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.

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.

Once John Doe awakens, he can rule out the possibility of it being Day 2 of his participation, and so can you. — Pierre-Normand

Why?

Once you have been assigned to John Doe, your credence (P(H)) 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. — Pierre-Normand

Yes, if you are randomly assigned an interview from the set of all interviews then the probability of it being a tails interview is greater than the probability of it being a heads interview.

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.

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.

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?

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

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.

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

Yes.

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?

Consider my extreme example. 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, 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.

We might want to measure which reasoning is proper by appeals to bets or expected values or success rate or whatever, but then there are two ways to reason on that:

1. $2\over3$ of all guesses are correct if every guess is that the coin landed heads 100 times in a row
2. $1\over2^{100}$ of all participants are correct if they all guess that the coin landed heads 100 times in a row

How do we determine which of these it is proper to apply?

So maybe there is no right answer as such, just more or less proper (or more or less compelling). And all I can say is that if the experiment is being run just once, and I am put to sleep and then woken up, my credence that the coin landed heads 100 times in a row would be $1\over2^{100}$.

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.

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.

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.

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 don’t quite understand this example. There are multiple coin flips and no amnesia?

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).

Then on the first day P = 1/2, the second day P = 1/4, the third day P = 1/8, etc.

Actually that’s not right (starting third day). Need to think about this. First two days are right though.

Not sure how this is at all relevant though.

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

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

First day P = 1/2, second day 1/4, third day 3/8, fourth day 5/16, etc.

Not sure what this is supposed to show?

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.

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. — Pierre-Normand

I don't think this is relevant to the Sleeping Beauty problem. It's a different experiment with different reasoning.

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.

Also it converges to 1/3 only as you repeat the coin tossing, whereas in the traditional problem the coin is only tossed once.

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:

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. — Pierre-Normand

That's the very point I disagree with, and is most evident with the example of tossing a coin 100 heads in a row. The possible outcomes have no bearing on my credence that the coin landed heads 100 times in a row. The only thing I would consider is that the coin landing heads 100 times in a row is so unlikely that it almost certainly didn't happen, and I think any rational person would agree.

This is less clear to see with the Sleeping Beauty problem given that heads and tails are equally likely, and so prima facie it doesn't matter which you pick, but given that there are two opportunities to win with tails there's no reason not to pick tails.

If you don't like to consider my extreme example because the numbers are too high then let's consider a simpler version. Rather than a coin toss it's a dice roll. If 1 - 5 then safehouse 1 with crocodiles for one day, if 6 then 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.

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 always reason as if it's tails).

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.

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. — Pierre-Normand

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, another coin toss each day determines if she will be woken (heads she will, tails she won't).

So the probability that the coin landed heads and she wakes on Monday is 1/4, the probability that the coin lands tails and she wakes on Monday is 1/4, and the probability that the coin lands tails and she wakes on Tuesday is 1/4. A simple application of Bayes' theorem is:

$\begin{aligned}P(Heads | Questioned) &= {{P(Questioned | Heads) * P(Heads)} \over P(Questioned)}\\&={{{1\over2}*{1\over2}}\over{2\over3}}\\&={3\over8}\end{aligned}$

As compared to the normal situation which would be:

$\begin{aligned}P(Heads | Questioned) &={{1*{1\over2}}\over1}\\&={1\over2}\end{aligned}$

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.

And with this variation, do you not agree that the probability of it being heads is 3/8? Would you not also agree that the probability of it being heads in this scenario must be less than the probability of it being heads in the traditional scenario, where being woken up on your assigned day(s) is guaranteed? If so then it must be that the probability of it being heads in the traditional scenario is greater than 3/8, i.e. 1/2.

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.

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. — Pierre-Normand

This is heading towards a betting example, which as I've explained before is misleading. There are three different ways to approach it:

1. The same participant plays the game 2¹⁰⁰ times. If they bet on 100 heads then eventually they will win more than they lose, and so it is rational to bet on 100 heads.

2. 2¹⁰⁰ participants play the game once. If they bet on 100 heads then almost everybody will lose, and so it is rational to not bet on 100 heads (even though the one winner's winnings exceed every losers' losses).

3. One participant plays the game once. If they bet on 100 heads then they are almost certain to lose, and so it is rational to not bet on 100 heads.

Given that the very premise of the experiment is that it is to only be run once, a rational person would only consider 3.

And I really don’t see any counterexamples refuting 3. I will never reason or bet that 100 heads in a row is more likely.

But even if the same participant were to repeat the experiment 2¹⁰⁰ 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.

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.