Among other things, there I address Elga's model and explain why it's valid but not sound. His proof is about a different problem which I call "No-Coin-Toss problem":
You are put to sleep and then awaken according to the outcome of a random number generator. If the generator outputs 0, you are awaken on Monday and a coin is put Heads. If 1 - you are also awaken on Monday but the coin is put Tails. If 2 - you are awakened on Tuesday and the coin is also put Tails.
It's kind of obvious in retrospect, because according to Elga's model:
P(Heads)=P(Heads&Monday)=P(Heads|Awake)=1/3 that is, non-conditional probability of allegedly fair coin being Heads is 1/3, which can only happen when there was no coin toss in the first place.
> Therefore, the odds of each event is equal, and there are 3 events, so your credence in the coin having come up heads and it being the first day should be 1/3.
This is true when the events are mutually exclusive so that sum of their probabilities adds up to 1. Like, once again, in No-Coin-Toss problem. In Sleeping Beauty, however, this is not the case because Tails&Monday and Tails&Tuesday are sequential. You can't treat them as individual outcomes and formally construct a probability space.
Ironically enough, this is exactly what Elga's paper shows - that you can't lawfully use probability theory with the framework of "centred possible works" which David Lewis came up with no mathematical justifications. Because if you do, it inevitably leads to fair coin having arbitrary probability.
The question you quoted is the one Elga posed, but not quite the one he solved. You summarized his solution, but without introducing the differences that explain it. The two should be equivalent, but those differences opened the door to the controversy over his answer.
The one he posed does not differentiate the "two days" in any way. The one he solved assumed that you will always be wakened on the "first day," which he called Monday, and that you will be wakened on Tuesday only if the coin came up Tails. And what does "first awakened" mean, since you cannot determine if it is "the first day?" Elga meant before you might be told either that the coin came up Tails (premise #1), or that it is the first day (premise #2).
The interesting part of his setup, is that it creates the situation that obfuscates the solution. But he also uses it to avoid addressing that situation. Which is, what is the status of Tuesday, after Heads, in the solution? This obfuscates the solution since halfers will try to rationalize a relationship between H&Tue and T&Tue. Specifically, since any effect due to H&Tue is ignored, they want to ignore T&Tue the same way. I'll return to this.
But there is a better setup. Once you are fully asleep, the researchers flip two coins; call them C1 and C2. Then this procedure is performed:
Procedure: If both coins are showing Heads, the procedure is ended without waking you. But if either coin is showing Tails, you are wakened and asked "To what degree ought you believe that coin C1 is showing Heads?" After you answer, you are put back to sleep with amnesia.
Once this procedure is ended, and you are again fully asleep, coin C2 is turned over and the procedure is performed again.
When you are awake, which will happen once or twice, you know that at the beginning of the procedure there were four equally-likely combinations for C1C2. They were HH, HT, TH, and TT. But since you were awakened in this iteration of the procedure, you know that the two coins are not showing HH. Of the three remaining possibilities, which must still be equally likely, only one has C1=H. The answer is 1/3.
Returning to Elga's solution, in it there are four possible situations where something happens in the experiment: {H&Mon, T&Mon, H&Tue, T&Tue}. What confuses halfers is that being left asleep counts as "something." The day still occurs, you just won't observe it. But in fact, it doesn't matter if you are awakened, as long as something different happens when you are..
And the "somethings" that might happen on different days count as independent outcomes to you, since you cannot connect it to another day due to the amnesia drug. The "new information" that you have is that the sample space is reduced to {H&Mon, T&Mon, T&Tue}, each with 1/3 probability. And I can't emphasize this enough: By saying you have no "new information," Halfers are implicitly claiming that Tuesday CEASES TO EXIST if you sleep through it. Because if it does exist, and you know it is not your current situation, that is "new information."
