If you follow your reasoning then you have to claim that your credence is 23. — Michael

I was assuming that was a typo, so I ignored it. Would you like to ask a question about what "my reasoning" would say? Because you have it horribly wrong. You don't even get the halfer logic right.

It's a false analogy because "not being woken up" is nothing like "playing football". It's a mistake to consider "Heads + Tuesday" at all. We can simplify the experiment as such: — Michael

It is true, because the information is what can happen on this day and days like it, vs. what can happen on other days that is different. It is true, because the model can be used to represent the experiment as you described it (and we all understand).

Her credence when asked is 1/3, because of the four possible cells in the 2x2 array, one is eliminated and only one of the remaining three is Heads.

If she is interviewed before playing football her credence that the coin landed on tails is not 1, but if she is interviewed after playing football her credence that the coin landed on tails is 1.

Playing football is additional information, but nothing like that is available in the traditional problem, and so your example is a false analogy. — Michael

You can't call it a "false analogy" if you don't actually address the analogy. Which you didn't. The point of asking a question is to get an answer to that question, not to have you re-arrange it to get an answer that avoids the question as asked.

So please, address the one with a random number selected from [1,M], M activities, and N days. The activities are distributed over an MxN calendar, with the one limitation that each activity appear at least once. And please note that an "activity" can be "just wake her up" or "don't wake her up." Finally, asking for a credence before the activity was not in my question, so it is irrelevant. But it doesn't break the analogy if you try it - there are just two activities on each waking day. One is "just wake her up" and one might be "tennis" or "football." Just not in your very, very specific distribution, in any distribution.

And yes, it is a proper analogy. The popular problem has M=N=2, A(1)="just wake her up," and A(2)="don't wake her up." A(1) is assigned to three cells in the 2x2 array, and A(2) is assigned to the fourth, Heads+Tuesday.

"New information" is an informal expression in probability theory. It does not mean "something not known before." That's a misrepresentation halfers use as an attempt to justify their answer. It means information about potential outcomes that eliminates some. The sample space is represented by the calendar in my question. Activity A(1) eliminates Heads+Tuesday, and so it is "new information."

You can't rule out that today is Monday or that today is Tuesday before playing either tennis or football, which is why your example is a false analogy. — Michael

So you are saying that if everything is the same as in the popular version but SB plays tennis before the interviews, that her credence is 1/3? But if the activity is "free day," it should be 1/2?

I'm sorry, but you seem to be confused about the problem. There is nothing false about the analogy.

An N-day experiment:

The days of the experiment are named D(1) through D(N).
D(0) is the night before the experiment begins, when SB is informed of all these details before going to sleep.
After she goes to sleep on D(0), an M-sided die will be rolled and preserved throughout out the experiment.
What occurs on each day of the experiment is predetermined using an M-row by N-column calendar. One of M distinct activities (that can all be differentiated from each other), A(1) through A(M), is assigned to each cell in the calendar, such that each appears at least once. The calendar is shown to SB on D(0).
With one variation about A(M) that will be explained below, SB will be awakened each day, and will participate in the activity assigned to that day's column and the row determined by the die roll.
After participating in a day's activity, except A(M), SB will be shown the calendar and asked for her credence in each possible die roll. After any waking activity, she will be put back to sleep with amnesia.


There are three possibilities for the variation:

For A(M), SB will be left asleep all day.
For A(M), SB will be taken to DisneyWorld.
For A(M), SB will be left asleep with probability 0<Q<1 or else taken to DisneyWorld. — JeffJo

I'm assuming that nobody can answer this is a way that is consistent with the canonical halfer answer.

Within the subject's knowledge, her waking experience is a random sampling of a certain number of possible combinations of a random integer within a range, and the day within the schedule. The combinations are all equally likely in the prior, but the rules of the experiment limit her waking to a subset of these combinations. Her "new information" is the subset that is consistent with what happens in this waking. The probability of each random number is the size of this intersection of this subset with that number, divided by the size of the subset.

An N-day experiment:

1. The days of the experiment are named D(1) through D(N).
2. D(0) is the night before the experiment begins, when SB is informed of all these details before going to sleep.
3. After she goes to sleep on D(0), an M-sided die will be rolled and preserved throughout out the experiment.
4. What occurs on each day of the experiment is predetermined using an M-row by N-column calendar. One of M distinct activities (that can all be differentiated from each other), A(1) through A(M), is assigned to each cell in the calendar, such that each appears at least once. The calendar is shown to SB on D(0).
5. With one variation about A(M) that will be explained below, SB will be awakened each day, and will participate in the activity assigned to that day's column and the row determined by the die roll.
6. After participating in a day's activity, except A(M), SB will be shown the calendar and asked for her credence in each possible die roll. After any waking activity, she will be put back to sleep with amnesia.

