A slightly different example. If the coin lands heads then Sleeping Beauty is woken on Monday and kept asleep on Tuesday. If the coin lands tails then on Monday and Tuesday a coin is tossed and Sleeping Beauty is only woken if the coin lands heads.

This gives us:


$\begin{aligned}P(Heads|Awake) = {{P(Awake|Heads) \cdot P(Heads)} \over P(Awake)}\end{aligned}$

Using halfer reasoning (each row is the outcome) we have:

$\begin{aligned}P(Heads|Awake) = {{1 \cdot {1\over2}} \over {7\over8}} = {4\over7}\end{aligned}$

Using thirder reasoning (each cell in the second and third columns is the outcome) we have:

$\begin{aligned}P(Heads|Awake) = {{{1\over2} \cdot {1\over2}} \over {1\over2}} = {1\over2}\end{aligned}$

Given that a fair coin toss is equally likely to be heads as tails and that she is less likely (not guaranteed) to wake if tails then if she does wake then she reasons that it's less likely to be tails. In waking she rules out TTT.

The answer of $4\over7$ seems correct, even though there will be an equal number of heads awakenings as tails awakenings after repeated runs.

We have two different experiments:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, A and B once each if tails

Your version of the experiment is comparable to the second experiment, not the first. — Michael

Um, no.

In "my experiment" I will literally and explicitly wake the single subject once if coin C1 lands on Heads, and twice if it lands on Tails. And there literally and explicitly is no second subject. So it is an exact implementation of 1, not 2.

What you are doing here, is confusing the change in the details that determine how this single subject will be awakened, with her being a different subject. You name the different subjects "subject A" and "subject B." Well, the same thing can be done with Elga's ("the most frequent") implementation:

3. Wake subject C once on Monday if Heads, or wake subjects C and D once each (on Monday and then Tuesday, respectively) if tails.

And the reason that neither 2, nor 3, fits the intended problem is that the subject has to know she will be wakened, but not remember whether it happens one or two times. So that she is the same person, but does not know if she is in "world" (I don't like this word here, but it is how philosophers approach the problem) A, B, C or D.

AGAIN:

The controversy created by Elga's ("the most frequent") implementation, of the same problem that I implement, is this:

• The incarnation of the single subject does not know if she is in world C or D.
• This prevents her from defining a sample space that applies to just her world.
• So she has to combine them into a single world
• Halfers do it by saying they are the same world as the world that contains both, so the single probability space that applies to the combination applies to each individually.
• Thirders do it by saying they are separate worlds, requiring a joint probability space and an assessment of which world she might be in.

And the way my implementation solves this is by creating a simple probability space that applies to just world A, or to just world B, so it no longer matters which world the subject is in.

In "my experiment" I will literally and explicitly wake the single subject once if coin C1 lands on Heads, and twice if it lands on Tails. And there literally and explicitly is no second subject. So it is an exact implementation of 1, not 2. — JeffJo

You toss two coins and don’t ask them their credence if both land heads. That’s what makes your experiment equivalent to my second example where B isn’t asked if heads.

To provide a proper analogy to Sleeping Beauty you must have it that the participant is guaranteed to be asked her credence at least once. That fact is why the answer is 1/2 and not 1/3.

The answer is only 1/3 when there’s a 1/4 prior probability of never being asked your credence.

But haven't you lost Sleeping Beauty's other constraint, that the chances of encountering one Italian or two Tunisians are equal? — Srap Tasmaner

In my original cosmopolitan analogy, the equal Italian and Tunisian populations mirrors the even likelihood of the coin landing on either side in the Sleeping Beauty problem. What makes it more likely to encounter a Tunisian—despite the equal population—is that Tunisians go for walks twice as often on average, increasing the odds of an encounter. This mirrors how Sleeping Beauty is woken up twice when the coin lands tails.

To fine-tune the analogy and preserve the feature of the Sleeping Beauty problem you've pointed out, we can assume that initially, you're equally likely to encounter an Italian or a Tunisian—perhaps because Tunisians walk in hidden pairs. When you meet a member of a Tunisian pair for the first time, their sibling ensures they are the next one you meet. Thus, when you have met an Italian, or a Tunisian for the second time in a row, your next encounter is equally likely to be with an Italian or a Tunisian, analogous to the Sleeping Beauty problem where a new coin toss (and a new Monday awakening) occurs after each heads or second tails awakening. Despite this, two-thirds of your encounters are with Tunisians, so the odds that any given encounter is with a Tunisian remain 2/3. (We can assume that the experiment begins with a small number of random "dummy" encounters to ensure that you lose track of the first "experimental" encounter.)

