Quote:
Originally Posted by pokerhsmtt
if we had 88 on the button we open then got 3 bet
Getting 3-bet increases the chance (unless they 3-bet 100%), but the calculation for
before they've acted is as follows:
2⋅36/1225 - [C(6,2)⋅6²/3 + 6]/C(50,4) = 267/4606 ≈ 5.8%
Again I started with what you did and then subtracted the chance of both players getting it.
C(6,2)⋅6²/3 is the #ways they can be dealt two different pairs. I divided by 3 because if the cards AAKK are dealt, we only want to count AA|KK cases as opposed to AK|AK. Only one out of 3 distributions is AA|KK. You can either do it like I did and divide by 3 on this step, or you can wait until the end and make the denominator C(50,4)⋅3 or C(50,2)⋅C(48,2)/2. But then you'd have to multiply the 6 same-pair combos by 3 so that its 3 cancels with the denominator's, since all 3 ways to distribute AAAA result in AA|AA. In other words that would look like [C(6,2)⋅6² + 6⋅3] / [C(50,4)⋅3]