There are three possibilities for the variation:

1. For A(M), SB will be left asleep all day.
2. For A(M), SB will be taken to DisneyWorld.
3. For A(M), SB will be left asleep with probability 0<Q<1 or else taken to DisneyWorld.

After participating activity A(m), for 1<=m<M, what should SB's credence for any die roll d be? I say that it is the number of times A(m) appears in row d, divided by the number of times it appears in the calendar. I say that it does not depend, in any way, on which variation is used.

The popular SB problem is a version of this with M=N=2 and using variation #1. A(1) is simply "interview" and appears in every cell except the HEADS row and the Tuesday column. So the answer, when she is interviewed, is that Pr(HEADS|INTERVIEW)=1/3.

The extra details DO NOT make this problem's solution inapplicable to the popular SB problem. In fact, any solution to the popular one has to be consistent with this problem's solution. All the added details do is illustrate why various rationalizations made by halfers, to avoid using simple conditional probability methods,are invalid. In particular, the three variations for A(M) are meant to show how A(M) cannot affect the solution.

I've been trying to get many questions answered for about a month now, and non have been addressed, except with unsupported assertions that I must be wrong. If that is true, there must be a solution to this problem I present today that differs from what I said.

If you disagree, please provide a solution to this problem, with specific detail about how the three variations affect the answer, not just an assertion that they must.

All I can say is that we aren't agreeing as to the semantics of the problem. Your sample space includes the counterfactual possibility (H, Tuesday), which isn't in the sample space of the experiment as explicitly defined. — sime

So, are you saying that the week skips from Monday to Wednesday if the coin lands on Heads? What it they wait to flip the coin until Tuesday Morning?

As I have described in several ways, and constructed questions to test, the experiment consists of two days, with two COIN+DAY potential-observation methods. But the amnesia means only one is "seen" at a time, and prevents them from being connected (for Pierre: this means no path information). But H+Tue, with no indication of waking or sleeping, is still a possibility, and a sample space must describe all possibilities.

You appeal to "if we awoke SB on tuesday on the event of heads" might be a perfectly rational hypothetical...

Please, what is inconsistent about it, when we ignore whether she is awake? Do you think she does not know it can happen?

But that hypothetical event isn't explicit in the problem description.

Why does the problem description have to explicitly say that something which obviously can happen, can actually happen? The description only says that SB will sleep through it, not that it is excluded from the realm of possibility.

But again, look at the version where, on H+Tue, SB is left asleep with probability Q and otherwise taken to DisneyWorld. The original problem is a sub-case of this with Q=0. the one you think is a different problem, but is not, has Q=1. Instead of thinking of me adding something, address this one as Q approaches, and then equals, 0.

Furthermore, the problem is worded as a philosophical thought experiment from the point of view of SB as a subject who cannot observe that tuesday occurred on a heads result,

But who still knows it can happen. All I'm saying is that your "philosophical thought experiment" does not "philosophically eliminate Tuesday from the week" if the coin lands Heads.

OR. we could just go back to the problem that spawned the SB problem, which was more of that "philosophical thought experiment."

• SB is put to sleep on day 0.
• A fair coin is flipped, and an N-sided die is rolled.
• Over the next N days, SB is woken, or not, based on what is currently showing on the coin and die.
• If she is woken, she is amnesia-ed before going back to sleep that night.
• Regardless of whether she is woken, the die is rotated down by 1 (resetting to N if it was a 1).

The conditions for waking her are if the coin is currently showing Heads, or the die is currently showing 1.

In this version, an exact implementation of the problem proposed by Arnold Zuboff, SB knows FOR A FACT that the experimenters look at the coin and die this morning. She knows, ALSO FOR A FACT, the there were 2*N equally-likely combinations they could have seen on this particular day. She knows, AGAIN FOR A FACT that she would not be awake for N-1 of them, she would be awake for N+1, and in only one of those is the coin showing HEADS. Her credence in Heads is 1/(N+1).

