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Old 12-11-2003, 01:50 PM   #4 (permalink)
rsl12
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I *don't* think the math is right. If you look at the problem *without* assuming that there is $X in cup A, then you could argue something like this:

There's two possible scenarios: cup A has $X and cup B has $2X, or else cup A has $2X and cup B has $X. The expected value in both the cups is $1.5X, so it doesn't matter which one I choose.

That makes TOTAL sense, right? So where's the mathmatical flaw in the original argument?
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