This is a fallacy:

If Monday, P(Monday-Heads) = P(Monday-Tails)
If Tails, P(Monday-Tails) = P(Tuesday-Tails)
Therefore, P(Monday-Heads) = P(Monday-Tails) = P(Tuesday-Tails)

The conclusion doesn't follow, because the first two equalities depend on the conditionals being true.

You can see this by observing that

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

Also satisfies the two conditional statements, without satisfying the conclusion — hypericin

The reason why the Double-halfer splits the probability P(Tails) = 1/2 between P(Monday-Tails) and P(Tuesday-Tails) is because they treat them as exclusive outcomes as if a second coin would be tossed to determine if Sleeping Beauty would be awakened on Monday or Tuesday, but not both.

Elsewhere, I made my argument more explicit. Let me rehearse it here by transposing it to the zoo variation:

I must first justify the inferences from:

1. P(Tiger|First) = P(Hippo|First)
2. P(Tiger|T-path) = P(Toucan|T-path)

to

1b. P(Tiger) = P(Hippo)
2b. P(Tiger) = P(Toucan)

The first inference is justified by the fact that placing more enclosures with different animals in them on either path doesn't alter the relative probabilities of the Tiger or Hippo outcomes since I will be seeing those new animals (such as toucans) in addition to seeing the tigers and hippos and not to the exclusion of them.

The second inference is justified by the fact that generating alternative timelines where I don't see either tigers or toucans doesn't alter their relative probabilities (but rather lowers both in equal proportion). The second inference is actually common ground between Thirders, Halfers and Double-halfers, which is presumably why Michael only challenged the first.

See also my most recent reply to Michael regarding this example.

Your claim that "The next enclosure is the toucan enclosure iff I first turned right at the fork (P = 1/2) and then passed the tiger enclosure," is an assumption that can't be put forward without begging the question against the Thirder. You need to substantiate, rather than presuppose, that when you're nearing an enclosure, there's a 1/2 chance the path you're on is a T-path. — Pierre-Normand

My wording may have been imprecise.

Both of these are true (note the tense):

1. To reach the toucon enclosure I must first turn right at the fork and then pass the tiger enclosure
2. The probability that I will turn right at the fork is $1\over2$

When I wake and consider my credence that the next enclosure is the toucon enclosure I consider what must have happened (or not happened) for the next enclosure to be the toucon enclosure. I know that I must have first turned right at the fork (A) and then passed the tiger enclosure (B).

P(A, B) = P(A) × P(B|A)

My claim is that the probability of having turned right at the fork is equal to the probability of turning right at the fork, i.e. $1\over2$.

Your claim is that the probability of having turned right at the fork is equal to the fraction of all encountered enclosures which are right-side enclosures, i.e. $2\over3$.

I don't think your claim makes any sense. The probability of the first event having happened isn't determined by what could happen after that first event happens. The probability of the first event having happened is determined only by the probability of that first event happening.¹

Your conclusion only applies if I'm dropped into an enclosure at random, perhaps via parachute. This is your so-called "episodic perspective" (where it’s not the case that I turned right at the fork; it’s only the case that I’m on the right-side path). But given that this isn't what happens to me, I shouldn't reason this way.

¹ _{Where no new relevant information is learned.}

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.

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. — JeffJo

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

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, and the prior probability that she will be woken on Monday and the coin landed on Heads is 1/2.

So when she "rules out" Pr(Heads & Day = Tuesday) she's "ruling out" some Pr = 0, not some Pr = 1/4. "Ruling out" some Pr = 0 after waking does nothing because it was already "ruled out".

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.

No, the prior probability that she will be woken on Tuesday, and the coin landed Heads, is 1/4. — JeffJo

The rules of the experiment say that she won’t be woken on Tuesday if the coin lands heads. That means that the prior probability that she will be woken on Tuesday and the coin lands heads is 0.

P(Heads, woken on Tuesday) = P(Heads) × P(woken on Tuesday|Heads) = 1/2 × 0 = 0.

Just look at how you calculated the probability of waking up in my experiment. It’s the same reasoning.

If a die rolls a 6 then Sleeping Beauty is woken six times otherwise she is woken once. When woken what is her credence that the die rolled a 6?

Halfers have to say $1\over6$ and thirders have to say $6\over11$.

