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.

But what determines the right question to ask isn't the statement of the Sleeping Beauty problem as such but rather your interest or goal in asking the question. I gave examples where either one is relevant. — Pierre-Normand

Are you referring to the safehouse and escape? That's a different scenario entirely.

I flip a coin. If heads then I flip again. If heads you win a car, otherwise you win nothing. If the first flip is tails then I flip again. If heads you win a motorbike, otherwise I flip again. If heads you win a motorbike, otherwise you win nothing.

I do this in secret and then tell you that you've won a prize. Given that you're more likely to win a prize if it's tails then it's reasonable to believe that it was most likely tails.

Now consider a similar scenario, but if heads then you win a car and if tails then you win two motorbikes. I do this is in secret and tell you that you've won at least one prize. It is not reasonable to believe that it was most likely tails.

Your safehouse and escape example is analogous to that first case. Your introduction of opportunities to escape changes the answer, and so it isn't comparable to the Sleeping Beauty case where there is nothing analogous to an escape opportunity with which to reassess the probability of the coin toss.

we are changing the structure of the problem and making it unintelligible that we should set the prior P(W) to 3/4. — Pierre-Normand

That's precisely the point, and why I suggested ignoring days and just saying that if heads then woken once and if tails then woken twice. P(W) = 1.

That P(Heads & Tuesday (or second waking)) consideration is a distraction that leads you to the wrong conclusion.

Even though the player is dismissed (as opposed to Sleeping Beauty, who is left asleep), a prior probability of P(Dismissed) = 1/4 can still be assigned to this state where he loses an opportunity to bet/guess. Upon observing the game master pulling out a ball, the player updates his prior for that state to zero, thus impacting the calculation of the posterior P(Red|Opp). If we assign P(Dismissed) = 1/4, it follows that P(Red|Opp) = 1/3. — Pierre-Normand

If you're going to reason this way then you also need to account for the same with blue. You reach in after the second blue and pull out nothing. So really P(Dismissed) > 1/4.

I just don't think it makes sense to reason this way.

If it helps, consider no tequila after the final ball. After being given a ball and asked your credence you're then dismissed if it's either red or second blue.

Your revised scenario seems to neglect the existence of a state where the player is being dismissed. — Pierre-Normand

It doesn't. You're dismissed after red or the second blue.

It is still the case that if I don't know whether this is Monday or Tails then I reason as if my ball is randomly selected from one of the two bags, such that P(R) = 1/2 and P(B1) = P(B2) = 1/4 (or just P(R) = P(B) = 1/2). It better reflects how the experiment is actually conducted.

I don't reason as if my ball is randomly selected from a pile such that P(R) = P(B1) = P(B2) = 1/3.

This scenario doesn't accurately reflect the Sleeping Beauty experiment. Instead, imagine that one bag is chosen at random. You are then given one ball from that bag, but you're not allowed to see it just yet. You then drink a shot of tequila that causes you to forget what just happened. Finally, you are given another ball from the same bag, unless the bag is now empty, in which case you're dismissed. The balls are wrapped in aluminum foil, so you can't see their color. Each time you're given a ball, you're invited to express your credence regarding its color (or to place a bet, if you wish) before unwrapping it. — Pierre-Normand

I didn't properly address this but I actually think it illustrates the point quite clearly. I'll amend it slightly such that the blue balls are numbered and will be pulled out in order, to better represent Monday and Tuesday.

If I'm told that this is my first ball (it's Monday) then P(R) = P(B₁) = 1/2.

If I'm told that this is a blue ball (it's Tails) then P(B₁) = P(B₂) = 1/2.

If I don't know anything then I should reason as if my ball was randomly selected from one of the two bags, and so P(R) = 1/2 and P(B₁) = P(B₂) = 1/4 (or just P(R) = P(B) = 1/2).

This contrasts with your reasoning as if my ball is randomly selected from a pile such that P(R) = P(B₁) = P(B₂) = 1/3.

At the very least this shows how halfers can be double halfers to avoid Lewis' P(Heads | Monday) = 2/3.

Would not a halfer say that they are equally as likely? — Pierre-Normand

Equally likely to happen, such that P(Monday & Heads) = P(Monday & Tails) = P(Tuesday & Tails) = 1/2, as per that earlier Venn diagram, but not equally likely that today is that interview, because if P(Heads) = 1/2 then P(Monday & Heads) = 1/2, and so P(Monday & Tails) + P(Tuesday & Tails) = 1/2, therefore if P(Monday & Tails) = P(Tuesday & Tails) then P(Monday & Tails) = P(Tuesday & Tails) = 1/4.

This was Lewis' reasoning in his paper.

Sleeping Beauty's inability to single out any one of those possible awakenings as more or less likely than another — Pierre-Normand

Well, that’s the very thing being debated. A halfer might say that a Monday & Heads awakening is twice as likely as a Monday & Tails awakening, and so it is a non sequitur to argue that because Tails awakenings are twice as frequent in the long run they are twice as likely.

