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.

She's certainly able to update it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. I'm saying that it's irrational of her to.

The only rational approach, upon waking, is to recognize that it landing heads 100 times in a row is so unlikely that it almost certainly didn't, and that this is her first and only interview. — Michael

She does recognize that for the coin to land 100 times in a row is unconditionally unlikely. But why would it not be rational for her to condition her credence in the truth of this proposition on her knowledge that her awakening isn't an event that she can distinguish in respect of its cause, and that a very large number of such indistinguishable awakening events can stem from such an unlikely cause?

The difference is that the unconditional probability of being called up is very low, and so just being called up at all affects one's credence. In the Sleeping Beauty case (both the normal and my extreme version), she's guaranteed to be awoken either way. — Michael

I can easily adjust my lottery study example such that I am guaranteed to be selected but, once selected, the very (unconditionally) low event that led to my selection (alongside a very large number of people) still is more likely than not to have been the cause of my selection. All that is needed is to shorten the waitlist by about 98%.

There's actually two spaces. See here. — Michael

Yes, I agree with your representation.

Then you have to say the same about my extreme example. Even when she knows that the experiment is only being run once, Sleeping Beauty's credence that the coin landed heads 100 times in a row is greater than here credence that it didn't.

And I think that's an absurd conclusion, showing that your reasoning is false. — Michael

I'm not sure why you think this is absurd. Compare again my lottery study example. Suppose there are one billion people on the waiting list. If a coin lands heads 20 times in a row, then, 100 million people get pulled from the list. Else, one single person gets pulled from the list. I am then informed that I got pulled from the list (but not whether I am alone or one from 100 million). Is it absurd for me to believe that the coin landed heads 20 times in a row? My credence in this proposition should be roughly 99%

In Sleeping Beauty's case, your intuition that her credence in the high probability of the sequence of heads ought to be absurd apparently stems from your unwillingness to contemplate the possibility of her being able to updating it on the basis of her knowledge that she might be awoken an even more absurdly large number of times as a consequence of this very unlikely event. The need for her to update her credence upon being awakened stems from her lacks of power of distinguishing those events in respect of their causes (i.e. the coin flip results). Another reason why you are neglecting the need she has for updating her credence might be a result of your characterizing the experiment that is being run once as starting with a sequence of coin flips. But from Sleeping Beauty's perspective, it really begins with a random awakening.

There is a space of possible awakening/interview events A that are being characterised by the day in which they occur ((M)onday or (T)uesday) and by the state of a coin that has been flipped prior to them occurring ((H)eads or (T)ails). P(H) = P(T) = 0.5. The rest of the conditional dependencies are part of the stipulation of the problem or can be inferred from them.

I never buy betting arguments unless the random variables are set up! — fdrake

They are!

They describe completely different approaches to modelling the problem. That doesn't immediately tell us which SB ought to model the situation as, or whether they're internally coherent. — fdrake

One clue to this is to let SB bet on the outcome that her credence is about and see if her betting behavior leads her to realize the EV she is anticipating.

1. If the experiment is run once, what is Sleeping Beauty's credence that the coin landed heads?

2. If the experiment is repeated several times, what is the probability that a randomly selected interview from the set of all interviews followed the coin landing heads?

Thirders answer the second question, which I believe is the wrong answer to the first question. The experiment doesn't work by randomly selecting an interview from a set of interviews after repeating the experiment several times and then dropping Sleeping Beauty into it. — Michael

I would rather say that the experience works by ensuring that Sleeping Beauty finds herself being awoken in circumstances that she knows to be twice as likely to occur (because twice a frequent) as a result of a coin having landed heads than as a result of this coin having landed tails. This formulation entails no metaphysical queerness.

