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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