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#9
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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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