Before she is first put to sleep she is to bet on whether or not the die will roll a 6 – paid out at the end of the experiment – and each time she is woken she is allowed to change her bet.

If she bets according to her credence then both halfers and thirders have to say that before she is first put to sleep she will bet that the die will not roll a 6.

Thirders then have to say that when woken she will change her bet and bet that the die did roll a 6.

Are thirders willing to commit to their position and change their bet?

Both of these are true (note the tense):

1. To reach the toucon enclosure I must first turn right at the fork and then pass the tiger enclosure
2. The probability that I will turn right at the fork is 1/2

When I wake and consider my credence that the next enclosure is the toucon enclosure I consider what must have happened (or not happened) for the next enclosure to be the toucon enclosure. I know that I must have first turned right at the fork (A) and then passed the tiger enclosure (B).

P(A, B) = P(A) × P(B|A)

My claim is that the probability of having turned right at the fork is equal to the probability of turning right at the fork, i.e. 1/2.

Your claim is that the probability of having turned right at the fork is equal to the fraction of all encountered enclosures which are right-side enclosures, i.e. 2/3.

I don't think your claim makes any sense. The probability of the first event having happened isn't determined by what could happen after that first event happens. The probability of the first event having happened is determined only by the probability of that first event happening. — Michael

It actually often makes sense that the probability of an event having happened is determined by what has been found to happen (as a consequence of it) after that first event happened. Take the simple example of a coin toss: The initial probability that a coin would land heads was 1/2. But suppose we have tossed the coin, and we now see it landed heads. Our updated probability that it landed heads is now 1. In this case, our current situation—our observing the fact of the coin having landed heads—has completely determined the probability of the previous event, the coin landing heads. This may seem trivial, but it is a similar principle at play in our journey through the zoo, and it is also key to the Sleeping Beauty problem. The probability of one finding oneself in a specific situation is not only determined by the initial probabilities of different outcomes (or paths taken) but also by the subsequent encounters or observations that are stipulated to occur as a result of those outcomes. Importantly, it is precisely when the proportion of these subsequent observations (or encounters) is dependent on the earlier outcome that those observations warrants a Bayesian updating of our credence.

Let's look back at our zoo journey. Right now, as we approach an enclosure, what would have had to happen for us to be approaching a hippo, tiger, or toucan enclosure? For a hippo enclosure, we must have taken a new fork after passing either a hippo or a toucan enclosure. For a toucan enclosure, we must have walked past a tiger enclosure. Every fork is equally likely to lead to a hippo or a tiger enclosure directly, so we can outline the possible scenarios as follows:

Hippo -> Hippo (1/2) or Tiger (1/2)
Toucan -> Hippo (1/2) or Tiger (1/2)
Tiger -> Toucan (1)

Now, let's consider (for the sake of argument) that we are equally likely to have just passed any of the three types of enclosures. This leads to the following scenarios with their probabilities:

1/3 of the time we passed a Hippo -> we're now approaching a Hippo (1/6) or Tiger (1/6)
1/3 of the time we passed a Toucan -> we're now approaching a Hippo (1/6) or Tiger (1/6)
1/3 of the time we passed a Tiger -> we're now approaching a Toucan (1/3)

This shows that, even if we start with equal chances of having just passed any kind of enclosure, and even if every new fork is equally likely to lead directly to a H-path or T-path, the equal probabilities of approaching each kind of enclosure at each new step are maintained. This refutes your argument that we should have a 1/2 chance of approaching a hippo enclosure based on the equal chances of taking a H-path or T-path at the previous fork. It is precisely because every new fork was equally likely to lead directly to a H-path or T-path that, whenever we are approaching an enclosure, the probability that it is a hippo enclosure is 1/3.

Let us now consider a closer analogy to the Sleeping Beauty problem where there only is one fork (and only one coin toss).

It's Christmas morning and young Leonard Shelby is seated beside a gleaming Christmas tree, eyes wide with excitement. Two of his favorite aunts, Jane and Sue, have come bearing gifts. Each year, it's a tradition that one aunt gifts a singular, unique and expensive present, while the other aunt bestows upon Leonard two less expensive, yet equally fun presents. Who brings the unique gift is decided by a flip of a fair coin - heads for Jane and tails for Sue.

