In that scenario, P(R|R or B1) would be 2/3 and P(B1|R or B1) would be 1/3. — Pierre-Normand

How do you get that?

However, if what you mean is that, from the bettor's perspective and in light of the evidence available to them at the time of betting, the bet (distinguished from other bets within the same experimental run, which from the agent's point of view, may or may not exist) is more likely to have been placed in circumstances where the coin landed tails, then I would argue that the inference is indeed warranted. — Pierre-Normand

I believe this response to PhilosophyRunner addresses this claim. Specifically:

Although it's true that most interviews follow the coin landing heads 100 times, every single one of those interviews belongs to a single participant, and for each participant the probability that they are that single participant is $1\over2^{100}$.

So although it's true that "any given interview is twice as likely to have followed the coin landing heads 100 times" it is false that "my interview is twice as likely to have followed the coin landing heads 100 times"

Does that procedure accurately represent how Sleeping Beauty understands her own epistemic situation when she is being awakened on a day of interview, though? — Pierre-Normand

It's not intended to. It's intended to show that this inference is not valid a priori:

P(A|A or B) = P(B|A or B)
∴ P(A) = P(B)

Elga's argument depends on this inference but he doesn't justify it.

His chain of entailments when applied to my counterexample leads to a false conclusion, and so it needs to be explained why this chain of entailments is valid for the Sleeping Beauty case.

Not necessarily so, just because we would do it does not mean that they would have the same motivations we do. — Sir2u

Certainly not necessarily so, but unless we're something special it stands to reason that at least one would.

First of all, why would they have to be more advanced than we are. True, there are many older galaxies out there that could have developed highly intelligent life forms along time ago, but there is also evidence that many galaxies have already died out. Anyone of the many galaxies could have life similar to our own at with the same level of technology, thus unable to come visiting. — Sir2u

Of course it's possible, and one explanation for the Fermi paradox is that we are one of the first intelligent species in the galaxy. But given that the oldest planet in the Milky Way is 12.7 billion years old and the Earth is only 4.5 billion years old, it would appear reasonable to infer that there were advanced civilisations long before us.

Second point, a million years ago when they set out it would have been impossible for them to even guess that we might appear on this planet. So why would they head in this direction instead of one of the other millions of possibilities in all of the other galaxies? — Sir2u

Just considering species born in the Milky Way, as I said before, the conjecture is that a species would explore all of it. Assuming the resources are available and they don't die out first, it's unclear why they wouldn't.

Last point, no one said that intelligent life is common. — Sir2u

Actually, lots of people do. It's called the mediocrity principle. Of course others also propose the Rare Earth hypothesis in opposition.

A given seeing of it is twice as likely to be tails. — PhilosophyRunner

This is an ambiguous claim. It is true that if you randomly select a seeing from the set of all possible seeings then it is twice as likely to be a tails-seeing, but the experiment doesn't work by randomly selecting a seeing from the set of all possible seeings and then "giving" it to Sleepy Beauty. It works by tossing a coin, and then either she sees it once or she sees it twice.

If we return to my example of tossing the coin 100 times, assume there are 2¹⁰⁰ participants. Each participant knows two things:

1. Of the 2¹⁰² interviews, 2¹⁰¹ follow the coin landing heads 100 times

2. Of the 2¹⁰⁰ participants, the coin landed heads 100 times for 1 of them

You are suggesting that they should ignore 2 and use 1 to infer a credence of $2\over3$.

I am saying that they should ignore 1 and use 2 to infer a credence of $1\over2^{100}$.

Although it's true that most interviews follow the coin landing heads 100 times, every single one of those interviews belongs to a single participant, and for each participant the probability that they are that single participant is $1\over2^{100}$.

So although it's true that "any given interview is twice as likely to have followed the coin landing heads 100 times" it is false that "my interview is twice as likely to have followed the coin landing heads 100 times".

And by the exact same token, although it's true that "any given interview is twice as likely to be tails" it is false that "my interview is twice as likely to be tails".

The likelihood of your interview being a tails interview is equal to the likelihood that the coin landed tails in your experiment, which is $1\over2$.

That I get to see something twice doesn't mean that I'm twice as likely to see it. It just means I get to see it twice.

Using frequencies over multiple games to argue for the probabilities in a single game is a fundamental way probabilities are calculated. — PhilosophyRunner

Only when it's appropriate to do so. It is in the case of rolling a dice, it isn't in the case of counting the number of awakenings.

