I accept 6 and reject 5. My credence that my current interview is a heads interview isn't equal to the fraction of interviews which are heads interviews.

My argument is:

P1. My credence is the degree to which I believe that a proposition is true
P2. My current interview is a heads interview iff I have been assigned one heads interview
C1. Therefore my credence that my current interview is a heads interview is equal to my credence that I have been assigned one heads interview (from P1 and P2)
P3. If I have been assigned at random by a fair coin toss either one heads interview or two tails interviews then the probability that I have been assigned one heads interview is 1/2
P4. I have been assigned at random by a fair coin toss either one heads interview or two tails interviews
C2. Therefore the probability that I have been assigned one heads interview is 1/2
(from P3 and P4)
P5. My credence that I have been assigned one heads interview is equal to the probability that I have been assigned one heads interview
C3. Therefore my credence that I have been assigned one heads interview is 1/2
(from C2 and P5)
C4. Therefore my credence that my current interview is a heads interview is 1/2
(from C1 and C3) — Michael

The issue arises from a conflation of two distinct ways of individuating events and counting probabilities. We can see this more clearly if we distinguish between the 'timeline perspective' and the 'episodic perspective'. Your propositions P1 through P4 and C1 though C4 above frequently shift between those two perspectives, which vitiates the validity of some inferences.

Let's consider the shift from the 'timeline perspective' to the 'episodic perspective'. In the timeline perspective, there are two possibilities ("possible worlds"): an H-timeline and a T-timeline, each with an equal chance of 1/2. The T-timeline, however, comprises two distinct awakening episodes ("centered possible worlds"). This does not create more exclusive events sharing the probability space; rather, it articulates the unique structure of the T-timeline.

Therefore, when shifting to the episodic perspective, it would be a mistake is to divide the probability of the T-timeline (1/2) between the two T-awakenings, suggesting each has a probability of 1/4. This line of thinking presumes these awakenings to be exclusive events within the T-timeline (as they might be if a second coin would be tossed to determine which one of the two would occur). However, that's not the case. In the T-timeline, the two awakenings are guaranteed to occur concurrently if at all; the only unknown is which of them Sleeping Beauty currently finds herself in.

The shift from the timeline perspective to the episodic perspective is not a straightforward Bayesian update on new information. Instead, it's a shift in how we count the alternatives. This shift happens automatically when Sleeping Beauty awakes, because she can't tell apart the two T-episodes and what were concurrent possibilities become exclusive possibilities.

Once we've dealt with the faulty reasoning that made it appear like the T-first-awakening and T-second awakening were lowered from 1/2 to 1/4 when shifting to the episodic perspective, we can now see how the equiprobability between the H-awakening and T-first-awakening must also be retained when we shift perspectives. When Sleeping Beauty wakes up and doesn't know whether it's Monday or Tuesday, this doesn't change the equiprobability of a H-awakening or a T-first-awakening that she would express her credence in were she to know that it's Monday. Instead, her ignorance about the day of the week introduces an additional possibility—that of her being in a T-second-awakening—which in turn increases the total probability of being in a T-awakening.

So, since the shift to the episodic perspective preserves both the equiprobabilities P(T-first-awakening) = P(T-second awakening) and P(T-first-awakening) = P(H-awakening), and all three outcomes are exclusive from this perspective, the probabilities must sum up to 1 and therefore must shift from 1/2, 1/2, and 1/2 to 1/3, 1/3 and 1/3.

Then before the experiment starts the thirder will say "since I now know that I will soon rationally infer that the coin will have landed heads with probability 1/3 (on the basis of no new information), I can already infer this right now, before the coin is tossed."

But I think this is wrong. — Michael

You are correct that this would be wrong. The entire aim of my variation (and the Leonard Shelby variation before it) was to highlight that there is indeed some new information available to Sleeping Beauty upon awakening, and that this information can be retained by her on Wednesday through a note. This information wasn't available to her before the experiment began, and isn't available to her on Wednesday when she doesn't receive a note.

The objective of my discussion was also to highlight the unique nature of this information. It's not a form of information indicating, consistent with this information, a higher proportion of possible worlds in which the coin landed tails. Indeed, the proportion of possible worlds remains exactly the same. It's rather information that is generated by placing Sleeping Beauty in a higher proportion of centered possible worlds (her distinct awakening episodes within a single experimental run) within the T-run timelines.

This type of information is the same as the information that is transmitted from her to her own future self (on Wednesday) when she awakens, by selecting twice as many future recipients of the note in the long run when it is a T-note. This is akin to the information you gained that someone you met was a Tunisian with a probability of 2/3, not because there are twice as many Tunisians as there are Italians (there were actually as many of each in the city), but because Tunisians venture outside of their homes twice as often, doubling your opportunities of meeting them. Likewise, the setup of the Sleeping Beauty experiment makes the coins that land tails twice as likely to "meet" her on the occasion of her awakenings.

They're not nearly identical. On Wednesday she knows that she only had the opportunity once. When she wrote the note she didn't know that it was her only opportunity. So contrary to the above, there is new information on Wednesday. — Michael

I would argue that the situations are nearly identical since this new knowledge is almost inconsequential. Suppose n = 100. On Wednesday, Sleeping Beauty knows that she only had a single opportunity and can thus rule out the miniscule 0.01% chance that she would have had two opportunities. The probability that the note she obtained was an H-note (produced on the occasion of an H-awakening) therefore is P(H-note|single)/(P(H-note|single)+P(T-note|single)) where

P(H-note|single) = P(single|H-note)P(H-note)/P(single) = (1)(0.5%)/(0.0149) ≈ 0.3356
(since P(single) = 0.5%+(1/2)*0.0198 = 0.0149, and we had calculated the 0.0198 before)

and

P(T-note|single) = P(single|T-note)P(T-note)/P(single) = (99%)(1%)/(0.0149) ≈ 0.6644

As expected, P(H-note|single) and P(T-note|single) sum up to 1, and the probability that the note Sleeping Beauty obtained on Wednesday was an H-note rises to 0.3356. This figure is slightly larger than 1/3 only because the rare cases of H-runs where two opportunities to write a note were present are discounted when she only receives one.