That doesn't mean that the credence isn’t transitive. My premises "fail" to account for it because it's irrelevant.

A iff B
P(B) = 1/2
Therefore, P(A) = 1/2 — Michael

While this kind of inference is often valid, it doesn't apply in the Sleeping Beauty problem.

Credences, or probabilities, can be thought of as ratios. My credence at a given time for an outcome O reflects the proportion of cases where O occurs in a similar situation S. In other words, it is the "ideal" or "long run" ratio O/S. For instance, my credence that a randomly drawn card from a shuffled deck is a spade, P(Spade) = 1/4, reflects my belief that the ratio of spade outcomes to card-drawing situations is 1/4.

The general validity of the inference you propose is based on the assumption that the mapping between O and S is constant. However, this mapping is contentious in the Sleeping Beauty problem, with Halfers and Thirders disagreeing, resulting in conflicting interpretations of P(Heads).

As long as Halfers and Thirders stick to their own definitions, this isn't a problem—though it can lead to miscommunication. Being aware of these divergent definitions also helps avoid invalid inferences.

Let's take A as Sleeping Beauty being in a H-awakening episode and B as her being in a H-run. While A iff B holds true, note that:

P(B) = 1/2 = B/O, where O represents a representative set of experimental runs.

P(A) = 1/3 = A/O', where O' represents a representative set of awakening episodes.

Equating P(B) and P(A) and inferring one from the other can only be valid if O remains constant—in other words, if the mapping from potential outcomes to potential situations doesn't change.

You toss two coins and don’t ask them their credence if both land heads. That’s what makes your experiment equivalent to my second example where B isn’t asked if heads. — Michael

I do not ask anybody (for) their credence if both coins landed on Heads. I don't ask anybody about coin C2 at all, although it has to be taken into account.

I ask for a credence in whether coin C1 is currently showing Heads. And the way the problem is set up, that is always the same event where coin C1 landed on Heads. Maybe you do not realize that coin C1 is the coin in the problem? And coin C2 is not?

C2 is just the coin that controls ordering. Since amnesia makes ordering irrelevant to the single subject, what coin C2 is showing cannot affect the answer.

That’s what makes your experiment equivalent to my second example where B isn’t asked if heads. — Michael

There is no B anywhere, as far as I can tell. You don't seem to want to explain the important details, like whether B is a person, a person in a different situation, or (as it seems here) if B is an event that is not a part of the experiment.

AGAIN: I use a single subject. That subject is always wakened, but may be wakened twice. That subject is always asked "if heads" about the only coin that the original problem is concerned with, (It's even possible we could pose the question while she sleeps, but we shouldn't expect an answer.) When she can answer, she knows that there are three equally likely combinations for what the two coins are showing, and in only one is coin C1 showing Heads.

This is not difficult. But you do need to stop trying to contrive a situation where it is wrong. So far, you have not even described a situation that applies.

I do not ask anybody (for) their credence if both coins landed on Heads. — JeffJo

Exactly. It is precisely because the prior probability of being asked at least once is 3/4 that the probability that the first coin landed heads is 1/3.

If the prior probability of being asked at least once is 1 then the probability that the first coin landed heads is 1/2.

This is why your example isn’t comparable to the traditional problem and is comparable to my second example where the prior probability of being asked at least once is 3/4.

My credence at a given time for an outcome O reflects the proportion of cases where O occurs in a similar situation S. — Pierre-Normand

I know that it is. I'm trying to show that it shouldn't be. Such reasoning is only correct where the situation is such that your outcome is randomly selected from the set of all outcomes, which isn't the situation with Sleeping Beauty. First her interview set is randomly selected from the set of all interview sets and then her interview is randomly selected from the set of all interviews in her set. This is how the experiment is actually conducted.

It just doesn't make sense to say that A iff B but that P(A) != P(B). And Bayes' theorem shows that P(A) = P(B).

I do not ask anybody (for) their credence if both coins landed on Heads. — JeffJo

Exactly. It is precisely because the prior probability of being asked at least once is 3/4 that the probability that the first coin landed heads is 1/3. — Michael

The original problem is about one coin, not two. Asking about two would make it a different problem. Asking about one is what makes it the same problem.