Again, it doesn't matter that if the coin lands heads 100 times in a row then I will be woken 2¹⁰¹ times. When I'm put to sleep, woken up, and asked my credence that the coin landed heads 100 times in a row – or my credence that my current interview is a 100-heads-in-a-row interview – the only thing that's relevant is the probability of a coin landing 100 heads in a row, which is $1\over2^{100}$. It simply doesn't matter that if the experiment were repeated 2¹⁰⁰ times then $2\over3$ interviews are 100-heads-in-a-row interviews.

If you want to say that it must still have to do with frequencies, then what matters is the frequency of a coin landing heads 100 times in a row, not the frequency of interviews that follow the coin landing heads 100 times in a row. You're using an irrelevant frequency to establish the probability.

@Pierre-Normand

Thought you might be interested in my short exchange with Elga:

Dear Professor Elga,

I've read your paper Self-locating belief and the Sleeping Beauty problem and hope you could answer a question I have regarding your argument. You state that "P(T₁|T₁ or T₂) = P(T₂|T₁ or T₂), and hence P(T₁) = P(T₂)" and by implication state that P(H₁|H₁ or T₁) = P(T₁|H₁ or T₁), and hence P(H₁) = P(T₁).

However I cannot see in the paper where this inference is justified, as it is not valid a priori.

If I have one red ball in one bag and two numbered blue balls in a second bag, and I pick out a ball at random and show it to you then P(R|R or B₁) = P(B₁|R or B₁) but P(R) = ½ and P(B₁) = ¼.

So the (double-)halfer can accept that P(H₁|H₁ or T₁) = P(T₁|H₁ or T₁) but reject your assertion that P(H₁) = P(T₁) follows. Is there something in your paper that I missed to justify this inference?

Thanks for your time. — Michael

Dear Michael,

Thanks for your interest in this stuff. The form of reasoning I had in mind was the following chain of entailments:

P(X|X or Y) = P(Y|X or Y)
P(X&(X or Y))/P(X or Y) = P(Y&(X or Y))/P(X or Y)
P(X)/P(X or Y) = P(Y)/P(X or Y)
P(X) = P(Y).

I wish you the best with your research. — Elga

Unfortunately I don't quite see how it addresses my counterexample, which seems to show that there must be a mistake with that chain of entailments, but I won't push him on it.

The main argument I see is not if aliens exist, but why would the come here? Any ideas about that? — Sir2u

I don’t think it’s specifically about coming here. One of the arguments is just that a sufficiently advanced civilisation would colonise their entire galaxy, even if just with unmanned probes, whether for research or to find resources.

At 10% the speed of light it would take a million years to cross the Milky Way. If intelligent life is common you’d have expected someone to have done it in the last few billion years.

Indeed, not only would their expected value (EV) be positive, but it would be positive because the majority of their individual bets would be winning bets. Michael, it seems, disagrees with the idea of individuating bets in this way. — Pierre-Normand

I disagree with the step from "the majority of winning bets are tails bets" to "tails is more probable".

It's either a non sequitur or affirming the consequent, where the implicit premise is "if tails is more probable then the majority of winning bets are tails bets".

In this case the majority of winning bets are tails bets only because you get to place more bets if it's tails.

This is why, as I have often said, betting examples just don't answer the question at all. They're a red herring. Betting on tails might be more profitable, but it is still the case that one's credence should be that P(Heads|Awake) = 1/2.

If you repeated the experiment a trillion times, and kept a note of whether you guess was correct or not each time, and I did the same. We would find that I got it correct more than you. By the law of large numbers that would mean the outcome I guessed for was more probable than yours. — PhilosophyRunner

More frequent but not more probable.

If the game is played once I wouldn't argue that the coin most likely landed heads 100 times in a row and that my interview is most likely a 100-heads-in-a-row interview. I would argue that the coin most likely didn't land heads 100 times in a row and that this is most likely my first and only interview.

I think using frequencies over multiple games to argue for the probability in a single game is a non sequitur.

Fair enough, but then a person betting that it did land on heads 100 times in a row will have a greater expected value for their winning (as long as the winnings for heads are greater than 2^100 than for tails). And their position would be the rational one. — PhilosophyRunner

It can be rational in the sense that it can be profitable to bet when the expected value is greater than the cost, much like a lottery that costs £1 with a prize of £2,000,000 and a probability of winning of 1/1,000,000.