So how does the thirder argue that they are equally likely if not by first reasoning as if an interview is randomly selected from the set of possible interviews?

If she opts to track her awakenings (centered possible worlds), her credence in heads is 1/3. — Pierre-Normand

How does one do this if not by reasoning as if one's interview is randomly selected from the set of possible interviews?

It seems quite counterintuitive that if my credence concerns the outcome of the experimental run I'm in, it is P(10) = 1/10, and if it's the outcome of the present awakening, it's P(10) = 10/19, and that both outcomes are perfectly correlated. — Pierre-Normand

So this goes back to what I said before. Either we reason as if we’re randomly selected from the set of all participants, and so P(10) = 1/10, or we reason as if our interview is randomly selected from the set of all interviews, and so P(10) = 10/19.

Given that the experiment doesn’t work by randomly selecting an interview from the set of all interviews, I don’t think it rational to reason as if it is. The experiment works by rolling a dice, and so it is only rational to reason as if we’re randomly selected from the set of all participants.

How we chose to bet just has no bearing on one’s credence that one is likely to win. With your lottery example we play even if we know that we’re most likely to lose (and in fact I play the lottery even though the expected value of winning is less than the cost). And with this example I might bet on 10 even if my credence is that it is less likely, simply because I know that I will win in the long run (or, if playing one game, I’m just willing to take the risk because of the greater expected value).

Yes, it is rational to believe that if you repeat the game enough times then you will win more than you lose, but it is still irrational to believe that if you play the game once then you are most likely to win.

For instance, suppose you offer me the opportunity to purchase a $100 lottery ticket that carries a one in a septillion chance of winning me $200 septillion. Despite the expected value being positive, it may not be reasonable for me to purchase the ticket. However, it would be a logical fallacy to extrapolate from this example and conclude that it would also be unreasonable for me to buy a $100 lottery ticket with a one in ten chance of winning me $2000. Given I'm not in immediate need of this $100, it might actually be quite unreasonable for me to pass up such an opportunity, even though I stand to lose $100 in nine times out of ten. — Pierre-Normand

It would be unreasonable of you to believe that you are most likely to win, even if it’s financially reasonable to play.

I did mention this. There are two ways to reason:

1. I should reason as if I am randomly selected from the set of possible participants
2. I should reason as if my interview is randomly selected from the set of possible interviews

I do the former, he does the latter.

My use of variants, such as that of tossing the coin 100 times, was to show that applying his reasoning leads to what I believe is an absurd conclusion (that even if the experiment is only done once it is rational to believe that P(100 Heads) = 2/3).

Although he accepts this conclusion, so at least he’s consistent.

But you’re right that this fundamental disagreement on how best to reason might make these arguments irresolvable. That’s why I’ve moved on to critiquing Elga’s argument, which is of a different sort, and to an application of Bayes’ theorem with what I believe are irrefutable terms (although we disagree over whether or not the result actually answers the problem).

I can’t see it

Or you could make it so that the Join discussion doesn't show if we're logged in, i.e. assigned to a category that only Guests can view?

I'm not sure what you mean by the sampling mechanism. There is one experiment with one coin toss. We both appear to agree on that.

Rather, it pointed out that your calculation of P(Heads | Monday or Tuesday) = 1/2 simply restates the unconditional probability P(H) without taking into account Sleeping Beauty's epistemic situation. — Pierre-Normand

As Elga says:

This belief change is unusual. It is not the result of your receiving new information — you were already certain that you would be awakened on Monday. (We may even suppose that you knew at the start of the experiment exactly what sensory experiences you would have upon being awakened on Monday.) Neither is this belief change the result of your suffering any cognitive mishaps during the intervening time — recall that the forgetting drug isn’t administered until well after you are first awakened. So what justifies it?

The answer is that you have gone from a situation in which you count your own temporal location as irrelevant to the truth of H, to one in which you count your own temporal location as relevant to the truth of H.

Sleeping Beauty's "epistemic situation" is only that her current situation is relevant. She doesn't learn anything new. All she knows is that her temporal location is either Monday or Tuesday. Before the experiment began this wasn't relevant, and so she only considers P(H). After being woken up this is relevant, and so she considers P(H | Mon or Tue).

That they both give the same answer (because Monday or Tuesday is trivially true) just suggests that Lewis was right. It really is as simple as (in his words) "Only new relevant evidence, centred or uncentered, produces a change in credence; and the evidence (H₁ ∨ H₂ ∨ H₃) is not relevant to HEADS vs TAILS".

The argument you've put forward could be seen as suggesting that the vast body of literature debating the halfer, thirder, and double-halfer solutions has somehow missed the mark, treating a trivial problem as a complex one. This isn't an argument from authority. It's just something to ponder over. — Pierre-Normand

Well, I would also think that my argument that P(A|A or B) = P(B|A or B) doesn't entail P(A) = P(B) is quite trivial. Maybe I've made a mistake (whether with this or my interpretation of Elga), or maybe Elga did. I'll admit that the former is most likely, but my reasoning appears sound.