But yes, it is indeed true that the prior probability of 3/4 is what makes the answer 1/3. You identified the wrong event for that prior probability (she is always asked), but it is the fact that this same prior probability applies to any waking, and not different prior probabilities depending on whether the subject is wakened on Monday or Tuesday, that makes it usable in a valid solution.

Thank you for stating, in your own words, why this is so.

The original problem is about one coin, not two. Asking about two would make it a different problem. Asking about one is what makes it the same problem.

But yes, it is indeed true that the prior probability of 3/4 is what makes the answer 1/3. But it is the fact that this same prior probability applies to any waking, and not different prior probabilities depending on whether the subject is wakened on Monday or Tuesday, that makes it usable in a valid solution.

Thank you for stating, in your own words, why this is so. — JeffJo

In the original problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1, which is why the answer is 1/2 and why your example isn't comparable.

In your example and my second example the prior probability of being asked one's credence at least once is 3/4 and the prior probability of being asked one's credence at least once if heads is 1/2, which is why the answer is 1/3.

In the original problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1, which is why the answer is 1/2 and why your example isn't comparable. — Michael

So now it isn't that I never asked about two coins ("You toss two coins and don’t ask them their credence if both land heads. That’s what makes your experiment equivalent to my second example where B isn’t asked if heads.")?

Do I need to explain "my version" to you again? You now say it is different because:

1. In the original problem the prior probability of being asked one's credence at least once is 1 ...
   
   • In mine, the prior probability of being asked ones credence (that coin C1 is showing Heads) at least once is also 1.
2. ... and the prior probability of being asked one's credence at least once if heads is 1 ...
   
   • In mine, the prior (?) probability of being asked one's credence at least once if Heads is also 1
   • But this doesn't appear to be a prior probability. That seems to be what "if Heads" means.

So you are still describing how it is the same problem, while claiming that it is different.

... which is why the answer is 1/2 and why your example isn't comparable.

Non sequitur.

Please, show me how

• The existence of two credence=1 events...
• ... which, while the facts that they are certain are valid, do exist as an issue in the problem,

... produce the conclusion "the answer is 1/2."

Do you want to try again? But do try to realize that probabilities used within a solution cannot affect whether the problems are the same. They can only affect the solutions, and credence=1 events do not do that.

These are two different problems:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, A and B once each if tails

In the first problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1. The answer to the problem is 1/2.

In the second problem the prior probability of being asked one's credence at least once is 3/4 and the prior probability of being asked one's credence at least once if heads is 1/2. The answer to the problem is 1/3.

In your problem two coins are tossed, and only if at least one coin is tails am I asked my credence that the first coin is heads. The prior probability of being asked one's credence at least once is 3/4 and the prior probability of being asked one's credence at least once if heads is 1/2. The answer to the problem is 1/3.

The first problem is the Sleeping Beauty problem. Your problem isn't analogous. Your problem is analogous to the second problem which gives a different answer.

It just doesn't make sense to say that A iff B but that P(A) != P(B). And Bayes' theorem shows that P(A) = P(B).

I have indeed conceded that the inference is valid (as are the applications of Bayes' theorem predicated on it) as long as we avoid equivocating the meaning of P(). To prevent this equivocation, we must be mindful of the specific ratio implicitly referred to when we discuss Sleeping Beauty's credence P(H). It's important to ensure that, when you lay out your demonstrations, you do not switch between two inconsistent definitions of P(), even within the same formula.

Consider again the pragmatic dice scenario where Sleeping Beauty is awakened six times in the East Wing if the die lands on 'six', and awakened once in the West Wing otherwise. It's rational for her to instruct her Aunt Betsy to wait for her at the West Wing exit, because once the experimental run concludes, the odds of her exiting there are P(not-'six') = 5/6. This also implies that P(not-'six'-awakening) is 5/6, if we understand it to mean that in five out of six potential runs of awakenings she awakens into, she finds herself in not-'six' runs (regardless of the number of times she awakens in that run). However, what does not logically follow is that P'(not-'six') = 5/6, if we interpret this to mean that in five out of six potential awakening episodes, she finds herself in not-'six' episodes. The relevant ratio in this context is P'(not-'six') = 6/11. This ratio is also the relevant one for her to predict from which wing she would likely exit from if she had a chance to escape during any given awakening episode.