I address this here. — Michael

Thank you! I'll respond to this separately.

And as it grows smaller, P(H) tends to 1. I don't understand the relevance of any of these three answers.

Why is the correct answer given by any of these situations, let alone by the situation where n is arbitrarily large? — Michael

The issue with making n small is that it allows Sleeping Beauty on Wednesday to decrease her credence P(H) regarding the origin of the single note. This is because (1) she did not receive two notes and (2) in a significant proportion of cases where a T-run occurs, two such notes are generated instead of one. This makes her epistemic situation dissimilar to her situation when she experiences a particular awakening episode. During such episodes, she can never know that there are two of them due to her amnesia. Making n large makes Sleeping Beauty's epistemic situation on Wednesday, when she receives a note, nearly identical to her situation when she wrote the note, since the Bayesian updating she can perform on the basis of the note being unique is negligible.

When evaluating P(H) on Wednesday, assuming n >> 1, the question Sleeping Beauty is asking is:

"What are the odds that this single note that I received was written by me during an H-awakening?"

The answer is approximately 1/3. However, the note could only have been written during an H-awakening if the coin landed on H. Therefore, P(H) is 1/3.

The second step in the reasoning is to consider that when Sleeping Beauty awakens and finds an opportunity to write a note, she knows that when she reads it on Wednesday (except on the very rare occasion when she finds two notes) she will be able to rationally infer that the odds that the note was written during an H-awakening are 1/3. Since it is now certain that she will read the note on Wednesday and will possess no more information regarding the circumstances of production of the current note than she currently has, she can already infer that this note is being written by her during an H-awakening with 1/3 odds.

A streamlined version of Sleeping Beauty's inference is: "Since I now know that I will soon rationally infer that this note was written during an H-awakening with probability 1/3 (on the basis of no new information), I can already infer this right now." (Here, I am making use of van Fraassen's reflection principle.)

The last step in the argument requires reflecting on the case where Sleeping Beauty doesn't find an opportunity to write a note. In that case, when she awakens, she can reason counterfactually:

"If I had had an opportunity to write a note to myself, I would then have inferred on Wednesday that P(H) (regarding the current awakening episode that is the source of the note) is 1/3, and hence known now that P(H) is 1/3. But the odds that I am currently experiencing an H-awakening are probabilistically independent of my finding an opportunity to write a note. Therefore, they are 1/3 and the only reason why I will be unable to know this when I awaken on Wednesday (and rather infer that P(H) = 1/2) is because I will have lost the special causal connection that I currently have to my present awakening episode.

Note that when Sleeping Beauty doesn't receive a note on Wednesday, her credence P(H) = 1/2 doesn't merely differ in value from her credence P(H) = 1/3 during awakenings; the predicates P() also have different meanings. During awakenings, P(H) refers to the odds that her current awakening episode is occurring during a coin toss that landed heads. On Wednesday, P(H) refers to the odds that the experimental run she is exiting from was an H-run. While in each case the biconditionals "I am now in an H-awakening iff I am now (and will be) in an H-run" or (on Wednesday) "I was in an H-awakening iff I am now in an H-run" hold, the probabilities don't necessarily match due to the two-to-one mapping between T-awakenings and T-runs.

To emphasize this last point, suppose Sleeping Beauty writes a note on each awakening occasion and the experiment is run many times. She ends up with a collection of identical notes, approximately two-thirds of which were written during T-awakenings. She now has lost track of the pairing between the notes. Two things can now be true at the same time:

(1) Since 1/3 of those notes are H-notes, Sleeping Beauty was right during the occasions where she wrote them to believe P(H-note) = 1/3 and hence that P(H) = 1/3.

(2) Since at the end of each experimental run, Sleeping Beauty received either one H-note or two T-notes with equal probabilities, the probability that the individual note(s) she received were T-notes (or H-notes) was 1/2 (or 1/2). In other words, in advance of counting how many notes there were on any given Wednesday, Sleeping Beauty could point at the note(s) and say that they were equally likely to be H-notes or T-notes.

This analysis again highlights how Halfers and Thirders can both be right at the same time but talk past each other when they fail to attend precisely to their respective definitions of P(H), and especially how H and T outcomes are to be individuated and counted.

Here, I've asked GPT-4 to summarise the argument and highlight the main points:

Though I don't see why I should accept your claim that if "she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note." — Michael

This is because, when the experimental protocol is expanded to enable Sleeping Beauty to hand notes to her future self in such a manner, the episodes of her receiving a note on Wednesday are produced twice as often in the long run when the coin has landed tails. On the occasion where she awakens and is offered the opportunity to write a note, Sleeping Beauty therefore is enabled to reason thus:

"When I will read again the note that I am currently writing, on Wednesday, I will be able to rationally infer that it is twice as likely that this note was written by me on the occasion of a T-awakening. But it can only be the case that it will have been written on the occasion of a T-awakening if I am now experiencing a T-awakening. Therefore, it is now twice as likely that I am experiencing a T-awakening."

Notice also that, since the probability that Sleeping Beauty would be offered an opportunity at any given awakening to write a note is the same regardless of whether it is an H-awakening or a T-awakening, being offered such an opportunity gives her no independent ground to update her credence.