My reasoning is that P(Awake) = 0.5 given that there are 6 possible outcomes and I will be awake if one of these is true:

1. Heads and I am 1
2. Tails and I am 2
3. Tails and I am 3 — Michael

If you mean P(Awake) to refer to the probability of your being awakened at all (on at least one day) then P(Awake) is indeed 0.5. But in that case P(Awake|Heads) should, consistently with this interpretation, refer to your being awakened at all conditioned on the case where the coin landed heads. This is (1/3+2/3)/2 = 0.5

I don't think it makes sense to say P(Awake) = 3/4. P(Awake) is just the probability that she will be woken up, which is 1. — Michael

But I'm not saying that. What I'm saying is that she is being awoken every Mondays and she is awoken half the time on Tuesdays. So, on average, on a random day, she is being awoken 3/4 times. I am then using Bayes' theorem to deduce the probability of a random awakening having occurred on a Tuesday. But the result was rather trivial.

(I woke up early)

The question which has been eating me is "What is the probability of the day being Tuesday?". I think it's necessary to be able to answer that question for the thirder position. But I've not found a way of doing it yet that makes much sense. Though I'm sure there is a way! — fdrake

P(Tuesday|Awoken) = (P(Awoken|Tuesday) / P(Awoken)) * P(Tuesday)

Sleeping Beauty is awoken with probability 3/4 on an average day (Monday or Tuesday). On Tuesdays, she is awoken with P = 1/2. Therefore, P(Awoken|Tuesday) / P(Awoken) = (1/2)/(3/4) = 2/3.

This (2/3) is the Bayesian updating factor. The unconditioned probability of the day being Tuesday is 1/2. The updated probability therefore is P(Tuesday|Awoken) = (2/3)*(1/2) = 1/3, as expected.

I think you numbers there are wrong. See this. — Michael

In the quoted post you say: "P(Awake|Heads) is just the prior probability that she will be woken up if the coin lands heads"

I think my lottery study analogy suggests a better interpretation of the P(Awoken|Heads)/P(Awoken) Bayesian updating factor. But I must go to sleep now. Thanks for engaging! I'll be back with equal probability on one of my next two awakenings.

Also this makes no sense. You can't have a probability of 2. — Michael

This is not a probability. It's a ratio of probabilities that I have expressed as a ratio of corresponding frequencies. The updated probability P(Heads|Awoken) is 2/3. The quoted ratio being larger than one just reflects the fact that Bayesian updating results in a probability increase in this case.

Being able to bet twice if it lands tails, and so make more money, doesn’t make it more likely that it landed tails; it just means you get to bet twice.

You might as well just say: you can place a £1 bet on a coin toss. If you correctly guess heads you win £1; if you correctly guess tails you win £2.

Obviously it’s better to bet on tails, but not because tails is more probable. — Michael

It makes it twice as likely that individual bets are winning bets. Right? Likewise in Sleeping Beauty's problem, the fact that she is being awoken twice when the coin lands heads makes it more likely that a randomly selected awakening is the result of a coin having landed heads. When she if afforded the opportunity to make one singe bet on any given awakening, her expected value when making this bet is conditioned on the probability that this particular awakening is the result of the coin having landed heads. Do you agree that her expected value for this single bet (in my scenario) is $120? If she would rather judge that the probability for the coin having landed heads is 1/2, she should expect the expected value of her bet to be $90 and would be rationally justified to decline waging $100.

This is a follow-up to my previous post.

How do you condition on such a thing? What values do you place into Bayes' theorem?

P(Heads|Questioned)=P(Questioned|Heads)∗P(Heads) / P(Questioned) — Michael

In the case of Sue's selection to participate in the lottery study, we have

P(Heads|Selected)=P(Selected|Heads)∗P(Heads) / P(Selected)

Since on each fair coin toss, 1.5 participants are being selected on average and when the coin lands on heads 2 participants are selected, P(Selected|Heads) / P(Selected) is 2/1.5 = 4/3.

