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.

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

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

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

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

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

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

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

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

I’m not asking about your shopping example. — Michael

But I am.

I’m asking about this example:

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

And the reason for the shopping example is pointing out that the four parts that I highlighted and labeled A, B, C, and D each have a prior of 1/4. And that what does or does not happen in C cannot change SB's credence in A, B, or D. All that matters is that when she knows she is in A, B, or D, she can "rule out" C. You are treating C as if it it isn't a point in the experiment, and it is.

But you would have to address my questions to discuss this, and any truthful answer to them would discredit all you have argued. So you continue to ignore them, and reply only with this same non sequitur.

Which is what? — Michael

That 1/4 chance that she would have been taken shopping.

In your experiment the prior probability P(HH) = 1/4 becomes P(HH) = 0 when she’s asked her credence. But there is no prior P(X) = 1/4 that becomes P(X) = 0 when she’s asked her credence in my simplified form of the experiment.

Hence your experiment is not equivalent and your solution doesn’t apply. — Michael

Tell me if you remember reading this before: In any experiment, measures of probability define a solution, not the experiment itself. The more you repeat this non sequitur (that your preferred solution can't be applied to my version of the experiment), the more obvious it becomes that you recognize that my experiment is correct.

BUT, Q4 thru Q6 show that there is a "prior P(X) = 1/4 that becomes P(X) = 0 when she’s asked her credence" in your experiment. There is a possibility that she can rule out, and it DOES NOT MATTER whether she would be awake, or asked anything, in that possibility. All that matters is that she knows it is not the current case.

Now, we could discuss all this if you would reply with anything except this same, tired non sequitur. Q1 thru Q3 are an easy start. But they don't involve mentioning prior probabilities.

The prior P(Heads & Second Time) = 0 as established by the rules of the experiment. She will never be asked a second time if the coin lands heads. So there's nothing for her to later rule out when she's asked her credence. — Michael

She cannot use ""First Time" and "Second Time" in her solution at all, except in the context of the Law of Total Probability. Because she cannot have the slightest clue which "time" (and you don't define if "second time" means the second possible waking, or the second actual waking) it is. So she can "rule out" this case as part of this Law's application.

The entire point of my suggested implementation is that "First Time" and "Second Time" look exactly the same. So the "prior" means the circumstances that govern "This Time," not "How this time relates to the other possible time."

But you still ignore that the details used anybody's solution do not establish whether the implementation of the problem is correct. They can only establish whether a solution is. Your argument here is a non sequitur. The experiment does not require the subject to distinguish between which "time" it is; and in fact, it prevents isolating such a difference.

Let me repeat "the rules of the experiment." From Elga's 2000 paper titled "Self-locating belief and the Sleeping Beauty problem" :

Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?

Q1: Do you agree, or disagree, that the procedure I have outlined (with two coins, turning coin C2 over, but asking only for credence in coin C1) correctly implements this?

Q2: Do you agree, or disagree, that the subject can be wakened, and asked for this credence, during the "second time" the waking+interviewing process could occur?

Q3: Do you agree, or disagree, that the subject's answer (when asked) can be based solely on the state of the coins between time A when were examined to see if she should be wakened, and time B when/if she is subsequently wakened and interviewed?

And you also ignored this:

1. Shopping 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 given amnesia and taken shopping.
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. — JeffJo

Q4: Do you agree, or disagree, that she can have a credence in Heads while shopping in step 3 of this variation of what you presented?

Q5: And that she even can be asked for it without affecting her credences at other times?

Q6: And that those credences can't be affected by whether they lied to her about step 3?

No it's not. She doesn't know that she's being asked a second time. She can't rule out heads. — Michael

Yes, it is. The Law of Total Probability says that

Pr(Heads) = Pr(Heads&First Time) + Pr(Heads&Second Time)
But "not ruing out" has not significance." Do this instead:

1. Shopping 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 given amnesia and taken shopping.
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.

If Shopping Beauty is taken shopping, she knows that Pr(Heads)=0, even tho she isn't asked.If she is asked for a credence, she know that one of four possibilities is ruled out.

She’s asked once in step 1 and then, optionally, again in step 4. — Michael

Okay, I missed that.

No prior probability is ruled out when asked. — Michael

When she is asked the second time, the "prior probability" of heads is ruled out.

That is what is wrong with your solution. You disagree with this criticism, but that does not invalidate these steps. In fact, nothing about anybody's solution invalidates anybody's set of steps.

But you have not addressed my steps, or my solution. You have only repeated your solution. A solution I claim is invalid.

OK, I understand your argument now, — Michael

No, you seem to understand the process finally, but your counterargument completely misses the point of the argument.

