I’ve spend sometime looking into the issue and I’m quite confident that this proof isn’t sound. Is it already a known fact?
[Question] Is Adam Elga’s proof for thirdism in Sleeping Beauty still considered to be sound?
I think a link or summary of the “proof”, and description of the parts of it that you’re wondering about, would go a long way toward making this a useful question.
I couldn’t find a concise summary of exactly which argument you’re asking about, so I’ll just say that all probability is in the agent’s map—it’s a prediction of future experiences. There may or may not be an underlying “reality”, but even if there is, there’s no access to any outside observer.
So 1⁄2 or 1⁄3 are valid answers to different questions.
I’ve reviewed again the argument in Section 2 of Elga’s paper (https://www.princeton.edu/~adame/papers/sleeping/sleeping.pdf) and it seems valid to me. It doesn’t have the same simplicity of some other arguments for 1/3, though, such as the betting arguments, and my argument in https://glizen.com/radfordneal/anth.abstract.html that the probability of anyone having your detailed experience on awakening is twice are large if you are woken twice.
There are many good arguments for 1⁄3. The arguments for 1⁄2 are all bad, and defending them forces people into contorted positions such as that there is no actual probability in this situation, or the probability depends on what class of people you see yourself as belonging to, or as seen in a comment below, that 1⁄2 and 1⁄3 are both valid answers. (Regarding the later, Beauty can decide to break down the door of the room, go out, and ask someone what day it is. There is nothing ambiguous about asking what the probability is that she will find out that it is Monday.)
For a contrary position, see https://philpapers.org/archive/BURSBR-2.pdf (which I just found by a search, and haven’t read yet), which however does acknowledge that 1/3 is the favoured position.
I stand by my “valid answers for different questions” position, but it’s pretty difficult to swallow “The arguments for 1⁄2 are all bad”, without being clear that you’re excluding the simplest of questions: if nothing changes from the setup, what is Beauty’s expected experience of being told on Wednesday what the results were?
The chance of a fair coin coming up tails is 1⁄2. Beauty has no evidence on waking up to alter that, as she’d experience that either way.
There are OTHER questions, and betting-odds considerations that lead to 1⁄3 being absolutely correct. I don’t dispute that in any way, but I do object to claims that it’s the only answer, without specifying further which questions she’s answering.
I’m not sure what you mean by “Beauty’s expected experience … on Wednesday”. There are two possibilities. On Sunday, she will expect that they have equal probabilities. But that says nothing more than that on Sunday she expects Heads and Tails to be equally likely. When woken on Monday or Tuesday, here expectation of what Wednesday will be like isn’t necessarily the same.
I also can’t figure out what you mean by saying that 1⁄3 is “absolutely correct”, while also saying that there could be another answer.
What’s Beauty credence for Heads when she wakes on Wednesday and doesn’t remember any of her awakenings on Monday/Tuesday? If it’s 1⁄2 what is the reason for the change from 1/3? The fact that she has forgotten the information about her awakening during the experiment?
Suppose SB wakes up on Wednesday. She is told that it’s Wednesday and then left alone to dress before she will be asked for the last time to guess Heads or Tails. She doesn’t remember any of her awakenings and her credence for Heads is once again 1⁄2. However she finds a piece of paper hidden in her pajamas from the past version of herself, experiencing an awakening. This past version managed to cheat and send a message to the future where she claims that her current credence for Heads is 1⁄3. Should she expect to successfully guess Tails with 2⁄3 probability?
Or suppose that SB is given memory loss drug only before she is awaken on Tuesday so that on Wednesday she always remembers her last awakenings though she doesn’t know whether it was on Monday or Tuesday. Is her credence for Heads still 1/3?
What’s Beauty credence for Heads when she wakes on Wednesday and doesn’t remember any of her awakenings on Monday/Tuesday? If it’s 1⁄2 what is the reason for the change from 1/3? The fact that she has forgotten the information about her awakening during the experiment?
