The Gordon Pair Principle
By Phil Gordon
Originally published on November 29, 2006 by ESPN.com
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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
| NUMBER OF PLAYERS REMAINING |
| | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 22 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 |
| 33 | 5.5 | 11 | 16.5 | 22 | 27.5 | 33 | 38.5 | 44 | 49.5 |
| 44 | 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 |
| 55 | 4.5 | 9 | 13.5 | 18 | 22.5 | 27 | 31.5 | 36 | 40.5 |
| 66 | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 |
| 77 | 3.5 | 7 | 10.5 | 14 | 17.5 | 21 | 24.5 | 28 | 31.5 |
| 88 | 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 |
| 99 | 2.5 | 5 | 7.5 | 10 | 12.5 | 15 | 17.5 | 20 | 22.5 |
| TT | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 |
| JJ | 1.5 | 3 | 4.5 | 6 | 7.5 | 9 | 10.5 | 12 | 13.5 |
| QQ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| KK | 0.5 | 1 | 1.5 | 2 | 2.5 | 3 | 3.5 | 4 | 4.5 |
Some examples:
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.
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