Specifically, you are saying that by not applying probability theory as you do, in the same way you do, that I am not implementing the problem correctly. I am saying - and nothing you have said challenges this - that:

1. I have implemented the original problem (not the variation of it that Elga solved) exactly.
   
   • I may have added a detail (coin C2), but that is how the specifics of the original problem (one or two wakings) is managed.
   • You (well, Elga) are adding a similar detail with Monday/Tuesday schedule. It also adds a detail, and is correct. But it obfuscates the solution instead of clarifying it.
2. Mine has a trivial solution that establishes what the answer to any correct implementation of the problem is.
3. So any solution that does not get that answer, to a correct implementation, is wrong.

in your experiment the prior probability P(HH) = 1/4 is ruled out when woken — Michael

Which is the entire point. You refuse to recognize that something must be "ruled out" due to the fact that your SB cannot be awake, after Heads, on Tuesday. The reason you do this is because the probability space that governs the situation on your Monday is different than that which governs Tuesday, and you disagree with how a joint probability space is calculated by Thirders.

My entire point is to create a single probability space that applies, unambiguously, to any waking.

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

This does not implement the original problem. She is wakened, and asked, zero tomes or one time.

From Elga:

The Sleeping Beauty problem:
Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking. When you are first awakened, to what degree ought you believe that the outcome of the coin toss is
Heads?

So if the coin combination is HH then the participant will be asked their credence during the second pass? If so then you are wrong when you said "she also knows that the fact that she is awake eliminates (H,H) as a possibility." — Michael

No. Please pay attention to the highlighted texts.

My implementation of the SB problem, the one I have been describing, is:

But the difference Elga introduced was unnecessary. So don't do it; do this instead:

Tell SB all the details listed here.
Put SB to sleep.
Flip two coins. Call them C1 and C2.
Procedure start:
If both coins are showing Heads, skip to Procedure End.
Wake SB.
Ask SB "to what degree do you believe that coin C1 is currently showing Heads?"
After she answers, put her back to sleep with amnesia.
Procedure End.
Turn coin C2 over, to show its opposite side.
Repeat the procedure.
Wake SB to end the experiment. — JeffJo

What matters is the probability that you will be asked for your credence at least once during the experiment. — Michael

0
What matters is not the question, but that you are awake for an interview. And in my implementation, you are. After careful double-checking I found that you not only linked to the wrong posting, you ignored the part that said:

Now, let the "further details" be that, if this is the first pass thru experiment, the exact same procedure will be repeated. Otherwise, the experiment is ended. Whether or not you were asked the question once before is irrelevant, since you have no memory of it. The arrangement of the two coins can be correlated to the arrangement in the first pass, or not, for the same reason. — JeffJo

I did forget to say that coin C2 is turned over, but that was said before. What I outlined in 3 posts above is identical to the SB problem. What I said in that post you dissected applies to one pass only, and the intent was to have two passes where, if there was no question in the first, there would be in the second.

The lab assistant only asks your credence if the coin combination isn't HH. — Michael

And in the original, on Tuesday after Heads, you are also not asked for a credence. The only difference is that in this subset of my implementation, you started out awake but forget the occurrence. Oh, yeah, it had only one possible waking, so it was not the full SB problem.

The point of extracting this from the full procedure was (A) the probability can be calculated based on the state of the current pass, and (B) only asking a question is important, not waking or sleeping.

My implementation of the SB problem, the one I have been describing, is:

But the difference Elga introduced was unnecessary. So don't do it; do this instead:

Tell SB all the details listed here.
Put SB to sleep.
Flip two coins. Call them C1 and C2.
Procedure start:
If both coins are showing Heads, skip to Procedure End.
Wake SB.
Ask SB "to what degree do you believe that coin C1 is currently showing Heads?"
After she answers, put her back to sleep with amnesia.
Procedure End.
Turn coin C2 over, to show its opposite side.
Repeat the procedure.
Wake SB to end the experiment.


When SB is awake, she knows that she is in the middle of the procedure listed in steps 4 thru 9. Regardless of which pass thru these steps it is, she knows that in step 5 of this pass, there were four equally-likely combinations for what (C1,C2) were showing: {(H,H),(H,T),(T,H),(T,T)}. This is the "prior" sample space. — JeffJo

What you were looking at was one pass thru step 4 to 9.

Probability=1 means "is certain." You said the probability of being asked was 1. That means certain. The subject in my implementation is always asked.

"Prior probability" means "prior to information being given," so "prior probability...if Heads" is vacuous.

I have no idea what you mean by "my example." My implementation is an exact version of your problem #1. I notice that you don't claim otherwise, you just try to say that #2 is different and I implemented #2. #3 is closer, and it is equivalent to #1.