Because of the memory erasure, I think it’s best to regard Beauty on Monday, on Tuesday (if woken), and on Wednesday as different people, all “descended” from Beauty on Sunday. But if you want to think of them as the same person, the change from 1⁄3 to 1⁄2 is a consequence of the new information from being told that it is Wednesday. The same shift happens if she’s told that it is Monday. (If she’s told it is Tuesday, she of course shifts to probability 1 for Tails.)
In the standard setup, by the way, I think it’s implicit that on Monday and Tuesday awakenings, Beauty is told that it is not Wednesday (ie, that it is either Monday or Tuesday).
Suppose SB wakes up on Wednesday. She is told that it’s Wednesday and then left alone to dress before she will be asked for the last time to guess Heads or Tails. She doesn’t remember any of her awakenings and her credence for Heads is once again 1⁄2. However she finds a piece of paper hidden in her pajamas from the past version of herself, experiencing an awakening. This past version managed to cheat and send a message to the future where she claims that her current credence for Heads is 1⁄3. Should she expect to successfully guess Tails with 2⁄3 probability?
Assuming that successfully hiding a piece of paper is an unlikely event, she should think the probability of Heads is 1⁄3 simply because that makes a hidden piece of paper twice as likely (since then her past self will have had two chances to do this). The message on the paper is irrelevant (assuming it doesn’t contain any new arguments that she hadn’t thought of on Sunday).
Or suppose that SB is given memory loss drug only before she is awaken on Tuesday so that on Wednesday she always remembers her last awakenings though she doesn’t know whether it was on Monday or Tuesday. Is her credence for Heads still 1/3?
That’s an interesting scenario. My argument for why Beauty when not told it is Wednesday should think Tails is twice as likely as Heads is that with Tails (hence two awakenings) the probability of anyone experiencing what she is (eg, exact position of fly seen on the wall, stray thought about how good omelettes taste, …) is twice what it is with Heads (one awakening), so standard Bayesian updating makes Tails twice as likely. When woken, told it is Wednesday, and having memories of her last awakening, however, the probability of her experience (including memories) is the same for Heads or Tails. The two awakenings with Tails don’t double the probability of her remembered experience being whatever it is (along with knowing it is Wednesday), since whatever she experienced on Monday is forgotten. So her probability for Heads should be 1⁄2.
How does it work? Especially in the scenario where SB doesn’t loose her memory of the last awakening. What kind of Bayesian update is it? Why are coins less likely to turn out to be Tails On Wednesday than on Monday or Tuesday? Does the universe retcon the result of the coin toss somehow?
Let’s look at the scenario where Beauty remembers her last awakening.
When Beauty is woken and not told it’s Wednesday, she should think Heads has probability 1⁄3, Tails 2⁄3. She knows that if the coin landed Heads, it is Monday, and that she will next wake up on Wednesday without forgetting anything. She knows that if the coin landed Tails, then with probability 1⁄2 it is Monday and her memory will soon be erased, and with probability 1⁄2 it is Tuesday and she will next wake up on Wednesday without forgetting anything.
So waking up on Wednesday without forgetting anything is twice as likely for Heads as for Tails. Hence, if that’s what happens, she should update her probability of Heads from 1⁄3 to to 1⁄2 (ie, odds for Heads change from 1⁄2 to 1, changing by a factor of two due to the likelihood ratio of two).
But what about if instead her memory is erased before a Tuesday awakening? In that case, this instance of Beauty effectively dies, and there’s nothing to say about what her probability of Heads is.
Compare with the situation where someone plays Russian roulette with a revolver that they know has probability 1⁄2 of being empty, and probability 1⁄2 of having bullets in 3 of the 6 positions. If they survive, they should update their probability of the revolver being empty to 2⁄3 (ie, odds for empty change from 1 to 2), since they are twice as likely to survive if it is empty as if it has 3 of 6 positions with bullets.
Similarly, if Beauty mistakenly thinks when woken and told it’s not Wednesday that Heads and Tails both have probability 1⁄2, then if she is woken on Wednesday without forgetting anything, she should by the rules of probability change her probability of Heads to 2⁄3. And that is obviously wrong.
Indeed. This I understand.
