Hi,You have AK of hearts, the flop shows two more hearts (say 2 and 10), what is the probability that you hit the flush if there only 2 players at the table (you and somebody else).I was thinking 9/45+9/44 (I either get another heart on the turn or on the river), but it is about 40% and it looks to be way to high... am I missing something? Also, I guess we must make an assumption that the other player has no hearts.Thanks.

I am not very familiar with Hold'em, but the flop is shared between you and the other player, isn't it ?So basically 2 more cards will be dealt (in the turn and river) and you ask the probability of having at least 1 more heart if the other player did not have a heart in his two first cards ? I think you made a small mistake in your computation, as your probability for the second draw does not contain the fact that you could have picked a heart in the first one.

QuoteOriginally posted by: bluetrinI am not very familiar with Hold'em, but the flop is shared between you and the other player, isn't it ?So basically 2 more cards will be dealt (in the turn and river) and you ask the probability of having at least 1 more heart if the other player did not have a heart in his two first cards ?Yes, that's correct.

With 4 cards known (and hearts) and 48 cards unknown, the probability of 0 hearts in the next 2 cards would be 1- (1-9/48)*(1-9/47) = 34.3%

Traden4Alpha, there are 45 cards unknown as 2 have been dealt to the each player and 3 are in the pot but that is a small detail.

QuoteOriginally posted by: bluetrinTraden4Alpha, there are 45 cards unknown as 2 have been dealt to the each player and 3 are in the pot but that is a small detail.Thanks for the correction I am not very familiar with Hold'em, either. The hidden cards in other player's hands don't count unless the other players' hands can affect how many cards get exposed. Only the visible pot and your hand matters. There's still 9 hearts among the unknown cards and a total of 47 unknown cards.1- (1-9/47)*(1-9/46) = 34.97%

X is hypergeometric(2, 47, 9).P(X >= 1) = 1 - P(X=0) = 1 - (9 over 0)(47-9 over 2-0)/(47 over 2) = 1 - 38*37/(47*46).That is, the probability to get the flush if we have not yet seen the flop is: [X is hyg(5, 50, 11)]P(X >=3) = 1 - P(X=0) - P(X=1) - P(X=2) = [computer] = 0.06399Step 2:If by "hit the flush" you mean actually seing the card that gives the flush we must consider the situation. I believe that it is likely that if one of the other two players stay in the game, then they have at least one heart. This reduces the probability of hitting the flush.Step 3:Model: p = a1*p1 + a2*p2 + a3*p3 + a4*p4Where ak is the probability that there are k other hearts out there and pk is the probability that we hit our flush in that situation.New problem: How do we estimate the a-probabilities?