Turn coin C2 over, to show its opposite side — JeffJo

OK, I understand your argument now, and it's what I said to you before: in your experiment the prior probability P(HH) = 1/4 is ruled out when woken; in the normal experiment where I am woken once if heads and twice if tails there is no prior probability P(X) = 1/4 that is ruled out when woken.

To make this fact clearer the experiment will be conducted in the simplest possible form:

1. Sleeping Beauty is given amnesia and asked her credence that the coin will or did land heads
2. The coin is tossed
3. If the coin lands heads then she is sent home
4. If the coin lands tails then she is given amnesia, asked her credence that the coin will or did land heads, and sent home

It said race could not be considered a reason to permit or deny admission into college under the Constitution. — Hanover

Unless it's a military academy, in which case they can.

GOP had evidence disproving Biden bribery claims in 2019, top Democrat says

The top Democrat on the House Oversight Committee has released evidence that casts significant doubt on GOP claims that the FBI ignored evidence that President Joe Biden accepted a bribe from a Ukrainian energy mogul during his time as vice president.

In a letter to House Oversight Committee chair James Comer, Maryland Representative Jamie Raskin reminded his GOP counterpart that Congress has had evidence “that directly contradicts the allegations” levied against Mr Biden in an FBI form which Republicans have claimed to be proof of alleged corruption on the part of the president.

“As part of the impeachment inquiry against then-President Trump, Congress learned that ... the Ukrainian oligarch and the owner of Burisma, whom Republican Committee Members appear to have identified as the source of the allegations memorialized in the Form FD-1023, squarely rebutted these allegations in 2019,” Mr Raskin said.

Prosecutors are prepared to hit Trump and his allies with new charges, sources say

The Department of Justice is prepared to seek indictments against multiple figures in former president Donald Trump’s orbit and may yet bring additional charges against the ex-president in the coming weeks, The Independent has learned.

According to sources familiar with the matter, the department has made preparations to bring what is known as a “superseding indictment” — a second set of charges against an already-indicted defendant that could include more serious crimes — against the ex-president in the Southern District of Florida.

But prosecutors may also choose to bring additional charges against Mr Trump in a different venue, depending on how they feel the case they have brought against him in is proceeding.

The Independent understands that prosecutors’ decision on whether to seek additional charges from a grand jury — and where to seek them — will depend in part on whether they feel the Trump-appointed district judge overseeing the case against him in the Southern District of Florida, Aileen Cannon, is giving undue deference to the twice-impeached, now twice-indicted former president.

The team of federal prosecutors working under Special Counsel Jack Smith is currently prepared to add an “additional 30 to 45 charges” in addition to the 37-count indictment brought against Mr Trump on 8 June, either in a superseding indictment in the same Florida court or in a different federal judicial district. In either case, they would do so using evidence against the ex-president that has not yet been publicly acknowledged by the department, including other recordings prosecutors have obtained which reveal Mr Trump making incriminating statements.

What I said in that post you dissected applies to one pass only, and the intent was to have two passes where, if there was no question in the first, there would be in the second. — JeffJo

So if the coin combination is HH then the participant will be asked their credence during the second pass? If so then you are wrong when you said "she also knows that the fact that she is awake eliminates (H,H) as a possibility."

And in the original, on Tuesday after Heads, you are also not asked for a credence. — JeffJo

What matters is the probability that you will be asked for your credence at least once during the experiment.

In the normal problem you are certain to be asked at least once during the experiment.

In your problem you are not certain to be asked at least once during the experiment.

Your problem isn’t equivalent and so the answer to your problem is irrelevant.

The subject in my implementation is always asked. — JeffJo

No they're not. I'll quote you:

1. Two coins will be arranged randomly out of your sight. By this I mean that the faces showing on (C1,C2) are equally likely to be any of these four combinations: HH, HT, TH, and TT.

2. Once the combination is set, A light will be turned on.

3. At the same time, a computer will examine the coins to determine if both are showing Heads. If so, it releases a sleep gas into the room that will render you unconscious within 10 seconds, wiping your memory of the past hour. Your sleeping body will be moved to a recovery room where you will be wakened and given further details as explained below.

