[AaronBrown]

I need to make a partial retraction, this particular example is not hard. Since the probability of drawing any rank of card from a well shuffled deck is equal, it is obvious that it doesn't matter how you choose the card to match.

The hard part is proving the theorem in general. Given two independent variables X and Y and some set of events E (that is some set of values of X and Y, such as the set where X > Y or the set such that X + Y < 1), prove that the probability of E is equal to the expected value over all possible values of X of Pr{E|X}, that is the probability of event E given the value of X.

For example, suppose I choose p from a uniform distribution over (0,1), then flip a coin that has probability p of coming up heads until I get a tail. I count the number of heads that come up before the first tail. The chance of getting n heads before the first tail, given p, is (1-p)*p^n. If I integrate this over p, I get 1/[(n+1)*(n+2)]. I want to say this is the probability of getting n heads before the first tail. But this has infinite expected value, despite the fact that n has finite expected value for any legal value of p. That bothered some people until we got a good measure theory.