[...]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.

She also knows that the fact that she is awake eliminates (H,H) as a possibility. This is a classic example of "new information" that allows her to update the probabilities. With three (still equally likely) possibilities left, each has a posterior probability of 1/3. Since in only one is coin C1 currently showing Heads, the answer is 1/3. — JeffJo

Your proposed scenario certainly provides an interesting variation, but it doesn't quite correspond to the structure of the situation typically discussed in literature, the one that seems to give rise to a paradox.

In your scenario, there are four potential outcomes from the experiment, each of which is equally probable:

HH (end) --> Never awakened
HT HH --> Awakened once
TH TT --> Awakened twice
TT TH --> Awakened twice

When Sleeping Beauty awakens, her credences corresponding to these four outcomes shift from {1/4, 1/4, 1/4, 1/4} to {0, 1/3, 1/3, 1/3}.

However, in the scenario most frequently discussed, entire experimental runs in which Sleeping Beauty is awakened once are just as common as those where she is awakened twice. Furthermore, since there isn't any experimental run where Sleeping Beauty is not awakened at all, it's debatable whether her experiencing an awakening provides new information that would cause her to adjust her initial probabilities (as Halfers are inclined to argue).

But there are two sources of randomness in this example, the die and the coin.

Similarly for all analyses that treat SB's situation as describable with two coin flips. We only have one. — Srap Tasmaner

Indeed, in my examples (labelled "First step" through "Fourth step"), there's only a single source of randomness, which consists in the random assignment of individuals to either population Pop-1 or Pop-2 (awakened once or twice with white or red tulips).

Halfers contend that Sleeping Beauty's awakening cannot serve as evidence indicating she is more likely to be part of Pop-2, as there's nothing that allows her to distinguish an awakening in Pop-1 from one in Pop-2. Yet, the same reasoning can be applied to the inability to distinguish a 'six' roll from a loaded die versus a 'six' roll from a fair die. Yet, the occurrence of a 'six' increases the likelihood that the die is loaded.

You're correct in stating that there's only one source of randomness in Sleeping Beauty's case, unlike the dice scenario. However, the two situations share a strong resemblance. The reason a 'six' outcome increases the probability that a die is loaded is because loaded dice generate more instances of players confronting a 'six'. Similarly, being part of Pop-2 in Sleeping Beauty's setup leads to more instances of self-aware awakenings. This is simply an analogy - for a more compelling argument, refer back to my cases 1 through 4 in the post you quoted.

The halfer position comes back to individuation, as you suggested some time ago. Roughly, the claim is that "this interview" (or "this tails interview" etc) is not a proper result of the coin toss, and has no probability. What SB ought to be asking herself is "Is this my only interview or one of two?" The chances for each of those are by definition 1 in 2.

Indeed, the choice between a Halfer (P(Pop-1) = 1/2) and a Thirder (P(Pop-1) = 1/3) updated credence is a matter of individuation. While I focused on the individuation of events, you had seemed to suggest that different (more or less extended) conceptions of self might lead people towards one stance or another. This struck me as insightful, although personal psychological inclinations don't provide valid justifications. Currently, I don't identify as a Thirder or a Halfer. Rather, I believe that Thirders and Halfers are talking past each other because they each focus solely on one of two possible types of outcome distributions that could be considered in Sleeping Beauty's credence update. My previous "pragmatic" examples aimed at highlighting this duality (not a dichotomy!) When Sleeping Beauty wakes and considers her situation, is she weighing the opportunities to either evade or confirm her current situation (facing lions or crocodiles)? In this case, she should reason as a Thirder. Or is she weighing the opportunity to end, or validate, the nature of her ongoing predicament (and be rescued by Aunt Betsy) at the end of her current series of awakenings? If so, she should reason as a Halfer. The root question of what her credence should be upon awakening is inherently ambiguous, and the thought experiment is tailored to create this ambiguity.

In the original scenario as I have described it, Ned reads the printout, but he only reads a part of it. And, importantly, he does not read a part of it where he is reading the printout -- that would be self-referentially problematic. Because there is no self-referentially in the parts of the printout that Ned does read, there is nothing necessarily theoretically vicious about Ned reading some parts of the printout. — NotAristotle

If Ned only reads a part of the printout that doesn't mention him reading it, his actions wouldn't contradict the thesis of determinism. For instance, suppose determinism entails that, given some initial conditions at time t1 (along with the laws of nature), Ned will drink a glass of water at time t2. The printout Ned reads must inform him of this prediction. In response, Ned decides to drink a glass of orange juice instead, fulfilling his intention to act contrary to the prediction.

We must consider when Ned's reading of the printout occurs. Does it occur before or after time t2 in the scenario you're envisioning?

