16/08/20 12:55:13.28 o5QeTUwB.net
>>121
つづき
There is a way to rephrase the paradox in which the axiom of choice is eliminated, and the difficulty is then shifted to the construction of product measure.
Suppose the warden can only assign a finite number of black hats, but is otherwise unconstrained. The warden therefore picks a configuration “uniformly at random” among all the configurations with finitely many black hats (I’ll come back to this later).
Then, one can again argue that each prisoner has only a 50% chance of guessing his or her own hat correctly, even if the prisoner gets to see all other hats, since both remaining configurations are possible and thus “equally likely”.
But, of course, if everybody guesses white, then all but finitely many go free. Here, the difficulty is that the group \lim_{n \to \infty} {\Bbb Z}_2^n is not compact and so does not support a normalised Haar measure.
(The problem here is similar to the two envelopes problem, which is again caused by a lack of a normalised Haar measure.)
引用おわり