These are two different problems:

1. A is woken once if heads, twice if tails
2. A is woken once if heads, A and B once each if tails — Michael

Yes they are.

#1 has two subjects, and #2 has two.

In #1, the only subject knows she will be wakened, just not how many times. Her credence in Heads is what we are asked about. Since both Heads and Tails are possible when she is awake, 0<Pr(Heads|Awake)<1.

In #2, A knows she will be wakened and that the coin is irrelevant. So Pr(Heads|Awake)=Pr(Heads)=1/2. B knows that she will only be wakened if Heads. So Pr(Heads/Awake)=1.

In my version, there is one subject who knows she will be wakened, just not how many times. Please, without referring to how you you would solve the problem once it is established, how is this different than #1? And how does the presence of B in #3 make it like mine?

In #2, A knows she will be wakened and that the coin is irrelevant. So Pr(Heads|Awake)=Pr(Heads)=1/2. B knows that she will only be wakened if Heads. So Pr(Heads/Awake)=1. — JeffJo

Neither participant knows if they are A or B.

In my version, there is one subject who knows she will be wakened, just not how many times. — JeffJo

She doesn't know that. If both coins land heads then she's not asked her credence.

However, what does not logically follow is that P'(not-'six') = 5/6, if we interpret this to mean that in five out of six potential awakening episodes, she finds herself in not-'six' episodes. The relevant ratio in this context is P'(not-'six') = 6/11. — Pierre-Normand

This has nothing to do with credence.

I am asked to place two bets on a single future coin toss. If the coin lands heads then only the first bet is counted. What is it rational to do? Obviously to bet on tails, even though my credence isn't that tails is more likely. The same principle holds in the Sleeping Beauty experiment where I'm put to sleep and woken up either once or twice depending on a coin toss.

In both cases the fact that there are twice as many T-wins as H-wins has nothing to do with the likelihood of the coin having landed tails. There are twice as many T-wins as H-wins because the experimental setup says that if it's heads then you get one bet and if it's tails then you get two bets, and heads and tails are equally likely.

That in many cases there are twice as many A-outcomes as B-outcomes because A is twice as likely isn't that in every case if there are twice as many A-outcomes as B-outcomes then A is twice as likely. There are plenty of other factors that can contribute to an outcome ratio that doesn't reflect the likelihood of each outcome. The Sleeping Beauty problem is one such example.

I have indeed conceded that the inference is valid (as are the applications of Bayes' theorem predicated on it) as long as we avoid equivocating the meaning of P(). — Pierre-Normand

I think there's only one meaning of P(), and it is such that P(HInterview) = P(HRun) = 1/2.

It is a mistake to reason that P(X) should reflect the ratio of X to not-X, as explained above.

This ratio is also the relevant one for her to predict from which wing she would likely exit from if she had a chance to escape during any given awakening episode.

If on each day the chance to escape is 1/2 then the prior probability of being given at least one chance to escape if the dice rolls 1-5 is 1/2 and if the dice rolls 6 is 63/64. Given the prior probability that I'm more likely to be given at least once chance to escape if the dice rolls 6 it is reasonable to infer that if I am given a chance to escape that the dice most likely rolled 6.

This is the same reasoning used here to infer that one's credence in heads should be greater in experiment 1 than in experiment 2 and used here to infer that one's credence should favour heads.

Neither participant knows if they are A or B. — Michael

I'm going to ignore the fact that neither A nor B is woken twice, so this isn't the SB problem. What you seem to mean is that the subject is woken once as A if Heads, and once each as A and as B if tails.

Then you have created a third problem that is the equivalent of #1. Literally, it just adds a certain detail about names that is significant in #2 when they apply to different people, but makes it identical to #1 when it is the same person that can have either name. And an event it uses to choose the name - Heads or Tails - is already mentioned in the problems:

1. A is woken once if Heads, twice if Tails.
2. A is woken once if Heads, A and B once each if Tails.
3. SB if woken once in room 100A if Heads, or once each in of the rooms 100A and 100B if tails.

My version is indeed equivalent to both #1 and #3. It is not equivalent to #2 if A and B are different people, which was not given in its definition. In the special case where that is specified, all three are the same problem.

What matters is that in the Sleeping Beauty problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1, and in your problem the prior probability of being asked one's credence at least once is 3/4 and the prior probability of being asked one's credence at least once if the first coin is heads is 1/2.