My explanation of how your additional constraints turned #2 into #3 is trivially true. As is how #3 is equivalent to #1.

What credence is asked for does not affect, in any way, how the events themselves might occur. You are arguing with non sequiturs.

All you have identified is how my solution differs from yours, not how my implementation of the problem differs from the actual problem. This is because it is trivially obvious that it does not. But you can't attack my solution, which is also trivially correct, so you attack the implementation.

What matters is that in the Sleeping Beauty problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1 — Michael

What has no significance to the SB problem, is what might be different if event X, or event Y, happens WHEN BOTH ARE CERTAIN TO HAPPEN.

• X = The prior probability of being asked one's credence at least once, equal to 1.
• Y = The prior probability of being asked one's credence at least once if heads, also equal to 1.

Now, you are either a troll, very confused, or expressing yourself poorly. But if you won't consider that I might be right (as I keep assuming about you, until I actually find something wrong) there is no way to discuss anything with you.

Neither participant knows if they are A or B. — Michael

I'm going to ignore the fact that neither A nor B is woken twice, so this isn't the SB problem. What you seem to mean is that the subject is woken once as A if Heads, and once each as A and as B if tails.

Then you have created a third problem that is the equivalent of #1. Literally, it just adds a certain detail about names that is significant in #2 when they apply to different people, but makes it identical to #1 when it is the same person that can have either name. And an event it uses to choose the name - Heads or Tails - is already mentioned in the problems:

1. A is woken once if Heads, twice if Tails.
2. A is woken once if Heads, A and B once each if Tails.
3. SB if woken once in room 100A if Heads, or once each in of the rooms 100A and 100B if tails.

My version is indeed equivalent to both #1 and #3. It is not equivalent to #2 if A and B are different people, which was not given in its definition. In the special case where that is specified, all three are the same problem.

These are two different problems:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, A and B once each if tails — Michael

Yes they are.

#1 has two subjects, and #2 has two.

In #1, the only subject knows she will be wakened, just not how many times. Her credence in Heads is what we are asked about. Since both Heads and Tails are possible when she is awake, 0<Pr(Heads|Awake)<1.

In #2, A knows she will be wakened and that the coin is irrelevant. So Pr(Heads|Awake)=Pr(Heads)=1/2. B knows that she will only be wakened if Heads. So Pr(Heads/Awake)=1.

In my version, there is one subject who knows she will be wakened, just not how many times. Please, without referring to how you you would solve the problem once it is established, how is this different than #1? And how does the presence of B in #3 make it like mine?

In the original problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1, which is why the answer is 1/2 and why your example isn't comparable. — Michael

So now it isn't that I never asked about two coins ("You toss two coins and don’t ask them their credence if both land heads. That’s what makes your experiment equivalent to my second example where B isn’t asked if heads.")?

Do I need to explain "my version" to you again? You now say it is different because:

1. In the original problem the prior probability of being asked one's credence at least once is 1 ...
   
   • In mine, the prior probability of being asked ones credence (that coin C1 is showing Heads) at least once is also 1.
2. ... and the prior probability of being asked one's credence at least once if heads is 1 ...
   
   • In mine, the prior (?) probability of being asked one's credence at least once if Heads is also 1
   • But this doesn't appear to be a prior probability. That seems to be what "if Heads" means.

So you are still describing how it is the same problem, while claiming that it is different.

... which is why the answer is 1/2 and why your example isn't comparable.

Non sequitur.

Please, show me how

• The existence of two credence=1 events...
• ... which, while the facts that they are certain are valid, do exist as an issue in the problem,

... produce the conclusion "the answer is 1/2."

Do you want to try again? But do try to realize that probabilities used within a solution cannot affect whether the problems are the same. They can only affect the solutions, and credence=1 events do not do that.

I do not ask anybody (for) their credence if both coins landed on Heads. — JeffJo

Exactly. It is precisely because the prior probability of being asked at least once is 3/4 that the probability that the first coin landed heads is 1/3. — Michael

The original problem is about one coin, not two. Asking about two would make it a different problem. Asking about one is what makes it the same problem.

But yes, it is indeed true that the prior probability of 3/4 is what makes the answer 1/3. You identified the wrong event for that prior probability (she is always asked), but it is the fact that this same prior probability applies to any waking, and not different prior probabilities depending on whether the subject is wakened on Monday or Tuesday, that makes it usable in a valid solution.

Thank you for stating, in your own words, why this is so.

You toss two coins and don’t ask them their credence if both land heads. That’s what makes your experiment equivalent to my second example where B isn’t asked if heads. — Michael

I do not ask anybody (for) their credence if both coins landed on Heads. I don't ask anybody about coin C2 at all, although it has to be taken into account.

