[meep_42]

I believe this is correct for 2 flushes on the same board, assuming only 3 of a suit on the board.

C(8,2)/C(45,2) = 2.83%, or about 1/35
(8 spades left, C(8,2) gets the number of 2 card combinations possible, 45 unknown cards left, C(45,2) gets the total possibility of hole cards your opponent has.)

I'm not sure if this needs to be multiplied by 2 to negate that we're not worried about the order of the cards, though. :\

(I'm trying to learn these C(x,y) functions. :P )

-d