I use "single day" because each day is an independent outcome to SB. SB does not know anything about any other days, only what is happening in this day. Even if it is possible that the same thing could happen on other days. She can place this day within the context of the set of all days, and the subset where this day's outcomes can happen. BUT SHE ONLY KNOWS ABOUT ONE DAY.

We do not need payout schemes to determine this. We do not need to know what SB expects over the row. We do not need to address "indexicals" The correct answers are:

• COUNT(CELL=X in row d)/COUNT(CELL=X in table).
• COUNT(CELL=X in column Y)/COUNT(CELL=X in table).
• No.

I didn't ignore your post. — Pierre-Normand

You didn't respond to a single point in it. You only acknowledged its existence, while you continued your invalid analysis about changing bets and expected runs. None of which can answer the questions I raised. Using this table

. M T W H F S
1 A D C F E B
2 F B B C C F
3 A B C D F D
4 F E B D B C
5 C D F C E E
6 E E F C C F

• On each single day, after activity X, what is the probability/SB's confidence that the die roll was d, for d=1 to 6?
• On each single day, after activity X, what is the probability/SB's confidence that the day is Y, for Y in {M,T,W,H,F,S}?
• Does it matter if E is "Extended sleep"? That is, the same as Tuesday&Heads.

I use "single day" because each day is an independent outcome to SB.

We do not need payout schemes to determine this. We do not need to know what SB expects over the row. We do not need to address "indexicals" The correct answers are:

• COUNT(CELL=X in row d)/COUNT(CELL=X in table)
• COUNT(CELL=X in column Y)/COUNT(CELL=X in table)
• No.

According to a standard Thirder analysis, prior to being put to sleep, SB deems the two possible coin toss outcomes to be equally likely.

According to the often-misrepresented, original Thirder analysis by Adam Elga, there are two independent random elements: the coin toss, and the day. They combine in four (not three) ways. But I suppose Elga suspected how obtuse halfers would be about them, so he only considered the two overlapping pairs of two that you think constitute the entire sample space.

When she awakens, she could be in either one of three equiprobable situations: Monday&Tails, Monday&Heads and Tuesday&Tails (according to Elga's sensible argument)

That's not the reasoning.

• If (upon awakening) you were to learn that the toss outcome is Tails, that would amount to your learning that you are in either T1 or T2. Since being in T1 is subjectively just like being in T2, and since exactly the same propositions are true whether you are in T1 or T2, even a highly restricted principle of indifference yields that you ought then to have equal credence in each. 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).
• If (upon awakening) you were to learn that it is Monday, that would amount to your learning that you are in either H1 or T1. Your credence that you are in H1 would then be your credence that a fair coin, soon to be tossed, will land Heads. It is irrelevant that you will be awakened on the following day if and only if the coin lands Tails — in this circumstance, your credence that the coin will land Heads ought to be 1/2. 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.

SB's credence in the truth of the statement "Today is Tuesday" is 1/3.

.
Wrong. Elga's credence in the truth of the statement "Today is Tuesday and the coin landed Tails" is 1/3. That's what Elga's T2 means.

You are doing exactly what I said halfers do - denying the existence of what you would call Tuesday&Heads, and that Elga would call H2. The prior pronbability, of a day that is "H2 or T2", is 1/2. Because H2 and T2 are independent results, and the prior pronbabilities are P(H2)=P(T2)=1/4.

But because she is awake, SB receives the "new information" [see note] that H2 is eliminated, leading to Elga's result.

Before the experiment began, SB could (correctly) reason that is was equally likely that she would be awakened once when the coin toss result is Heads and twice when the coin toss result is Tails.

As far as I know, SB can tell that this day is just one of those three possibilities. Please explain if you think otherwise. And before the experiment begins (this is called the "prior" to those who understand probability), she also knows that the experiment exists on Tuesday&Heads, even though she will not be awake to observe it. So when she is awake, she eliminates that possibility.

Note: The way Baysean Updating works is that you define a sample space (called the prior) comprising all possible outcomes of a procedure. This procedure is doing something with SB on a single day based on a coin flip and the index of the day - the 1 or 2 in Elga's notation. So the red herring about indexicals does not apply. The prior probabilities of these outcomes should sum to 1.