4. But if either coin is showing tails, a lab assistant will come into the room and ask you a probability question. After answering it, the same gas will be released, your sleeping body will be moved the same way, and you will be given the same "further details."

The lab assistant only asks for my credence if the coin combination isn't HH.

If you take away this condition and so I am always asked my credence then the answer is 1/2.

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

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

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

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

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

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

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

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

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

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

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

That's not what I said.

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

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

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

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

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

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

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

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

This has nothing to do with credence.

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

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

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

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

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

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

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

Neither participant knows if they are A or B.

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

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

These are two different problems:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

This gives us:


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

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

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

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

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

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

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

The "elsewhere", e.g. anything outside my frame of reference, is incoherent to be talking about as it doesn't exist for me. — Benkei

This is where a realist would disagree. We can't know what's happening in the Andromeda Galaxy right now, but something is happening. Either "intelligent alien life exists in the Andromeda Galaxy" is (unknowably) true or "intelligent alien life doesn't exist in the Andromeda Galaxy" is (unknowably) true.

If we consider this in the context of presentism, the presentist would claim that only objects that exist in the present (including the present elsewhere) are real. But if B exists in A's present (elsewhere) and C exists in B's present (elsewhere) then C is real, even if C exists in A's future (elsewhere). And if special relativity is true then something like this is the case. Therefore presentism is false (as is the growing block universe theory). Or, given this fact, presentism and eternalism amount to the same theory.

Or are you an antirealist about events outside your (past?) light cone?

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. — Pierre-Normand

Because of what I said before:

A1. there are twice as many Tunisian walkers as Italian walkers
A2. therefore, if I go out and meet at random one of the walkers then I am twice as likely to meet a Tunisian walker
A3. therefore, there will be twice as many Tunisian-walker meetings

In Sleeping Beauty's case:

B1. Sleeping Beauty wakes up twice as often if the coin lands tails
B2. the coin is equally likely to land tails
B3. therefore, there will be twice as many tails awakenings

This argument is sound and fully explains the betting outcome. Your conclusion that "therefore, if Sleeping Beauty wakes up then the coin is twice as likely to have landed tails" just doesn't follow.

What would follow is "therefore, if I pick at random one of Sleeping Beauty's awakenings then it is twice as likely to be a tails awakening," but given that the experiment isn't conducted this way it doesn't make sense for Sleeping Beauty to reason this way to determine her credence.

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. — Pierre-Normand

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

The conclusion has to follow.

See also what I said before:

$\begin{aligned}P(TInterview | TRun) &= {{P(TRun | TInterview) * P(TInterview)} \over P(TRun)}\\\\P(TInterview) &= {{P(TInterview|TRun) * P(TRun)} \over P(TRun|TInterview)}\\\\P(TInterview) &= {{1 * P(TRun)} \over 1}\\\\P(TInterview) &= P(TRun)\end{aligned}$

(1) Relativity of simultaneity + all observers’ 3D worlds are real at every event = block universe

The argument on that page accepts that relativity of simultaneity is true but claims that "all observers’ 3D worlds are real at every event" is false because "Our intuitions don’t really know how to deal with “elsewhere”; it’s neither fixed and certain, since we can’t predict what happens there with certainty based only on the data in our past light cone, nor changeable since we can’t causally affect what happens there; we can only causally affect events in our future light cone."

This is a non sequitur. That an event cannot be predicted with certainty isn't that the event isn't certain. Or to phrase it another way, even if we cannot know (with certainty) whether or not "there is intelligent alien life in the Andromeda Galaxy" is true, it doesn't follow that it isn't true (or false).

And nobody is suggesting that it's changeable. In fact if the block universe is true then nothing is changeable; it just is what it is.

So the Andromeda Paradox is the claim that if "aliens are leaving Andromeda en route to Earth" is (unknowably) true in some reference frame then "aliens will leave Andromeda en route to Earth" is (unknowably) true in my reference frame.

And furthermore, that for every proposition "X will happen" either there is some reference frame A in my present elsewhere such that "X is happening" is (unknowably) true, and so "X will happen" is (unknowably) true in my reference frame, or there is no reference frame A in my present elsewhere such that "X is happening" is (unknowably) true, and so "X will happen" is (unknowably) false in my reference frame.

