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, [...] — Michael

Would not a halfer say that they are equally as likely?

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

If by "reasoning as if one's interview is randomly selected from the set of possible interviews" you're referring to Sleeping Beauty's avowal of her inability to single out any one of those possible awakenings as more or less likely than another, then yes, she is reasoning in that way. However, this form of reasoning is guided by her need to assess her credence in a heads-awakening outcome (her centered possible world), not by the setup of the experiment.

The design of the experiment allows participants to assess both their centered and non-centered credences. What they decide to track can be influenced by various factors. For instance, they might be trying to optimize their expected value or their chances of winning; they could be trying to maximize their chances of escaping alive versus maximizing their chances of being rescued. Their choice could also be guided by their interests, such as whether they want to track the frequency of heads-awakenings or the frequency of heads-experimental-runs. In other words, the choice is driven by the participants' individual needs and interests, not dictated by the structure of the experiment itself.

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. — Michael

Your assertion that I'm reasoning "as if" someone selected an interview from all possible interviews isn't quite accurate. Sleeping Beauty is fully aware that the current experimental run comprises either one or multiple awakenings, depending on the outcome of a coin toss. She doesn't assume that someone arbitrarily selected an awakening of hers. Rather, she knows she's currently experiencing an awakening and isn't certain whether it's the only one in the present run, or one among many.

She's asked about her credence regarding the outcome of the coin toss. Given the known setup of the experiment, she can choose to calculate her credence in either the centered or non-centered heads outcome. She could even calculate both. If she opts to track her awakenings (centered possible worlds), her credence in heads is 1/3. If she decides to track her experimental runs (non-centered), her credence in heads is 1/2. These are credences in distinct outcomes, and both can serve different purposes.

For example, in my hostage scenario, one approach maximizes one's chance of survival during an escape, while the other maximizes one's chances of being rescued by the police when communicating probable location. (I believe I may have made a mistake in my earlier analysis of the police rescue situation).

Perhaps it's a misnomer to call them correlated, because there's no meaningful notion of a joint event of both occurrences within the same sample space. — fdrake

What I mean is that whenever the coin landed heads during a particular awakening, then it also landed heads during the particular experimental run this awakening is a part of, and vice versa. But Sleeping Beauty's credences in those two different outcomes (centered = 1/3 and uncentered = 1/2) is different, they are both right, and they both perfectly tack the frequencies of those corresponding outcomes when the experiment is repeated.

Yes, 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. — Michael

Indeed, it's true that if you play the game once, it's not most likely that you'll win during that particular experimental run. However, when I suggest it's more likely than not that the outcome is 10 during the current awakening, I'm not claiming that it's most likely that the current experimental run will result in a win for me. These are two distinct statements, despite their truth values being perfectly correlated. This seems to be the point that was made in the conclusion of the paper that @fdrake mentioned. Furthermore, these claims hold true irrespective of whether the experiment is run once or multiple times.

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. But that's just because the one gerrymanders the other.

It would be unreasonable of you to believe that you are most likely to win. — Michael

Consider this scenario in the context of the Sleeping Beauty problem where a ten-sided dice is thrown. In this situation, it would be rational for me to believe that I am more likely to win an even money bet placed on the outcome being a 10 during the current awakening. Despite the fact that in 9 out of 10 experimental runs I would lose my only bet, in the remaining run, I would win ten times my bet by the time the experiment concludes. Thus, on average, I'd win 10 out of 19 times. And this is because my credence that the outcome is 10, P(10) = 10/19, would exactly match, in the long run, the proportion of times the outcome indeed was 10 when I awoke.

I don’t think any reasonable person would believe this. I certainly wouldn’t. — Michael

Perhaps a rational individual might not believe in the plausibility of being woken up and put back to sleep 2^101 times. But even if this extreme scenario makes the consideration of such a highly unlikely albeit extraordinary event unreasonable, it does not follow that thirder-like solutions to a less extreme version of the Sleeping Beauty problem would be equally unreasonable.

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 nine times out of ten.

I ended up in a state of confusion in the calculations, having a few contradictions in reasoning, which this paper elevates into framings of the experiment (including SB's setting within it) having inconsistent sample spaces between the centred and non-centred accounts. Thus yielding a "dissolution" of the paradox of the form; it's only a paradox when centred and non-centred worlds are equated. — fdrake

Thanks! I'm going to read this paper. I like the conclusion: "Thus, we suggest that although it is true that Beauty is in a heads-awakening if and only if the coin landed heads, Beauty, upon awakening, should assign probability 1/3 to the former and probability 1/2 to the latter."

Is that not closely analogous to my distinction of case-2 and case-1 (respectively) in my analysis of the prisoner scenario? My analysis also suggests a pragmatist interpretation of the choice between centered and non-centered accounts. The choice depends not on metaphysical preferences, and not either on arbitrary stipulations regarding sampling mechanisms, but rather on the specification of the use one intends to make of one's credence in the outcome.