Note that, as you said yourself, if the probability of her writing a note is 1/2 then if she finds exactly one note then her credence in Heads is 1/2.

That was only in the specific case where n = 2. As n grows larger, P(H) tends towards 1/3.

If tails then:

The probability of her writing on Monday is 1/100
The probability of her writing on Tuesday is 1/100 — Michael

Yes.

The probability of her writing on both Monday and Tuesday is 1/100 * 1/100 = 1/10,000
The probability of her writing on neither Monday or Tuesday is 1 - (1/100 * 1/100) = 9,999/10,000

The latter is actually 99/100*99/100 = 0.9801 ("both" and "neither" aren't complementary cases.)

The probability of her writing on Monday or Tuesday but not both is (1/100 + 1/100) / 2 = 1/100

It's actually 1 - "both" - "neither" = 1 - 0.0001 - 0.9801 = 0.0198 ≈ 2%, which is roughly twice the probability of writing a note in the H case.[/quote]

If heads and n = 100 then the probability of writing a note is 1/100

If tails and n = 100 then the probability of writing exactly one note is 1/100.

So if she finds exactly one note her credence in heads is 1/2. — Michael

Her probability of writing a note is 1/100 on each occasion she awakens. Since she awakens twice when the coin lands tails, her probability of writing a note is 2/100 when a T-experimental run occurs (discounting the 1/10000 cases where she writes two notes).

[...]But notice that as the probability of writing a note each time approaches 1 the "greater likelihood" of it having been tails gets smaller, approaching 1.[...] — Michael

Yes, indeed, which is why I edged my specification by stipulating that the occasions to write a note were rare.

If Sleeping Beauty would receive two notes on Wednesday, she'd be able to infer that there were two awakenings and hence that the coin didn't land heads. On the earlier occasions when she was writing those notes, by contrast, she wasn't able to know this. When the probability that she would be able to write a note on each awakening occasion is exactly 1/2, the overlapping cases are just numerous enough to enable her to infer on Wednesday, when she receives one single note, that P(H) = 1/2.

As the occasions to write a note become rarer (e.g. 1/n with n >> 1), the frequency of those overlapping notes become negligible (n times as many single notes are received as double notes) and Sleeping Beauty's epistemic state (i.e. the value of her credence) approaches asymptotically her epistemic state as she was writing the note. And, as I had suggested in my previous post, this is because when she receives a single note on Wednesday, Sleeping Beauty comes to be causally and epistemically related to the coin result in the exact same manner as she was when she originally wrote the note.

I'll address the other cases and analyses you have presented separately.

Nothing is ruled out when woken or asked her credence that wasn’t already ruled out before the experiment started.

Even Elga understood this: — Michael

I believe Elga was mistaken about this. There actually is some information that becomes available to Sleeping Beauty when she awakens, though the nature of this information is rather peculiar. I discussed the nature of this information with GPT-4 in this earlier post.

What informs Sleeping Beauty about the likelihood that the coin landed (or will land) tails, allowing her to update her credence from 1/2 to 2/3, is the fact that she awakens and that, whenever she awakens, the coin landed (or will land) tails two times out of three. After the experiment is over, and she is awoken on Wednesday (assuming she always receives the amnesia-inducing drug after each interview), this information is lost to her, and her credence reverts back to 1/2. The reason why she can't retain the information available to her during each awakening is that this information pertains specifically to the state of the coin in relation to her current episode of awakening. Upon awakening on Wednesday, she loses this information because she loses the ability to refer deictically to her past episodes of awakening (not even knowing how many of them there were).

This loss of information can be emphasized further by modifying the experiment in such a way that the information is not lost by her on Wednesday. Suppose we update the protocol so that on rare occasions, which present themselves with equal probability on each awakening episode, Sleeping Beauty is able to write down a note saying "I have now been awakened and interviewed." She can retain this note and read it again on Wednesday. Upon rereading the note on Wednesday, she can reason that it is twice as likely that such a note was produced if the coin landed tails since she would have been twice as likely to write it during such an experimental run. Therefore, she can retain her updated credence P(T) = 2/3 through ordinary Bayesian reasoning.

The key realization is that the same information that allows Sleeping Beauty to update her credence P(H) from 1/2 to 2/3 upon receiving the note, is inherent in every awakening she experiences due to the causal structure of the experiment. Each awakening serves as an implicit notification of her being in one of the two potential kinds of awakening episodes, which are twice as likely to occur if the coin landed tails. This causal relationship between coin toss results and awakenings, established by the experimental setup, provides information that is available to her in every awakening, even when she doesn't have the opportunity to physically write it down. Essentially, the note merely serves to extend this causal relationship to her Wednesday state, providing her with twice as many opportunities to receive the note if the coin landed tails.

Pierre-Normand is saying that P(X) refers to the ratio of Xs to non-Xs in some given reference class.

I'm saying that P(X) refers to the degree to which I believe X to be true.

If P(X) refers to the degree to which I believe X to be true, and if I believe that A iff B, then P(A) = P(B). — Michael

Actually, I suggested that P(X) could be understood as referring to the ratio of |{X}| to (|{X}| + |{not-X}|) in epistemically identical situations with respect to X. There is some flexibility in defining what the relevant situations are.

In the case where Sleeping Beauty can say "I am experiencing an H-awakening iff I am experiencing an H-run", and there is a one-to-one mapping between H-awakenings and H-runs, we still can't logically infer that P(H-awakening) = P(H-run). This is because one can define P(H-awakening) as |{H-awakening}|/|{awakenings}| and similarly define P(H-run) as |{H-run}|/|{run}| where {awakenings} and {run} are representative sets (and |x| denotes cardinality). For the inference to hold, you would also need a one-to-one mapping between the sets of T-runs and T-awakenings.