If it takes place after t2, then Ned's actions wouldn't actually contravene the prediction. Indeed, by then, he would have already drunk the glass of water, hence aligning with the prediction. However, if he reads the printout prior to t2 to pre-empt the predicted outcome, we run into the self-referential issue. In this case, the computer would need to forecast what occurs as a consequence of Ned reading the printout. This would invariably involve predicting Ned's reaction to the prediction, triggering the problematic self-reference loop I had mentioned.

I suppose we could stipulate that Ned has enough information about his immediate environment to make an accurate prediction about how he will act. It doesn't really concern us whether this sort of information can, as a matter of practicality, be acquired; the concern is whether in principle, if this information were acquired, could Ned act in opposition to it. And the answer to that seems to be yes. — NotAristotle

I concur that we must accept, in principle, that this information can be acquired by Ned in order for his contrarian behavior to make any sense. Would Ned's decision to act contrary to his prediction be an arbitrary or unexplainable one? No, we cannot attribute his behavior to the occurrence of miracles and then use it as evidence against determinism.

But, if we consider that it is Ned's reaction to the acquired information (provided by a computer, for example) that prompts him to act differently, the computer's prediction must also account for the possibility that Ned will read and respond to its output. In other words, the computer's predicted output becomes a component of its own predictive behavior loop, transforming Ned into a sort of "contrarian cog" within the deterministic system.

This issue can be reduced to the challenge of programming a computer capable of modeling its own deterministic algorithmic process in such a way that it will output "yes" if it predicts it will output "no" and vice versa. This is a conundrum that can't be resolved in principle, yet this doesn't impact the deterministic nature of the computer's behavior. It rather highlights the impossibility of creating a self-referential prediction system within the boundaries of determinism.

But historically it has. There are a multitude of multilateral treaties that prove even enemies will agree on all sorts of things. WTO, UN, Geneva and the Hague conventions, Vienna Convention on the laws of treaties, Vienna Convention on diplomatic relations, etc. — Benkei

And the Montreal Protocol that has been ratified by all member states of the United Nations.

But she is only asked a question once in the whole year. One of the wakings is randomly selected to be the one where she is asked the question. On this randomly selected waking, she is asked the question "what is the probability that this randomly selected waking shows a heads." The answer is 1/3, as per Problem A in my previous post. — PhilosophyRunner

A Halfer might argue that Sleeping Beauty being posed such a question, along with the provided context of the question's delivery (i.e., through a random selection among all awakenings), indeed provides the grounds for Sleeping Beauty to update her initial credence P(H) from 1/2 to 1/3. However, they might also assert that this type of questioning doesn't exist in the original setup. Therefore, they might insist that, in the absence of such randomly assigned questioning, Sleeping Beauty should maintain her credence of 1/2.

A Thirder might counter-argue by saying: The crucial element that turns the questioning into information, enabling Sleeping Beauty to update her credence, is the fact that it results from randomly selecting an awakening from all possible awakenings. Given that there are twice as many awakenings under the 'tails' condition than under 'heads,' a random selection is twice as likely to yield a 'tails' awakening. We must recognize that Sleeping Beauty doesn't necessarily require external assistance to isolate her current awakening in a manner that is both random and statistically independent of the coin toss result.

Imagine an alternative method where an external agent, let's call her Sue, randomly selects awakenings from the complete set. Sue could examine a list of all scheduled awakenings, roll a die for each, and mark the awakening as evidence-worthy if the die lands on 'six'. The selected participants would then be equipped to update their credence P(H) to 1/3 after being presented with the evidence of their selection by Sue.

Now, it doesn't matter who performs the die-rolling selection; what's important is that any awakening marked as evidence-worthy is selected randomly by a method independent of the coin toss outcome. The participants themselves, not Sue, could roll the die and, if it lands on 'six,' consider their current awakening to have been randomly selected (as it would indeed have been!) from the entire set of awakenings. This random selection allows Sleeping Beauty to single out the fact of her current awakening as evidence for updating her credence P(H) to 1/3.

If the die doesn't land on 'six,' has Sleeping Beauty squandered an opportunity to identify her current awakening as a valuable piece of evidence? Actually, if the convention had been reversed to select awakenings by a die not landing on 'six', the chosen sample would still statistically represent all scheduled awakenings (with 1/3 of those being 'tails' awakenings). The Halfer's error is assuming that the mere occurrence of an awakening doesn't provide sufficient evidence for Sleeping Beauty. The participants' selection method, which involves identifying awakenings with the indexical expression "I am currently experiencing this awakening," is actually the most representative of all methods as it encompasses the entire population of awakenings!