I've been reading along, I have a meta question for you both Pierre-Normand@Michael - why is it helpful to discuss variants which are allegedly the same as the original problem when you both don't seem to agree what the sampling mechanism in the original problem is? — fdrake

I introduced the prisoner variant to satisfy Michael's insistence that Sleeping Beauty's credence only be evaluated within a single-run experimental context. It aimed to show that it is rational for the prisoner to act under the assumption that their safehouse is surrounded by crocodiles with probability 6/11 whenever they get an opportunity to act of the basis of their credence, and this in spite of the fact that they had initially been brought to a safehouse surrounded by lions with probability 6/11. It also highlighted how the credence of the participant (prisoner) might interact with the different credence of an external agent (the police). I think my distinction of case-1 and case-2 also helped diagnose the source of Michal's halfer intuition. It stems from confusing the standard Sleeping Beauty problem with a different but closely related one (that indeed may be construed as stemming from a different sampling mechanism).

The blue and red balls mechanism was introduced by Michael but I proposed to refine it in order to better mirror the Sleeping Beauty protocol.

Regarding the sampling mechanism, I believe the centered possible world approach offers a compelling interpretation, although Milano suggests that there may still be residual disagreement about setting priors. This approach is compelling because it allows the participant to reason about her epistemic situation in a natural way without the need to incorporate any weird metaphysical conjectures about how she finds herself in her current situation. However, I'm not entirely confident about this. Milano's chosen values for the parameters α and β seem well-argued to me, grounded in both the stipulations of the problem and uncontroversial rationality assumptions. But there may be something I've overlooked.

So ChatGPT is saying that P(Heads | Monday or Tuesday) = 1/2 is trivially true. Doesn't that just prove my point? — Michael

GPT-4 wasn't endorsing your conclusion. 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.

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.

Just for fun, I asked GPT-4 to comment on your Bayesian analysis.

I think this is a better way to consider the issue. Then we don't talk about Heads & Monday or Tails & Monday. There is just a Monday interview and then possibly a Tuesday interview. It's not the case that two thirds of all interviews are Tails interviews; it's just the case that half of all experiments have Tuesday interviews. Which is why it's more rational to reason as if one is randomly selected from the set of possible participants rather than to reason as if one's interview is randomly selected from the set of possible interviews. — Michael

It appears that you are suggesting a false dichotomy. Logically, both (1) two-thirds of all interviews being Tails interviews, and (2) half of all experiments having Tuesday interviews can simultaneously hold true. Both (1) and (2) are in fact logical implications from the problem's stipulations. The terms 'heads', 'tails', 'Monday', and 'Tuesday' merely serve as convenient labels for the spectrum of possibilities. Solving a problem doesn't involve selectively ignoring one of its stipulations or dismissing its acknowledgement as irrational. If you intentionally disregard a stipulation of the problem (or a logical consequence thereof), you are effectively addressing a different problem.

And I think it's even better to not consider days and just consider number of times wakened. So first she is woken up, then put to sleep, then a coin is tossed, and if tails she's woken again. Then we don't get distracted by arguing that her being asleep on Monday if Heads is part of the consideration. It doesn't make sense to say that she's asleep during her second waking if Heads.

I don't think anyone said that she experiences anything on Tuesday when she sleeps. Milano addresses the related concern that it is not an epistemic possibility that Sleeping Beauty will experience the centered possible world in which it is Tuesday and she is asleep. That doesn't mean that this possibility can't be considered in Bayesian conditionalization. You can ascribe non-zero probabilities to states that you will not experience, which happens for instance when you buy life insurance.

That's exactly the implication of Elga's reasoning. — Michael

Lewis's treatment yields an incredible result. Elga's treatment yields an unsurprising result by means of a controversial method. This is why I prefer the treatments by Stalnaker and (especially) Milano. They both demystify the method and Milano, in addition, shows the self centered world method to be consistent with Bayesian conditionalization, and both to satisfy van Fraassen's principles of reflection.

My example proves that this doesn't follow where P is the credence function I ought to have after being explained the rules of my game. Elga doesn't explain why it follows where P is the credence function I ought to have upon first awakening. — Michael

I apologize. I misunderstood this part of Elga's argument. Although P(T1) and P(H1) are the credences Sleeping Beauty should have upon being first awakened, they are not conditional on her being in a centered world where she is first awakened. It is rather P(H1|H1 or T1) that is thus conditioned.

I'm going to study the argument more closely and comment later.

Incidentally, the fact that Sleeping Beauty's credence that the coin landed (or soon is going to land) heads, upon first awakening, is being increased by 1/6 after she learns that today is Monday, is a point of agreement between Lewis and Elga.