So, the grounds for a Thirder's credence P(H-awakening) being 1/3 (where it is defined as |{H-awakening}|/|{awakenings}|) simply comes from the propensity of the experimental setup to generate twice as many T-awakenings as H-awakenings.

You argue that the number of T-awakenings is irrelevant to the determination of Sleeping Beauty's credence P(H-awakening) because her having multiple opportunities to guess the coin toss result when it lands tails doesn't impact the proportion of tails outcomes. However, while it doesn't impact |{H-run}|/|{run}|, it does impact |{H-awakening}|/|{awakenings}|, which is why I argue that Halfers and Thirders are talking past each other.

Consider an alternative experiment setup where Sleeping Beauty awakens less often rather than more often when the coin lands tails. For instance, we could eliminate Tuesday awakenings altogether and ensure that Sleeping Beauty awakens once on Monday when the coin lands heads, and only half the time when it lands tails (by tossing a second coin, say). This setup would, in the long run, reverse the ratio of H-awakenings to T-awakenings compared to the original setup. In that case, when Sleeping Beauty awakens, would you still argue that her credence P(H-awakening) remains 1/2? A Thirder would argue that her credence should now increase to 2/3, based on the same frequency-ratio reasoning.

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.

@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.

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.

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.

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.

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.

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.)

If you want a closer analogy with pedestrians, it's Tunisians walking around in pairs. If the chances of meeting an Italian or a pair of Tunisians are equal, then the chances of meeting *a* Tunisian are either nil, since you can't meet just one, or the same as meeting a pair.

Look at how hang-around times affect the pedestrian-encountering odds. Roughly, if you miss a short walker, you've missed him, but if you miss a long walker you get another chance. That's not how Sleeping Beauty works at all. There's no way to miss your first tatils interview but still catch the second one. — Srap Tasmaner

If I were to adjust the analogy, suppose that meeting a Tunisian pedestrian guarantees that you have met or will meet their sibling either in the previous or next encounter. In this scenario, would your credence that the pedestrian you're encountering is a Tunisian change? As long as you meet Tunisians twice as often as Italians, your credence P(Tunisian encounter) should remain 2/3 at the time of each individual encounter, regardless of the pairing situation.

This is an ambiguous claim. If there are half as many Tunisians but they go out four times as often but are only out for 10 mins, whereas Italians are out for 20 mins, then it would be that Tunisians are around equally as often as measured by time out. The only way you could get this to work is if the argument is set out exactly as I have done above:

A1. there are twice as many Tunisian walkers as Italian walkers (out right now)
A2. if (right now) I meet a walker at random from a random distribution of all walkers (out right now) then I am twice as likely to meet a Tunisian walker

But there's nothing comparable to "if (right now) I meet a walker at random from a random distribution of all walkers (out right now)" that has as a consequent "then my interview is twice as likely to be a T-interview". — Michael

To fine-tune the analogy, let's assume that there are an equal number of Tunisians and Italians, that they are out for the same duration, and that Tunisians go out twice as frequently. Importantly, there's no need for an extraneous process of random selection to generate an encounter with a citizen, or a tossed coin, in either example. In the cosmopolitan encounter case, the random distributions of citizens in the street at any given time (with, on average, twice as many Tunisians out) directly result in twice as many encounters with Tunisians. It is a straightforward causal relationship between the distribution of wanderers and the distribution of encounters. Similarly, in the Sleeping Beauty case, the setup guarantees that Sleeping Beauty will encounter twice as many coins having landed tails when she awakens, simply by ensuring that she is awakened twice as often when the coins land tails.

P1. If I am assigned at random either a T-interview set or a H-interview set then my interview set is equally likely to be a T-interview set
P2. I am assigned at random either a T-interview set or a H-interview set
P3. My interview is a T-interview iff my interview set is a T-interview set
C1. My interview is equally likely to be a T-interview

The premises are true and the conclusion follows, therefore the conclusion is true. — Michael

The conclusion doesn't follow because, while the biconditional expressed in P3 is true, this biconditional does not guarantee a one-to-one correspondence between the set of T-interviews and the set of T-runs (or "T-interview sets"). Instead, the correspondence is two-to-one, as each T-run includes two T-interviews. This is a central defining feature of the Sleeping Beauty problem that your premises fail to account for.

In the case of the meetings we have:

*P1) there are twice as many Tunisian walkers
*P2) if I meet a walker at random then I am twice as likely to meet a Tunisian walker (from *P1)
*P3) I meet a walker at random
*C) I am twice as likely to have met a Tunisian walker (from *P2 and *P3)

In Sleeping Beauty's case we have:

P1) there are twice as many tails interviews
P2) ?
P3) I am in an interview
C) I am twice as likely to be in a tails interview

What is your (P2) that allows you to derive (C)? It doesn't follow from (P1) and (P3) alone. — Michael

You are introducing premises *P2 and *P3 in an attempt to emphasize a perceived disanalogy between the cosmopolitan meeting scenario and the Sleeping Beauty problem. Both *P1 and *P2 seem to imply that there exists a pre-determined set of potential encounters (many Tunisians and half as many Italians strolling around), from which a random selection process subsequently generates an encounter. There indeed is no analogous situation in the Sleeping Beauty problem, as there isn't a pre-determined set of pre-tossed coins from which Sleeping Beauty randomly encounters one upon awakening. However, I would argue that this misrepresents the cosmopolitan meeting scenario.

In the cosmopolitan situation, the probability of meeting a Tunisian doubles because Tunisians are around twice as often. There is no need for a random selection from a pre-existing set of potential encounters. Similarly, in the Sleeping Beauty problem, coins that have landed on tails "hang around" twice as long (i.e., for two sequential awakenings instead of one), which makes it twice as likely for Sleeping Beauty to encounter this outcome each time she is awakened and interviewed throughout the experiment.

