Sorry, you lost me completely. I didn't prove that P(Heads | Monday) > 1/2 at all.
Could you say which step (1-6) is wrong, if I am Beauty, and I wake up, and I reason as follows?
The experiment is unchanged by delaying the coin flip until Monday evening.
If the current day is Monday, then the coin is equally likely to land heads or tails, because it is a fair coin that is about to be flipped. Thus P(Heads | CurrentlyMonday) = 1/2.
By Bayes' theorem, which is applicable because it cannot currently be both Monday and Tuesday:
P(Heads) = P(CurrentlyMonday) P(Heads | CurrentlyMonday) + P(CurrentlyTuesday) P(Heads | CurrentlyTuesday)
P(Heads | CurrentlyTuesday) = 0, because if it is Tuesday then the coin must have landed tails.
Thus P(CurrentlyMonday) = 2 * P(Heads) by some algebra.
It may not currently be Monday, thus P(CurrentlyMonday) != 1, thus P(Heads) < 1/2.
Sorry, you lost me completely. I didn't prove that P(Heads | Monday) > 1/2 at all.
You had said:
This is in contrast to the standard halfer position, where P(Heads | Monday) > 1/2
Neither of your links to the halfer position shows anyone claiming that. So I assumed you tried to deduce it from the halfer position. The obvious way to deduce it is wrong for the reason I stated.
Could you say which step (1-6) is wrong, if I am Beauty, and I wake up, and I reason as follows?
"CurrentlyMonday" as you have defined it is a per-awakening p...
If it's worth saying, but not worth its own post, then it goes here.
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