Elga's interpretation that her credence P(H) gets updated from 1/3 to 1/2 is easily accounted for by the fact that, after she can rule out today being Tuesday, the possible worlds [H1, asleep] and [T1, T2] remain equiprobable. Since the setup of the experiment doesn't even require that anyone look at the result of the toss before Monday night, nothing changes if the toss is actually performed after Sleeping Beauty's awakening. In that case the credences expressed on Monday are about a future coin toss outcome rather than an already actualized one. My previous remark that Sleeping Beauty and the "game master" here are in the same centered possible world and share the same evidence applies.

Lewis's halfer argument, though, commits him to claim that Sleeping Beauty's credence P(H) gets updated from 1/2 to 2/3. In other words, after she is first awakened and told that today is Monday, it becomes rationally warranted for Sleeping Beauty to believe that the probability that the coin landed heads (or will land heads when it is being tossed later tonight) is 2/3. Lewis bites this bullet and explains this as a peculiar form of knowledge about the future. It would have been intriguing to know if Lewis would have extended this line of reasoning to the game master's credence, given that they share the same epistemic situation as Sleeping Beauty. If so, this would suggest a highly unusual implication - that one could acquire knowledge about future events based solely on the fact that someone else would be asleep at the time of those events.

That’s not accurate. There is a difference between these two assertions:

1. P(R|R or B1) = P(B1|R or B1)
2. P(R) = P(B1)

The first refers to conditional probabilities, the second to unconditional probabilities, and in my example the first is true but the second is false. — Michael

It looks like you may have misinterpreted Elga's paper. He doesn't define P as an unconditional probability. In fact, he expressly defines P as "the credence function you ought to have upon first awakening." Consequently, P(H1) and P(T1) are conditional on Sleeping Beauty being in a centered possible world where she is first awakened. The same applies to P(R) and P(B1), which are conditional on you being in a centered possible world where you are presented with a ball still wrapped in aluminum foil before being given a tequila shot.

To understand what P(R) entails, let's look at the situation from the perspective of the game master. At the start of the game, there is one red ball in one bag and two blue balls in the other. The game master randomly selects a bag and takes out one ball (without feeling around to see if there is another one). They hand this ball to you. What's the probability that this ball is red? This conditional probability is represented as P(R). This also represents how you should update your credence about the color of the ball upon learning that this is the first (or the only) ball given to you before the tequila shot. In this scenario, both you and the game master occupy the same centered possible world and share the same evidence.

Incidentally, I think Stalnaker and Milano's paper both produce arguments that are easier to follow and, it seems to me, more rigorous than Elga's valiant first attempt.

There is a red ball in one bag and two numbered blue balls in a second bag. You will be given a ball at random. — Michael

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.

According to Elga's reasoning:

1. P(B1|B1 or B2) = P(B2|B1 or B2), therefore P(B1) = P(B2)

2. P(R|R or B1) = 1/2, therefore P(R) = P(B1)

3. Therefore, P(R) = P(B1) = P(B2) = 1/3

The second inference and so conclusion are evidently wrong, given that P(R) = 1/2 and P(B1) = P(B2) = 1/4.

So his reasoning is a non sequitur. — Michael

Here, P(R|R or B1) is the probability that the ball you've just received is red, conditioned on the information (revealed to you) that this is the first ball you've received in this run of the experiment. In other words, you now know you haven't taken a shot of tequila. Under these circumstances, P(R) = P(B1) = 1/2.

Incidentally, since we started this discussion, I've read Elga's and Lewis's papers. But I've also read most of Robert Stalnaker's "Another Attempt to Put Sleeping Beauty to Rest" (2013) and Silvia Milano's "Bayesian Beauty" (2020), both of which are illuminating. Stalnaker modifies Lewis' centred-world approach, and Milano reconciles such approaches (which interpret indexical content) with Bayesian principles.

And this is precisely why the betting examples that you and others use don’t prove your conclusion. — Michael

The betting examples serve to illustrate that the credences held by thirders, unlike those held by halfers, align with the frequencies of outcomes in such a way that if they were to place bets based on these credences, their expected values would satisfy the equation EV = Σ(P(i)*$i), where P(i) represents the probabilities of outcomes and $i the corresponding payouts. This illustration does not necessitate that Sleeping Beauty actually seeks to maximize her expected value. Rather, it reveals that if this were her goal, her credences would guide her actions effectively. Moreover, if numerous participants were placed in identical situations to Sleeping Beauty's, their credences would correspond to the frequencies of the outcomes they anticipate, which aligns closely with the intuitive definition of a credence.

On the other hand, it's challenging to reconcile a statement such as: 'My current credence P(H) is 1/2, but if I were placed in this exact same situation repeatedly, I would expect the outcome H to occur one third of the time.' This perspective, typically associated with halfers, seems to introduce an incongruity between stated credence and expected frequency of outcomes.