If you think I changed something by adding (if N>2) more days, or that changing the order makes it different, please explain how instead of avoiding the issue by simply saying it is. If you think SB knowing that some combinations could not be observed means that they must be excluded from the sample space, consider that she could be awoken on all days, observe the coin and die herself, and then a sleep+amnesia gas it released if she sees HEADS and DIE>1.

I am simply interpreting the thrux of your position in terms of an extended sample space. — sime

You are inserting details into the description of the outcomes, that provide no additional information. It has nothing to do with the [crux? thrust?] of my position. You are obfuscating the sample space in order to suggest an omission.

Look at it this way: a sample space is a set of distinct outcomes that include all possibilities. An event is a set of outcomes. These concepts often get confused. You are confusing them, by including the name of an event (awake, asleep), as your "new variable" that you claim helps to define the outcome.

There is a simple model for how SB views each waking in an NxN sleeping-beauty calendar. There are N^2 cells in the calendar. Each has an N^(-2) prior probability at the start of an amnesia'ed day, before SB is awakened. When an awake SB observes activity X, her credence in any row or column of the calendar is the number of times activity X occurs in that row or column, divided by the number of times it occurs in the entire calendar.

The only complication in this model, is what happens when one of the activities does not permit an observation. But since the crux of SB's observation is recognizing the activities that did not happen, that cannot be an issue. For there to be an effect, that calendar cell has to cease to exist, not be unobservable. So I'm trying to get that difference addressed.

This step is methodological and not about smuggling in new premises, except those that you need to state your intuitive arguments, which do constitute additional but reasonable premises.

That's just rationalization. I have proposed a model, that I claim represents the SB problem. Whether or not it is "smuggling in new premises" (it isn't, it is extending a premise that already exists), the issues here are only (A) Is the SB problem an example of my model, (B) does my solution apply to the model in general, and (C) how does an unobservable activity affect the solution.

In other words, when you claim that I am "smuggling in a new premise" you are looking at the changes I suggest add details. What I am trying to get you to address is how the SB problem is an example of my model with fewer details.

Look at this way: It is certainly is the case that according to the Bayesian interpretation of probabilities, one can speak of a joint probability distribution over (Coin State, Day State, Sleep State), regardless of one's position on the topic.

What you are doing is creating an event, a subset of the sample space, and applying it as a "new variable" in every outcome member of that event. If you were adding new descriptors, you should add all to each outcome. Yes, some will end up with zero probability, like H+Tue+Awake.

But the Sleeping Beauty Problem per-se does not assume that the Sleeping Beauty exists on tuesday if the coin lands heads, because it does not include an outcome that measures that possibility.

And you confuse "measuring the possibility" with "the possibility exists as an outcome." But I devised specific questions to address this exact issue, which have gone completely ignored.

• If were to wake SB on H+Tue, and take her to DisneyWorld (see how I don't need to add DISNEYWORLD asa descriptor in the sample space?), what is her credence in the event (H+Mon or H+Tue) if she observes the event (H+Mon or T+Mon or T+Tue)? It is:
• Pr(((H+Mon or H+Tue) and (H+Mon or T+Mon or T+Tue)) / Pr(H+Mon or T+Mon or T+Tue)
• = Pr(H+Mon) / Pr(H+Mon or T+Mon or T+Tue) = (1/4)(/(3/4) = 1/3.

[You] do require the introduction of a third variable to the sample space in order to express your counterfactual intuition that I called "sleep state" (which you could equally call "the time independent state of SB").

Except, you didn't add one. You applied a name that always applies to combination of the other variables. And pardon me for suggesting this, but it seems you are using it to not address my very specific questions, that your "new variable" adds nothing to. If you think it does, then use it as part of your answer to those questions.

And it is not "counterfactual." The outcome H+Tue can occur. The prior sample space is comprises every possibility before the random elements are determined, and before an observation is made. And I'm tired of all this repetition, but all this can be explored by addressing the questions I have asked.

As I understand it, your proposal is essentially the principle of indifference applied to a sample space that isn't the same as the stated assumptions of the SB problem, namely your sample space is based on the triple

{Coin,Day,Wakefulness}

upon which you assign the distribution Pr(Heads,Monday,Awake) = Pr(Tails,Monday,Awake) = Pr(Heads,Tuesday,Asleep) = Pr(Tails,Tuesday,Awake) = 1/4. — sime

No.

