16/08/20 12:54:00.36 o5QeTUwB.net
>>120
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Terence Tao Says:
September 13, 2007 at 9:58 pm | Reply
This paradox is actually very similar to Banach-Tarski, but involves a violation of additivity of probability rather than additivity of volume.
Consider the case of a finite number N of prisoners, with each hat being assigned independently at random. Your intuition in this case is correct: each prisoner has only a 50% chance of going free.
If we sum this probability over all the prisoners and use Fubini’s theorem, we conclude that the expected number of prisoners that go free is N/2. So we cannot pull off a trick of the sort described above.
If we have an infinite number of prisoners, with the hats assigned randomly (thus, we are working on the Bernoulli space {\Bbb Z}_2^{\Bbb N}),
and one uses the strategy coming from the axiom of choice, then the event E_j that the j^th prisoner does not go free is not measurable,
but formally has probability 1/2 in the sense that E_j and its translate E_j + e_j partition {\Bbb Z}_2^{\Bbb N} where e_j is the j^th basis element, or in more prosaic language, if the j^th prisoner’s hat gets switched, this flips whether the prisoner gets to go free or not.
The “paradox” is the fact that while the E_j all seem to have probability 1/2, each element of the event space lies in only finitely many of the E_j. This can be seen to violate Fubini’s theorem ? if the E_j are all measurable. Of course, the E_j are not measurable, and so one’s intuition on probability should not be trusted here.
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