Except the experiment is only conducted once. — Michael

I've also been working under the assumption that the experiment is conducted only once. Sleeping Beauty's calculation that P(H) = 1/3 doesn't hinge on her participation in the experiment being repeated. She's aware that if the coin lands heads, she will be awakened once, but if it lands tails, she will be awakened twice. If we run this experiment once with three participants, and all three of them bet on T every time they are awakened, they will be correct 2/3 of the time on average, which aligns with their credences.

In the same vein, in your dice roll variation, Sleeping Beauty knows that if the die didn't land on six, she will be awakened once, but if it did land on six, she would be awakened six times. This information is sufficient to justify her credence P(6) = 6/11. If we have eleven participants each participating in this experiment once, and they all bet on 'six' every time they are awakened, they will be correct 6/11 of the time on average.

But even if the same participant were to repeat the experiment 2^100 times, they don't bet on 100 heads because they think it's more likely, they bet on 100 heads because they know that eventually they will win, and that the amount they will win is greater than the amount they will lose. — Michael

Sleeping Beauty isn't asked to place a bet on the outcome on day zero, before she's put to sleep for the first time, with payouts occurring on each subsequent awakening based on her initial prediction. Instead, she's asked about her credence in the outcome of the 100 tosses on each individual awakening (and given an opportunity to place a bet). Most of her awakenings occur on the rare occasion when 100 tosses yield heads, which forms the basis for her credence P(100H) being greater than 1/2 on the occasion of a particular awakening. This same reasoning would also be the ground for her betting on that outcome, assuming her primary goal is to maximize her expected value for that single bet.

However, the Sleeping Beauty problem specifically inquires about her credence, not about the rationality of her attempt to maximize her expected value, or her preference for some other strategy (like maximizing the number of wins per experimental run rather than average gain per individual bet).

Even if she were to endorse your perspective on the most rational course of action (which doesn't seem unreasonable to me either), this wouldn't influence her credence. It would simply justify her acting in a manner that doesn't prioritize maximizing expected value on the basis of her credence.

Then this is a different scenario entirely. If we consider the traditional problem, it would be that after the initial coin toss to determine which days she could be woken, a second coin toss determines if she will be woken (heads she will, tails she won't). — Michael

It is fundamentally the same scenario, except we're adding external threats like crocodiles or lions to the test environment. In the hostage situation, the captive finding himself still held hostage upon waking up is parallel to Sleeping Beauty finding herself still under experimentation when she wakes up, as opposed to being dismissed on the seventh day.

The rational means that increase the captive's survival chances in our scenario (from 5/11 to 6/11) are comparable to those Sleeping Beauty would use to enhance her correct retrodictions of the die roll result being in the [1, 5] range (from 5/11 to 6/11).

The only significant divergence lies in the frequency of opportunities: the hostage can't be provided with frequent chances to escape without invalidating the analogy, whereas Sleeping Beauty can be given the chance to guess (or place a bet) every single day she awakens without undermining the experiment.

However, we can further refine the analogy by allowing the hostage to escape unharmed in all instances, but with the caveat that he will be recaptured unknowingly and re-administered the amnesia-inducing drug. This would align the scenarios more closely.

Also, as an aside, if you correctly reason that it's tails then you escape on the first day, and so you can rule out today being the second day (assuming you understand that you would also reason as if it's tails). — Michael

Yes, that's a valid point. This is precisely why I introduced the concept of a small, constant probability ε, representing the chance for the hostage to find the means of escape on any given day. By doing so, we can marginalize the potential inferences the hostage could make from the fact that he has not yet escaped (or perished in the attempt). The key aspect of the Sleeping Beauty problem, and indeed the comparison to our hostage scenario, is the existence of multiple actionable opportunities to express one's credence - be it through betting, verification, rational action, and so on. So, while your observation uncovers a minor discrepancy, I do not believe it fundamentally undermines my overall argument.

If you don't like to consider my extreme example because the numbers are too high then let's consider a less extreme version. Rather than a coin toss it's a dice roll. If 1 - 5 then woken once (or safehouse 1 with crocodiles for one day), if 6 then woken six times (or safehouse 2 with lions for six days). Any rational person would take the wooden plank, and 5 out of every 6 kidnapped victims would survive. — Michael

The analysis you provide would hold true if the hostage was guaranteed in advance to have exactly one opportunity to escape during the entirety of the experiment (case-1). I agree that in this context, upon being provided the means to escape, the prisoner's updated credence would match his initial credence, leading him to select the wooden plank and expect to survive five out of six times (or one out of two times in the original problem).