P(Heads|Selected) therefore is 4/3 * P(Heads) = (4/3)*(1/2) = 2/3

Likewise, in the case of Sleeping Beauty

P(Heads|Awoken) =(P(Awoken|Heads) / P(Awoken)) * P(Heads) = 2/1.5 * 1/2 = 2/3

The simplest "experiment" is just to imagine yourself in Sleeping Beauty's shoes. — Michael

Wasn't that rather the Cinderella problem?

You're inviting us to imagine ourselves in Sleeping Beauty's shoes to support the halfer position. However, there are other ways to imagine her situation that can be argued to support the thirder position. Consider the following thought experiment:

Suppose we introduce a character, Sue, who signs up for a lottery study. She joins a waitlist where participants are selected one or two at a time based on a coin toss: two participants when it lands heads and one when it lands tails. Upon being selected, they're given the option to buy a ticket for $100 that could potentially be worth $180 if the coin had landed heads (or else is worth nothing).

The expected value of the ticket, and whether Sue should purchase it, depends on her credence about the coin toss. If Sue holds the halfer position, her credence is 1/2, and the expected value of the ticket is $90. Hence, she shouldn't buy it. However, if Sue holds the thirder position, her credence in the proposition that the coin landed heads is 2/3, making the ticket's expected value $120. Hence, she should buy it.

Sue could argue for the thirder position as follows: if she has been selected from the waiting list, it is twice as likely that she has been chosen (together with another participant) as a result of the coin landing heads. As a frequent participant in the study, Sue would find, over time, that she profits if she always buys the ticket (an average net gain of $20 per participation), which corroborates the thirder position.

To make this scenario more directly analogous to the original problem, let's imagine that Sleeping Beauty, upon each awakening, can not only express her belief about the coin toss but also place a bet on it. In the long run, she would profit from taking the bet as a thirder, further reinforcing this position.

The pitfall of the 'imagine-yourself-in-her-shoes' argument lies in conflating Sue's perspective with the experimenter's by focusing only on Sue's situation before the coin toss. Eventually, everyone on the waitlist will be selected, just as Sleeping Beauty is guaranteed to be awoken at least once. Her credence that the coin will land heads is 1/2 before being put to sleep, and the same is true for the study participants before they're selected. However, once the coin has been tossed and they've been chosen, their credence about the value of their tickets being $180 (and that the coin landed on heads) should be updated to 2/3. The same applies to Sue's credence that her current awakening was due to a coin landing heads up.

There's the question of whether the "Bivariate Distribution Specification" reflects the envelope problem. It doesn't reflect the one on Wiki. The reason being the one on the wiki generates the deviate (A,A/2) OR (A,2A) exclusively when allocating the envelope, which isn't reflected in the agent's state of uncertainty surrounding the "other envelope" being (A/2, 2A).

It only resembles the one on the Wiki if you introduce the following extra deviate, another "flip" coinciding to the subject's state of uncertainty when pondering "the other envelope": — fdrake

In the Wikipedia article, the problem is set up thus: "Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?"

Your setup for the bivariate distribution specification is a special case of the problem statement and is perfectly in line with it. Let's call our participant Sue. Sue could be informed of this specific distribution, and it would represent her prior credence regarding the contents of the initially chosen envelope. If she were then to condition the Expected Value (EV) of switching on the hypothetical situation where her initially chosen envelope contains $10, the EV for switching, in that particular case, would be positive. This doesn't require an additional coin flip. She either is in the (5, 10) case or the (10, 20) case, with equal prior (and equal posterior) probabilities in this scenario. However, this is just one hypothetical situation.

There are other scenarios to consider. For instance, if Sue initially picked an envelope containing $5, she stands to gain $5 with certainty by switching. Conversely, if she initially picked an envelope with $20, she stands to lose $10 with certainty by switching.

Taking into account all three possibilities regarding the contents of her initially chosen envelope, her EV for switching is the weighted sum of the updated (i.e. conditioned) EVs for each case, where the weights are the prior probabilities for the three potential contents of the initial envelope. Regardless of the initial bivariate distribution, this calculation invariably results in an overall EV of zero for switching.