The reason I keep asking for specific answers to specific questions, is that I find that nobody addresses "my sample space." Even though I keep repeating it. They change it, as you did here, to include the parts I am very intentionally trying to eliminate.

There are two, not three, random elements. They are COIN and DAY. WAKE and SLEEP are not random elements, they are the consequences of certain combinations, the consequences that SB can observe.

There are two sampling opportunities during the experiment, not two paths. The random experiment, as it is seen by SB's "inside" the experiment, is just one sample. It is not one day on a fixed path as seen by someone not going through the experiment, but one day only. Due to amnesia, each sample is not related, in any way SB can use, to any other.

Each of the four combinations of COIN+DAY is equally likely (this is the only application of the PoI), in the prior (this means "before observation") probability distribution. Since there are four combinations, each has a prior ("before observation") probability of 1/4.

In the popular problem, SB's observation, when she is awake, is that this sample could be H+Mon, T+Mon, or T+Tue; but not H+Tue. She knows this because she is awake. One specific question I ask, is what happens if we replace SLEEP with DISNEYWORLD. Because the point that I feel derails halfers is the sleep.

Halfers seem to think SLEEP means H+Tue cannot be sampled. So, I change the problem so it can be sampled. SB's observation is now that she is in an interview. So this sample could be H+Mon, T+Mon, or T+Tue; but not H+Tue. She knows this because she is in an interview, not at DisneyWorld. There is no difference it her utilization of the mechanics of the experiment, nor of what her observation means.

By contrast, the probability space for the classical SB problem is that of a single coinflip C = {H,T}, namely (C,{0,H,T,{H,T}},P) where P (C = H) = 0.5 .

This probability space includes a distribution says that a single sampling is happening on two days at the same time. Halfers convince themselves that there is no contradiction; after TAILS the "other" awakening is identical, and after HEADS it is not observed. But that doesn't work if we change to DISNEYWORLD.

But what makes your argument incorrect [is] the use of a non-permitted sample space.

What is non-permitted? It is a functionally equivalent one. Consequence #1 occurs with three of the four combinations, and Consequence #2 occurs with the fourth. The question is only asked with Consequence #2. I'm sorry, but this is a rationalization.

But its use is why I asked the specific question about my Camp Sleeping Beauty version. There is a 6x6 calendar for the six days of camp, and a six-sided die is used to pick one row (the columns are days of the week). If campers partake of, say, activity C then how do we deduce the conditional probability that the die roll was, say, 5?

I claim that the 36 cells each have a prior probability of 1/36 (PoI again). And that the conditional probability of a 5, given activity C, is the number of times C appears in row 5, divided by the number of times C appears in the calendar.

AND, this does not, and cannot, change if one of the "activities" is "sleep through the entire day." This is what addresses you concern here, and I'd love to hear why D="DARTS" produces a different result than D="DOZE."

There is nothing "non-permitted" about including either D in the sample space. In fact, since either D is part of the camp counselors' planning, they must be functionally equivalent. And it applies to any NxN isomoprhic experiment, even if N=2.

We could discuss this conclusion, but that discussion will need to include answers.

Your bell is just a label for the event {Monday OR Tuesday} — sime

It's supposed to turn the continuous passage of time into a discrete outcome. And yes, I have had halfers try to make that an issue.

What I was pointing out is that this application of the principle of indifference isn't consistently applied to SB.

The Principle of Indifference is applied to single bell-rings, not to SB. It does not take SB into account. That's the point.

P(Monday,Heads) = P(Monday, Tails) = P(Tuesday,Tails) = 1/3

You keep abbreviating it, to insert SB into the PoI application. And I think you made a typo, that I corrected (if you intended to include H+Tue, but not T+Mon, I'd love to hear that explained).

What I'm saying is that the PoI applies to the prior probabilities, which are:

• Pr(H+Mon) = Pr(T+Mon) = Pr(H+Tue) = Pr(T+Tue) = 1/4

Note that there is no "application to SB" here. So I can't be doing it wrong. You are doing it wrong, since you try to include SB somehow.

The solution, is that when SB hears the bell ring, she can eliminate H+Tue. The event AWAKE is (H+Mon) or (T+Mon) or (T+Tue) with prior probability 3/4. And:

• Pr(H+Mon|AWAKE) = Pr((H+Mon) and AWAKE)/Pr(AWAKE) = (1/4)/(3/4)= 1/3.
• Pr(T+Mon|AWAKE) = Pr((T+Mon) and AWAKE)/Pr(AWAKE) = (1/4)/(3/4)= 1/3.
• Pr(T+Tue|AWAKE) = Pr((T+Tue) and AWAKE)/Pr(AWAKE) = (1/4)/(3/4)= 1/3.

