[Dave H.]

Sorry, but that's too vague a question. Do you mean something like:

What's the probability, if it's heads up, that I will have exactly one ace and you won't have any ace in the pocket?

I would determine the probability that I have an ace as

(4*48)/C(52,2)

Then I would calculate the probability that you do NOT have an ace as C(47,2)/C(50,2)

Now you would just multiply those two probabilities. I don't know if this is what you wanted, but you should probably start a new post and get some of the real probability gurus to look at it.