However, consider a different scenario where the hostage has a small, constant probability ε of discovering the means of escape each day (case-2). In this scenario, stumbling upon this means of escape would provide the hostage with actionable evidence that he could use to update his credence. Now, he would believe with a probability of 6/11 that he's in safehouse #2, thereby justifying his decision to pick up the torch. Consequently, given that 6 out of 11 kidnapped victims who find the means to escape are surrounded by lions, 6 out of 11 would survive.

In my previous post, I made a point to explain why the original Sleeping Beauty setup mirrors case-2 more closely than it does case-1. The crucial aspect in the Sleeping Beauty problem is the opportunity for her to express her credence, which, as in the case-2 scenario, is not a single guaranteed occurrence, but something that happens with different frequency depending on the result of the coin toss.

In this case you're in safehouse 1 if not tails yesterday and not tails today and you're in safehouse 2 if either tails yesterday or tails today. Obviously the latter is more likely. I think only talking about the preceding coin toss is a kind of deception. — Michael

The original Sleeping Beauty problem does indeed hinge on a single coin toss, but it's crucial to understand the unique nature of this coin toss within the context of the experiment. When we consider the result of 'the' coin toss, we're not considering a generalized coin toss in a vacuum; we're considering the specific toss that determined the course of the experiment that Sleeping Beauty currently finds herself in.

Sleeping Beauty's question - 'What is the probability that the coin shows heads?' - is not a generalized question about the inherent probability of a coin toss result. Instead, it's a targeted question about the outcome of the specific coin toss that dictated the structure of her current experience. Given this context, I argue that Sleeping Beauty is justified in updating her prior credence P(T) from 1/2 to 2/3.

Let's modify the safehouse example to more closely mirror the Sleeping Beauty problem. Imagine you are kidnapped and held hostage. The captor flips a coin only once. If the coin lands heads, you are blindfolded and transported to safehouse #1, held captive for one day, and then released. If the coin lands tails, you are taken to safehouse #2, where you're held for two days and then released. In the latter case, however, an amnesia-inducing drug is administered at the end of the first day, such that you forget the events of the day and wake up on the second day with no memory of the previous day.

Just like in the Sleeping Beauty problem, you are unable to distinguish between the first and second day of captivity in safehouse #2. Now, when you find yourself in a safehouse and try to guess the outcome of the coin toss, your best bet would be to assign a 1/3 probability to being in safehouse #1 (and hence the coin toss resulting in heads) and a 2/3 probability to being in safehouse #2 (and hence the coin toss resulting in tails). This is not a reflection of the inherent probabilities of a coin toss, but rather an acknowledgment of your unique epistemic position in the situation, and the contextual information you have access to as a captive. Your credence in each possibility is based on the number of ways in which you could find yourself in your current situation given the possible outcomes of the specific coin toss.

To further bolster the point, imagine that you find a hidden cellphone and have a brief window to send a text message to the police. However, due to character limitations, you can only provide the address of one of the two safehouses. Given your current epistemic position, would it not make sense to communicate the address of safehouse #2? This decision is based on your updated credence in the outcome of the coin toss: you believe there is a 2/3 chance you're in safehouse #2 and a 1/3 chance you're in safehouse #1. So, despite the inherent probabilities of a fair coin toss being 1/2 for each outcome, your unique context and the information you have access to lead you to favor safehouse #2 as the most likely location for your captivity.

Let us now consider the perspective of the police and the information available to them. If they have knowledge of the captor's coin-flip protocol, then prior to receiving any communication, they would have a 1/2 credence of the hostage being in either safehouse.

How does their epistemic position changes after they receive your message? I would argue that we now need to distinguish two cases.

In the first case, if the police have themselves provided the communication means in both safehouses, the message does not convey any new information about the coin toss nor about the hostage's location. Given that the hostage would have been equally likely to find and use the device in either location, the message itself doesn't change the police's prior beliefs.

In the second case, however, if the hostage independently finds a way to communicate, the longer duration spent at safehouse #2 makes it twice as likely for the hostage to send a message from there. This makes a crucial difference. If a message is sent, the police now have evidence that is more likely to occur if the coin landed tails, which should update their credences accordingly, both for the location of the hostage and the result of the coin toss.

The message, in this context, acts as evidence that bears differently on the hypotheses of interest (the coin toss and the location), given the known differences in the likelihoods of being able to send the message from the two locations.

In case-1, since the hostage has an equal chance to communicate from either safehouse, the police are not able to update their credences based on the received message. Here, the police's prior beliefs remain the same, P(H) = P(T) = 1/2, and their strategy should reflect these priors.

In case-2, however, the hostage's ability to communicate is influenced by the duration of captivity. If they get a message, the police can update their beliefs because the message is more likely to be sent if the coin landed tails and the hostage was in safehouse #2 for longer. Their updated beliefs should be P(H) = 1/3, P(T) = 2/3.

