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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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up vote 1 down vote accepted

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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I keep getting it wrong:… -- this is large than the probability of A getting 1 ace, which will result in a number larger than one after I divide for the conditional. – iDontKnowBetter Oct 14 '11 at 3:11
"Divide by the conditional" doesn't enter into it. There are 47 cards (after 5 have been dealt to A). Exactly 3 of them are aces. What is the probability that B gets 2 or more aces among her 5 cards? – Gerry Myerson Oct 14 '11 at 3:27
thanks for the help. Probability is kicking my arse. – iDontKnowBetter Oct 14 '11 at 4:16
You're welcome. If my answer helped you, you could vote it up. If my answer helped you a lot, you could even tick the box to accept it. – Gerry Myerson Oct 21 '11 at 10:54

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