The random process is fully specified by the equal distribution of coin toss outcomes (over the long run) and the longer "hang around" times of tails outcomes. Likewise, in the cosmopolitan meeting case, the process is fully specified by the equal distribution of Italians and Tunisians in the city and the increased frequency of encounters generated by Tunisians due to their longer "hang around" times. In neither case are additional random selection processes from a pre-determined set of possible encounters necessary.

Your argument is that: if 1) there are twice as many T-awakenings and if 2) I randomly select one of the awakenings then 3) it is twice as likely to be a T-awakening.

This is correct. But the manner in which the experiment is conducted is such that 2) is false. — Michael

Your point (2) doesn't factor into my argument. I've consistently held to the premise, as dictated by the problem statement, that Sleeping Beauty awakens once when the coin lands heads and twice when it lands tails. There's no necessity for an external agent to select an awakening, just as there's no need for someone to choose a street encounter. Instead, Sleeping Beauty, upon each awakening (or encounter), should consider the long-term distribution of these awakenings (or encounters) to formulate a rational belief about the current situation.

If we were to use the meetings example then:

1. A coin is tossed
2. If heads then 1 Italian walks the streets
3. If tails then 2 Tunisians walk the streets
4. Sleeping Beauty is sent out into the streets

What is the probability that she will meet a Tunisian? — Michael

In the Sleeping Beauty problem, she isn't asked to estimate the probability of being awakened in the future with the coin having landed heads. Instead, she's awakened and then questioned about her current belief regarding the coin's outcome. To maintain this structure in the street encounter example, we should consider Sleeping Beauty meeting a wanderer and then being asked to consider the probability that this wanderer is an Italian. If, over time, the setup leads to twice as many Tunisian encounters (perhaps because Tunisians wander about twice as long as Italians), then Sleeping Beauty's rational credence should be P(Italian) = 1/3.

"there are twice as many Tunisian-meetings" isn't biconditional with "there are half as many Tunisians and Tunisians go out four times more often" and so A doesn't use circular reasoning. — Michael

However, you seem to agree that in this scenario, one is twice as likely to encounter a Tunisian. The conclusion that there are twice as many Tunisian-meetings emerges from the premises: (1) there are half as many Tunisians and (2) Tunisians venture out four times more often. This inference is simply an intermediate step in the argumentation, providing an explanation for why there are twice as many Tunisian-meetings. Analogously, the Sleeping Beauty setup explains why there are twice as many T-awakenings. If the reason for twice as many Tunisian-meetings is that Tunisians venture out twice as often (assuming there are an equal number of Tunisians and Italians), then the analogy with the Sleeping Beauty scenario is precise. The attribute of being Tunisian can be compared to a coin landing tails, and encountering them on the street can be paralleled to Sleeping Beauty encountering such coins upon awakening. In the Sleeping Beauty setup, coins that land tails are 'venturing out' more often.

This is just repeating the same thing in a different way. That there are twice as many T-awakenings just is that Sleeping Bauty is awaked twice as often if tails — Michael

You accepted the validity of the reasoning when probability was deduced from frequencies in the Tunisian-meetings scenario. Why is this reasoning acceptable for people who were born Tunisian but questionable for coins that landed tails?

In this case:

1. there are twice as many Tunisian-meetings because Tunisian-meetings are twice as likely
2. Tunisian-meetings are twice as likely because there are half as many Tunisians and Tunisians go out four times more often

This makes sense.

So:

1. there are twice as many T-awakenings because T-awakenings are twice as likely
2. T-awakenings are twice as likely because ...

How do you finish 2? It's circular reasoning to finish it with "there are twice as many T-awakenings". — Michael

To fill in your number 2 with no circularity, we can draw a parallel to the first example:

2a. Tunisian-meetings are twice as likely because there are half as many Tunisians and Tunisians go out four times more often (and thus, Tunisian-meetings are twice as frequent as Italian-meetings)

Likewise:

2b. T-awakenings are twice as likely because, based on the experiment's design, Sleeping Beauty is awakened twice as often when the coin lands tails (and thus, T-awakenings are twice as frequent as H-awakenings)

Starting here you argued that P(Heads) = 1/3.

So, what do you fill in here for the example of one person woken if heads, two if tails? — Michael

I see. I was filling up a template that you had provided where P(Monday) = 2/3, thus making it clear that we were quantifying awakening episodes.

In that case P(Monday|Heads) = 1, and P(Heads) = 1/3 since one third of the awakenings are H-awakenings.

Therefore P(Heads|Monday) = P(Monday|Heads)∗P(Heads)/P(Monday) = (1)*(1/3)/(2/3) = 1/2.

Likewise, P(Heads|Awake) = P(Awake|Heads)∗P(Heads)/P(Awake) = (1)*(1/3)/(1) = 1/3

Note that when we quantify awakening episodes, P(Awake|Heads) = 1 since all H-awakenings are awakenings.

What wouldn't make sense is just to say that Tunisian-meetings are twice as likely because there are twice as many Tunisian-meetings. That is a non sequitur. — Michael

But why wouldn't it make sense? For example, if you're an immigration lawyer and your secretary has arranged for you to meet with twice as many Tunisians as Italians in the upcoming week, when you walk into a meeting without knowing the client's nationality, isn't it logical to say that it's twice as likely to be with a Tunisian?