Now, consider the question whether the police acting based on the hostage's updated credence rather than their own increases their chances of success, it seems that this depends on the case. In case-1, if the police act based on their own beliefs, they're expected to be right half the time. If they act based on the hostage's belief, they'd be misled half the time (when the coin lands heads). So, in case-1, the police need not update their credence.

In case-2, however, acting based on the hostage's beliefs (which coincide with the updated beliefs of the police if they consider the message as evidence) increases their chances of success. This is because the hostage's ability to communicate is correlated with the coin toss and their location, unlike in case-1. So, in case-2, the police are enabled to update their credence to align them with the hostage's.

The ambiguity stems from the fact that whether the police should act based on the hostage's belief depends on the specific circumstances of each case. A complete answer should specify the case under consideration and fully specify the epistemic perspectives of all the agents.

On edit: I foresee a possible objection to the previous discussion, particularly that it draws upon an analogy with a situation involving a communication device, which has no direct parallel in the original Sleeping Beauty problem. To address this, I propose another scenario that may mirror Sleeping Beauty's epistemic situation even more closely:

Not sure what this is supposed to show? — Michael

It's worth noting that your provided sequence converges on 1/3. If the captive is not keeping track of the date, their credence should indeed be exactly 1/3. The crucial detail here is that the captive gets to guess twice regarding the same coin toss when the result is tails. This very fact is what explains why their credence in being presently in safehouse #1 (and thus, the preceding coin toss resulting in heads) is 1/3 rather than 1/2.

I'd like to draw your attention to the use of the terms "presently" and "preceding" in the proposition's statement. These indexicals are vital as they give us insight into the captive's perspective on the previous coin toss. This subjective epistemic perspective must be distinguished from an external point of view that would consider the next coin toss.

Actually that’s not right. Need to think about this. — Michael

Take you time. I'm being moved to a new safehouse until tomorrow.

I don’t quite understand this example. There are multiple coin flips? — Michael

Yes, my first sentence was wrong. There is a new coin flip every day when the captor must decided on a new (or the same) safehouse. In the case where the coin lands tails, the new decision happens after two days have elapsed (which is analogous to waking up and interviewing Sleeping Beauty twice).

I’ve explained the error with betting examples before. Getting to bet twice if it’s tails doesn’t mean that tails is more likely. — Michael

Suppose you've been kidnapped. Each morning, your captor flips a coin. If it lands on heads, you're blindfolded and taken to safehouse #1 (or simply walked around and returned there). If it lands on tails, you're blindfolded and taken to safehouse #2 (or similarly walked around and returned) for two consecutive days. Both safehouses are indistinguishable, and you're informed of this procedure.

On any given day, what would be your credence in the proposition "I am at safehouse #1"? Would it not be P(#1) = 1/3? You could argue that it's 1/2, with the rationale that "getting to guess twice doesn't mean that tails is more likely". But wouldn't this conflate the epistemic perspective of the person making the guess at the time of guessing with the epistemic perspective of the person flipping the coin at the time of flipping?

I would say that both are true, but also contradictory. Which reasoning it is proper to apply depends on the manner in which one is involved.

For the sitter, his involvement is determined by being randomly assigned an interview, and so I think the first reasoning is proper. For the participant, his involvement is determined by tossing a coin 100 times, and so I think the second reasoning is proper. — Michael

Let us stick with the normal Sleeping Beauty scenario for now, if you don't mind, as I think the lessons drawn will generalize even to your extreme variation but the analysis is simpler.

You're arguing that Sleeping Beauty and the sitter can have different, yet warranted, credences of P(T) at 1/2 and 2/3 respectively. Let's consider this: they could both be regular participants in this experiment, occasionally meeting again by chance. Whenever they meet, they might agree to wager $1 on the outcome—with the sitter betting on tails and John Doe betting on heads.

Under this arrangement, Sleeping Beauty would expect to win two-thirds of the time, and John Doe would expect to lose two-thirds of the time, correct? However, if John Doe's credence P(T) truly is 1/2, should he not expect to break even in the long run? If he still expects the outcome to be tails two-thirds of the time, despite his 1/2 credence, would he not refuse to place a wager? His actions would reveal his true expectations about the probability of the outcome.

They ruled that out before the experiment begun. You might as well say that they can rule out it being the case that the coin landed heads and that this is day 3. — Michael

Before the experiment begins, neither John Doe nor the sitter can rule out a possible future in which we are on Day2 of the experiment and John Doe remains asleep. As soon as John Doe wakes up, they can both rule out the proposition "Today is Day2 of the experiment and the coin landed tails". This is one crucial change to their epistemic situation. They now are able to refer the the current day at issue with the indexical "today" and rationally update their credences by means (self-)reference to it.

