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.
