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The following hand occurred during the last five minutes of the first day of the $10,000 championship no-limit holdem event. It is a good hand to illustrate proper no-limit thought processes as well as a simple trick to estimate "all-in" odds.

With eight players at the table, the ante was $50 and the blinds were $100-$200. I was the small blind and had the moderate amount of $15,000 in front of me. A well-known tournament player had the big blind and about $5,000. I held pocket sevens and everyone folded to me. I made it $1,200 total. The player in the big blind moved in $3,800 more. Now what?

Obviously, this problem combines both reading hands and mathematics. At this point I was getting 6,600-to-3,800 odds. I need to win 3,800/10,400 of the time to have an even bet about 36.5 percent. To know what my chances are, I need to know not only what my opponent's possible holdings are, but also how to turn that into an overall price.

After some thought, I came to the conclusion that my well-known opponent did not need a great hand to move in in this spot. I know he hates to risk all of his chips on close gambles, but three factors told me this situation was an exception: namely, my position, his relatively short stack, and the fact that he probably felt an aversion to returning the second day with very few chips. It was better to double up or get out.

On the other hand, this opponent would not go overboard here. I thought he would choose only hands that had pretty good chances in a "race" even if I called him with a premium hand.

I postulated that he probably had a pair of fives or better, or A-K, or A-Q. Given that, would the pair of sevens win often enough to make a call correct?

The reason why this question is not simple is because not all hands are equally likely to be dealt. A-K and A-Q can be dealt 16 ways, while pairs can be dealt six ways (except for two sevens, for which my opponent had only one way). So, even if I know the chances the sevens have against each of my opponent's possible holdings, I can't simply average them. I must instead get a "weighted" average, which is time-consuming. But here's a trick.

First, I count the total number of combinations my opponent (in my opinion) could have. A-K plus A-Q is 32. Aces down to fives is 55 (9 x 6 plus 1). That's 87. Now I add up how many of these 87 possibilities figures to win for me. When he has one of the 32 nonpair combinations, I should win about 17 of them. When he has one of the 12 combinations of pairs smaller than mine, I should win about 10 of them. When he has one of the 42 combinations of pairs above mine, I should win about 7. That's 34. I then add a half for when we both have sevens. Altogether, my chances are about 34.5 out of 87, just short of 40 percent.

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