Re: The envelope problem, and a possible solution
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It "makes sense" to switch if the conditional probability, p, that the other envelope is larger, satisfies p>1/3. The value p is always a well-defined number between 0 and 1.
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Actually, that's incorrect. It makes sense to switch if the average expected value (EV) of the other envelope is larger than the value of the held envelope.
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If x is the value of the held envelope and p is as I defined it, then the EV of the other envelope is
2xp + 0.5x(1 - p) = x(0.5 + 1.5p),
which will be greater than x if and only if
1 < 0.5 + 1.5p,
which is true if and only if p>1/3.
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