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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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    I keep getting it wrong: http://www.wolframalpha.com/input/?i=1-3%28+%2844+choose+3%29%283+choose+2%29+%2B+%2844+choose+2%29%283+choose+3%29%29%29%2F%2847+choose+5%29 -- 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.2011-10-14
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    "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?2011-10-14
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    thanks for the help. Probability is kicking my arse.2011-10-14
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    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.2011-10-21