This year, all three gifts are packaged identically, and their order of presentation to Leonard is randomized. Due to the condition Leonard has had since his childhood (unlike what happens in the Memento movie), he forgets about the gifts as soon as he unwraps them. This makes every unwrapping experience as exciting as the first one. A small note accompanies each gift, indicating which aunt has brought it. Given the symmetries of this setup, before unwrapping any gift, Leonard's initial, or prior, probabilities are quite straightforward: there's a 1/2 chance the gift is from Jane, a 1/2 chance the coin landed heads, and a 1/3 chance that the gift is the unique one.

Now, let's consider a scenario where Leonard reads the attached card and learns that the gift is from Jane. What does this tell him about the coin toss? Well, if the coin landed heads, Jane would be the one to give the unique gift. But if it was tails, Jane would have two gifts to give. Knowing this, Leonard updates his belief about the coin toss. Now that he knows the gift is from Jane, the probability of the coin having landed heads, P(H|Jane), is reduced to 1/3.

This key conclusion is supported by Bayes' theorem, allowing Leonard to update his beliefs in light of the new evidence. Bayes' theorem here shows that the probability of heads, given that the gift is from Jane, is equal to the initial probability of Jane being the one to give the gift, given heads (1/3), times the prior probability of heads (1/2), divided by the prior probability of the gift being from Jane (1/2). This gives us a revised probability of 1/3 for heads, given that the gift is from Jane.

In short: P(H|Jane) = P(Jane|H)P(H)/P(Jane) = (1/3)(1/2)/(1/2) = 1/3.

Okay, so now imagine a similar scenario, but instead of gift-giving aunts and Christmas, it involves a scientific experiment Leonard is participating in. In this experiment, if a coin lands heads, Leonard is interviewed once in a room in the West Wing (let's call it Jane's Wing) and twice in a room in the East Wing (Sue's Wing). If the coin lands tails, the number of interviews in each wing is reversed. Similar to the Christmas scenario, Leonard is interviewed three times exactly and his priors before any interview are: P(West Wing) = 1/2, P(H) = 1/2, and P(Unique Interview) = 1/3. (In more details, his priors are: P(West-HU) = P(East-T1) = P(East-T2) = P(East-HU) = P(West-T1) = P(West-T2) = 1/6 where interviews ("awakenings") rather than gifts are labeled as unique ("U"), first ("1") or second ("2").

But now, let's say that Leonard finds out he's being interviewed in the West Wing. This new information allows Leonard to update his belief about the coin toss, similar to what happened in the Christmas scenario. Using Bayes' theorem again, Leonard finds that P(H|West Wing) = 1/3. In other words, given that he's in the West Wing (where the structure of the experiment is identical to the original Sleeping Beauty problem), Leonard's credence in the coin having landed heads is 1/3.

And there you have it! We've demonstrated that when considering the structure of the experiment and the new information that Leonard possesses, the probabilities he assigns to the coin having landed heads or tails should be updated. The key insight here is that the information Leonard gains in the revised scenario—namely that he's being interviewed in the West Wing—doesn't actually provide him with any new facts that he wasn't already aware of in the original Sleeping Beauty problem. Instead, learning he's in the West Wing simply makes explicit to Leonard the fact that he is in a situation which perfectly mirrors the original Sleeping Beauty setup. This underlines the fact that in the original scenario, Sleeping Beauty already has all the information necessary to update her beliefs in line with the Thirder position.

I appreciate the reply but I’ve run out of motivation and am going to end my involvement with this. I wouldn’t change my bet, so I’m a committed halfer. I don’t think any number of analogies are going to convince me to change my bet in that specific case.

If a die rolls a 6 then Sleeping Beauty is woken six times otherwise she is woken once. When woken what is her credence that the die rolled a 6?

Halfers have to say 1/6

and thirders have to say 6/11.

Before she is first put to sleep she is to bet on whether or not the die will roll a 6 – paid out at the end of the experiment – and each time she is woken she is allowed to change her bet.

If she bets according to her credence then both halfers and thirders have to say that before she is first put to sleep she will bet that the die will not roll a 6.

Thirders then have to say that when woken she will change her bet and bet that the die did roll a 6.

