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#1
01-16-2004, 01:50 PM
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Re: Odds of pockets aces being dealt twice in the same round
[ QUOTE ]
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. [/ QUOTE ] 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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#2
01-16-2004, 06:38 PM
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Re: Odds of pockets aces being dealt twice in the same round
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?
Yes, that is a very different probability than what he asked for. Of course it is much more likely that 2 players have aces if you already have aces. That's like the guy who takes his own bomb on a plane because he figures the odds of there being two bombs is astronomical. If you have aces, the probabilty that one of your 9 opponents has has aces with you is 9/C(50,2) = 9/1225 or 135-to-1.
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