But it's not rational to believe that (especially when playing once) that I am most likely to win betting that it landed heads 100 times. You're not most likely to win. The odds of winning are $1\over2^{100}$.

Following Pradeep Mutalik's argument, according to the Bayesian "Dutch Book argument", "a degree of certainty" or "degree of belief" or "credence" is essentially your willingness to wager. Specifically, if you have a "degree of certainty" of 1/n, then you should be willing to accept a bet that offers you n or more dollars for every dollar you bet.

In that case, it's not merely the expected value of the bet that determines the credence. Rather, it's your degree of certainty, 1/n, in the outcome being wagered on that makes you rationally justified in accepting a bet with such odds. — Pierre-Normand

Then apply this to my case of tossing the coin one hundred times, and where the experiment is only run once.

Will you bet that the coin landed heads 100 times in a row? I wouldn't. My credence is that it almost certainly didn't land heads 100 times in a row, and that this is almost certainly my first and only interview.

That I would be woken up a large number of times if it did land heads 100 times in a row just doesn't affect my credence or willingness to bet that it did at all.

And the same if I was one of 2¹⁰⁰ participants taking part. One person might win, but I will almost certainly not be the winner.

Only if I got to repeat the experiment 2¹⁰⁰ times would I bet that it did. But not because my credence for any particular experiment has increased; it's because my credence is that I'm likely to win at least once in 2¹⁰⁰ attempts, and that the winnings for that one time will exceed the sum of all my losses.

Yes, an individual tails interview event is twice as probable. A tails interview where Monday ans Tuesday interviews are grouped together is equally likely as a heads interview. it comes back to the language of the question and interpretation. — PhilosophyRunner

That I believe is a bad interpretation of probability.

The probability of the coin landing heads is 1/2, leading to one interview.
The probability of the coin landing tails is 1/2, leading to two interviews.

The probability that there will be a heads interview is 1/2.
The probability that there will be a tails interview is 1/2.

This is the correct interpretation.

Take away the amnesia. Does it follow that because there are two interviews after every tails that a tails interview is twice as probable?

Throwing in amnesia doesn't convert the increased frequency into an increased probability.

There is a twist that comes from the fact that a biconditional holds between the two propositions "E1 is now occurring" and "E2 is now occurring". How can they therefore have different probabilities of occurrence? This puzzle is solved by attending to the practical implications of establishing effective procedures for verifying their truths, or to the means of exploiting what such truths afford. — Pierre-Normand

I don't see the connection between credence in an outcome and practical implications. Proving that the optimum betting strategy over multiple games is to bet on tails doesn't verify that P(Tails|Awake) = 2/3 is true.

If there's a lottery where the probability of winning is 1/1,000,000 but the cost is £1 and the prize is £2,000,000 then it can be profitable to play 1,000,000 times, but it is still the case that for each lottery one's credence in winning should be 1/1,000,000.

Given that E1 iff E2 as you say, I would say that P(E1) = P(E2), and that the most appropriate credence for E2 (and so also E1) is 1/2, irrespective of practical considerations.

So, from Sue's perspective (based on the exact same evidence she shares with the participant), she concludes that the coin landed tails with a 2/3 probability, despite the coin having a 1/2 propensity to land tails. Sue's credence that the coin landed tails is a consequence of both the initial propensity of the coin to land tails and the propensities of the experimental setup to place her in a room that corresponds to a tails outcome. — Pierre-Normand

I think that part in bold is key.

There are two Sleeping Beauties, A and B. If the coin lands heads then A will be woken once and B twice, otherwise B will be woken once and A twice.

When woken each is asked their credence that they have been or will be woken twice.

Each of the three sitters is also asked their credence that their participant has been or will be woken twice.

Now it may be tempting to argue that the sitters ought to reason as if their interview has been randomly selected from the set of all interviews, but that might not be technically correct. The only way I can think of this to be done is for one of the sitters to be randomly assigned an interview from the set of all interviews (of which there are 3), for one to be randomly assigned an interview from the set of all remaining interviews (of which there are 2), and then one to be assigned the final interview.

So rather the sitters should reason as if they are randomly selected from the set of all sitters, and so their credence that their participant has been or will be woken twice will be 2/3.

