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08-11-2005, 11:54 AM
could someone tell me what the chances of someone having KK if i hav AA in a 10 handed game i would appreciate it if u could explain how u arrived at the answer
thanks mprom

Tom1975
08-11-2005, 12:16 PM
4/50 *3/49 *9=0.044081633

There are 50 unkown cards, so some has to first be dealt one of the four kings; next they have to get one of the three remaining kings out of 49 unknown cards. Finally there are nine people so multiply by nine.

08-11-2005, 12:59 PM
To put it another way if you are dealt KK 1000 times at a 10-hand table its likely 956 of those times you won't be up against someone with pocket Aces.

LetYouDown
08-11-2005, 01:12 PM
If you're looking for the probability that a specific opponent has K-K, then it's simply C(4,2)/C(50,2) = 0.48979591%.

If you're looking for the probability that at LEAST ONE opponent has K-K, then it's:

9 * 6/C(50,2) - C(9,2)/C(50,4) = ~4.39%

You need the inclusion/exclusion principle to get this one exactly right.

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