Once you have that, you make an observation about an outcome - WHICH IS WHAT HAPPENS ON A SINGLE DAY. This is called "new information" because some of those outcomes in the prior sample space can be eliminated. Not because what you know about what did happen was a surprise. This is another red herring halfers use. Once you "eliminate" any outcomes in the prior that are inconsistent with the observation, you update the prior probabilities so those that are consistent sum to 1.

In the SB experiment, there are four things that can happen ON A SINGLE DAY. They are H1, T1, H2, and T2. Prior probabilities are 1/4 each. When SB is awake, she knows that she is in A SINGLE DAY but it must be H1, T1, or T2. Not H2. So she updates these probabilities to 1/3 each.

This is a trivial conditional probability problem. The reason I posed the "Camp Sleeping Beauty" version, is that it exposes the red herrings. And I assume that is the reason you ignore it, and how the red herrings are exposed.

SB has no unusual "epistemic relationship to the coin," which is what the point of my new construction was trying to point out. That fallacy is based on the misconception that Tuesday somehow ceases to exist, in her world, if the coin lands on Heads. It still exists, and she knows it exists when she addresses the question.

But she also knows that it is not the current situation when she is asked the question. In this new construction, there are N random values (N=2 for the original, N=6 at camp) that determine which row of N days in the schedule is used. So there are N^2 equally likely entries in the schedule, with N possible ways an entry could be observed.

The probability for each random value is the number of times the observed activity appears in the row for that random value, divided the number of times it appears in the schedule. The probability for each day is the number of times the observed activity appears in the column for that day, divided the number of times it appears in the schedule. What the other observations are - even "I wouldn't be able to observe" - are irrelevant since SB knows what was observed, and that the opportunity for the other observations exist regardless of whether the activity is observable.

There is no need to debate a "payout schedule" since the probabilities apply to what SB knows at the time, which depend only which schedule entries are consistent and which are not. To argue otherwise, you have to defend why E="Egg Hunt" gives different answers than E="Extended Sleep," and what that change would be.

Sorry to resurrect. But I recently thought of a way to explain exactly how the halfers are misinterpreting the problem. It is based on how Marilyn vos Savant tried to make the answer to the Monty Hall Problem more intuitive by using more doors. I call it "Camp Sleeping Beauty," and it uses more days, more random choices, and more things that can happen.

SB arrives for camp orientation on Sunday Night and is informed of these details:

• Every day during the week (Monday thru Saturday), campers will take part in an activity that is determined by a six-sided die that will secretly be rolled after they are dismissed on Sunday. The activities are Archery, Bowling, Canoeing, Dodge Ball, Fishing, and an Extra Activity that is unnamed, but not one of those other five.
• The activities are selected using a 6x6 table that is randomly filled out. This can occur either after they are dismissed, or during orientation. (This shouldn't matter, but I suspect that some will try to argue that it does so I leave the option open.)
• This table includes a column for each day, and a row for each possible roll of the die.
• At the end of each day, campers will be shown the table and asked to assign a probability to each day, and to each die roll, based on what activity they participated in that day.
• After providing these probabilities, campers will be put to sleep with an amnesia drug that makes them forget everything that happened that day.

Here is sample of the table. I did make sure that each activity is represented at least once by placing them in order along the diagonal, again for those who think it should make a difference (it can't).

Mon Tue Wed Thur Fri Sat
1 A D C F E B
2 F B B C C F
3 A B C D F D
4 F E B D B C
5 C D F C E E
6 E E F C C F

So, for example, if the die roll was a 6, the campers would do the Extra Activity of Monday and Tuesday, go Fishing on Wednesday and Saturday, and go canoeing on Thursday and Friday.

It is my claim that this is a trivial conditional probability problem. Say the campers went fishing. An "F" appears eight times in this table. Since it appears once in each row except the "2" and "6" rows, there is a 1/8 probability for each die roll except 2 and 6, and a 1/4 probability for each of 2 and 6. Since "F" never appears in the Tuesday column, it can't be Tuesday. But it there is a 1/7 probability for Thursday and for Friday, and a 2/7 probability for Monday, Wednesday, and Saturday.

But by the halfer solution for the SB problem, each die roll has a 1/6 probability. Even a 5, if the campers went Bowling. Or will halfers want to use conditional probability incompletely, and claim Pr(5|B)=Pr(6|B)=0 and Pr(1|B)=Pr(2|B)=Pr(3|B)=Pr(4|B)=1/4, since there are four rows (this is applying conditional probability!) where it appears, but ignoring that there are two where it appears only once and two where it appears twice (this is not applying conditional probability!).