This approach also underlines the flaw the popular argument that, if sound, would generate the paradox. If we consider an initial bivariate distribution where the potential contents of the larger envelope range from $2 to $(2^m) (with m being very large) and are evenly distributed, it appears that the Expected Value (EV) of switching, conditioned on the content of the envelope being n, is positive in all cases except for the special case where n=2^m. This would suggest switching is the optimal strategy. However, this strategy still yields an overall EV of zero because in the infrequent situations where a loss is guaranteed, the amount lost nullifies all the gains from the other scenarios. Generalizing the problem in the way I suggested illustrates that this holds true even with non-uniform and unbounded (though normalizable) bivariate distributions.

The normalizability of any suitably chosen prior distribution specification (which represents Sue's credence) is essentially a reflection of her belief that there isn't an infinite amount of money in the universe. The fallacy in the 'always switch' strategy is somewhat akin to the flaw in Martingale roulette strategies.

You can conclude either strategy is optimal if you can vary the odds (Bayes or nonconstant probability) or the loss function (not expected value). Like if you don't care about amounts under 20 pounds, the optimal strategy is switching. Thus, I'm only really interested in the version where "all results are equally likely", since that seems essential to the ambiguity to me. — fdrake

If we assume that all results are equally likely, the EV of switching given that the chosen envelope was seen to contain n is (2n + n/2)/2 - n = 1.5n. Hence whatever value n might be seen in the initially chosen envelope, it is irrational not to switch (assuming only our goal is to maximize EV). This gives rise to the paradox since if, after the initial dealing, the other envelope had been chosen and its content seen, switching would still be +EV.

As I wrote, the prior probabilities wouldn't be assigned to the numbers (5,10,20), they'd be assigned to the pairs (5,10) and (10,20). If your prior probability that the gameshow host would award someone a tiny amount like 5 is much lower than the gigantic amount 20, you'd switch if you observed 10. But if there's no difference in prior probabilities between (5,10) and (10,20), you gain nothing from seeing the event ("my envelope is 10"), because that's equivalent to the disjunctive event ( the pair is (5,10) or (10,20) ) and each constituent event is equally likely — fdrake

I did indeed first assigned priors to the two cases—(5, 10) and (10, 20)—and only derivatively calculated priors regarding the possible contents of the first chosen envelope (or of the other one).

Edit: then you've got to calculate the expectation of switching within the case (5,10) or (10,20). If you specify your envelope is 10 within case... that makes the other envelope nonrandom. If you specify it as 10 here and think that specification impacts which case you're in - (informing whether you're in (5,10) or (10,20), that's close to a category error. Specifically, that error tells you the other envelope could have been assigned 5 or 20, even though you're conditioning upon 10 within an already fixed sub-case; (5,10) or (10,20).

The conflation in the edit, I believe, is where the paradox arises from. Natural language phrasing doesn't distinguish between conditioning "at the start" (your conditioning influencing the assignment of the pair (5,10) or (10,20) - no influence) or "at the end" (your conditioning influencing which of (5,10) you have, or which of (10,20) you have, which is totally deterministic given you've determined the case you're in).

I am not conditioning within an already fixed sub-case. Seeing that there is $10 in one envelope still leaves it open that there might be $5 or $20 dollars in the other one. To be clear, you seems to be emphasizing the role of conditioning and timing, arguing that my observation of the $10 should not affect my decision about whether I'm in the (5,10) or (10,20) case. You appear to argue that, by the time I've observed the $10, the case is already determined, and I can't switch between cases based on my observation.

However, it seems that you've misunderstood my use of Bayesian updating. I am not arguing that observing the $10 allows me to switch between cases. Rather, I'm saying that, given an observation of $10, I can update my beliefs about the probability of being in the (5,10) case or the (10,20) case.

(Some further discussion with GPT-4)