To make this comparable to the Sleeping Beauty problem; there are two Sleeping Beauties, one will be woken if heads, two will be woken if tails. When woken, what is their credence in heads? In such a situation the answer would be 1/3. Bayes' theorem for this is:

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

=(1/2)∗(1/2) / (3/4)=1/3

=1/3

This isn't comparable to the traditional problem. — Michael

Why isn't it comparable? I had proposed an identical version earlier. One effective way to erase Sleeping Beauty's memory without any side effects from an amnesia-inducing drug might be to switch her with her identical twin for the second awakening. They would each only experience one awakening at most as part of a team. Their epistemic perspectives regarding the coin toss would remain the same, and therefore so should their rational credences.

Incidentally, what is your version of Bayes' theorem for this where P(Heads) = 1/3?

I am unsure what it is that you are asking here.

If you want to be very precise with the terminology, the Andromeda Paradox shows that some spacelike separated event in my present is some spacelike separated event in some other person's causal future even though that person is also a spacelike separated event in my present. I find that peculiar. — Michael

In essence, you're saying that even though a distant event currently lies beyond your ability to influence it (due to the fact that any influence you exert cannot travel faster than light), someone else, presently positioned closer to the event, can influence it.

Some event (A1) in my (A0) future is spacelike separated from some event (B0) in someone else's (B1) past, even though this person is spacelike separated from my present. It might be impossible for me to interact with B1 (or for B1 to interact with A1), but Special Relativity suggests that A1 is inevitable, hence why this is an argument for a four-dimensional block universe, which may have implications for free will and truth. — Michael

If we let c approach infinity, Galilean spacetime converges with Lorentzian spacetime. In this case, the "absolute elsewhere" of an event (the region outside of the light cone) shrinks into a unique simultaneity hyperplane. In Galilean spacetime, an observer at a given time views any event in its (absolute) past as "inevitable." In Lorentzian spacetime, an observer deems "inevitable" any event that resides either in its (absolute) past light cone or in its (also absolute) elsewhere region. The "inevitability" relation between observers-at-a-time (events) and other observers-at-a-time becomes intransitive.

This intransitivity means that even if

1. A1 is inevitable by B1, and
2. B1 is inevitable by A0,
it does not follow that (3) A1 is inevitable by A0.

This inference is invalid because the inability of A0 to affect A1 indirectly by influencing B1 does not mean that A0 can't influence A1 directly.

the edge of the visible universe is receding from us faster than the speed of light. Although individual galaxies are much slower than light their apparent movement adds up radially away from us. Over billions of years we would see fewer galaxies spread further apart in ever darkening space. — magritte

This is due to the expansion of the universe, which is a general relativistic effect. It is unrelated to the shifting of the simultaneity plane due to the substitution of inertial reference frames is special relativity.

Very interesting, I had always heard that all the whites were descendents of slave owners, and ispo facto, all racists. — Merkwurdichliebe

I've also heard James Lindsay and Tucker Carlson claim that woke leftists generally believe this, but I've never heard a leftist actually say it.

Which of these are you saying?

1. There are twice as many T-awakenings because tails is twice as likely
2. Tails is twice as likely because there are twice as many T-awakenings

I think both of these are false.

I think there are twice as many T-awakenings but that tails is equally likely.

The bet's positive expected value arises only because there are twice as many T-awakenings. — Michael

I am not relying on 1, but it would be a valid inference if we assume that P(T) = 2*P(H). This assumption would hold if we define P(T) as P(T) =def |{T-awakenings}| / |{awakenings}| (and similarly for P(H)).

Your disagreement with 2 appears to stem from an assumption that (at least in the context of the Sleeping Beauty problem) the probability of an outcome must solely be a reflection of the realization of an object's intrinsic properties, such as a fair coin. However, we frequently talk about probabilities of (types of) events that depend on how we interact with objects and that only indirectly depend (if at all) on the propensities of those objects had to actualize their properties. For instance, if there are twice as many Italians as Tunisians in my city (and no other nationalities), but for some reason, Tunisians go out four times more often than Italians, then when I go out, the first person I meet is twice as likely to be a Tunisian.

Here's another example. The management of the Sleeping Beauty Experimental Facility organizes a cocktail party for the staff. The caterers circulate among the guests serving drinks and sandwiches. Occasionally, they flip a coin. If it lands heads, they ask a random guest to guess the result. If it lands tails, they ask two random guests. The guests are informed of this protocol (and they don't track the caterers' movements). When a caterer approaches you, what are the odds that the coin they flipped landed heads? This situation is similar to the case of encountering a Tunisian in my previous example. Upon meeting someone, you could rationally say, "The probability that this person is a Tunisian is 2/3," even though people don't inherently possess a propensity to be born a different nationality than the one they were actually born into.

In the cocktail party scenario, the probability that "this" coin landed tails is a function of both the coin's propensities (its fairness) and the method used to engineer interactions between guests and coins based on the outcomes of the tosses.

I think you're confusing two different things here. If the expected return of a lottery ticket is greater than its cost it can be rational to buy it, but it's still irrational to believe that it is more likely to win. And so it can be rational to assume that the coin landed tails but still be irrational to believe that tails is more likely. — Michael

The rationality of Sleeping Beauty betting on T upon awakening isn't because this bet has a positive expected value. In fact, it's the other way around. The bet's positive expected value arises because she is twice as likely to win as she is to lose. This is due to the experimental setup, which on average creates twice as many T-awakenings as H-awakenings. It's because her appropriately interpreted credence P(T) =def P(T-awakening) = 2/3 that her bet on T yields a positive expected value, not the reverse. If she only had one opportunity to bet per experimental run (and was properly informed), regardless of the number of awakenings in that run, then her bet would break even. This would also be because P(T) =def P(T-run) = 1/2.

