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#1
09-24-2004, 08:08 PM
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Question
If you are dealt AK, what are the odds that another player at the table will have AA or KK assuming a full ten-handed table? This question is not designed to test anybody, as I realize it is fairly simple, but rather so that I can verify my results as my math is not as advanced as many posters here.
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#3
09-24-2004, 11:26 PM
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Re: Question
[ QUOTE ]
See this thread: Answer [/ QUOTE ] That post only computed the probability for 1 specific player having AA or KK, which is 6/C(50,2) = 6/1225. For the probability that someone holds these hands with 9 opponents, the probability is 9*6/1225 - C(9,2)*6/C(50,2)*3/C(48,2) =~ 4.36% by the inclusion-exclusion principle. This and similar problems have been discussed many times before. If you are interested in more probabilities like this, you can find them in this post, though the method I used there was not as elegant as in the more recent posts.
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