The question, then, is whether or not the participant should reason as if they are randomly selected from the set of all participants, and so their credence that they have been or will be woken twice is 1/2. I will continue to say that they should, given the propensities of the experimental setup to place them in the position to be woken twice.

That the participant who is woken twice will be right twice as often (if she guesses that she has been or will be woken twice) isn't that each participant's credence should be that they are twice as likely to be woken twice.

Your suggestion that a thirder expects to gain from choosing amnesia would depend on her conflating the probability of making a correct prediction upon awakening with the frequency of the actual payout from the initial bet. — Pierre-Normand

I think a distinction needs to be made between the probability of making a correct prediction and the frequency of making a correct prediction. That a correct prediction of tails is twice as frequent isn't that a correct prediction of tails is twice as probable – at least according to Bayesian probability.

Maybe thirders who use betting examples are simply frequentists?

Perhaps there also needs to be a distinction made between the probability of making a correct prediction and the probability of the coin having landed tails. It could be that the answers are different. This might be especially true for frequentists, as the frequency of correct predictions is not the same as the frequency of coins landing tails (there can be two correct predictions for every one coin that lands tails).

I wonder like if it has four foot pedals and a steering wheel that requires twelve hands or something like that. — Hanover

Or it has writing in some gibberish like Chinese or Russian.

For some reason the UFO stories started gaining popularity on FoxNews and in conservative circles. I guess it goes along with the government conspiracy theory thing. — Hanover

What's interesting about this case (at least according to this), is that he "has given Congress and the Intelligence Community Inspector General extensive classified information about deeply covert programs that he says possess retrieved intact and partially intact craft of non-human origin."

Maybe that entire report is rubbish, or maybe he's given them a bunch of stuff that he erroneously believes to show evidence of aliens.

Trans Man Brutally Assaulted For Using Women’s Restroom at Campground

A transgender man camping in Ohio was arrested for disorderly conduct after he claims he was assaulted by a group of men for using the women's restroom on July 3.

Noah Ruiz, 20, told local Fox affiliate WXIX he was using the women's restroom at the Cross's Campground in Camden at the direction of the camp owner, Rick Cross, when a woman camper became very upset.

"She was like, 'No man should be in this bathroom. Like, if you're a man you need to use a man's bathroom,'" Ruiz told the outlet. "And I was like, 'I'm transgender. Like, I have woman body parts, and I was told to use this bathroom.'"

As he and his girlfriend left the bathroom, Ruiz said he was jumped by three large men who lifted him off the ground and choked him out, all the while using anti-LGBTQ+ slurs and threatening to kill him.

Alternatively, she loses $1 each time she wakes. What is her credence that she will lose $2?

It’s not the case that she loses $2 2/3 of the time, although it is the case that 2/3 of the time she wakes up she’s in an experiment where she’s going to lose $2.

I think thirders conflate the two. The latter is obvious but not what is asked about when asked her credence that the coin landed tails.

I wonder if the variation I considered here highlights the difference in perspective, with the question forcing you to consider it from the halfer perspective:

There are two Sleeping Beauties; one will be woken on Monday and one on both Monday and Tuesday, determined by a coin toss.

What is their credence that they have been or will be woken twice?

We can even consider it with just one Sleeping Beauty, as with the traditional problem.

Does it make sense for her credence to be that she's twice as likely to be woken twice simply because she's woken twice if she is woken twice compared to once if woken once? I don't think it does. The very question requires that you think of both wakings as being part of the same outcome, with your credence in that outcome being what is questioned.

But given that you're woken twice iff the coin landed tails, it stands to reason that the answer to "what is your credence that you have been or will be woken twice" is the same as the answer to "what is your credence that the coin landed tails".

I think you would benefit from reading Groisman. — Pierre-Normand

I've just read it. Seems to be saying what I said here:

There are two ways to reason:

1. $2\over3$ of all interviews are 100 heads in a row interviews, therefore this is most likely a 100 heads in a row interview
2. $1\over2^{100}$ of all participants are 100 heads in a row participants, therefore I am most likely not a 100 heads in a row participant

I would say that both are true...

That's correct since events that happen in the world don't come flagged with sign posts that say: "the current event begins here" and "the current event terminates here." How credences in the probabilities of events are assessed depend on the way those events are individuated and this can be dictated by pragmatic considerations. — Pierre-Normand

I don't think this is right. The experiment is only being conducted once and Sleeping Beauty is asked "what is your credence that the coin I tossed on Sunday landed heads?" I think there's only one appropriate way to reason, which above you seemed to accept gives an answer of $1\over2$.