Note that the PoI is not used here, either.

To verify that you are happy with this credence assignment, you need to check the hypothetical credences that this credence implies. In the case of P(Monday | Tails) we get

P(Monday | Tails) = P(Monday, Tails) / P(Tails) = (1/3) / (1/2) = 2/3.

No,

• Pr(Mon|Tails) = Pr((T+Mon) and ((T+Mon) or (T+Tue))) / Pr((T+Mon) or (T+Tue))
• Pr(Mon|Tails) = (1/4) / (1/2) = 1/2

Are you happy with this implied conditional credence?

This one? Yes. Since yours included observing H+Tue, and pulled the event "Tails" out of SB's observation, that one was bogus.

+++++

Now, replace "Leave SB asleep" on H+Tue with "Wake SB and go to DisneyWorld with probability Q, or leave her asleep with probability 1-Q."

How does SB's credence in Heads, in an interview that can't be on H+Tue, depend on Q?

No, the Halfer position doesn't consider SB to have any information that she could utilize when awakened, — sime

"Gee, what do I know? Well, if the coin landed Tails then there is another waking I have to go to, and I have to split the prior probability of Heads between today and that day. But if it landed Heads, the full probability weight of Heads is applied to today."

"So If I knew which way the coin landed, I could get different answers about today. BUT I DON'T KNOW. I have to moderate these possibilities in a way that makes the probability of TODAY equal for all."

Of course, all of this could be made clearer if we compare two versions of the experiment, one where SB sleeps through H+Tue, and one where she is awakened but taken to DisneyWorld instead of being interviewed. If the answer changes, she is using information about the "other" day that she is supposed to not know. If the answer is the same, she is not using it. But if it is the same, it can't be 1/2 for the COIN, and 1/4 for the COIN+DAY, since the probabilities addd up to more than 1.

Yes, an awakened SB doesn't know which of the possible worlds she inhabits and is indifferent with regards to which world she is in and rightly so.

So don't use that as a model, use the well-established methods of conditional probability. Ring a bell at noon of both days. An awake SB hears it, but a sleeping SB is unaffected in any way.

The prior probabilities of a specific bell-ring being on any member of {H+Mon, T+Mon, H+Tue, T+Tue} is 1/4. If SB hears it, H+Tue is eliminated. Conditional probability says:

Pr(H+Mon|Bell) = Pr(H+Mon)/[Pr(H+Mon)+Pr(T+Mon)+Pr(T+Tue)] = 1/3.

No, this doesn't imply that she should assign equal probabilty values for each possible world: For example, we have already shown that if an awakened SB assigns equal prior probabilities to every possible world that she might inhabit, then she must assign unequal credences for it being monday versus tuesday when conditioning on a tails outcome.

???? The probability of Tuesday, conditioned on a Tuesday outcome, is 1.

But SB's sample space (the set of possible outcomes) does not include "Tuesday." That is an event that is the union of the discrete outcomes "H+Tue" and "T&Tue." The significance is that she can only observe one.

To recap, if P(Monday) = 2/3 (as assumed by thirders on the basis of indifference with respect to the three possible awakenings), and if P(Tails | Monday) = 1/2 = P(Tails) by either indiffererence or aleatoric probability, then

????

Thirders never say Pr(Monday)=2/3. Since SB can only observe one combination of COIN+DAY at a time, they say that Pr(MONDAY|I AM OBSERVING ONE COMBINATION) = 1/3. A dreaming SB might say PR(MONDAY|I'M DREAMING)=1, but any evidence she actually has includes only one of these outcomes.

quote="Pierre-Normand;1022039"]Before Beauty sleeps, the attendants lay out a garden she knows everything about:

At the gate there's a fair fork: Heads-path or Tails-path (that's the aleatoric 1/2 vs 1/2).

Along the paths they place the stopping spots (interviews): one lit spot on Heads, two lit spots on Tails.

Amnesia just means each lit spot looks the same when she stands under it.[/quote]

The difference here is that she seems to get a second chance at finding a lamppost along one path, but not the other. Your version pre-loads the probabilities based on which path is taken, which is what I mean when I say you use information that SB cannot possess.

