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Old 07-08-2003, 06:30 AM
mongeron mongeron is offline
Join Date: Sep 2002
Posts: 97
Default Pocket pair probability.

What is the probability in Hold'em of one player having a pocket pair, and running into a higher pocket pair? I need just the idea to calculate it for different ranks of pairs myself.

- mongeron
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Old 07-08-2003, 11:55 AM
doormat doormat is offline
Join Date: Dec 2002
Posts: 84
Default Re: Pocket pair probability.

This is a good question but a little tricky. I assume you want to know the chance of someone else at the table having a higher pair than yours. The chance of being dealt any pair is (4*3*13)/(52*51), or 1 in 17. If you are dealt a pair, the chance of the next person to act having a higher pair is (4*3*n)/(50*49), where n is the number of cards of higher rank than yours. Let's say you have a pair of 9's, so 5 pairs can beat you. So the chance of a particular hand beating you is (4*3*5)/(50*49), or .0245. Therefore the chance of you having a higher pair than the next guy is .9755 in this case. If there are 6 players left to act after you, the chance of your 9's being higher than any pair they may have is .9755 to the power of 6, or .8617. You would have to work this out for all pairs and all number of players left to act, and it would still be just approximate because if players folded in front of you, it implies higher cards behind you, and if players called or raised before you their chance of beating you is higher. Also, it ignores non-paired hands that rank higher than some pairs, etc. but I think it should be close enough for government work, as they say.

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