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After handing out 5 cards to each player, what is the probability that, given that Mr. A has exactly 1 ace, none of the other 3 players have more aces than Mr. A?

I figured the probability of handing him one ace is $\binom{52-4}{4}$ divided by the sample space $\binom{52}{5}$. But everything I've tried after that only fails.

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Note that at most one other player can have more aces than A (since there are only 4 aces in the deck). So you just have to find the probability that B has 2 or more aces, multiply that by 3, and then subtract the result from 1.

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