Are thirders willing to commit to their position and change their bet? — Michael

Thirders wouldn't change their bet in this scenario. Although it's true that in this setup, a bettor aware of P("6") = 6/11 (i.e., the proportion of "H-awakenings" to "All-awakenings" equals 6/11) might be tempted to bet on the outcome "6". They're also aware that a successful bet will be rewarded only once at the end of the experimental run, no matter how many times they've made that assessment.

Here's where the nuance comes in: over the course of six experimental runs, they would, on average, place a winning bet 6 times out of 11. This is if we imagine each individual bet as standing alone. However, due to the rules of this experiment, the six winning bets will only result in a single even-money payout. The lack of profitability in this scenario doesn't fault the credence in the outcome. Instead, it reveals a unique aspect of this setup where multiple winning bets are not individually rewarded. Therefore, the "bet" one ought to make doesn't straightforwardly track one's credence in the outcome of the die roll, but rather, it must take into account the rules of payout in this specific experimental setup.

I had an idea for a rather dark version of this:

Beauty gets cloned with all her memories on a given flip, such that each Monday and Tuesday has a 50% chance of resulting in a new clone being created.

On Wednesday, all the Beauties wake up in an identical room with identical memories. The real Beauty is given breakfast and allowed to leave her room and enjoy the magical castle until Sunday, when the experiment will be rerun.

The clone Beauties are killed by the first person who comes to the door.

The real Beauty retains her memories of her free weekends.

Now, let's say the experiment has been running for a long time, three years. A lot of other Beauties have likely been cloned and killed by this point. But, if you're the real Beauty, and you consistently think you are the real Beauty when you wake up, then you have indeed been right about that fact for three years straight. So, when you wake up next time, how worried should you be that you'll be killed?

Based on (an admittedly simple) Bayesian take, Beauty should be increasingly confident that she is the real Beauty with each passing week. The whole idea is that repeated trials should move the dial in our probability estimates. And yet, this doesn't seem right, no?

Beauty gets cloned with all her memories on a given flip, such that each Monday and Tuesday has a 50% chance of resulting in a new clone being created. — Count Timothy von Icarus

Are you suggesting that one new clone is always created, but the coin flip determines on which day? Furthermore, are Sleeping Beauty and her clones only awakened on Wednesdays? Regardless, I believe that your experiment could completely ignore the element of randomness introduced by the coin flip, and Sleeping Beauty's uncertainty on Wednesday would still exist, solely because she doesn't know if she's the original Beauty or a clone.

Based on (an admittedly simple) Bayesian take, Beauty should be increasingly confident that she is the real Beauty with each passing week. The whole idea is that repeated trials should move the dial in our probability estimates. And yet, this doesn't seem right, no?

It doesn't seem to me that Bayesian principles would allow Sleeping Beauty to grow increasingly confident with each passing week that she isn't a clone. That's because the design of the experiment ensures that she survives if she is the original Beauty. She can recall (or inherit once the "fake" memory) that every previous week, upon awakening, she had (from her viewpoint) a 1/2 chance of being a clone. However, her past survival events weren't probabilistically independent events, given that the survival of the original is always ensured by the fact that she is the original. It remains true, however, that until she leaves the room and survives yet again, her belief that she is currently the original is reduced to 1/2.

On edit: After asking GPT-4 (always the sycophant) to check my English, I discussed another issue with it.

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.

Therefore, the "bet" one ought to make doesn't straightforwardly track one's credence in the outcome of the die roll, but rather, it must take into account the rules of payout in this specific experimental setup. — Pierre-Normand

If each outcome has the same reward then it is rational to bet on the most probable outcome.

Therefore, if her credence that the die landed on a 6 is $6\over11$ then she will change her bet. Therefore, if she doesn't change her bet then her credence that the die landed on a 6 isn't $6\over11$.

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

Seems like a trick question.

The probability remains the same every time they flip the coin after she wakes.

But if she wrote the tally down, then she'll see heads written for the tally 100% of the time. But since it's the coin toss that's all that matters.

So if the last coin flip doesn't effect the outcome, doesn't that mean it's always 50/50? If i understand correctly, getting heads the first time means it's over, which balances getting tails twice in a row, and the last one is 50/50.

It is a confusing issue, do there exist methods for keeping sleeping beauty from remembering the other days in the experiment?

If each outcome has the same reward then it is rational to bet on the most probable outcome.