Whether or not that makes a difference can be seen if you address the Camp Sleeping Beauty version, where she may or may not "find" a spot along the indicated path. Specifically, how does, say, Friday's lamp being lit or unlit change her credence on Monday? You are taking a property of the path she is on, that she cannot know, as evidence she can use.

The only information she makes use of about the run she's in is that the fair coin decided it. — Pierre-Normand

So she doesn't make use of the fact that her "run" might include another waking, which is supposed to be beyond her worldview? Specifically, that there might be a connection that she cannot see?

It's the exact same information that she's also making use of in your own "four-cards" variation since it's on the basis of this information that she knows that getting a T-Monday, a T-Tuesday or a H-Monday

You seem to have left out H-Tuesday. By doing so, you are utilizing information about one "run" that you ignore in another. And this omission is the exact reason I'd like to see my questions, which are designed to demonstrate how you are utilizing this information inconsistently, addressed. If you feel you are not utilizing this information, there should be no reason to not address them.

Sime's piece of the puzzle: The grounding of SB's credence is aleatoric. The fair coin doesn't just draw the map, it drives the frequencies the map will realize across many runs (or, justify expectations over one single run) — Pierre-Normand

When SB N is awake, while she is aware of the map, she has no information that she can use to place herself in that map. IT IS IRRELEVANT.

In any way that SB can assess her credence, that does not reference her position in the map, the answer is 1/3.

• Using four volunteers, where each sleeps though a different combination in {H&Mon, T&Mon, H&Tue, T&Tue}? On any day, the credence assigned to each of the three awake volunteers cannot be different. and they must add up to 1. The credence is 1/3.
• Use the original "awake all N days, or awake on on one random day in the set of N" problem? N+1 are waking combinations, only one corresponds to "Heads." The credence is 1/(N+1).
• Change the "sleep" day to a non-interview day? It is trivial that the answer is 1/3.

I'm sure there are others. The point is that the "halfer run-based" argument cannot provide a consistent result. It only works if you somehow pretend SB can utilize information, about which "run" she is in, that she does not and cannot posses.

But to consider a halfer's objections, I have to get that halfer to address the direct questions I have asked about how a halfer argument applies in other scenarios that must use similar arguments. So far, I can't get that to happen.

I definitely would not say that my credence is 50/50, because any statistic computed with that credence would not be reflective of the physical information that you have provided. — sime

Then what would you say it is? If you say Q, then your credence in Tails must be 1-Q, and you have a paradox.

The SB problem is a classic illustration of confusing what probability is about. It is not a property of the system (the coin in the SB problem), it is a property of what is known about the system. That is, your credence in an outcome is not identically the prior probability that it will occur. Example:

• I have a coin that I have determined, through extensive experimentation, is biased 60%:40% toward one result. But I am not going to tell you what result is favored.
• I just flipped this coin. What is your credence that the result was Heads?

Even though you know that the probability-of-occurrence is either 60% or 40%, your credence in Heads should be 50%. You have no justification to say that Heads is the favored result, or that Tails is. So your credence is 50%. To justify, say, Tails being more likely than Heads, you would need to justify Tails being more likely to be the favored result. And you can't.

And the reason I have not responded to many of Pierre's comments, is that they try to justify answers that directly contradict the answers to the questions I have asked, but have gone unanswered. Because he is trying to convince me with unsupported logic that would be dismissed if he answers mine. Since there is no end in sight to the carousel of unanswered questions, I am going to assert the answers to mine.

I'm going to describe several alternate scenarios that encompass my point. All include amnesia at the end of each day. What I would like to see, is either agreement with these assertions; or disagreement, with reasons. And if reasons are given, I will respond to them.

• Heads & Monday: Wake SB and interview her in conference room A.
• Tails & Monday: Wake SB and interview her in conference room B.
• Heads & Tuesday: Wake SB and take her to Disney World.
• Tails & Tuesday: Wake SB and interview her in conference room D.
• When SB is interviewed, she is asked fro her credence in each line item.

Note: the conference rooms will be indistinguishable to SB.

Assertion #1: These are different outcomes, regardless of SB's ability to distinguish them.

Assertion #2: As she wakes up, SB's credence in today being each line item should be 25%.

Assertion #3: SB cannot make use of runs. That is, if she is being interviewed in conference room A there will be a trip to Disney World. If she is being interviewed in D, she can't make use of the fact that she had been interviewed in C. Such knowledge is of no use to her.