But if you want to insist that it's ambiguous, I'll just fall back to that previous application of Bayes' theorem (despite ChatGPT's objection):

P(Heads | Mon or Tue) = P(Mon or Tue | Heads) * P(Heads) / P(Mon or Tue)
P(Heads | Mon or Tue) = 1 * 1/2 / 1
P(Heads | Mon or Tue) = 1/2

As Elga says, she learns nothing new (and Lewis would agree); she only recognises that her temporal location (of which all she knows is it's either Monday or Tuesday) is relevant. But contrary to Elga's reasoning (which I believe I showed to be faulty) this doesn't change her credence.

It might just be as simple as this.

The issue of optimum betting strategies over multiple games is irrelevant to the problem. There's nothing to debate on that; it's obviously better to bet on tails. I'm sure Lewis and every other halfer would agree.

But as this debate has gone on long enough and I don't think I have the energy to continue it much more, I'm happy to just say that both $1\over2$ and $1\over3$ are correct answers to distinct but equally valid interpretations of the question.

So thanks for the enlightening discussion.

Since on my approach probabilities track frequencies, even if there is just one kidnapping event, the hostage's chances of survival are 5 in 11 whenever an escape attempt occurs. — Pierre-Normand

So then there are two different ways to reason with nothing to prove that one or the other is the "right" way?

So, she should be carrying a plank and end up being eaten by lions on 6 out of 11 escape attempts? — Pierre-Normand

5 out of every 6 victims escape. I count by participants, not by escape attempts. I think it's more reasonable.

The manner in which (1) is stated suggest that Sleeping Beauty is referring to the wide centered possible world spanning the whole experiment run. In that case, her credence in H should be 1/2.

The second one makes it rational for her to rely on her credence regarding narrow centered possible worlds spanning single awakening episodes. There indeed isn't any entailment from the suitability of one framing of the question from (1) to (2) or vice versa. The two sentences concern themselves with different questions. — Pierre-Normand

And that first question is the premise of the problem. Sleeping Beauty is asked her credence that the current experiment's coin toss landed heads. That's it. She's not being asked to consider the most profitable betting strategy for multiple games, which even Lewis (I assume) would agree is to bet on tails.

However, in the scenarios with Sleeping Beauty and the prisoner, merely being presented with an opportunity to bet or escape does not give them any new information about the outcome of the coin toss (or throw of the die). They must decide how to take advantage of this opportunity (by choosing to carry the torch or the plank, or choosing what safehouse address to communicate to the police) before gaining any knowledge about the success of the attempt. The offering of the opportunities carry no information and provide no ground for updating credences. — Pierre-Normand

She can and should use known priors to condition her credence, and one such prior is that she is more likely to win a prize/have the opportunity to escape if tails/a dice roll of 6. As such, if she wins a prize or has the opportunity to escape she should condition on this and her credence should favour tails/a dice roll of 6, otherwise she should condition on not winning a prize or having the opportunity to escape and her credence should favour heads/a dice roll of 1-5.

And if she's guaranteed the opportunity to escape each day, or she never has the opportunity to escape each day, then her credence should favour a dice roll of 1-5.

But honestly, all this talk of successes is irrelevant anyway. As I said before, these are two different things:

1. Sleeping Beauty's credence that the coin tossed on Sunday for the current, one-off, experiment landed heads
2. Sleeping Beauty's most profitable strategy for guessing if being asked to guess on heads or tails over multiple games

It's simply a non sequitur to argue that if "a guess of 'tails' wins 2/3 times" is the answer to the second then "1/3" is the answer to the first.

In your scenario, the nature of the prize is conditioned on the coin toss results. — Pierre-Normand

Then forget the nature of the prize.

If I know that I’ve won a prize my credence favours the first coin toss having landed tails, otherwise it favours heads. If I see that the tulip is red my credence favours the coin toss having landed tails, otherwise it favours heads.

The main point is that seeing Rex Harrison being featured (while knowing that 1% of the movies randomly being shown in this theater feature him) doesn't impact your credence in this movie being part of a double feature. — Pierre-Normand

That's only because I walk into one film. If I'm given amnesia and walk into the second film (if there is a second film) then it affects my credence.