Yes. — Michael

So, as they await the interviewer, John Doe and the sitter contemplate the probability that the coin landed tails. The coin might be right there on a nightstand with piece of cardboard covering it. Both John Doe and his sitter are equally informed of the full details of the protocol. After John Doe has woken up, they both have access to exactly the same relevant information. There credences target the proposition: "This coin landed tails". Since they are evaluating the exact same proposition from the exact same epistemic perspective, why don't they have the same credences on your view?

The question is whether or not it is rational for the participant to reason this way. Given that the experiment doesn't work by randomly assigning an interview to them from the set of all interviews, I don't think it is. The experiment works by randomly assigning an interview set from the set of all interview sets (which is either the head set or the tail set), and so I believe it is more rational to reason in this way. — Michael

Do you agree that from the sitter point of view, the probability that the coin landed tails is 2/3?

Why? — Michael

Sorry, I meant to say that he can rule out it being the case the the coin landed heads and that this is Day2.

A more comparable example would be if there are four doors, two containing a goat and two containing a car. You pick a door (say 1) and then Monty opens one of the two doors that contain a goat (say 2). What is the probability that your chosen door contains a car? What is the probability that the car is behind door 3, or door 4? — Michael

In that case, the probability that my chosen door contains a car remains 1/2. The probabilities that a car is behind door 3 or behind door 4 get updated from 1/2 to 3/4 each.

To apply this to the traditional problem: there are two participants; one will be woken on Monday, one on both Monday and Tuesday, determined by a coin toss.

I am one of the participants. What is the probability that I am woken twice?

Do I reason as if I am randomly selected from the set of all participants, and so that I am equally likely to be woken twice, or do I reason as if my interview is randomly selected from the set of all interviews, and so that I am more likely to be woken twice?

Halfers do the former, thirders the latter.

Which is the most rational?

Given the way the experiment is conducted I think the former (halfer) reasoning is the most rational. — Michael

The halfer and thirder responses, as you frame them here, correspond to different questions answered from different epistemic perspectives.

Consider this: let's say you are being hired as an observer in these experiments. Each observer is tasked to attend one session with one participant. The participants are woken up and interviewed once on Day 1, or twice on Day 1 and Day 2, which is determined by a coin toss. As an observer, you are assigned to a randomly chosen participant, and you don't know whether this is their only awakening or one of two.

In the experiment facility, there are, on average, twice as many rooms occupied by participants waking twice (due to tails on the coin toss) as there are rooms with participants waking once (heads on the coin toss). Now suppose you had access to the participant registry where all active participants are listed. You spot the name 'John Doe.' What are the chances he will be woken up twice? You credence is 1/2, and this would also be the case for John Doe's credence before he undergoes the first sleep session.

Now, let's say that by a stroke of luck, you are assigned to John Doe on that particular day. Your job is to measure his vitals as he awakens and get him breakfast as he waits for the interview. You arrive in John's room and wait for him to wake up. What are the chances that the coin toss resulted in tails, indicating this could be one of two awakenings rather than the only one?

Once you have been assigned to John Doe, your credence (P(T)) in this proposition should be updated from 1/2 to 2/3. This is because you were randomly assigned a room, and there are twice as many rooms with participants who wake up twice as there are rooms with participants waking up once. Once John Doe awakens, he can rule out the possibility of it being Day 2 of his participation, and so can you. His credence (P(T)) then aligns with yours because both your credences are targeting the exact same proposition, and both of you have the same epistemic access to it.

I think this is basically a Monty Hall problem. I would say that the probability that I will be put to sleep is 1/2, that the probability that the person to my left will be put to sleep is 1/2, that the probability that the person to my right will be put to sleep is 1/2, and that the probability that one of the other two will be put to sleep is 1/2. — Michael

Your variation of the problem indeed appears to me to contain elements reminiscent of the Monty Hall problem, but with a key difference in the selection process.

In the original Monty Hall scenario, Monty's action of revealing a goat behind a door you didn't choose is non-random. This non-random revelation influences your strategy, prompting you to switch doors. (The probabilities of your initial choice and the remaining door hiding the car that were 1/3 and 1/3, respectively, get updated to 1/3 and 2/3.

Your scenario can be compared to a modified Monty Hall scenario. In this modified scenario, imagine Monty picks one of the unchosen doors at random and reveals a goat by chance. Because this revelation is a random event, it doesn't provide the same targeted information as in the traditional Monty Hall scenario. Consequently, the initial probability estimates (1/3) for your chosen door and the remaining unchosen door hiding a car are updated to 1/2 each. Therefore, you would be indifferent about whether to stick with your original door or switch to the remaining door.

