Strictly mathematical flush question.

[ChromePony]

Take a hypothetical situation where you hold suited cards in you hand, lets say hearts, you play through to the river and hit the flush so there are three hearts on board. There are 45 cards left unaccounted for, 8 of them are hearts, what are the chances that someone else was dealt two of them, the odds that they also make it to the river arent important here, just the pf deal. Assume 10 person table. Thanks.

[BugsBunny]

C(45,2) = 990
C(8,2) = 28

28/990 * 9 = 0.2545454545

This isn't exact because there's also a chance that up to 4 people (other than yourself) have the flush.

2 people will have the flush:
C(45,4) = 148995
C(8,4) = 70
C(9,2) = 36
70/148995 * 36 = 0.0169133192

3 people:
C(45,6) = 8145060
C(8,6) = 28
C(9,3) = 84
28/8145060 * 84 = .00028876398701

4 people:
C(45,8) = 215553195
C(8,8) = 1
C(9,4) = 126
1/215553195 * 126 = .00000058454248382

That means the exact answer (using inclusion/exclusion) =
0.2545454545 - 0.0169133192 + .00028876398701 - .00000058454248382 = 0.2379203148

Or about 23.79% of the time at least 1 other person will have the flush along with you (assuming they play any 2 suited and go to the river).

[ChromePony]

Exactly what I was looking for, thanks a lot.