Donald Trump Fantasized About Having Sex With Ivanka, New Book Says

“Aides said he talked about Ivanka Trump’s breasts, her backside, and what it might be like to have sex with her, remarks that once led [former Chief of Staff] John Kelly to remind the president that Ivanka was his daughter,” Taylor, who served as a Department of Homeland Security chief of staff under Trump, wrote in his book.

“Afterward, Kelly retold that story to me in visible disgust,” Taylor writes. “Trump, he said, was ‘a very, very evil man.’

Surprising no-one.

We have two different experiments:

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

Your version of the experiment is comparable to the second experiment, not the first. Your case of HH is equivalent to the case that the coin landed heads and I am B. The very fact that I'm being asked my credence allows me to rule out one of the non-zero prior probabilities, i.e. P(HH) = 1/4 and P(Heads and B) = 1/4.

The second experiment is not equivalent to the first experiment. In the first experiment there is no non-zero prior probability that I can rule out when being asked my credence.

Given that I'm guaranteed to be woken up if heads in the first experiment but not the second (and guaranteed to be woken up if tails in both experiments) it makes sense that my credence in heads is greater in the first experiment than in the second. If I'm less likely to be woken up if heads then it's less likely to be heads if I'm woken up.

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

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

However, consider:

P4. If my sitter is assigned at random either a H-interview or a T-interview then his interview is half as likely to be a H-interview
P5. My sitter is assigned at random either a H-interview or a T-interview
P6. My interview is a H-interview iff my sitter's interview is a H-interview
C2. My interview is half as likely to be a H-interview

Prima facie the premises are true and the conclusion follows, therefore prima facie the conclusion is true. However, C1 and C2 are contradictory, therefore one of the arguments must be unsound.

Let's say that my sitter happens to be John:

P7. If John is assigned at random either a H-interview or a T-interview then his interview is half as likely to be a H-interview
P8. John is assigned at random either a H-interview or a T-interview
P9. My interview is a H-interview iff John's interview is a H-interview
C3. My interview is half as likely to be a H-interview

The issue is with P9. My interview is not biconditional with John's interview given that he is not guaranteed to be my sitter. That second argument commits a fallacy. P4 and P5 are true only under a de re interpretation of "my sitter" and P6 is true only under a de dicto interpretation.

This is why the participant shouldn't update his credence to match his sitter's.

I introduce the additional premise(s) because this is a non sequitur:

A1. there are twice as many Tunisian walkers as Italian walkers
A2. I am twice as likely to meet a Tunisian walker

If all the Tunisian walkers are in one area of the town but I'm walking in another then the conclusion is false. If, unknown to me, all the Tunisian walkers are wearing red and all the Italians wearing blue and I'm told to meet someone wearing blue if a coin lands heads or red if tails then the conclusion is false. If I don't go out to meet anyone then the conclusion is false. The experiment needs to be set up in such a way that the walkers are randomly distributed throughout the town and that I meet at random any one of the walkers. Only when this setup is established as a premise will the conclusion follow:

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

Similarly, this is a non sequitur:

B1. there are twice as many T-interviews as H-interviews
B2. my interview is twice as likely to be a T-interview

You would instead need something like:

C1. there are twice as many T-interviews as H-interviews
C2. if my interview is randomly assigned from the set of all interviews then my interview is twice as likely to be a T-interview

But this reasoning doesn't apply to Sleeping Beauty because her interview isn't randomly assigned from the set of all interviews.

For Sleeping Beauty the correct argument is the one that properly sets out the manner in which the experiment is conducted:

D1. there are an equal number of T-interview sets as H-interview sets
D2. 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
D3. I am assigned at random either a T-interview set or a H-interview set
D4. my interview is a T-interview iff my interview set is a T-interview set
D5. my interview is equally likely to be a T-interview

B is fallacious, C is inapplicable, and D is sound, hence why P(Heads|Awake) = 1/2 is the only rational conclusion. The fact that there are twice as many T-interviews as H-interviews is irrelevant. It's a premise from which no relevant conclusion regarding credence can be derived. It's only use is to explain betting outcomes.