Therefore, if her credence that the die landed on a 6 is 6/11
6
11
then she will change her bet. Therefore, if she doesn't change her bet then her credence that the die landed on a 6 isn't 6/11
6
11
. — Michael

She doesn't change her bet because although her credence changes, the specific payoff structure doesn't warrant her doing so. Before the experiment begins, she's given an opportunity to bet on the outcome of the coin toss while this singular bet will be paid off at the end of the experiment. As the experiment proceeds, when she is first awakened (as well as during any other awakening episode) she is being offered an opportunity to change her bet. Although her credence that the die landed on a 6 is updated to 6/11, she has no incentive to change her bet. That's because she knows that the payout will occur only once at the end of the current experimental run regardless of how many awakenings will have occurred during the whole run. So, her choosing to bet that the die will not have been found to have landed on a 6 at the end of the current experimental run doesn't reflect her (updated) credence that is has landed on a 6 on the occasion of her current awakening episode.

Sorry to resurrect. — JeffJo

It's not a resurrection. It's just a reawakening :wink:
Yes, yours is a nice variation that highlights the motivation for the thirder reasoning.

You have it backwards. It's not that her credence changes and she bets against it when she can only place a single bet; it's that her credence doesn't change and she bets against it when she can place multiple bets.

Her reasoning after waking up when she can only place a single bet is:

A1. If I keep my bet and the die didn't land on a 6 then I will win £100
A2. If I change my bet and the die did land on a 6 then I will win £100
A3. The probability that the die landed on a 6 is $1\over6$
A4. Therefore, the expected return if I keep my bet is £83.33
A5. Therefore, the expected return if I change my bet is £16.67
A6. Therefore, the expected return if I keep my bet is $5\over6$
A7. Therefore, I will keep my bet

And her reasoning after waking up when she can place multiple bets is:

B1. If I always bet that the die didn't land on a 6 and it didn't then I will win £100
B2. If I always bet that the die did land on a 6 and it did then I will win £600
B3. The probability that the die landed on a 6 is $1\over6$
B4. Therefore, the expected return if I always bet that the die didn't land on a 6 is £83.33
B5. Therefore, the expected return if I always bet that the die did land on a 6 is £100
B6. Therefore, the expected return if I always bet that the die did land on a 6 is $6\over11$
B7. Therefore, I will bet that the die did land on a 6

Whereas your argument appears to be that her reasoning after waking up when we can only place a single bet is:

C1. If I keep my bet and the die didn't land on a 6 then I will win £100
C2. If I change my bet and the die did land on a 6 then I will win £100
C3. The probability that the die landed on a 6 is $6\over11$
C4. Therefore, the expected return if I keep my bet is £83.33
C5. Therefore, the expected return if I change my bet is £16.67
C6. Therefore, the expected return if I keep my bet is $5\over6$
C7. Therefore, I will keep my bet

Which makes no sense at all. If she truly believes C3 then she would have calculated different expected returns and changed her bet, just as you or I would if we came to learn that the die is loaded in favour of landing on a 6. But waking up doesn't function like learning that the die is loaded in favour of landing on a 6, and so her credence in the outcome doesn't change. Her credence is always that the probability that the die landed on a 6 is $1\over6$, consistent with common sense and explaining why she bets the way she does in both betting scenarios.

Thirder reasoning appears to conflate B6 and C3, which is both a mistake and a contradiction given B3.

@Michael

I am happy to grant you all of this (except for the last comment regarding credence updating), not as a resolution of the paradox however, but rather as a sensible explanation of the reasoning that motivates the halfer response to the SB problem. I am not myself either a halfer or a thirder. I am rather arguing that both the halfer and the thirder (or "sixther", in our current example) advocates present incompatible but equally valid and complementary interpretations of the epistemic situation Speeping Beauty finds herself in, and of her credence in the "outcome". This is possible because the outcome (i.e. the coin having landed heads or tails) is incompletely specified in the statement of the problem. This outcome purports to represent an event that SB has incomplete information about, but the precise way in which this event is meant to be individuated is unspecified which opens the room for two equally plausible interpretations.