Assertion #4: Since her credence in each outcome is 25%, and she cannot utilize "runs," when she is interviewed her credence in each of the three "interview" outcomes updates to 33%.

• Heads & Monday: Wake SB and interview her in conference room A.
• Tails & Monday: Wake SB and interview her in conference room A.
• Heads & Tuesday: Wake SB and take her to Disney World.
• Tails & Tuesday: Wake SB and interview her in conference room A.

Assertion #5: There is no difference, that can affect SB's credence, in this scenario. Whatever "identifies" an interview has nothing to do with the room where it occurs, it is the circumstances under which it occurs. But not "runs."

• Heads & Monday: Wake SB and interview her.
• Tails & Monday: Wake SB and interview her.
• Heads & Tuesday: Leave SB asleep
• Tails & Tuesday: Wake SB and interview her.

Assertion #6: There is no difference, that can affect SB's credence in an interview, in this scenario. Not being able to observe H&Tue does not remove it from the set of outcomes she knows can happen. Those are determined by the plan described on Sunday, not SB being able to observe it.

Assertion #7: Not being able to observe H&Tue does not make allow SB to utilize the difference between a "Heads run" and a "Tails run."

• The Camp Sleeping Beauty setup, with six distinguishable activities named A, B, C, D, E, and F. Each day in the six-day-by-six-die-rolls camp calendar is randomly assigned one.
• After participating in each day's activity, SB is asked for her credence about the possible die rolls.

Assertion #8 (Thirder version): Her credence in die roll D should be the number of days that today's activity occurs in row D, divided by the number of times it appears on the calendar.

Assertion #8A (Impossible Halfer Version): Her credence in each die roll should be 1/6, even for die rolls where today's activity does not appear. (It is impossible since SB knows the die roll can't be for a row where today's activity does not appear.)

Assertion #8B (Inconsistent Halfer Version): Her credence in each die roll where today's activity does not appear has to be 0. Fore those where it does, it should be 1/N, where N is the number of rolls where it appears. (It is inconsistent since it contradicts the halfer concept that her credence can't be updated.)

Assertion #9: It does not matter if one of the activities is "sleep all day and skip the question."

Assertion #10: The halfer logic is inconsistent. The correct answer is the thirder's.

You repeatedly claimed that I'm disallowed to make reference to any awakening opportunity Sleeping Beauty isn't currently experiencing. — Pierre-Normand

You can refer to any part of the experiment you want. Sleeping Beauty knows all of the parts (*), but has no means to relate her current awake period to any others. You are saying halfers base their answer on doing that. They can't.

But how do you yourself arrive at a credence of 2/3 without making reference to the fact that there are three possible awakening opportunities in total and not just the single one that she is experiencing?

Are you really that obtuse? As I indicated with the (*), she knows all of the parts. That's what establishes the prior sample space. All four possibilities, with equal probabilities. Since she is awake, she eliminates the one she sleeps through.

And as I have said, betting arguments don't work because you have to agree on how many bets are placed. But there is no logical fallacy in a direct probability analysis, as I have done.

There is nothing vague about my questions, unless you refuse to understand it.

• Compare two versions of the popular problem; one where she stays asleep on H+Tue, and one where she is awakened but taken to Disney World instead of being interviewed. In the halfer, two-runs model, does her credence in Heads change between these two versions? What is her credence in Heads when she goes to DisneyWorld?
• In my Camp Sleeping Beauty version, is her credence in die roll D (# times today's activity appears in row D)/(# times today's activity appears in table), as thirders would claim, or is it 1/6 as halfers would claim? How does the halfer's answer change if today's activity does not appear in all rows?

No. I just mean that when she awakens she isn't able to tell if she's in a T-run anymore than she can tell if she's in a T-Monday-awakening or any other possible awakening. — Pierre-Normand

And I'm saying that this is the exact reason why she cannot base credence on what may, or may not, be the other part(s) of the "run" she is in. I'm saying that all she can base credence on is the one day she can see. And this is trivial to confirm, by addressing the questions you refuse to acknowledge.

SB doesn't have the magical power to make the other awakenings, or their mutual causal relationships, drop out of existence on the occasion where she awakens. — Pierre-Normand

Exactly. That is the opposite side of the ability you claim she could have, to make one "other awakening" selectively pop into significance based on know;edge she does not possess. That is, to treat the "other Tails awakening" when the coin landed Tails differently than the "Heads awakening."