The same logic applies in my 'escape scenario', where Sleeping Beauty's room is surrounded by crocodiles (and she awakens once) if the die doesn't land on 'six', and is surrounded by lions (and she awakens six times) if the die does land on 'six'. Given a rare chance to escape (assuming opportunities are proportional to the number of awakenings), Sleeping Beauty should prepare to face lions, not because of the relative utilities of encounters with lions versus crocodiles, but because she is indeed more likely (with 6/11 odds) to encounter lions. Here also, this is because the experimental setup generates more encounters with lions than it does with crocodiles.

Would you not agree that this is a heads interview if and only if this is a heads experiment? If so then shouldn't one's credence that this is a heads interview equal one's credence that this is a heads experiment? — Michael

Indeed, I have long insisted (taking a hint from @fdrake and Laureano Luna) that the following statements are biconditional: "The coin landed (or will land) heads", "I am currently experiencing a H-awakening", and "I am currently involved in a H-run".

However, it's important to note that while these biconditionals are true, they do not guarantee a one-to-one correspondence between these differently individuated events. When these mappings aren't one-to-one, their probabilities need not match. Specifically, in the Sleeping Beauty problem, there is a many-to-one mapping from T-awakenings to T-runs. This is why the ratios of |{H-awakenings}| to |{awakenings}| and |{H-runs}| to |{runs}| don't match.

If so then the question is whether it is more rational for one's credence that this is a heads experiment to be 1/3 or for one's credence that this is a heads interview to be 1/2.

Rationality in credences depends on their application. It would be irrational to use the credence P(H) =def |{H-awakenings}| / |{awakenings}| in a context where the ratio |{H-runs}| / |{runs}| is more relevant to the goal at hand (for instance, when trying to be picked up at the right exit door by Aunt Betsy) or vice versa (when trying to survive potential encounters with lions/crocodiles).

What the Andromeda Paradox implies is that the observed universe apparently shifts in its entirety towards a moving observer. Which means that in the forward moving direction many more of the most distant galaxies come into possible view and we lose some distant galaxies from possible view behind us. This is all pretty absurd, yet it is demonstrably true. — magritte

The galaxies you are moving towards would have come into view regardless of your motion, only at a later time as measured by your clock. Similarly, the galaxies you are moving away from will also come into view, but at a later time. In a flat spacetime, you cannot indefinitely outrun light rays. Interestingly, as you acquire more velocity relative to both sets of galaxies, they "move" closer to you due to the effects of Lorentz contraction.

However, as you gain speed, the reason why the light from the galaxy behind you doesn't catch up to you sooner (despite the contracted distance) is due to the recalibration of your plane of simultaneity. As this occurs, the photons that were a distance D away before you started moving suddenly "jump" back in time and are "now" less advanced on their journey towards you!

No. This has nothing to do with what one person sees. There are distant events happening in my present that I cannot see because they are too far away. According to special relativity some of these events happen in your future even though they are happening in my present. This is what I find peculiar. — Michael

In Special Relativity, an observer can be identified with an inertial reference frame in which they are at rest, and relative to which they make all their space and time measurements. This can be likened to a set of co-moving rods and clocks, all synchronized by light signals. For instance, Clocks A and B can be synchronized through a light signal sent back and forth between them, with the time at Clock B being set at the mid-point interval between the departure and return times at Clock A. Consequently, two observers can have different simultaneity planes due to the fact that events are timed with reference to distinct sets of clocks that have been synchronized differently. @Benkei's train example above illustrates this concept well.

Actual "observers" (such as human beings) are free to choose whatever reference frames they want, and can translate space-time coordinates of events using the Lorentz transformations. So long as they reside outside of their light cones, the issue of events being located in their "relative past" or "relative future" doesn't have any physical or metaphysical significance. It only has relevance to the degree that it challenges certain presentist or "growing block universe" conceptions of time. This was Putnam's main point in his paper, "Time and Physical Geometry".

Previously you've been saying that P(Heads) = 1/2. — Michael

In earlier messages? Yes. I shouldn't have used this prior in the context of the Thirder intepretation of P(H). I was unclear between the individuation of events as they related to the two possible interpretations of P(H) for Sleeping Beauty. So, some of my earlier uses of Bayes' theorem may have been confused or inconsistent. It is, I now think, the very fact that P(H) appears intuitively (but misleadingly) to reflect Sleeping Beauty's epistemic relation to the coin irrespective of the manner in which she tacitly individuates the relevant events that generates the apparent paradox.

I think Bayes’ theorem shows such thirder reasoning to be wrong.

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

If P(Unique) = 1/3 then what do you put for the rest? — Michael

P(Heads|Unique) = 1 and P(Heads) = 1/3 (since 1/3 of expected awakenings are H-awakenings)

P(Unique|Heads) is therefore 1, as expected.

Similarly:

P(Heads|Monday)=P(Monday|Heads)∗P(Heads)P(Monday)

If P(Monday) = 2/3 then what do you put for the rest?

P(Monday|Heads) = 1 and P(Heads) = 1/3.

P(Heads|Monday) = 1/2, as expected.

This is a non sequitur. — Michael

My argument follows a consistent line of reasoning. Given Sleeping Beauty's understanding of the experimental setup, she anticipates the proportion of indistinguishable awakening episodes she will find herself in, on average (either in one or in many experimental runs), and calculates how many of those will be H-awakenings given the evidence that she will presently be awakened.

What we can say is this:

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

We know that P(Unique | Heads) = 1, P(Heads | Unique) = 1, and P(Heads) = 1/2. Therefore P(Unique) = 1/2.

Therefore P(Unique|W) = 1/2.

And if this experiment is the same as the traditional experiment then P(Heads|W) = 1/2.