So, when SB awakens, she could be wondering what the chances are that the experimental run her present awakening is a part of is a 6-run, regardless of how many times she is being awakened in this run. That is your favorite interpretation. And the answer is 1/6 precisely on the basis of the reasonings that you present in your previous post. But she can also wonder what the chances are that her present awakening episode is a 6-awakening regardless of the manner in which such events are stringed into continuous experimental runs. In this case, your analysis of the thirder reasoning in your previous post is inaccurate (and an elucidation of the sort recently proposed by @JeffJo is more apposite).

The reason why SB can take a thirder rather than a halfer stance regarding her current awakening episode is because she may care about the long-term average frequency of such events (6-awakenings) rather than the average frequency of such experimental runs (6-runs). The reason why she doesn't change her bet is because, as I've explained, although in the long run she finds herself having been (on each occasion where she had been given an opportunity to change her bet) in a 6-awakening 6/11 of the times, she only is being awarded £100 once for each series of six awakening episodes where she consistently changes her bet, but is awarded the full £100 every single time when she changes her bet on the occasion of the comparatively rarer 6-awakenings.

Of course, given this specific payout structure (that is not being provided as part of the statement of the original SB problem), which has been conceived precisely to highlight a rationale for the halfer reasoning, the halfer stance appears more intuitive and the thirder stance appears more convoluted. But you only have to change the payout structure, such that even-money bets are paid on the occasion of each awakening episode rather than once after each run, to reverse the situation and make the halfer stance appear unnecessarily convoluted.

During the last round of SB discussions that had taken place two years ago, I had devised a few variation (hostage taking and escape opportunities, zoo visit scenario, Christmas gift scenario, etc.) that were meant to highlight how pragmatic considerations and specific payout structures can motivate either a halfer or a thirder stance as the most natural interpretation of the problem. The original SB problem just strips away the required interpretative or pragmatic context in such a way as to leave SB's "credence" regarding the "event" (are two tails-awakening within a single run two events or two parts of the same event?) indeterminate and open to interpretation.

The reason why SB can take a thirder rather than a halfer stance regarding her current awakening episode is because she may care about the long-term average frequency of such events (6-awakenings) — Pierre-Normand

She isn't being asked "what is the long-term average frequency of being woken up when the die did land on a 6?" Her answer to that question is the same both before being put to sleep and after being woken up, and so there wouldn't be a problem to solve.

The problem only exists because there is the counter-intuitive suggestion that her credence in the outcome of a die roll changes, comparable to being told that the die is loaded, after being woken up despite prima facie not being provided with any new information.

The actual question she is being asked is "what is the probability that the die did land on a 6?" which is the same as being asked for the value of $n$ below:

1. If I correctly bet that the die didn't land on a 6 then I will win £$a$
2. If I correctly bet that the die did land on a 6 then I will win £$b$
3. The probability that the die did land on a 6 is $n$
4. Therefore, the expected return if I bet that the die didn't land on a 6 is £$x$
5. Therefore, the expected return if I bet that the die did land on a 6 is £$y$

There is only one correct value for $n$ and that value is the value that gives the correct values for $x$ and $y$ – which is $1\over6$ both before being put to sleep and after being woken up.

Whereas if she were told that the die is loaded then the values of $n$, $x$, and $y$ all change and with it her actual credence in the outcome of the die roll.

She isn't being asked "what is the long-term average frequency of being woken up when the die did land on a 6?" — Michael

That's right. If this had been the question there would be no apparent paradox and the answer would unambiguously be 6/11. But notice that she isn't being asked "what is the long-term average frequency of experiencing a series of six successive awakenings when the die did land on a 6?" either. If this had been the question, then the answer would be unambiguously 1/6. This is your favored interpretation.

Typical formulations of the SB problem are ambiguous, however. For instance, it is being asked, on the occasion of her next awakening, what is her credence that the die landed on a six? (or that the coin landed tails?). What this "occasion" precisely is is left unstated. Ostensibly, it is an event whereby SB encounters a specific die in a determinate albeit unknown state. It's a reference to an event. However, is this "event" consisting of one among a set of undistinguishable awakening episodes or is it one among a set of undistinguishable experimental runs?

You seem to think that the mere reference to the state of a specific (fair) die, and incomplete knowledge of the process whereby that die ended up in a specific state, is sufficient for establishing her credence. But the question isn't being asked about a credence derived from the perspective of someone running the experiment, but rather from SB's own epistemic situation when she is being awakened (and, as stated, this is ambiguous).