What Elga did, in your first two premises, was reduce the sample space further by eliminated both H&Tue, which is already eliminated, and one additional situation. So he IGNORED H&Tue, while Halfers imply it cannot exist.
https://www.lesswrong.com/posts/SjoPCwmNKtFvQ3f2J/lessons-from-failed-attempts-to-model-sleeping-beauty
Among other things, there I address Elga's model and explain why it's valid but not sound. His proof is about a different problem which I call "No-Coin-Toss problem":
You are put to sleep and then awaken according to the outcome of a random number generator. If the generator outputs 0, you are awaken on Monday and a coin is put Heads. If 1 - you are also awaken on Monday but the coin is put Tails. If 2 - you are awakened on Tuesday and the coin is also put Tails.
It's kind of obvious in retrospect, because according to Elga's model:
P(Heads)=P(Heads&Monday)=P(Heads|Awake)=1/3 that is, non-conditional probability of allegedly fair coin being Heads is 1/3, which can only happen when there was no coin toss in the first place.
> Therefore, the odds of each event is equal, and there are 3 events, so your credence in the coin having come up heads and it being the first day should be 1/3.
This is true when the events are mutually exclusive so that sum of their probabilities adds up to 1. Like, once again, in No-Coin-Toss problem. In Sleeping Beauty, however, this is not the case because Tails&Monday and Tails&Tuesday are sequential. You can't treat them as individual outcomes and formally construct a probability space.
Ironically enough, this is exactly what Elga's paper shows - that you can't lawfully use probability theory with the framework of "centred possible works" which David Lewis came up with no mathematical justifications. Because if you do, it inevitably leads to fair coin having arbitrary probability.
I comprehensively talk about it here:
https://www.lesswrong.com/posts/gwfgFwrrYnDpcF4JP/the-solution-to-sleeping-beauty
The question you quoted is the one Elga posed, but not quite the one he solved. You summarized his solution, but without introducing the differences that explain it. The two should be equivalent, but those differences opened the door to the controversy over his answer.
The one he posed does not differentiate the "two days" in any way. The one he solved assumed that you will always be wakened on the "first day," which he called Monday, and that you will be wakened on Tuesday only if the coin came up Tails. And what does "first awakened" mean, since you cannot determine if it is "the first day?" Elga meant before you might be told either that the coin came up Tails (premise #1), or that it is the first day (premise #2).
The interesting part of his setup, is that it creates the situation that obfuscates the solution. But he also uses it to avoid addressing that situation. Which is, what is the status of Tuesday, after Heads, in the solution? This obfuscates the solution since halfers will try to rationalize a relationship between H&Tue and T&Tue. Specifically, since any effect due to H&Tue is ignored, they want to ignore T&Tue the same way. I'll return to this.
But there is a better setup. Once you are fully asleep, the researchers flip two coins; call them C1 and C2. Then this procedure is performed:
Procedure: If both coins are showing Heads, the procedure is ended without waking you. But if either coin is showing Tails, you are wakened and asked "To what degree ought you believe that coin C1 is showing Heads?" After you answer, you are put back to sleep with amnesia.
Once this procedure is ended, and you are again fully asleep, coin C2 is turned over and the procedure is performed again.
When you are awake, which will happen once or twice, you know that at the beginning of the procedure there were four equally-likely combinations for C1C2. They were HH, HT, TH, and TT. But since you were awakened in this iteration of the procedure, you know that the two coins are not showing HH. Of the three remaining possibilities, which must still be equally likely, only one has C1=H. The answer is 1/3.
Returning to Elga's solution, in it there are four possible situations where something happens in the experiment: {H&Mon, T&Mon, H&Tue, T&Tue}. What confuses halfers is that being left asleep counts as "something." The day still occurs, you just won't observe it. But in fact, it doesn't matter if you are awakened, as long as something different happens when you are..
And the "somethings" that might happen on different days count as independent outcomes to you, since you cannot connect it to another day due to the amnesia drug. The "new information" that you have is that the sample space is reduced to {H&Mon, T&Mon, T&Tue}, each with 1/3 probability. And I can't emphasize this enough: By saying you have no "new information," Halfers are implicitly claiming that Tuesday CEASES TO EXIST if you sleep through it. Because if it does exist, and you know it is not your current situation, that is "new information."
What Elga did, in your first two premises, was reduce the sample space further by eliminated both H&Tue, which is already eliminated, and one additional situation. So he IGNORED H&Tue, while Halfers imply it cannot exist.