I've been reading the Sleeping Beauty problem wiki. The contradicting answers, to me, appear to stem from frequentist and Bayesian interpretations:
The "thirder" solution refers to a limit in an infinite number of identical experiments. If Sleeping Beauty says tails, she will be correct in $1/3$ of such experiments.
The "halfer" solution is that Sleeping Beauty has no information with which to update her Bayesian prior, which ought, therefore, to remain at $1/2$.
This is paradoxical. I can't understand, however, why in the "halfer" solution, $1/2$ was a sensible prior? If it is simply insufficient reason, it isn't that surprising that the results contradict each other. I could by insufficient reason guess that the probability of two disjoint outcomes is $1/2$. If the outcomes are rolling $1$-$5$ or a $6$ on a dice I would be wrong. But that is not a paradox.
If Sleeping Beauty chooses $1/2$ because she knows that $p(h)=p(t)=1/2$, i.e. because she is cognizant of the details of the experiment, so why can't she also follow the "thirder" reasoning and have her prior as $1/3$?
EDIT: I now see that before the experiment SB is asked "What is your belief in heads?", not "What would be your belief in heads upon waking?" So my objection doesn't make sense. But it's not clear how from being woken SB can update her prior. I now fully see the paradox!
Imagine that the Sleeping Beauty Problem is repeated with one minor change: she is always wakened, but the room she is in depends on the coin and the day. The room is painted red unless Heads was flipped and it is Tuesday. Then she is wakened in a room that is painted blue. And she knows these details. What should she say the probability of Heads is if she finds herself in a red room?
Nobody has a problem with her claiming answer is 1/3. There originally were four equally-likely situations, one is eliminated, and the remaining three are still equally likely. But some refuse to accept that when she can have no knowledge of any other situation because of the amnesia drug, it doesn't matter whether she can observe them (because she is awake) or not (because she is asleep). Each other situation- even if she is asleep - still represents an observation opportunity that she knows is different from her current one.
This variation is identical to the classic Sleeping Beauty Problem, and the answer is 1/3. To both.
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Here's another variation that proves the answer quite nicely. Imagine you are one of four people to volunteer for the “Let’s Make A Sleeping Beauty Deal” game show. Three of you will play the game on Monday, and a different set of three on Tuesday, as determined by the flip of a fair coin Sunday night. With the customary amnesia drug, of course, administered on Monday night to those who will play twice. Each of you was assigned – and told – a unique combination of Monday or Tuesday, and Heads or Tails, that represent the combination where you will not play. Your combination is Tuesday+Heads.
The game is quite simple: You are asked for your confidence that you will play the game once, or twice. Note that since your combination is Tuesday+Heads, “once” corresponds identically to Heads, and “twice” identically to tails. Can you logically deduce what your confidence should be, or is it a toss up?
You can logically deduce it. Whenever you play, you are one of three contestants with functionally equivalent information. Exactly one of you will play once. Your confidence can only be 1/3.