And here I stopped following you. There is exactly one person in her epistemic situation (Waking up on Wednesday remembering the previous awakening) in both Heads and Tails worlds. According to both SSA and SIA no update has to happen.
There are two ways one might try to figure out these probabilities. One is that, in whatever final situation is being considered, you figure out the probability from scratch, as if the question had never occurred to you before. The other is that as you experience things you update your probability for something, according to the the likelihood ratio obtained from what you just observed, and in that way obtain a probability in the final situation.
When figuring out the probability of Heads from scratch on Wednesday, I think everyone agrees that it should be 1⁄2. The only issue from my perspective is that one might think that my argument for Heads having probability 1⁄3 when Beauty is woken before Wednesday still applies, but as I explained above, it doesn’t.
So the question now is whether this is consistent with updating the probability of Heads from a value of 1⁄3 when Beauty is woken before Wednesday, assuming that this instance of beauty does not have her memory erased before being woken on Wednesday. Not having her memory erased and then being woken on Wednesday is twice as likely for Heads as for Tails, so yes, the probability does get updated to 1⁄2, consistent with the “from scratch” method.
Note that it makes no sense to ask how an instance of Beauty whose memory has been erased will “update” her probability of Heads—she doesn’t even remember what “her” (really someone else’s) previous probability of Heads was.
As far as I can tell, you don’t seem to be updating based on the ratio of likelihoods from what is observed, but I don’t know what method you are using. SSA and SIA aren’t usually phrased as methods for updating probabilities over time as new information arrives. And even if they were, it would certainly be controversial to say that they should take precedence over standard Bayesian reasoning.
But in any case, I don’t agree that “there is exactly one person… waking up on Wednesday remembering their previous awakening”. There is of course only one person who is actually woken on Wednesday, but I take it you mean potential persons, who might have been the one woken on Wednesday. There is Beauty on Sunday, who if the coin lands Heads will not have her memory erased at any point and will then be woken on Wednesday. Alternatively, when the coin lands Tails, there is Beauty as woken on Tuesday, who from that point on does not have her memory erased and is woken on Wednesday. The Beauty who is woken on Monday and then has her memory erased is effectively dead.
Ape in the coat: “What’s Beauty credence for Heads when she wakes on Wednesday and doesn’t remember any of her awakenings on Monday/Tuesday?”
The snarky answer is “irrelevant.” The assessment of her credence, that she gives on Monday or Tuesday, is based on the information that it is either Monday or Tuesday within the waking schedule described in the experiment. She is not responsible for incorporating false information into her credence.
One thing that seems to get lost in many threads about this problem, is that probability and/or credence is a function of your knowledge about the state of the system. It is not a property of the system itself. You are trying to make it a property of the system.
Halfers are most guilty of not understanding this, since their arguments are based on there being no change to the system despite an obvious change to the knowledge (I’m ignoring how to treat that knowledge). Thirders recognize the change, but keep trying to cast the difference in terms of episiotomy when it really is much simpler.
“However she finds a piece of paper hidden in her pajamas from the past version of herself, experiencing an awakening. This past version managed to cheat and send a message to the future where she claims that her current credence for Heads is 1⁄3. Should she expect to successfully guess Tails with 2⁄3 probability?”
No, since her knowledge is different now than it was then.
“Or suppose that SB is given memory loss drug only before she is awaken on Tuesday so that on Wednesday she always remembers her last awakenings though she doesn’t know whether it was on Monday or Tuesday. Is her credence for Heads still 1/3?”
If I understand this correctly, she knows that it is Wednesday. Her credence is 1⁄2. She can, at the same time on Sunday or Wednesday, make an assessment of the state of her knowledge on Monday or Tuesday, and recognize how it is different. In other words, “my credence now is 1⁄2, but my credence then should be 1⁄3.” The paradox you seem to be fishing for does not exist.