Again: the prior sample space comprises FOUR combinations of Coin+Day. In the prior, each is equally likely to apply at the moment the lab techs decide whether or not to awaken her. If they do, to entirety of her information about it is that it is one of the THREE combinations that correspond to an awakening. To her, there is no more, or less, of a connection to the "other" day in this two-day run that indicates, to her, whether it is Monday or Tuesday, if the coin landed Heads or Tails, or which "run" she is in. If you think otherwise, I'd be glad to hear why. An explicit reason why, on T&Mon, she could be more or less likely to think it is H&Mon. This requires knowledge of whether she is in a Heads or Tails run, not the knowledge that such runs are possibilities.

When SB, as a Halfer, says that the odds that the coin landed tails are 1/2, what she means is that her current awakening episode is part of a set of indistinguishable runs that, in the long run, will turn out to have been T-runs one half of the time. — Pierre-Normand

Indistinguishable? You contradict yourself here, because in the long run you do distinguish them.

But you use this argument to once again evade answering the direct questions I have asked several times. One of them is "If the sleeping day is changed to a non-interview waking day, what should her answer be on an interview waking day?" It can't be 1/2, because that would not allow here to have 100% credence in Heads in the non-interview waking day. So she must answer 1/3. But if she answers 1/3, what is different in her knowledge on an interview waking day in the original version?

And if you try to hand-wave a difference, how does in work in the Camp Sleeping Beauty version when each run can contain a different number of waking days?

But I've given up the silly notion that you will address these questions. Which probably means you can't.

The Halfer's run-centered measure just is a way to measure the space of probabilities by partitioning the events that Sleeping Beauty's credence — Pierre-Normand

<Sigh.> I can repeat this as often as you ignore it.

The experiment, when viewed from the outside, consists of two possible runs. The experiment that SB sees is one day, from one run, and to her that one day is independent of whichever run she is in. Since she cannot know which run she is in, that is not information that is useful to her. Inside the experiment, an outcome consists of one "day" only. The only point that is significant to SB is that she can tell that an interview day is not a sleeping day. This constitutes "new information" in probability.

In fact, "new information" is not defined in probability. The information that allows for probability updates is whatever eliminates outcomes that exist in the prior sample space, but are inconsistent with that information. Yes, this usually means a positive fact about the outcome, that does not apply to all, hence some call it "new information." But being "new" isn't what is important, it is the elimination. H&Tue is a member of the prior sample space. It is eliminated when she is awoken and interviewed.

And you can check this is several ways, all of which you ignore. One, you can change the sleeping day to one where she is awakened, but not interviewed. I'll stick with the example that you take her to DisneyWorld. Now one "day" in what you call the "Heads run" is eliminated when she is interviewed. Since the "Heads run" has a 50% probability, and she is can't be in all of the probability-weight of the "Heads Run,", her credence in Heads must be less than 50%/

But it cannot matter what happens on H&Tue. What affects SB's credence is that she knows that the current "day" is not H&Tue. Which she knows whatever happens on H&Tue.

Or you could address the Camp Sleeping Beauty version with more than just "it illustrates the thirder view." You could try to apply the "six day run" theory to Camp Sleeping Beauty. And you will not be able to do so consistently.

• Is SB's credence in each die roll the number of times today's activity appears in that row, divided by the number of times it appears in the 6x6 calendar. If you disagree, please say what it is based on the "six day run" theory.
• If one of the activities is replaced with "sleep through this day," does that change her credence in any way? HOW, and TO WHAT?

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

Because her current knowledge and existence in the experiment is fully limited to one day. Knowing that she will always be awakened on Monday does not change that. See above.

I do think this related to the Monty Hall problem where information affects probabilities. Information does affect probabilities, you know. — ssu

This is called conditional probability.

It's easier indeed to understand the Monty Hall when there's a lot more doors

What that does is make it more intuitive. Since there is a 99.9999% chance Monty Hall picked that one door for the specific reason that it has the car, and a 0.0001% chance that he picked a goat door randomly, it makes sense top go with the 99.9999%. This is harder to wee then the numbers are 66.7% and 33.3%.

So yes, there is similarity in that the information that allows conditional probability to be used is hard to see. But the reasons are quite different. In Sleeping Beauty, it is because philosophers want to propose inconsistent ways to view information.

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.