Finally, it cannot matter what "E" means in the table. All that matters is how many times the day's actual activity does. Even if "E" means "don't wake the campers up." Each activity that is demonstrated to not have happened is eliminated the same way.

The classic SB problem is Camp Sleeping beauty with a 2x2 table. Three cells are populated with "Wake and Interview," and one with "sleep." The answer for the "Heads" row is the number of times "Wake and Interview" appears in that row, divided by the number of time is appears in the table. That is 1/3.

This is a Veritasium video on the Cinderalla problem.

It's known as the Sleeping Beauty Problem.

I think it suggests that a fair coin flip can have odds other than 50/50

No. It "suggests" that the conditional probability of an outcome depends on any information that is obtained about that outcome. For example, if I pick a random card the probability that it is the Ace of Spades is 1/52. If I tell you it is a black card, the conditional probability is 1/26. If I tell you it is an Ace, the conditional probability is 1/4.

None of these answers changed anything. They evaluated different sets of information about the card. The issue in the Sleeping Beauty Problem is what, if any, information is gained? And two things seem confuse both the halfers and thirders described in the video:

1. Both seem to think that SB being left asleep on Tuesday, if the coin landed on Heads, is not a result of the experiment. They think this because SB can't observe it. I'm sorry, but it is a result, just one that can only be observed from the outside. If you doubt this, ask yourself if anything changes, on the days she is awakened in the video, if she is awakened on Heads+Tuesday, but not asked the question.
2. The memory wipe between Monday and Tuesday makes the observations SB can make on those two days independent to her, but not an outside observer.

MY ASSERTION: The correct solution is that SB will participate in four equally-likely outcomes that are independent to her, but not to an outside observer, because of the memory wipe. One of these outcomes is eliminated due to the fact that she observes it. So the answer is 1/3.

ONE PROOF BY DEMONSTRATION: There actually are four possible variations of the same experiment. She could be wakened always on Monday, and optionally on Tuesday; or optionally on Monday, and always on Tuesday. With either schedule, she could be wakened on the optional day after Heads, or after Tails; with this variation, she will be asked about the coin result where it is optional.

Regardless of what you think the answer is, that answer applies to each of the four variations. So, use all four at the same time. Assign a different variation to each, and use the same coin flip for them all.

On each day, three volunteers will be wakened. Two of these will be wakened both days, and one will be wakened on this day only. The question each addresses is, in essence, whether she is the "one wakening" volunteer. Yet none have any information to support the probability being different for any of the three. So the answer must be 1/3.

This is true regardless of whether each was told what combination she was assigned, and regardless of whether they can discuss it with each other (as long as they don't share assignments). So it also must be true if the other three exist, or SB merely supposes it. Which is the original problem. Essentially, the "missing" volunteer represents the "missing" observation opportunity.

+++++

Sailor's Child problem
The Sailor's Child problem, introduced by Radford M. Neal, is somewhat similar. It involves a sailor who regularly sails between ports. In one port there is a woman who wants to have a child with him, across the sea there is another woman who also wants to have a child with him. The sailor cannot decide if he will have one or two children, so he will leave it up to a coin toss. If Heads, he will have one child, and if Tails, two children. But if the coin lands on Heads, which woman would have his child? He would decide this by looking at The Sailor's Guide to Ports and the woman in the port that appears first would be the woman that he has a child with. You are his child. You do not have a copy of The Sailor's Guide to Ports. What is the probability that you are his only child, thus the coin landed on Heads (assume a fair coin)?

This is actually not the quite same thing. To see that, imagine what the answer would be if you do have a copy of The Sailor's Guide to Ports.

• If your port is the last one listed, you know that the coin landed on Tails. That is, Pr(Heads|Last)=0.
• If your port is the first one listed, you have no information about the coin. So the conditional probability is the same as the unconditional probability, Pr(Heads|First)=1/2.

But if you don't know what Q is, the answer must be between these two answers; that is, 1/2 can't be correct. So, say the fraction of the other entries that come after your port is Q, where 0<=Q<=1, then Pr(Heads|Q)=Q/(1+Q). Note that this agrees with the other two answers, and that it is 1/3 if Q=1/2.

