[BruceZ]

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Say you are holding KK or some other hand (not important as long as an ace isn't part of it)

Then I have read on the forum that the probability of EXACTLY one holding AA is:

9*C(4,2)/C(50,2) - C(9,2)/C(50,4)

[/ QUOTE ]

This is the probabilty of at least 1 player having AA, which is the the case for which I have always given this probabilty.