How to calculate runner-runner odds in your head

[k3nt]

I've been thinking about this one ever since my [9c] ran into [7c] on a flop of [Tc][4c]. We were all-in on the flop and the jerkwad sucked out running 7s for the win.

In general, when two hands are all-in on the flop and one hand needs running cards to win, here's how you calculate their chances of winning.

First, you need to know that in a case like this, there are always 990 different boards that can come down. (45 different cards can hit on the flop, * 44 on the turn, then divide by 2 because it makes no difference what order the cards hit in: [4c] is the same as [5d]. 45 * 44 / 2 = 990.) Your opponent's chance of winning is always x / 990, where x is the number of winning boards.

In a runner-runner situation, your opponent does not have any "outs" per se. He has "outs to have outs." He has to hit one of y outs on the turn, and then he also has to hit one of z outs on the river.

The math is super simple. x = y * z / 2

x = number of boards that win for your opponent (out of 990 total)
y = number of "outs" on the turn
z = number of "outs" on the river

.....

Take the simplest case. You have AA and your opponent has [6c] and the board is AA . Your opponent has to hit the and the to win. On the turn, he has 2 outs (either the 9 or the 8) and on the river he has 1 out (the other one). 2 * 1 = 2. 2/2=1. His chances are 1 in 990, or almost exactly 0.1%.

.....

In the case of my hand, things are a bit more complicated. My opponent needed to catch running QQ 77 Q7, or else KJ. But if either the K or the J were a club, that would give me the flush, so the and the are not cards that help him.

So on the flop, he has 6 outs to catch either a Q or a 7. Then if he catches one of those, he has 5 outs to catch another one on the turn. 6 * 5 = 30. 30/2 = 15 boards that win.

In addition to that, he has 6 outs to catch either a K or a J (non-club) on the turn. If one of those hits, he has another 3 outs to finish the straight on the river (if he hit a K, he needs a J, and vice-versa). 6 * 3 = 18. 18/2 = 9 boards that win.

So he has a total of 15 + 9 = 24 winning boards. His chances were 24 / 990. Because 990 is so close to 1000, just round it to 2.4 percent.

You can go to twodimes.net to confirm this: 24 winning boards out of 990.

[Ichoi]

[redhouse]