Yes, I fully concur with this calculation. It interprets Sleeping Beauty's credence P(Unique|W) = P(H|W) upon awakening as the proportion of complete experimental runs in which Sleeping Beauty expects to find herself in an H-run ('unique awakening run'). However, this doesn't negate the Thirder interpretation, which becomes relevant when Sleeping Beauty is focusing on the individual awakening events she is expected to experience, rather than on the individual experimental runs. This interpretation distinction is highlighted in various practical examples I've provided: for example, preparing to face lions or crocodiles while escaping relies on the Thirder interpretation, whereas being picked up by Aunt Betsy at the East Wing at the end of the experiment follows the Halfer interpretation, and so on.

It may still be that the answer to both is 1/3, but the reasoning for the second cannot use a prior probability of Heads and Tuesday = 1/4, because the reasoning for the first cannot use a prior probability of Heads and Second Waking = 1/4.

But if the answer to the first is 1/2 then the answer to the second is 1/2. — Michael

For the first case, we can use priors of P(H) = 1/2 and P(W) = 3/4, given that there are three awakenings in the four possible scenarios (H&Mon, H&Tue, T&Mon, T&Tue) where Sleeping Beauty can be. P(W|H) = 1/2, as she is only awakened on Monday when the coin lands heads.

Consequently, P(H|W) = P(W|H)P(H)/P(W) = (1/2)(1/2)/(3/4) = 1/3.

In the second case, we can set up a similar calculation: P(Unique|W) = P(W|Unique)*P(Unique)/P(W)

P(Unique) is the prior probability that an awakening will be unique rather than part of two. P(Unique) = 1/3, as one-third of the experiment's awakenings are unique. P(W) is now 1, as Sleeping Beauty is awakened in all potential scenarios.

We then find that P(Unique|W) = P(W|Unique)P(Unique)/P(W) = (1)(1/3)/(1) = 1/3.

This second case calculation is straightforward, but the similarity between the two cases is illuminating. Bayes' theorem works for updating a belief in an outcome given new evidence, P(O|E), by increasing the prior probability of the outcome, P(O), in proportion to the ratio P(E|O)/P(E). This ratio quantifies how much more likely the outcome becomes when the evidence is known to be present.

In both cases, Sleeping Beauty's evidence is that she is currently awake. In the first case, the relevant ratio is (1/2)/(3/4), which reflects how much more likely the coin is to land heads when she is awake. In the second case, the relevant ratio is (1)/(1), indicating how much more likely a unique awakening situation (due to the coin landing heads) is when she is awake. Both cases yield the same result (1/3), aligning with the ratio of possible H-awakenings ('unique awakenings') to total possible awakenings produced by the experimental designs.

Another interpretation of P(H) is the proportion of entire experimental runs in which Sleeping Beauty ends up in an H-run ('unique awakening run'). According to this interpretation, the Halfer solution P(H) = 1/2 is correct. The choice between Thirders or Halfers' interpretation of P(H) should depend on the intended use: during individual awakenings (Thirders) or throughout the experiment (Halfers).

I'll throw in one last consideration. I posted a variation of the experiment here.

There are three beauties; Michael, Jane, and Jill. They are put to sleep and assigned a random number from {1, 2, 3}.

If the coin lands heads then 1 is woken on Monday. If the coin lands tails then 2 is woken on Monday and 3 is woken on Tuesday.

If Michael is woken then what is his credence that the coin landed heads?

Michael's credence before the experiment is P(1) = 1/3, so if woken he ought to continue to have a credence of P(1) = 1/3 since he gains no new relevant evidence if he wakes up during the experiment. — Michael

In this variation, it seems to me that being awakened does provide Michael with relevant evidence. Given that the coin landing on tails results in one person being awakened, and the coin landing on heads results in two persons being awakened, on average, 1.5 out of three persons are awakened. Therefore, the prior probability that Michael will be awakened is P(MA) = 1/2. The conditional probabilities are P(MA|H) = 1/3 and P(MA|T) = 2/3 (and these are the same for Jane and Jill).

Hence, when Michael awakens, it's more probable that the coin landed tails.

P(T|MA) = P(MA|T)*P(T) / P(MA) = (2/3)*(1/2)/(1/2) = 2/3.

And given that if woken he is 1 iff the coin landed heads, he ought to have a credence of P(Heads) = 1/3.

Do we accept this?

Yes, we do.

If so then the question is whether or not Sleeping Beauty's credence in the original experiment should be greater than Michael's credence in this experiment. I think it should.

I'd be curious to understand why you think so.

Recently, I contemplated a similar variation wherein candidates are recruited as part of a team of two: Jane and Jill, for example. On Monday, either Jill or Jane is awakened (selected at random). On Tuesday, if a coin lands on tails, the person who wasn't awakened on Monday is now awakened. If the coin lands on heads, the experiment ends. (In this variation, as in yours, there's no need for an amnesia-inducing drug. It's only necessary that the subjects aren't aware of the day of their awakenings.)

Just like in your variation, tails generates two awakenings (for two different subjects), while heads generates only one. On average, 1.5 out of two persons are awakened. Jane's prior is P(JA) = 3/4, and the conditional probabilities are P(JA|H) = 1/2 and P(JA|T) = 1.

As before, Jane's awakening provides her with evidence that the coin landed tails.

P(T|JA) = P(JA|T)*P(T) / P(JA) = (1)*(1/2)/(3/4) = 2/3.

I would argue that this case is structurally identical to the one discussed in the original post (as well as in Lewis and Elga), with the sole exception that the relevant epistemic subjects are members of a team of two, rather than a single identifiable individual potentially being placed twice in the "same" (epistemically indistinguishable) situation. You could also consider a scenario where Jill falls ill, and her team member Jane volunteers to take her place in the experiment. In this instance, the amnesia-inducing drug would be required to maintain the epistemic separation of the two potential awakenings in the event that the coin lands heads.