Because of this your problem is not equivalent. The answer to your problem is 1/3 only because of those prior probabilities. The Sleeping Beauty problem has different prior probabilities and so a different answer of 1/2.

This has nothing to do with credence.

I am asked to place two bets on a single coin toss. If the coin lands heads then only the first bet is counted. What is it rational to to? Obviously to bet on tails. Even though my credence isn't that tails is more likely. The same principle holds in the Sleeping Beauty experiment where I'm put to sleep and woken up either once or twice depending on a coin toss. That it's rational to bet on tails isn't that my credence is that it's most likely tails; it's that I know that if it is tails I get to bet twice.

The same principle holds with the dice roll and the escape attempts. — Michael

I find it unusual that you maintain that when faced with a potential outcome O in a situation S, your credence P(O) should only reflect the intrinsic propensity of an object to generate O, disregarding how O affects the likelihood of you being in this situation.

Consider a scenario where fireflies are equally likely to have gene XYZ, which makes them brighter and thus twice as likely to be seen from the same distance. If you happen to be in proximity to such a firefly, it is twice as likely to catch your attention when it has the XYZ gene. Therefore, from a population where half of the fireflies have this gene, you witness twice as many flashes from the ones carrying XYZ. According to your logic, your credence about any given firefly flash should remain P(XYZ) = 1/2 (because the firefly generating it had a 50% chance of inheriting this gene), despite the fact that you would have twice as many betting opportunities on fireflies with the XYZ gene. You seem to consider this increase in betting opportunities irrelevant to your credence P(XYZ), even though your encounters with such fireflies are twice as frequent.

This line of reasoning appears to be an ad hoc restriction on the common understanding of credence, primarily designed to disqualify the Thirder interpretation of the Sleeping Beauty problem from the outset. This restriction seems to have limited applicability outside of this specific problem. In most cases, we focus more on the overall frequency of the outcomes in proportion to the relevantly similar situations, rather than on the intrinsic propensities of the objects generating these outcomes.

I find it unusual that you maintain that when faced with a potential outcome O in a situation S, your credence P(O) should only reflect the intrinsic propensity of an object to generate O, disregarding how O affects the likelihood of you being in this situation. — Pierre-Normand

That's not what I said.

In the Sleeping Beauty problem I am guaranteed to wake up at least once if tails and guaranteed to wake up at least once if heads. The coin toss does not determine the likelihood of me waking up. It only determines the number of times I'm woken up. But this frequency is irrelevant given the guarantee.

Only if there is no guarantee is the frequency relevant. Again, see this and this.

What matters is that in the Sleeping Beauty problem the prior probability of being asked one's credence at least once is 1 and the prior probability of being asked one's credence at least once if heads is 1 — Michael

What has no significance to the SB problem, is what might be different if event X, or event Y, happens WHEN BOTH ARE CERTAIN TO HAPPEN.

• X = The prior probability of being asked one's credence at least once, equal to 1.
• Y = The prior probability of being asked one's credence at least once if heads, also equal to 1.

Now, you are either a troll, very confused, or expressing yourself poorly. But if you won't consider that I might be right (as I keep assuming about you, until I actually find something wrong) there is no way to discuss anything with you.

In your example being asked your credence isn't certain. In Sleeping Beauty's it is. That's why your example isn't equivalent.

Pierre-Normand also tried to explain this to you here.

Probability=1 means "is certain." You said the probability of being asked was 1. That means certain. The subject in my implementation is always asked.

"Prior probability" means "prior to information being given," so "prior probability...if Heads" is vacuous.

I have no idea what you mean by "my example." My implementation is an exact version of your problem #1. I notice that you don't claim otherwise, you just try to say that #2 is different and I implemented #2. #3 is closer, and it is equivalent to #1.

My explanation of how your additional constraints turned #2 into #3 is trivially true. As is how #3 is equivalent to #1.

What credence is asked for does not affect, in any way, how the events themselves might occur. You are arguing with non sequiturs.

All you have identified is how my solution differs from yours, not how my implementation of the problem differs from the actual problem. This is because it is trivially obvious that it does not. But you can't attack my solution, which is also trivially correct, so you attack the implementation.

That's not what I said.