Remember the variation I had suggested whereby someone knocks at a random door in a village. Some of the doors belong to a singular dwelling that has two doors. And some of the doors belong to a dwelling that had only one door. We also assume that the probability that each singular dwelling in this village would have been built with one or two doors has been established by a coin toss. On the occasion where you knock at a door, the probability that this is one door from a two-door dwelling is 2/3 since two thirds of the doors in this village belong to two-door dwellings. Yet, the probability that this dwelling would have had two doors rather than one was 1/2 since it was the result of a coin toss. What accounts for the different credence of the visitor is their specific epistemic situation in relation with the dwelling that they come to face on that occasion, and not just the event that led to this dwelling being constructed with either one or two doors. Although the specific dwelling that they now face was equally likely to end up being a two-door dwelling or a single-door dwelling, your credence that it is a two-door dwelling is 2/3 owing to you encountering such dwellings twice as often. (And similarly for Sleeping Beauty more often encountering dice that have landed on a six, or coins that landed tails, under the thirder interpretations of the problem.)

This is your favored interpretation. — Pierre-Normand

My "favoured" interpretation is the literal interpretation; she is being asked about the probability that a die rolled a six.

She isn't being asked about the long-term average frequency of being woken up when the die did land on a 6 and she isn't being asked about the long-term average frequency of experiencing a series of six successive awakenings when the die did land on a 6. Either of these two questions gives the same answer when asked before being put sleep as when asked after being woken up, and so there would be no problem to solve.

The problem only exists when the question being answered before being put sleep is the same question being answered after being woken up, and where the answer (allegedly) changes despite (apparently) no new information.

If the Thirder's answer before being put to sleep is $1\over6$ and if their answer after being woken up is $6\over11$ then either they are not answering the same question or one of their answers is wrong. And it is obvious in context that the correct answer to the question being asked before being put to sleep is $1\over6$.

My "favoured" interpretation is the literal interpretation; she is being asked about the probability that a die rolled a six. — Michael

In most formulations of the problem, she is not being asked "What is the probability that a die landed (past tense) tails" from some sort of a disengaged perspective. She always is being asked a question that makes specific reference to her peculiar epistemic situation such as: "what her degree of belief that “the outcome of the coin toss is Heads” ought to be when she is first awakened". If you're just ignoring the peculiarities of her queer epistemic situation, then, of course, the paradox evaporates.

The problem only exists when the question being answered before being put sleep is the same question being answered after being woken up, and where the answer changes despite no new information.

Yes, that's the root to the paradox. Since she acquired no new information about the specific coin that was tossed, how could her credence regarding the same event change? Clue: owing to the change in her epistemic situation, she is not referencing the "same" event in the same way.

If the Thirder's answer before being put to sleep is 1/6 and if their answer after being put to sleep is 6/11 then either they are not answering the same question or their answer is wrong.

...or, thirdly, their epistemic relation to the target outcome changed. This is the alternative that you seem to have a persistent blind spot for. You haven't addressed the village/doors example. When a house is built, the number of doors giving access to it is set at random to be one to two according to the toss of a fair coin. When a traveller in that village picks a door at random, the house that this door belongs to was equally likely (when it was constructed) to be a two-door dwelling or a one-door dwelling.

However, owing to the fact that the traveller must establish their credence on the occasion of encountering one among a set of indistinguishable doors, and 2/3rds of such doors belong to two-door dwellings, their credence that this house that they now are facing is a two-door dwelling is 2/3. This is true in spite of the fact that the traveller can also reason: "this house that I am now facing was equally likely when it was built to become a two-door or a one-door dwelling". What makes those two statements compatible is the fact that owing to the traveller's peculiar epistemic situation, their present encounter is twice as likely to be an encounter with a two-door dwelling.

When the traveller came to face a particular house, they didn't acquire any new information regarding this particular house. That's because they don't know what house it is that they are now facing. But they do know that they now are twice as likely to be facing a two-door dwelling. The question is no longer about a specific house that they never encountered before but rather about a specific encounter with a house that has already been built.

