[Cobra]

In order to answer this type of question you need to make certain assumptions and start at a given point. I will find a slightly different number. I am going to solve the problem that given you start with a pair.

1. You find your self in a set/set/set situation after the flop.

2. One or more of you end up with four of a kind.

Obviously you must assume all people dealt a pocket pair stay to see the flop and all people with a set on the flop stay to the end.

Given that you have a pocket pair!

The flop gives you sepecifically a set

=2*(12c2)*4*4/(50c3) = 10.7755% or 1 in 9.28 flops

Prob two of your five opponents also hold a set

=18/(50c2)/(48c2)*(5c2) = .0139% or 1 in 7,676.67 times

Prob one or more of your opponents ends up with four of a kind

=(3*40+3)/(43c2) = 13.6213% or 1 in 7.34 times

So given you flop a pocket pair the probability of seeing a set/set/set flop is:

=10.7755%*.0130% = .0014037% or 1 in 71,242 times

So given that you have a pair the probability that three of you flop a set and one or more end up with a four of a kind is:

=.0014037%*13.6213% = .0001911% or 1 in 523,019 times

Now if you wanted to start from scratch and say what is the probability that on the next deal I get a pair and the above happens you will be dealt a pair 1 in 17 times so multiply the above numbers by 17.

This does not answer the posters question of how often this will happen at a six person table. It answers how ofter will I be involved in the above in a six person table.

Cobra