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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