Thirders then claim that:

P(6|Monday)=6/11

P(¬6|Monday)=5/11 — Michael

Unless my memory is faulty, the variation we had discussed (two years ago) was one where Sleeping Beauty was awakened only once, on Monday, unless the die lands on 6, in which case she is being awakened six times from Monday through Saturday. In that case, thirders would claim that

P(6|Monday)=1/6 (Since one sixth of Monday-awakenings are Six-awakenings)

P(¬6|Monday)=5/6 (Since five sixths of Monday-awakenings are Non-six-awakenings)

Right?

Sorry, I deleted that post because it's late and I'm tired and I may have messed up the specific numbers. The general gist is what I said before. Your argument is that her reasoning after being woken up is:

A1. If I keep my bet and the die didn't land on a 6 then I will win £100
A2. If I change my bet and the die did land on a 6 then I will win £100
A3. My credence that the die landed on a 6 is $6\over11$
A4. Therefore, the expected return if I keep my bet is £83.33
A5. Therefore, the expected return if I change my bet is £16.67

But A3, A4, and A5 are inconsistent. If A3 really was true then she would calculate different values for A4 and A5, concluding that it is profitable to change her bet. But she doesn't do this.

You can't have it both ways. Either she genuinely believes it to be more likely that the die landed on a 6, and so she changes her bet, or she continues to believe it to be more likely that it didn't, and so she keeps her bet.

However, owing to the fact that the traveller must establish their credence on the occasion of encountering one among a set of indistinguishable doors, and 2/3rds of such doors belong to two-door dwellings, their credence that this house that they now are facing is a two-door dwelling is 2/3. — Pierre-Normand

That doesn't follow. It depends on the manner in which the door is chosen. Compare with a red bag containing 100 balls and a blue bag containing 50 balls. You "encounter" a ball. What is the probability that it came from the red bag? Is it $2\over3$ because $2\over3$ of the balls come from the red bag? Not if one "encounters" a ball by putting one's hand in a bag at random, as the probability that one picks the red bag is $1\over2$.

So you need to first specify the mechanism by which one has "encountered" a door, and this mechanism must be comparable to the Sleeping Beauty scenario for it to be an apt analogy.

Sorry, I deleted that post because it's late and I'm tired and I may have messed up the specific numbers. The general gist is what I said before. Your argument is that her reasoning after being woken up is:

A1. If I keep my bet and the die didn't land on a 6 then I will win £100
A2. If I change my bet and the die did land on a 6 then I will win £100
A3. My credence that the die landed on a 6 is 6/11
A4. Therefore, the expected return if I keep my bet is £83.33
A5. Therefore, the expected return if I change my bet is £16.67

But A3, A4, and A5 are inconsistent. If A3 really was true then she would calculate different values for A4 and A5, concluding that it is profitable to change her bet. But she doesn't do this. — Michael

A thirder will not agree with A4 or A5. If SB is allowed to change her bet when she awakens, she must do do consistently as a matter of policy since she can't distinguish between different occasions of awakening (i.e. day of the week). She knows that a policy of changing her bet lowers her expected return since there only is one payout per experimental run. Although her systematically betting on a six would result in her being right on six out of eleven occasions when she is being given the opportunity to do so, in accordance with her credence, she only is being paid £100 once at the end of the experimental run when she does so (and the die landed 6) but this policy also makes her forfeit the full prize on the five occasions out of eleven where the die didn't land on six. All this shows is that the lopsided payout structure makes it irrational for her to bet on the most likely outcome.

A thirder will not agree with A4 or A5. — Pierre-Normand

Correct. Her reasoning would be:

A1. If I keep my bet and the die didn't land on a 6 then I will win £100
A2. If I change my bet and the die did land on a 6 then I will win £100
A3. My credence that the die landed on a 6 is $6\over11$
A4. Therefore, the expected return if I keep my bet is £45.45
A5. Therefore, the expected return if I change my bet is £54.55

And yet you say she doesn't change her bet even though she has calculated that changing her bet is more profitable? There's something amiss with your reasoning.

Either she does change her bet or her credence that the die landed on a 6 continues to be $1\over6$.

All this shows is that the lopsided payout structure makes it irrational for her to bet on the most likely outcome. — Pierre-Normand

Again, you have it backwards. The most likely outcome is always that the die didn't land on a 6, but when she is allowed to place multiple bets it is irrational to bet on the most likely outcome.