Similarly, in your variation, if a participant is revealed in random way to have been selected to be put to sleep, much like Monty randomly revealing a goat, the probabilities for the remaining three participants (including yourself) would need to be updated similarly. If it's known that exactly one of the remaining three participants had been selected, then the probabilities for each of those three would increase to 1/3.

The double halfer approach does entail:

P(Heads & Monday) = P(Tails & Monday) = P(Tails and Tuesday) = 1/2 — Michael

From the external point of view of the experimenter, it makes sense that the tree probabilities add up to more than one since the three outcomes are non exclusive. The events (Tails & Monday) and (Tails and Tuesday) can (and indeed must) be realized jointly. The intersection zone in the Venn diagram highlights the fact that those two events aren't mutually exclusive. The perspective (and epistemic position) of Sleeping Beauty is different. When she is awoken, those three possible states are mutually exclusive since it is not possible for her to believe that, in her current state of awakening, the current day is both Monday and Tuesday. Her credences in the three mutually exclusive states that she can be in therefore must add up to 1.

Finding out that today is Monday just removes the blue circle. — Michael

I agree with the idea that Sleeping Beauty's credence in H is updated to 1/2 after she learns that her current awakening is occurring on a Monday. The question is what was this credence updated from, before she learned that the current day was Monday. At that point she could not rule out that the current day was Tuesday. Does that not imply that her credence in H was lower than 1/2 before she learned that the day was Monday? Before she learned that, all three mutually exclulsive areas of the Venn diagram represent possible states for her to be in. She therefore should have non-zero credences for each one of them.

I now realize that in the OP's stipulation of the problem, and in line with most discussions of it, it is when the fair coin lands tails that Sleeping Beauty is awoken twice, and else she is awoken once. I all my examples I had assumed the opposite. This is just an arbitrary convention but I hope my unfortunate reversal of it didn't generated any confusion.

In any case, in the future, I'll revert back to the conventional practice.

This isn't comparable to the Sleeping Beaty problem because being a participant isn't guaranteed. That makes all the difference. — Michael

I appreciate your viewpoint, but I can modify my analogies to meet the condition in your first scenario.

Scenario 3 (Lottery study)

Imagine that tickets numbered from one to 2^100 are randomly mixed. We have a waitlist of 2^101 participants. The experimenter selects tickets one by one from the mixed pile. When they draw the 'winning' ticket numbered 2^100, they select (2^100)+1 participants from the waitlist and assign them this ticket. Otherwise, they select just one participant and allocate the losing ticket drawn. This protocol guarantees that the waitlist is exhausted, and everyone gets a chance to participate.

As one of the 2^101 participants, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, as half the participants (plus one) have been assigned this ticket.

Scenario 4 (Sleeping Beauty)

Let's again take tickets numbered from one to 2^100, randomly mixed. We have a waitlist of 2^100 participants. Each participant is assigned a ticket, put to sleep, and the participant with the 'winning' ticket is awoken and interviewed (2^100)+1 times, while everyone else is awoken and interviewed only once.

As one of the 2^100 participants who has just been awoken, your credence that you've been assigned the winning ticket numbered 2^100 should be equal to (or slightly greater than) the probability that you haven't, given that half the participant interviews (plus one) are with participants assigned the winning ticket. Or so I still would argue; here I am just addressing your most recent objection regarding guaranteed participation.

Scenario 5 (Bonus for @fdrake)

Returning to the original Sleeping Beauty problem: upon awakening and awaiting her interview, Sleeping Beauty reflects on her credence that the coin landed on heads. As she checks her smuggled-in cellphone and learns that today is Monday, her halfer position would require her to maintain her credence at 1/2, despite ruling out the possibility of it being Tuesday. The thirder position, on the other hand, allows her to adjust her credence from 2/3 to 1/2 after acquiring this new piece of information. This seems more plausible, and I would like the halfer to explain why ruling out the Tuesday scenario doesn't affect their credence in the coin toss outcome at all.

Regarding betting, expected values, and probability:

Rather than one person repeat the experiment 2^100 times, the experiment is done on 2^100 people, with each person betting that the coin will land heads 100 times in a row. 2^100 - 1 people lose, and 1 person wins, with the winner's winnings exceeding the sum of the losers' losses. The expected value of betting that the coin will land heads 100 is greater than the cost, but the probability of winning is still 1/2 ^100

Even though I could win big, it is more rational to believe that I will lose. — Michael

For sure, but your new variation doesn't mirror the Sleeping Beauty problem anymore. You earlier version was better. We must rather imagine that in the unlikely event that the coin lands heads 100 times in a row, 2^101 persons get pulled from the waitlist (or else, only one person does). In that case, if you are one random person who has been pulled from the waitlist, it is twice as likely that your have been so pulled as a result of the coin having landed heads 100 times in a row than not. It is hence rational for you to make this bet and in the long run 2/3 of the people pulled from the waitlist who make that bet with be right.