[BruceZ]

Hi Qui-Gon,

Welcome to the forum.

The probability that none has a pocket pair is then (16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17 * 16/17) = 0,6157.

This means that the probability that at least one player has a pocket pair is (1-0,6157) = 0,3843!

There you go, a much easier way for the calculation.


It is much easier, but it is only an approximation, though often a good one. It is only approximate because the event of one player not having a pair is not independent of another player not having a pair, so you cannot simply multiply these probabilities to get an exact answer. For example, if the first player does not have a pair, the probability of the second player not having a pair is no longer 16/17 = .941176, but instead it changes to 1 - (11*6+3+3)/(50*49/2) = .941224. Since these are close, the approximation is fairly good.

Another approximation which is sometimes good is to simply multiply the number of players by the probability for 1 player. This approximation assumes that the events can be approximated as mutually exclusive rather than independent, but it is not accurate in this problem. Often one or the other of these approximations is good, or both, so it is good to mention them.

The poster wanted a solution which would work for all such problems, and that required the inclusion-exclusion method. This gives an exact answer. If the elementary methods were exact, I certainly would have used them.

-Bruce