In the Sleeping Beauty problem I am guaranteed to wake up at least once if tails and guaranteed to wake up at least once if heads. The coin toss does not determine the likelihood of me waking up. It only determines the number of times I'm woken up. But the frequency is irrelevant. The only thing that matters is the guarantee. — Michael

When Sleeping Beauty awakens, she could potentially be experiencing either a guaranteed awakening (i.e. T-Monday or H-Monday) or an optional awakening (i.e. T-Tuesday). Since she cannot definitively rule out the possibility of experiencing an optional awakening, this uncertainty should affect her credence P(T), as P(T) = P(T-Monday) + P(T-Tuesday), and P(T-Monday) is always equal to P(H-Monday) regardless of the value of P(T-Tuesday). Therefore, P(T) should be greater than 1/2 whenever Sleeping Beauty cannot conclusively rule out the possibility of it being T-Tuesday.

If we apply your reasoning to the example here then we conclude that P(Heads|Awake) = 1/2, which I think is wrong.

I'm less likely to wake if tails and so if I do wake it's less likely to be tails, and so P(Heads|Awake) > P(Tails|Awake).

This is satisfied only if we use halfer reasoning and conclude that P(Heads|Awake) = 4/7.

@Michael

Let me adjust my previous firefly case to meet your objection.

We can assume that half of the fireflies have gene XYZ, which causes them to flash twice every five minutes. The other half, lacking gene XYZ, flash once every five minutes.

A witness can see every flash and thus is guaranteed to see the first flash of every firefly. The second flash, however, is optional as it depends on the firefly having gene XYZ. This mimics the guaranteed and optional awakenings in the Sleeping Beauty problem.

When the witness sees a flash, they know it could either be a first flash (which is guaranteed from every firefly) or a second flash (which is optional and only comes from the fireflies with gene XYZ).

Just like in the Sleeping Beauty problem, every flash is an 'awakening' for the witness. The presence of gene XYZ is akin to a coin landing tails (T), leading to an optional second flash (analogous to the T-Tuesday awakening).

Upon witnessing a flash, the observer's credence that they're seeing a firefly with gene XYZ should be more than 1/2, as the witness cannot conclusively rule out that it's a second, optional flash. This aligns with the reasoning that P(T) > 1/2 for Sleeping Beauty when she cannot rule out the possibility of T-Tuesday.

This analogy illustrates how an increased frequency of a particular event (the witnessing of a second flash, or T-Tuesday) can impact overall credence.

In your scenario there are a bunch of flashes going off in a forest and me, a passer-by, randomly sees one of them. This is comparable to a sitter being assigned a room.

But this isn't how things work for Sleeping Beauty. It's not the case that her interview is randomly selected from the set of all interviews. It's the case that her interview set is randomly selected from the set of all interview sets. That's just how the experiment works and so is how she should reason.

In your scenario there are a bunch of flashes going off in a forest and me, a passer-by, randomly sees one of them. This is comparable to a sitter being assigned a room. — Michael

The passer-by sees all of the flashes and does not know the genetic status of the fireflies producing them. This is analogous to Sleeping Beauty experiencing all of her awakenings but not knowing if they're unique (generated by a coin having landed heads) or one of a series of two (generated by a coin having landed tails). The witness's credence P(XYZ) = 2/3 applies to all of the flashes they are witnessing, just like Sleeping Beauty's credence P(T) = 2/3 applies to all the awakenings she is experiencing, not merely to random samplings of them.

What is true of a random sampling of these awakening episodes (or flash sightings), due to the fact that the sampling would represent the relevant frequencies, is even more applicable to the total population of awakening episodes. However, in the latter case, no additional sampling method (nor the presence of a randomly assigned sitter) is required.

The passer-by sees all of the flashes and does not know the genetic status of the fireflies producing them. This is analogous to Sleeping Beauty experiencing all of her awakenings but not knowing if they're unique (generated by a coin having landed heads) or one of a series of two (generated by a coin having landed tails). — Pierre-Normand

The passer-by can see a mix of single and double-flashing fireflies. Sleeping Beauty can't. She either sees one firefly flash once or she sees one firefly flash twice (forgetting if she's seen the first flash).

All you've imagined here is that we collect several iterations of the Sleeping Beauty experiment, mix up all the interviews, and then have Sleeping Beauty "revisit" them at random. That's not at all equivalent. This is equivalent to a sitter being assigned a room and I already accepted that for the sitter the answer is 1/3.