It's exactly like my scenario with the coin toss and prizes. If heads then the car is the possible prize, otherwise the motorbike is the possible prize. If a car then a single coin toss determines if I get it (if heads), if a motorbike then two coin tosses determine if I get it (one head is enough to win).

If I'm told that I've won a prize then my credence favours that it is a motorbike. If I'm told that I didn't win a prize then my credence favours that the first coin toss was heads.

Knowing that I've won (or not won) a prize is the same as (not) seeing a red tulip or Harrison. It conditions my credence of that first coin flip.

Sorry, misunderstand the movie example. It’s a different answer if I only get to walk into one film, which would be comparable to Sleeping Beauty only waking on Monday (or Tuesday) if tails.

In such a case I can’t say it’s more likely a double feature, and thirders would say Sleeping Beauty can’t say it’s more likely tails.

Consider this analogy: you're entering a movie theater where there's an even chance of a double feature being shown. There's a one percent chance that any given film will feature Rex Harrison. Suppose you see Harrison featured in the first film. Does that increase your credence that there will be a subsequent feature? — Pierre-Normand

No, but if I walk in not knowing if it’s the first or second film then my credence favours it being part of a double feature.

If you think about it, Lewis's notion—that Sleeping Beauty can conclude from knowing it's Monday that a future coin toss is more likely to yield heads with a 2/3 probability—is already rather puzzling. — Pierre-Normand

I agree, which is why I offered the example of taking balls from a bag which is a double halter interpretation of the problem.

What if there is a 1% chance that the tulip is red on any given awakening day? Would that make any difference? — Pierre-Normand

Yes. The probability of it being red on a waking day if heads is 1%. The probability of it being red on a waking day if tails is 1 - 0.99^2 = 1.99%.

The probability of it being red on a waking day if tails is greater than the probability of it being red on a waking day if heads, therefore if it's red on a waking day then it is more likely tails and if it's not red on a waking day then it is more likely heads.

This might be clearer if we say that the probability of it being red on a waking day if heads is 1.5%. Sleeping Beauty should still reason that it is more likely tails. But if we increase it further to 2% then she should reason that it is more likely heads.

Introducing these additional probabilistic events upon which to condition her credence changes the answer and so doesn't accurately represent the problem without them.

Suppose there is a 0.01% chance to find an opportunity to escape on any given day held captive regardless of that day being the only one or one among six in a kidnapping event. Finding such opportunities doesn't yield any updating of credence. — Pierre-Normand

It does.

Dice roll 1-5 (safehouse #1): day 1, 0.01% opportunity to escape
Dice roll 6 (safehouse #2): day 1, 0.01% opportunity to escape; day 2, 0.01% opportunity to escape; day 3, 0.01% opportunity to escape; day 4, 0.01% opportunity to escape; day 5, 0.01% opportunity to escape; day 6, 0.01% opportunity to escape

If I'm doing my maths right, the probability of being offered an opportunity to escape from safehouse #2 is 0.058, so you're almost 6 times more likely to be offered the opportunity to escape from safehouse #2, so upon being offered the opportunity to escape, you can condition on this and determine that P(Dice roll 6|opportunity to escape) > P(Dice roll 1-5|opportunity to escape).

Conversely, upon not being offered the opportunity to escape, you can condition on this and determine that P(Dice roll 6|no opportunity to escape) < P(Dice roll 1-5|no opportunity to escape).

And it would be strange to say that P(Dice roll 6|no opportunity to escape) < P(Dice roll 1-5|no opportunity to escape) where there is a 0.01% chance of an opportunity to escape each day but then to say P(Dice roll 6) > P(Dice roll 1-5) where there is a 0% chance of an opportunity to escape each day.

Likewise, enabling Sleeping Beauty to bet on H on each awakening provides no information to her, provided only the payouts are delivered after the experiment is over. — Pierre-Normand

You may have missed my edit above:

I don't understand the connection between Sleeping Beauty's credence that the coin landed heads and the tracked frequency of heads-awakenings. It's a non sequitur to claim that because tails-awakenings are twice as frequent over repeated experiments then a coin toss having landed tails is twice as likely in any given experiment.

Sleeping Beauty is being asked "in this current, one-off experiment, what is the probability that the coin I tossed on Sunday evening landed heads?".

She's not being asked to guess if it's heads or tails and then being rewarded for each successful guess.

Her choice of guess in the latter has nothing to do with what her answer would be to the former.