Since you have no information about Q, you can treat it as as a uniformly distributed random variable. As the number of entries grows large, Pr(Heads|you were born) approaches 1-ln(2) ~= 0.0307,

Pr(Heads & Day = D) = 1/2 * 1/N. — Michael

Fixed; I was in a hurry, and that didn't affect the answer. All the probabilities I gave were off by that factor of 1/2.

That aside, using your above reasoning, in the normal problem the prior probability that she will be woken on Tuesday and the coin landed on Heads is 0

No, the prior probability that she will be woken on Tuesday, and the coin landed Heads, is 1/4. The prior probability that she is awake and the coin landed Heads is 0. "Will be woken" and "is awake" are not the same events.

So when she "rules out" Pr(Heads & Day = Tuesday)

And for about the tenth time, "rules out" is not a valid expression. I only use it since you can't stop using it, and then only when I really mean a valid one. The conditional probability of event A, given event C, is defined to be:

Pr(A|C) = Pr(A&C)/Pr(C).

I gave the correct realization of this definition in my answer. The fact that it differs from yours, in "ruling out" certain probabilities, can only prove that one is wrong. Not which. The fact that mine is valid, and ours is not, proves which.

It is valid, because it looks at what could have happened, not what could not. The prior probability of reaching an interview on any given day, after Tails is flipped, is twice that of Heads. Always. On any day in your experiment.

It's not. You say:

"If the coin landed on Heads, then an N-sided die is rolled, where N>=2. She is woken on day D1 - that is, D1 days after day 0 - where D1 is the result of this roll, and asked her credence. Then she is put back to sleep." — Michael

It's irrelevant (it refers to occurrences after she has answered). But I did intend to take that one out.

I did adjust the second one to match what you said, so your next point is not only just as irrelevant, it is incorrect when you say I "put her back to sleep" after the second interview:

In my example she isn't put back to sleep. The experiment just ends. The same with her second tails interview. So we have no idea how many days the experiment will last. It could be anywhere between 1 and N days.

But it is also incorrect where you claim you identified when she is sent home (you didn't). Or the implication that the experiment's length makes any difference whatsoever. But since these are the only differences you could find, and they make no difference, I can easily correct them to match the version you now say you want. Which, by your standards, is different than what you said before:

1. She is put to sleep on day 0.
2. A coin is flipped.
3. If the coin landed on Heads, then an N-sided die is rolled, where N>=2. She is woken on day D1 - that is, D1 days after day 0 - where D1 is the result of this roll, and asked her credence.Then she is sent home.
4. If the coin landed Tails, then two N-sided dice are rolled. If they land on the same number, repeat the roll until they are different. She is woken on day D1 and day D2, and asked her credence. Then, on the first of these days, she is put back to sleep with amnesia. On the second, she is sent home.

This is the exact procedure you asked for, except: (A) It lasts between 1 and N>=2 days, not between 1 and 14 days. And (B) the selection of the two random "TAILS" days in that period is uniform, instead of weighted toward the earlier days.

On each day in the range 1 to N, the prior probability that she will be woken on that day AND the coin landed on Heads is 1/N. And the prior probability that she will be woken on that day AND the coin landed on Tails is 2/N.

We can proceed two different ways from here. The first is easier, but relies on the statement "the probabilities are the same regardless of which day it is, so we can treat the day D as a known constant.:

• Pr(Heads&Day=D) = (1/2)*(1/N) = 1/(2N).
  
  • This is the prior probability from above.
  • All events of the form (Heads&Day=d), where d is not equal to D, are "ruled out" because it is day D.
• Pr(Tails&Day=D) = (1/2)*(2/N) = 1/N.
  
  • All events of the form (Tails&Day=d), where d is not equal to D, are "ruled out" because it is day D.
  • But I will reiterate that "ruled out" is not a definition that is ever used in probability theory. It is one you made up. What is done, is what follows next.
• The conditional probability of event A, given that it is day D, is found by this definition:
  
  • Pr(A|C) = Pr(A&C)/Pr(C)
  • Pr(Heads|Day=D) = Pr(Heads&Day=D)/[Pr(Heads&Day=D)+Pr(Tails&Day=D)
  • Pr(Heads|Day=D) = (1/(2N))/(1/(2N)+1/N) = 1/3.
  • In other words, what is done is to only use the prior probabilities of events that are consistent with the condition. In this case, with a specific day.

