Originally Posted by NineLions;hing. I 512430
At this point I'm still wondering what the chances are he's on KK or AA.
This is an article from Expert Insight.com written by Phil Gordon that will help answer this question for you. I didn't include the chart that he has along with this so check out the website if you want to see it.
I was playing in a sit and go tournament at Full Tilt a few days ago with my fiancee looking on. We were down to three-handed, all the stacks were about the same, though I was the short stack. The blinds were very high -- the average stack was about 12 big blinds. I had 2-2 on the button. I raised all-in and was called by 6-6. I went broke.
"That was a really bad play, Phil. How can you go all-in there?" she said.
I protested vigorously: "Honey, it is well against the odds that either of my opponents will have a higher pocket pair. With only 12 big blinds, I'm either all-in or I fold in this situation. Doing anything else is just crazy, I think. Especially because we're already in the money, and the difference between second and third place isn't very significant."
"Well, I think it's much more likely for them to have a pocket pair. What are the exact odds?" she asked.
I didn't know off the top of my head, which just seemed to give her more ammunition for her argument. It is hard to argue odds when you don't know them. So, I set off to do some math so I could "prove" to her that I was right. In the process, I "discovered" a general mathematical formula that everyone can use when arguing with a significant other.
I'm calling this rule the "Gordon Pair Principle" or GPP. I've always wanted a theorem named after me, and so here it is. A few years back, I got zero credit for naming the "Rule of 4 and 2," and I'm a little on tilt about it. Now, I'm not claiming that I discovered the "Rule of 4 and 2," but I do claim naming it and referring to it in print as such for the first time (see my book "Poker: The Real Deal").
So, here goes.
The Gordon Pair Principle
Let C = percent chance someone left to act has a bigger pocket pair Let N = number of players left to act Let R = number of higher ranks than your pocket pair (i.e., if you have Q-Q, there are two ranks higher. If you have 8-8, there are six ranks higher)
Then, C = (N x R) / 2
You have pockets 10s and there are six players left to act. Someone will have a bigger pocket pair about 12 percent of the time.
You have pocket kings under the gun in a 10-handed game. You'll be up against pocket aces (and probably broke) about 4.5 percent of the time.
Now, this formula isn't exact, but it is a damned close approximation. It's definitely close enough to use when arguing with your significant other. Of course, I showed her this calculation after about an hour of work and she still thinks I made a stupid play despite the fact that my 2-2 is the best hand there 88 percent of the time.
Good luck at the tables. Better luck arguing the subtleties of no-limit with your significant other.