![]() ![]() I hope you'll forgive me if I just do the probability of getting exactly a pair, including hands that also form a straight or flush. Using that formula you get the following for all situations.įor those not familiar with the hold 'em terminology, you are asking for the probability of at least a pair in six cards, given that the first two are (the hole cards) of different ranks. For example, with a pair of jacks in a 10 player game the probability of a higher pocket pair is 3*9/2 = 13.5%. That is the percentage probability that there will be at least one higher pair. ![]() Take the number of higher pairs, multiply by the number of other players, and divide by 2. After this column appeared one of my fans, Larry B., wrote a brute force combinatorial program to solve this problems. So my approximation of the probability of at least one higher pocket pair is 1-e -n*r*(6/1225). The table below shows those probabilities. For examle if you have pocket queens and there are 9 other players the expected number of players with a higher pocket pair is 0.0882, so the probability of at least one player having a higher pocket pair is 1-e -0.0882 = 8.44%. Given that assumption the probability that at least one player will beat you is 1-e -µ, where µ is the mean. To get the probability that at least one player will beat you I will make the not entirely correct assumption that the number of players with a higher pocket pair is a Poisson random variable with a mean in the above table. ![]()
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