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Morendil comments on Updating, part 1: When can you change your mind? The binary model - Less Wrong
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Hold on - p(D|H) and P(D) are not point values but probability distributions, since there is yet another variable, namely what day it is.
The other variable has already been marginalized out.
So long as it is not Saturday. And the ideas that p(H) = ½ comes from Saturday.
But marginalizing over the day doesn't work out to P(D)=1 since on some days Beauty is left asleep, depending on how the coin comes up.
Here is (for a three-day variant) the full joint probability distribution, showing values which are in accordance with Bayes' Law but where P(D) and P(D|H) are not the above. We can't "change the values" willy-nilly, they fall out of formalizing the problem.
Frustratingly, I can't seem to get people to take much interest in that table, even though it seems to solve the freaking problem. It's possible that I've made a mistake somewhere, in which case I'd love to see it pointed out.
I was just talking about the notation "p(D|H)" (and "p(D)"), given that D has been defined as the observed data. Then any extra variables have to have been marginalized out, or the expression would be p(D, day | H). I didn`t mean to assert anything about the correctness of the particular number ascribed to p(D|H).
I did look at the table, but I missed the other sheets, so I didn`t understand what you were arguing.
It seems to say that p(heads|woken) = 0.25. A whole new answer :-(
That's in the three-day variant; it also has a sheet with the original.
It has three sheets. The respective conclusions are: p(heads|woken) = 0.25, p(heads|woken) = 0.33 and p(heads|woken) = 0.50. One wonders what you are trying to say.
That 1/3 is correct in the original, that 1/2 comes from allocating zero probability mass to "not woken up", and the three-day version shows why that is wrong.