I ask for a credence in whether coin C1 is currently showing Heads. And the way the problem is set up, that is always the same event where coin C1 landed on Heads. Maybe you do not realize that coin C1 is the coin in the problem? And coin C2 is not?

C2 is just the coin that controls ordering. Since amnesia makes ordering irrelevant to the single subject, what coin C2 is showing cannot affect the answer.

That’s what makes your experiment equivalent to my second example where B isn’t asked if heads. — Michael

There is no B anywhere, as far as I can tell. You don't seem to want to explain the important details, like whether B is a person, a person in a different situation, or (as it seems here) if B is an event that is not a part of the experiment.

AGAIN: I use a single subject. That subject is always wakened, but may be wakened twice. That subject is always asked "if heads" about the only coin that the original problem is concerned with, (It's even possible we could pose the question while she sleeps, but we shouldn't expect an answer.) When she can answer, she knows that there are three equally likely combinations for what the two coins are showing, and in only one is coin C1 showing Heads.

This is not difficult. But you do need to stop trying to contrive a situation where it is wrong. So far, you have not even described a situation that applies.

We have two different experiments:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, A and B once each if tails

Your version of the experiment is comparable to the second experiment, not the first. — Michael

Um, no.

In "my experiment" I will literally and explicitly wake the single subject once if coin C1 lands on Heads, and twice if it lands on Tails. And there literally and explicitly is no second subject. So it is an exact implementation of 1, not 2.

What you are doing here, is confusing the change in the details that determine how this single subject will be awakened, with her being a different subject. You name the different subjects "subject A" and "subject B." Well, the same thing can be done with Elga's ("the most frequent") implementation:

3. Wake subject C once on Monday if Heads, or wake subjects C and D once each (on Monday and then Tuesday, respectively) if tails.

And the reason that neither 2, nor 3, fits the intended problem is that the subject has to know she will be wakened, but not remember whether it happens one or two times. So that she is the same person, but does not know if she is in "world" (I don't like this word here, but it is how philosophers approach the problem) A, B, C or D.

AGAIN:

The controversy created by Elga's ("the most frequent") implementation, of the same problem that I implement, is this:

• The incarnation of the single subject does not know if she is in world C or D.
• This prevents her from defining a sample space that applies to just her world.
• So she has to combine them into a single world
• Halfers do it by saying they are the same world as the world that contains both, so the single probability space that applies to the combination applies to each individually.
• Thirders do it by saying they are separate worlds, requiring a joint probability space and an assessment of which world she might be in.

And the way my implementation solves this is by creating a simple probability space that applies to just world A, or to just world B, so it no longer matters which world the subject is in.

So we have two different versions of the experiment: — Michael

I'm not quite sure why you quoted me in this, as the two version you described do not relate to anything I've said.

To reiterate what I have said, let me start with a different experiment:

You volunteer for an experiment. It starts with you seated at a table in a closed room, where these details are explained to you:

1. Two coins will be arranged randomly out of your sight. By this I mean that the faces showing on (C1,C2) are equally likely to be any of these four combinations: HH, HT, TH, and TT.
2. Once the combination is set, A light will be turned on.
3. At the same time, a computer will examine the coins to determine if both are showing Heads. If so, it releases a sleep gas into the room that will render you unconscious within 10 seconds, wiping your memory of the past hour. Your sleeping body will be moved to a recovery room where you will be wakened and given further details as explained below.
4. But if either coin is showing tails, a lab assistant will come into the room and ask you a probability question. After answering it, the same gas will be released, your sleeping body will be moved the same way, and you will be given the same "further details."

So you sit in the room for a minute, the light comes on, and you wait ten seconds. A lab assistant comes in (so you weren't gassed, yet) and asks you "What is the probability that coin C1 is showing Heads?

The answer to this question is unambiguously 1/3. Even tho you never saw the coins, you have unambiguous knowledge of the possibilities for what the combinations could be.

Now, let the "further details" be that, if this is the first pass thru experiment, the exact same procedure will be repeated. Otherwise, the experiment is ended. Whether or not you were asked the question once before is irrelevant, since you have no memory of it. The arrangement of the two coins can be correlated to the arrangement in the first pass, or not, for the same reason.

The point of my "alternate version" that I presented above is that it creates what is, to your knowledge, the same probability space on each pass. Just like this one.It exactly implements what Elga described as the SB problem. He changed it, by creating a difference between the first and second passes. The first pass ignores the coin, so only the second depends on it.

What you describe, where the (single) coin isn't established until the second pass, is manipulating Elga's change to emphasize that the coin is ignored in the first pass. It has nothing to do with what I've tried to convey. The only question it raises, is if Elga's version correctly implements his problem.