Reply to thread by Casino.us

Message: [QUOTE="ginNjuice, post: 514230, member: 17325"] 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. [B]The Gordon Pair Principle[/B] 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, [B]C = (N x R) / 2[/B] 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. [/QUOTE]

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