The more formal method is to use that calculation separately for each day d in the range [1,N], add them all up, and divide the result by N. It still gets 1/3.

There is one unorthodox part in all this. The random variable D in the prior probabilities is different from the random variable d used by SB when she is awake. This is because D can take on two values in the overall experiment, but only one to SB when she is awake. So (D=1) and (D=1) are not independent events in the prior, while (d=1) and (d=2) are independent to SB when she is awake

Finally, I'll point out that if N=2, this is the two-coin version you have ignored.

I can't prove a negative. — Michael

Yet that is the basis of your argument. You even reiterate it here. And it is part of your circular argument, which you used this non sequitur to divert attention from:

1. M's solution is right.
2. J's solution gets a different answer, so it must be wrong.
3. J's solution "rules out" an event.
4. M's does not. In fact, it it says that event doesn't exist (the "negative" you claim).
5. That must be the error, since there must be an error (see #2).

If there is some prior probability that is ruled out when woken then tell me what it is.

I have. You ignore it. But this is a fallacious argument. Claiming I did something different does not prove the way you handles the different thing is right and mine was wrong.

If you can’t then I have every reason to accept that there isn’t one.

Quite an ultimatum, from one who never answers questions and ignores answers he can't refute. Since you haven't proven why the event "Heads&Tuesday" doesn't exist - and in fact can't, by your ":can't prove a negative" assertion, I have every reason to accept that it does exist.

+++++

Here is a version of your latest procedure. The only non-cosmetic change is that I made it easier to determine the prior probabilities.The "prior probabilit[ies] that [are] ruled out" are easily identified. I just need you to agree that it is equivalent to your procedure first. IF YOU DON'T, I HAVE EVERY REASON TO ACCEPT THAT IT IS.

1. She is put to sleep on day 0.
2. A coin is flipped.
3. If the coin landed on Heads, then an N-sided die is rolled, where N>=2. She is woken on day D1 - that is, D1 days after day 0 - where D1 is the result of this roll, and asked her credence.Then she is put back to sleep.
4. If the coin landed Tails then two N-sided dice are rolled. If they land on the same number, repeat the roll until they are different.She is woken on day D1 and day D2, and sked her credence. On the first of these days, she is put back to sleep with amnesia.

Notes on the changes; all but one are cosmetic only:

• By numbering day 0, you allow numbering all days.
• You left out the coin flip, but we all know where it has to go.
• The singular of "dice" is "die."
• Your tenses don't mesh with stating when the coin is flipped.
• There is nothing special about 14. Any number works - it doesn't even have to be even - that allows two wakings.
• Rolling two N-sided dice at the same time, rather than two N/2 dice sequentially, makes the distribution of wakings over the N days uniform rather than a complicated function. So it simplifies any calculations based on it, if needed.

Whether the coin landed on Heads or Tails she can wake on any day between Day 1 and day N.

Something is indeed is ruled out when she wakes. You just don't recognize how the sample space you described doesn't recognize it.

When I said that the only things that matter are:

1. She has either one or two interviews determined by a fair coin toss and
2. She doesn’t know if she’s already had one

I was referring to her just waking up, not being told any further information. — Michael

And the only point of mentioning new information, was to show that the information you ignore has meaning. Not to solve the problem or alter the problem. But you knew that.

No prior probability is ruled out here when woken so your example isn't equivalent. — Michael

And again, you keep using circular logic. You deny that events with non-zero prior probability are "ruled out" in your solution. So you claim that my solution, which does "rule out," must be wrong. This is a fallacy; your presumption that you are right is your only defense. You have never argued for why you think they aren't events.

So, lets try this variation of what you most recently proposed. The only changes are to make the conditions you require, as in that quote above, easier to describe with both words and probabilities:

1. SB is put to sleep on day 0, and a coin is flipped.
2. The counting of days is facilitated by a tear-away calendar that is posted on the door to her room at midnight. It starts at Day 1, and as the top page is torn off at midnight every day thereafter, the day numbers are increased by 1.
3. If the coin landed Heads, then an N-sided die (N>=2) is rolled and placed on a shelf by the door.
4. If the coin landed Tails, then two N-sided dice are rolled and placed on the shelf. (If they both land on the same number, the roll is repeated until they are different.)
5. Over the next N days, if the calendar number matches a die number, she is woken and asked for her credence.