If I were Sleeping Beauty I would answer "1/2" and guess tails.

Indeed, which is basically the 'thirder' solution (in this case, the 5/11er solution). — Pierre-Normand

Yes, but it's only P(Dice roll 6|opportunity to escape) > P(Dice roll 1-5|opportunity to escape) because of the opportunity to escape (and where the %s are set up a certain way). Without that opportunity to escape it's P(Dice roll 1-5|no opportunity to escape) > P(Dice roll 6|no opportunity to escape), which is comparable to the Sleeping Beauty problem.

Which is why I had included the proviso that the (rare) opportunities be proportional to the number of days the hostage is held captive. Under those conditions, they carry no information to the hostage. — Pierre-Normand

Can you give actual numbers? Because that determines the answer. If there's a 90% opportunity to escape on day 1 in safehouse #1 but a 1% opportunity to escape on each day in safehouse #2 then P(Dice roll 6|opportunity to escape) < P(Dice roll 1-5|opportunity to escape).

Or if you set the %s up right then it can work out as P(Dice roll 6|opportunity to escape) = P(Dice roll 1-5|opportunity to escape), and so it's equally likely.

To set out the scenario:

Dice roll 1-5 (safehouse #1): day 1, 50% opportunity to escape
Dice roll 6 (safehouse #2): day 1, 50% opportunity to escape; day 2, 50% opportunity to escape; day 3, 50% opportunity to escape; day 4, 50% opportunity to escape; day 5, 50% opportunity to escape; day 6, 50% opportunity to escape

It's quite straightforward that P(Dice roll 6|opportunity to escape) > P(Dice roll 1-5|opportunity to escape), but P(Dice roll 1-5|no opportunity to escape) > P(Dice roll 6|no opportunity to escape).

And so when there's a 0% opportunity to escape on each day, as with the traditional problem, P(Dice roll 1-5) > P(Dice roll 6).

The opportunity to escape just enables the prisoner to put their credence to good use, and to chose how to most appropriately define the states that those credences are about. It doesn't change their epistemic situation. — Pierre-Normand

It does change the epistemic situation. It's exactly like the scenario with the coin tosses and the prizes, where in this case the prize is the opportunity to escape.

You're more likely to win a prize if it's tails, therefore upon being offered a prize you reason that tails is more likely.

You're more likely to win the opportunity to escape if in safehouse #2, therefore upon being offered the opportunity to escape you reason that safehouse #2 is more likely.

Introducing the concept of escape possibilities was intended to illustrate that what is at stakes in maximizing the accuracy of the expressed credences can dictate the choice of the narrow versus wide interpretations of the states that they are about.

In the safehouse and escape example: if the prisoner's goal is to maximize their chances of correctly predicting 'being-in-safehouse-#1' on any given awakening day, they should adopt the 'thirder' position (or a 5/11 position). If their goal is to maximize their chances of correctly predicting 'being-in-safehouse-#1' for any given kidnapping event (regardless of its duration), they should adopt the 'halfer' position (or a 6/11 position). — Pierre-Normand

Introducing the concept of escape possibilities simply changes the answer. You're more likely to have an opportunity to escape in safehouse #2, and so if given an opportunity to escape then you are more likely in safehouse #2. P(Safehouse #2|escape opportunity) > P(Safehouse #1|escape opportunity)

This answer isn't relevant to a scenario where there are no opportunities to escape, where P(Safehouse #1|no escape opportunity) > P(Safehouse #2|no escape opportunity).

If the agents expression of their credences are meant to target the narrow states, then they are trying to track frequencies of them as distributed over awakening episodes. If they are meant to target the wide states, then they are trying to track frequencies of them as distributed over experimental runs (or kidnaping events). — Pierre-Normand

I don't understand the connection between Sleeping Beauty's credence that the coin landed heads and the tracked frequency of heads-awakenings. It's a non sequitur to claim that because tails-awakenings are twice as frequent over repeated experiments then a coin toss having landed tails is twice as likely in any given experiment.

Sleeping Beauty is being asked "in this current, one-off experiment, what is the probability that the coin I tossed on Sunday evening landed heads?".

She's not being asked to guess if it's heads or tails and then being rewarded for each successful guess.

Her choice of guess in the latter has nothing to do with what her answer would be to the former.

If I were Sleeping Beauty I would answer "1/2" and guess tails.