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Old 06-04-06, 04:18 AM
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Join Date: Feb 2005
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I almost posted the other day and said exactly what you said, but I realized later he's actually right. It's a question of conditional probability.

"What's the probability of the fourth Ace in play GIVEN there are three Aces on the board".

A = three aces on the board
B = exactly one hole card ace

P(B|A) = P(A and B)/P(A)
=1/1353 / 1/553
=553/1353
=1/2.45
which is about 40%.

My calculations assume 10 players, therefore 20 hole cards, and the three Aces being anywhere on the board - not just on the flop.

I guess we'll have to agree to disagree...
 


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