In the cosmopolitan situation, the probability of meeting a Tunisian doubles because Tunisians are around twice as often. — Pierre-Normand

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".

We start with the mutually agreeable premise:

P1) there are twice as many T-awakenings

Your conclusion is:

C) T-awakenings are twice as likely

Obviously this is a non sequitur. We need the second premise:

P2) if there are twice as many T-awakenings then T-awakenings are twice as likely

This is something that I disagree with and that you need to prove.

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.

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. — Pierre-Normand

I believe this credence is based on fallacious reasoning as explained here.

Her reasoning is: if 1) there are twice as many Tunisian walkers and if 2) I randomly meet one of the walkers then 3) it is twice as likely to be a Tunisian walker.

Given the manner in which the experiment is conducted (2) is false and so this isn't the correct reasoning with which to determine one's credence.

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. (3) doesn't follow from (1) alone.

(2) is true for the sitter assigned an interview but not for the participant.

For the participant it is the case that 1) there are twice as many T-awakenings, 2) I randomly select one of the awakening sets, 3) it is equally likely to be a T-awakening set, and so 4) it is equally likely to be a T-awakening. (1) it turns out is irrelevant.

You can't just ignore (or change) the manner in which Sleeping Beauty participates in the experiment, which is what your various analogies do.

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. — Pierre-Normand

This goes back to my distinction between:

1. One should reason as if one is randomly selected from the set of all participants
2. One should reason as if one's interview is randomly selected from the set of all interviews

In the case where I go out and meet someone on the street it is certainly comparable to 2, and this is why when we consider the sitters it is correct to say that the probability that they are assigned a heads interview is 1/3.

But Sleeping Beauty isn't assigned an interview in the same way. It's not the case that there is one heads interview, two tails interviews, and she "meets" one of the interviews at random (such that P(T interview) = 2/3); instead it's the case that there is one heads interview, two tails interviews, and first she is assigned one of the interview sets at random (such that P(T interviews) = 1/2) and then she "meets" one of the interviews in her set at random.

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? That there are twice as many Tunisians isn't that her meeting a Tunisian is twice as likely.

We have two different experiments:

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

Given that I'm guaranteed to wake up if heads in the first experiment but not guaranteed to wake up if heads in the second experiment (and guaranteed to wake up if tails in both experiments) I think it only reasonable to conclude that P(Heads|Awake) in the first experiment is greater then P(Heads|Awake) in the second experiment.

And given that P(Heads|Awake) = 1/3 in the second experiment I think it only reasonable to conclude that P(Heads|Awake) > 1/3 (i.e. 1/2) in the first experiment.

These are two different sets of claims:

A1. there are twice as many Tunisian-meetings because Tunisian-meetings are twice as likely
A2. Tunisian-meetings are twice as likely because there are half as many Tunisians and Tunisians go out four times more often

B1. there are twice as many T-awakenings because T-awakenings are twice as likely
B2. T-awakenings are twice as likely because Sleeping Beauty is woken twice as often if tails

"there are twice as many T-awakenings" is biconditional with "Sleeping Beauty is woken twice as often if tails" and so B uses circular reasoning.

"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.

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 — Pierre-Normand

This is just repeating the same thing in a different way. That there are twice as many T-awakenings just is that Sleeping Beauty is awakened twice as often if tails. So your reasoning is circular.

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? — Pierre-Normand

I've since edited my post to make my point clearer. To repeat:

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".

I am unsure what it is that you are asking here. — Pierre-Normand

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?

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\\\&={{P(Awake | Heads)*{1\over3}}\over{P(Awake)}}\end{aligned}$

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. — Pierre-Normand

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".

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? — Pierre-Normand

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:

$\begin{aligned}P(Heads | Awake) &= {{P(Awake | Heads) * P(Heads)} \over P(Awake)}\\&={{{1\over2}*{1\over2}}\over{3\over4}}\\&={1\over3}\end{aligned}$

This isn't comparable to the traditional probem.

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

I'm not seeing it. Light cones concern causal past and causal future. The Rietdijk–Putnam argument and Andromeda Paradox concern events outside the light cone.

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.

And let's take it further and consider this:



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.