Question

[1p0kerb0y]

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.

[uuDevil]

See this thread: Answer (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1040786&page=&view=&s b=5&o=&vc=1)

[BruceZ]

[ QUOTE ]
See this thread: Answer (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=1040786&page=&view=&s b=5&o=&vc=1)

[/ 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 (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Board=probability&Number=417 383).

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 (http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Number=213933), though the method I used there was not as elegant as in the more recent posts.