The first paper on the sleeping beauty problem, to the best of my knowledge, was by Adam Elga. In it, Elga argued that thirding is the right answer. His paper is worth reading in full, being clear and direct, numbering at only six pages. Elga’s argument seems demonstrably correct and yet isn’t discussed much these days in the discussion of the sleeping beauty problem. Elga begins his paper by presenting the problem:

The Sleeping Beauty problem: Some researchers are going to put you to sleep. During the two days that your sleep will last, they will briefly wake you up either once or twice, depending on the toss of a fair coin (Heads: once; Tails: twice). After each waking, they will put you to back to sleep with a drug that makes you forget that waking.2 When you are first awakened, to what degree ought you believe that the outcome of the coin toss is Heads?

He argues that you should think the odds are 1/3. There are two basic premises:

The odds that it is the first day and the coin came up tails = the odds that it is the second day and the coin came up tails.

The odds that it is the first day and the coin came up tails = the odds that it is the first day and the coin came up heads.

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.

The first premise is quite trivial. If you know the coin came up tails, you have no evidence for it being either the first or second day, so you should be indifferent between the two of them. This is just like, if you know it’s the weekend but don’t know what day it is, you should think it being Saturday and Sunday are equally likely.

The second premise is also straightforward enough. Suppose that whether the coin was flipped heads or tails, you’d only be awoken once. Well then upon waking up, you should assign each result a 50% probability. But what will happen the second day shouldn’t affect your credence today. Finding out, after learning it’s the first day, that you’ll be awoken again tomorrow shouldn’t affect your credence in the coin having come up heads or tails. The presence of future wakings shouldn’t influence your credence in current wakings.

Here’s another way to see it—suppose that the coin flip will happen at the end of the first day. You wake up and don’t know it’s the first day. You’re asked to estimate the odds that the coin that is flipped at the end of the first day comes up heads (not you don’t know whether this has already happened or whether it will happen). On the halfer view, you should be 50/50. Then you learn it’s the first day. Conditional on the hypothesis that the coin flipped at the end of the first day comes up heads, the odds it would be the first day are 50%. So, therefore, you should now think the odds the coin will come up heads are 66.66666666666%. But this means that your credence in the coin coming up heads, that you know hasn’t been flipped yet, depends on its consequences. If instead you learned that there wouldn’t be a second waking conditional on the coin coming up tails, then you should go back to being 50/50. But this is crazy! Your credence in a future event shouldn’t depend, in this way, on its consequences. If someone is about to flip a coin, the odds that it will come up heads shouldn’t depend on what it will do if it comes up heads.

Seriously imagine just how weird the other view is on credences. After waking up, you should be 50/50 on the coin that will be flipped at the end of the day coming up heads and tails. But then, after learning that you’ll be woken up tomorrow if it comes up tails, you should now think it will probably come up heads. That’s crazy!

I basically think this settles the issue.

edited Mar 22https://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.