Your knowledge about the state of the system can either correctly represent the state of the system or not. If you believe that the probability that the coin is Tails is 2⁄3, while on a repeated experiment about 50% times the coin is Heads that means that your probability estimate is not correct. Of course you can declare that your credence is answering some other question and this is fine. But there is a potential for misunderstanding—people can mistakenly think that they can actually guess the result of the coin toss better than chance as if they received some relevant evidence.
I explore this dynamics in my recent post.
Whether your knowledge correctly represents the state of the system, or not, is irrelevant. Your credence is based on your knowledge. If they lie to SB and wake her twice after Heads, and three times after Tails? But still tell here it is once or twice? A thirder’s credence should still be 1⁄3.
And those who guess would not be considered the rational probability agent that some versions insist SB must be.
The correct answer is 1⁄3. See my answer to this question.
Well, when you explicitly acknowledge that “2/3 credence for Tails” doesn’t correspond to Tails being the result of the coin toss in 2⁄3 of experiments where the beauty is thinking this way, we have cleared the possible misunderstanding. You are talking about the motte not the bailey. We can still argue about whose definitions are better, and I’d like to make a few points about it. But essentially the crux is resolved.
In such mind experiments the possibility of lies is usually stated explicitly. Like in the riddle of three idols. And people do not usually claim that if the person trusted the lying idol their belief is “correct” and it’s not their fault they got lied to.
When the possibility of a lie wasn’t explicitly stated but the lie happened, then your point is valid. We can say that the credence corresponded to a different experiment setting, thus in this sense can still be correct, even if it’s not correct in the new experimental setting.
However, this is absolutely not the case here. No one is lying to the beauty. The experiment is going exactly is stated. So if beauty’s knowledge doesn’t represent the state of the system, it means she made some incorrect inference from her evidence which is her failure as a rational agent.
When you declare that map representing the territory is irrelevant it’s not clear how “correct” is a meaningful term anymore. I suppose you mean that it corresponds to your mathematical model. This is true but you can always find some mathematical model that your answer corresponds to. The interesting question is whether this model corresponds to reality.
My point is that SB must have reason to think that she exists in the “Monday or Tuesday” waking schedule, for her to assign a credence to Heads based on that schedule. If she is awake, but has any reason to think she is not in that situation, her credence must take that into account.
You told her that she would be asked for her credence on Monday and maybe on Tuesday. “What’s Beauty credence for Heads when she wakes on Wednesday and doesn’t remember any of her awakenings on Monday/Tuesday?” is irrelevant because you are allowing for the case where that is not true yet you want her to believe it is. That is lying to her, in the context of the information you want her to use.
But she can from an opinion. “My credence should be halfer/thirder answer if Wednesday has not yet dawned, or 1⁄2 if it has. Since the cue I was told would happen—being asked for my credence—has not yet occurred, I am uncertain which and so can’t give a more definitive answer.” And if you give that cue on Wednesday, you are lying even if you promised you wouldn’t.
And yes, my mathematical model corresponds to SB’s reality when she is asked for her credence. That is the entire point. If you think otherwise, I’d love to hear an explanation instead of a dissertation that does not apply.
No lies are necessary. I didn’t have to tell her beforehand that she will be asked any questions at all. Or I could have told her that she will be asked on Monday and Tuesday (if she is awake) without knowing which day it is and then she will be asked on Wednesday, knowing that it is indeed Wednesday. None of it changes the experiment. And even if I didn’t ask her about her credence that the coin is Heads on Wednesday, she still has to have some probability estimate doesn’t she?
The point of the Wednesday question is to highlight, that, what you mean by “credence”, isn’t actually a probability estimate that the coin is Heads. What you are talking about, is the probability that the coin is Heads, weighted by the number of times this question is asked. Which is a meaningful category. But confusing it with the probability that the coin is Heads can be extremely misleading
AINC: “What’s Beauty credence for Heads when she wakes on Wednesday and doesn’t remember any of her awakenings on Monday/Tuesday?”
If she has no reason to think this is not one of her awakenings on Monday/Tuesday, then her credence is the same as it would have been then: 1⁄2 if she is a halfer, and 1⁄3 if she is a thirder.
AINC: “If it’s 1⁄2 what is the reason for the change from 1/3?”
