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 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.
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.