Four flush cards on the board....

[BugsBunny]

There are 46 cards left in the deck. 11 cards per suit for the suits you have in hand, and 12 cards persuit of the other 2 suits. The question here is how often will a 4 (or 5) flush of one of the suits that you don't have in hand be dealt.

4 flush = C(12,4) * (46-12) = 495 * 34 = 16830 combinations
5 flush = C(12,5) = 792 combinations
16830 + 792 = 17622 combinations

C(46,5) = 1370754 total 5 card combinations

17622/1370754 = 0.0128556984
But there are 2 possible suits that this can happen with so:
2 * 0.0128556984 = 0.0257113968 or about 2.57% of the time you'll lose to a flush this way.

To get an exact figure you'd have to subtract some straight flush combinations (split pots) (4 per suit) and factor in those times when they get their flush but you get a boat to win (4 flush that includes an A with the 5th card pairing one of the other cards).
4 flush that includes an A = C(11,3) = 165 (per suit)
There is 1 A + 3 of each of the other 3 cards for a total of 10 cards so
165 * 10 = 1650 combinations where they hit the flush but still lose (per suit)

17622 - 4 - 1650 = 15968

Per suit you'll end up losing to a flush:
15968/1370754 = 0.0116490632

0.0116490632 * 2 = 0.0232981264 or about 2.3%


Of course you could also lose to a straight etc. Your total equity in this situation is, per pokerstove, 95.249%

[IggyWH]

Good math Bugs :

http://twodimes.net/h/?z=1140205

or

http://twodimes.net/h/?z=1140208