The only way it could change from 1⁄3 to 1⁄2, is if she is a thirder and you tell her that it is Wednesday. And the reason it changes is that you changes the state of her information, not because anything about the coin itself has changed.
But if you think her credence should be based on the actual day, even when you didn’t tell her that is was Wednesday, then you have told an implicit lie. You are asking her to formulate a credence based on Monday/Tuesday, but expecting her answer to be consistent with Wednesday.
AINC: The point of the Wednesday question is to highlight, that, what you mean by “credence”, isn’t actually a probability estimate that the coin is Heads.
And the point of my answer, is that it is actually a conditional probability based on an unusual state of information.
The “valid answers for different questions” position is a red herring, and I’ll show why. But first, here is an unorthodox solution. I don’t want to get into defending whether its logic is valid; I’m just laying some groundwork.
On Sunday (in Elga’s re-framing of the problem he posed—see my answer), the sample space for the experiment seems to be {T,H} with equal probability for each outcome. Both Monday and Tuesday will happen, and so cannot be used to specify outcomes. This is one of your “different questions.”
But when SB is awake, only one of those days is “current.” (Yes, I know some debate whether days can be used this way—it is a self-fulfilling argument. If it is assumed to be true, you can show that it is true.) Elga was trying to use the day as a valid descriptor of disjoint outcomes, so that in SB’s world the outcomes T&Mon and T&Tue become distinct. The prior probabilities of the three disjoint outcomes {T&Mon, T&Tue, H} are each 1⁄2, and the probability for H can be updated by the conventional definition:
Pr(H|T&Mon or T&Tue or H) = Pr(H)/[Pr(T&Mon)+Pr(T&Tue)+Pr(H)]
Pr(H|T&Mon or T&Tue or H) = (1/2)/[(1/2)+(1/2)+(1/2)] = 1⁄3
If you think about it, this is exactly what Elga argued, except that he didn’t take the extreme step of creating a probability that was greater than 1 in the denominator. He made two problems, that each eliminated an additional outcome. I suggest, but don’t want to get into defending, that this is valid since SB’s situation divides the prior outcome T into two that are disjoint. Each still has the prior probability of the parent outcome.
But this is predicated on that split. Which is what I want to show is valid. What if, instead of leaving SB asleep on Tuesday after Heads, wake her but don’t interview her? Instead, we do something quite different, like take her on a $5,000 shopping spree at Saks Fifth Avenue? (She does deserve the chance of compensation, after all). This gives her a clear way to distinguish H&Mon&Interview from H&Tue&Saks as disjoint outcomes. The use of the day as a descriptor is valid, since this version of the experiment allows for it to be observed.
The actual sample space, on Sunday, for what can happen in an “awake” world for SB is {T&Mon, H&Mon, T&Tue, H&Tue&Saks}. Each has a prior probability of 1⁄4, for being what will apply to SB on a waking day. When she is actually interviewed, she can eliminate one.
And now I suggest that it does not matter, in an interview, what happens differently on H&Tue. SB is interviewed in the context of an interview day, so the question is placed in that context. Claiming it has the context of Sunday Night ignores how the Saks option can only be addressed in a single-day context, and can affect her answer.
Much of the debate gets obscured in the question of outside (of SB’s perception) information. That’s really the purpose of the thought experiment—if the memory wipe is complete and SB has literally no way of knowing whether it’s monday or tuesday, why are they considered (by her) to be different experiences? For purposes of sampling, there is only one waking, the one she’s experiencing.
Alternate formulations (especially frequentist models where there is some “truth” to probability, as opposed to it being purely subjective) can be different, but only by introducing some distinction of experience to make them into two wakings, somehow.
My view is that to an outside observer, the probability is 1⁄2 before the flop, 1 or 0 after the flip. To Beauty, it remains 1⁄2 until she receives evidence, in the form of something she can observe that would be different with heads than with tails. She should WAGER 1⁄3 in some formulations (for instance, if the wagers for Monday and Tuesday are summed, rather than being one evaluation).
