Originally Posted by metski
yeah but lets say villains both call flop and i assign them 200 of the same combos
70 of those combos contain Qx
35% of their range is Qx
since their continuing range has both Qx and other non Qx random stuff i cant substract the Qx combos in player 2 range right?
You count how many combinations player 2 can have that are not Qx when player 1 doesn't have Qx. See example below.
what is the probability either have Qx when they both call
As I said, we need the ranges for that, and you won't tell us what they are. So I'll make up my own problem and solve that.
For simplicity, suppose both players will play any 2 cards T or higher, and nothing else. When the flop comes Qyy where y is < T, what is the probability that at least one of them has Qx, where x can be a Q or non-queen?
There are C(19,2) = 19*18/2 = 171 hands that they would see the flop with. Of these, 51 contain a Q. The probability that the first player has Qx is 51/171 or almost 30%. The probability that he doesn't have Qx then is 120/171. If player 1 doesn't have Qx, there are 3*14+3 = 45 ways that player 2 can have Qx out of C(17,2) = 17*16/2 = 136 possible hands. The probability that he has Qx when player 1 does not have Qx is 45/136,. The probability that he does have Qx when player 1 doesn't have Qx then is 91/136. So the probability that either one of them has Qx is 1 minus the probability that neither has Qx is
1 - (120/171)*(91/136)
Now if you had simply computed 1 - (120/171)2
as some people have so egregiously suggested, you would have gotten 50.75%.