Re: Odds of pockets aces being dealt twice in the same round
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C(10,2)/C(52,4) = 6015-to-1.
1/C(52,4) is the proability of 2 specific players having the 4 aces, and we multiply this by the number of ways to choose 2 players out of 10, or C(10,2). Since no more than 2 players can have 4 aces, this method is exact.
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It could be he's asking, if * I * have aces, what are the odds anyother player has aces as well. Is that a different probability?
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