I’m sure you’d agree that, if the room has a calendar, Beauty should assign 1⁄2 on Sunday and Monday, and 0 on Tuesday. Never 1⁄3, right?
Two points…
Like a number of people (including Elga), you’re converting an almost-doable thought experiment into one that may be impossible in principle. If the experiment is done with some not-yet-invented but plausible memory-erasing drug, but is otherwise realistic, Beauty will not have the same experiences when woken Monday and when woken Tuesday. Various aspects of her sensed environment, as well as internal thoughts, will be different for the two awakenings. We just assume that none of these differences allow her to infer the day of the week.
Second, if she should wager as if the probability of Heads is 1⁄3, in what sense is the probability of Heads not actually 1/3? The only point of probabilities is to influence actions. (Also, it is not just “some” situations where she should wager on the basis that the probability of Heads is 1⁄3. That’s always what she should do.)
No, much of the debate gets obscured by trying to ignore how SB’s information, while it includes all of the information that will be used it separate the experiment into two observations, also is limited to the “inside information” and she is in just one of those parts. That’s the purpose of the shopping spree I suggested. Correcting what you wrote, if the memory wipe is complete and SB has literally no way of knowin what information might apply to what is clearly a distinct part of the experiment, but she knows there is a different part than the current one, how can it not be “considered (by her) to be a different experiences.”
For the purposes of sampling, there are two parts to the experiment. What you call them is not relevant, only that SB knows that she is in one, and only one, of two parts. In the classic version, as Elga modifided it from the actual problem he proposed, one part must have a waking and one part has either a sleeping or a waking. Are you really claiming that the part where she is left asleep is not a part of the experiment? One that she knows is a possibility that is ruled out by her current state? And so fits the classic definition of “new evidence” that you deny? But if you really believe that, what about my version where she is taken shopping? It is even better as a classic example of evidence.
Or you could look at my answer. There, I explain how the problem you are solving actually is an “alternate formulation.” And, while there are still two parts, they are equivalent so we do not need to treat them separately. The answer is unequivocally 1⁄3.
Or try this: Instead of a coin, roll two six-sided dice. On Monday, ask her for the probability that the resulting sum is 7. On Tuesday, if the sum is odd, ask her the same question. But if it is even, ask her for the probability that the sum is 8.
If the room has a calendar, Beauty should certainly say Pr(S=7)=1/6 on Sunday and Monday. But on Tuesday, in answer to “what is the probability of 7,” she should say Pr(S=7)=1/3. But without a calendar, Pr(S=7)=1/6 can’t be right. Because it might be Tuesday, and Pr(S=7)=1/3 can’t be right. And it might be Monday, when Pr(S=7)=1/6 can’t be right. It has to be something in between, and that something is (2/3)*(1/6)+(1/3)*(1/3)=4/18=2/9. Yes, even though this is never the answer if she knows the day. That is how conditional probability works.
I can’t tell you whether it is considered to be sound. In my opinion, it is. But I do know that the issues causing the controversy to continue for 23 years are not a part of the actual problem. And so are unnecessary.
If anyone thinks that sounds odd, they should go back and read the question that Elga posed. It does not mention Monday, Tuesday, or that a Tails-only waking follows a mandatory waking. Those were elements he introduced into the problem for his thirder solution.
In the problem as posed, the subject (I’ll call her SB, even though in the posed problem it is “you”) is woken once, or twice, based on the outcome of a fair coin flip (Heads=once, Tails=twice). Elga enacted that description with a mandatory waking on Monday, and an optional one on Tuesday. This way, could create two partial solutions by revealing two different bits of information that removed one of the three possibilities. The unfortunate side effect of this was that the conditions surrounding Monday and Tuesday are different; thirders require Tuesday to be part of a different outcome, and halfers insist it is the same outcome as Monday+Tails.
And that is what is unnecessary. Instead of that Monday/Tuesday schedule, simply flip two coins (call them C1 and C2) after SB is first put to sleep. Then:
If both coins are showing Heads, skip to step 6.
Wake SB.
Ask SB for her credence that coin C1 is showing Heads.
