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Possible Duplicate:
Card doubling paradox

I came across the following preferences paradox:

Suppose you have two identical boxes $A$ and $B$. One of them contains an unspecified number of coins. Another contains twice this amount. The first one you touch is the one you open. Which box do you choose?

Suppose I choose Box $A$. Suppose $E(A) = x$. Then the value of box $B$ is $x$ or $\frac{1}{2}x$. Thus $E(B) = x(\frac{1}{2})+(2x)(\frac{1}{2}) = 1.25x$. One can do a similar analysis when starting with Box $B$. It seems that both both boxes should have the same expected value a priori. Is it wrong to assume that $E(A) = x$ or $E(B) = x$? The value of these boxes may be $x$, but that doesn't imply that their expected values if $x$?

Or maybe our definition of expected value if flawed? Should we look at log utility? This seems similar to the Monty Hall Problem.

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marked as duplicate by Qiaochu Yuan Jul 18 '11 at 15:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I'm unable to follow your calculations at all. Could you please go into more detail and spell out the paradox? – t.b. Jul 18 '11 at 15:31
And you haven't even gotten into which of the boxes contains the more valuable coins! :-) But seriously: Why is the value of box $B$ $x$ or $\frac12x$? And how does this lead to $E(B) = x(\frac{1}{2})+(2x)(\frac{1}{2})$? – joriki Jul 18 '11 at 15:32
Added the "puzzle" tag: Use when posting a question whose answer is already known. – GEdgar Jul 18 '11 at 15:39

See Wikipedia for a detailed discussion on the Two envelopes problem.

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