Whether Heads or Tails, she can wake once or twice over the next N days. The prior probability of a waking is the same on every day (your method didn't make this true). And she will never know if it is the first waking, the second waking, or the only waking.

Do you agree that this does the same things your latest experiment does (if N=14) with the unimportant exception that the prior probability distribution over the N days varies in your version?

I most certainly can. You just won't try to understand it. Or answer any question placed before you, if you don't want to accept the answer. So once again, just answer mine first and I'll answer yours.

So tell me what prior probability is ruled out in my experiment above. — Michael

This is getting tiresome. The answer follows trivially from what I have said before - have you read it? But I will not continue to dangle on your string. I'll point it out after you tell me:

1. Whether you agree that procedure I just described implements "The only things that matter are that she has either one or two interviews – determined by a fair coin toss – and that she doesn’t know if she’s already had one [or will have another]."
2. If not, what "thing that matters" is not implemented.
3. What her credence in Heads should be.

The specific days she’s woken or kept asleep are irrelevant. — Michael

I agree, with a caveat. The specific details of whether she is woken at a specific point ("day") in the experiment do matter. You can test this in Elga's solution. If she is told that the current day is Tuesday, then she knows Pr(Heads) must decrease to 0. If she is told that it is not Tuesday, the Law of Total Probability actually requires it to go up. It goes from 1/3 to 1/2 in the Thirder soluton, and from 1/2 to 2/3 in the Halfer solution. This is quite relevant, even if you think it is wrong AND CAN PROVIDE A VALID REASON

But it isn't the day name that matters, it is the details associated with that day name. And since these details are different on different "days," we can track them by naming the "days."

Now, you could argue about why those details might matter; but so far you have refused to. You have just asserted they don't (in spite of evidence like I just presented). But naming them cannot affect the correct answer, no matter how those details affect it. SO THERE IS NO REASON TO NOT NAME THE "DAYS." And even the possibility that they might have an effect makes them relevant.

In other words, you are proffering the red herring here. You are insisting that we must ignore a piece of potential information, because you think it has no affect. If so, there is no harm in including it.

The only things that matter are that she has either one or two interviews – determined by a fair coin toss – and that she doesn’t know if she’s already had one. Everything else is a red herring. — Michael

I agree (well, if you add "or if she will have one later") about "only things." But the "red herring" you claim exists, is trying to utilize information that you say has no affect while refusing to address affects that are clearly present.

But if you truly believe in this, then it should be easy to address this:

1. She is put to sleep. A coin, call it C1, is tossed to determine if "she has either one or two interviews."
2. But then a second coin is tossed, call it C2.
   
   1. Call the state of the coins at this moment S1.
   2. If either coin is showing Tails in S1, wake her with amnesia and interview her.
   3. If both are showing Heads, do SOMETHING ELSE.
   4. If she isn't asleep at this point, put her to sleep again.
3. Turn coin C2 over to show its other side.
   
   1. Call the state of the coins at this moment S2.
   2. If either coin is showing Tails in S2, wake her with amnesia and interview her.
   3. If both are showing Heads, do SOMETHING ELSE.
   4. If she isn't asleep at this point, put her to sleep again.
4. Wake her and send her home.

This procedure creates what, in your words, are "the only things that matters." She has either one, or two, interviews and does not know if another will/did happen. The only thing that is different, is that your possible two interviews occur under different circumstances; one is mandatory, and one is optional. What we disagree about is whether the part that is missing - the non-interview when the option is not taken - matters.

Here they occur under identical circumstances. That is, either steps 2.1 thru 2.4, or steps 3.1 thru 3.4. And those circumstances can be used to answer the question. I did "name" the details by calling them state S1 or S2, but since they are identical to SB she can call them state S.

There are three possible combinations of the two coins in state S, and they are equally likely. Her credence in state S=(H,T) is 1/3.

This has nothing to do with what may or may not get "ruled out" in a solution to your version of the experiment. That difference is the red herring in the "most frequent" presentation of the problem. This is a self-contained experiment with a trivial answer.

But it's an answer you don't like. So you will either ignore it, or repeat the non sequitur that it includes the "ruling out of a 1/4 probability" that we are debating about above, which is circular logic.

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.

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.

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.

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.

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.

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.

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.