Wait for her answer.
Put her back to sleep with amnesia.
End this stage of the experiment.
Now turn coin C2 over to show its other face, and repeat these steps.
When SB is woken, she knows that: (A) She is in step 2 of a pass thru these six steps. (B) In step 1, there were four equally-likely states for the two coins: HH, HT, TH, and TT. (C) Since she is awake, the state HH is eliminated but the other three remain equally likely. (D) In only one of those states is C1 showing Heads. So she can confidently state that her credence is 1⁄3.
The difference between this, and how Elga enacted the problem he posed, is that here there is no ambiguity about how she arrived at her current, awake situation.
I originally upvoted your answer as it presented an interesting version of the SB which I didn’t see before.
However, it has similar problems to the original and it’s even more convoluted, you may notice that there is no need to throw the second coin other than to confuse everyone. You can just put it tails and then turn over to get the exact same results.
HH, HT, TH and TT are not four elementary outcomes of the experiment, as there is causal connection between them, even if such formulation makes it less obvious and harder to talk about it.
The “need to throw the second coin” is to make the circumstances underlying any awakening the same. Using a random method is absolutely necessary, although it doesn’t have to be flipped. The director could say that she is choosing her favorite coin face. As long as SB has no reason to think that is more likely to be one result than the other, it works. The reason Elga’s version is debated, is because it essentially flips Tails first for the second coin.
What the coins are showing at the moment are elementary outcomes of the experiment-within-the-experiment. “Causal connection,” whatever you think that means, has nothing to do with it since we are talking about a fixed state of the coins from the examination to the question.
Which statement here to you disagree with?
Just before she was awakened, the possible states of the coins were {HH, HT, TH, TT} with known probabilities {1/4, 1⁄4, 1⁄4, 1⁄4}.
SB knows this with certainty. That is, she did not have to be awake at that moment to know that this was true at that time.
With no change to the coins, they were examined and the next part of the experiment depends on the actual state.
SB knows this with certainty.
Since she is awake, she knows that the possible change is that the state HH is eliminated.
SB knows this with certainty.
The probabilities in step 1 constitute prior probabilities, and her credence in the state is the same as the posterior probabilities
Pr(HT|Awake) = Pr(HT)/[Pr(HT)+Pr(TH)+Pr(TT)] = 1⁄3.
But there are other ways, that have whatever “causal connection” you think is important. That is, they match the way Elga modified the experiment. That’s another point you seem to ignore, that the two-day version differs from the actual question by more than mine does.
Try using four volunteers instead of one, but one coin for all. Each will be wakened on both Monday and Tuesday, except:
SB1 will be left asleep on Tuesday, if the coin landed on Tails. This is Elga’s SB, with the same “causal connection.”
SB2 will be left asleep on Tuesday, if the coin landed on Heads.
SB3 will be left asleep on Monday, if the coin landed on Tails.
SB4 will be left asleep on Monday, if the coin landed on Heads.
This way, on each of Monday and Tuesday, exactly three volunteers will be wakened. And unless you think the specific days or coin faces have differing qualities, the same “causal connections” exist for all.
Each volunteer will be asked for her credence that the coin landed on the result where she would be wakened only once. With the knowledge of this procedure, an awake volunteer knows that she is one of exactly three, and that their credence/probability cannot be different due to symmetry. Since the issue “the coin landed on the result where you would be wakened only once” applies to only one of these three, this credence is 1⁄3.
I assume you mean from https://www.princeton.edu/~adame/papers/sleeping/sleeping.pdf ? The one about “if you knew it was Monday it would be 0.5, surely adding Tuesday as an option causes an update,” followed by a standard claim that symmetry implies 1/3?
Both parts are independently correct. But I think there’s probably not enough step-by-step talk about symmetry for this to serve the social role of a proof for confused people. I’ve found it’s important, when explaining anthropics, to really focus on “centered possible worlds” (or hypotheses that reproduce your experiences, to frame it more like Solomonoff induction), and why a question like “what is the universe like outside my head?” can be answered by considering hypotheses anew when you have different memories.