I don't buy this.
Call the amounts in the two envelopes "A" and "B". A is known, B is not. So you guess at B and pick the number "B'".
Now, If B' > A, you will pick B, and you will be right if B > A. B can be anything, so the probability of this is 50%. Alternately, if you pick B' such that A > B', your chance of being correct with respect to B is also 50%. So, 2 out of 4 possibilities of being correct. Even odds overall.

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Ryan Koppenhaver, Aspiring Perl Hacker

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"I ask for so little. Just fear me, love me, do as I say and I will be your slave."
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