Floped full house Vs floped quads 6 handed

[BogusPomp]

Ok- i'm going to take a stab as well

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So it's two events, firstly two people dealt any unique pocket pair, the odds of this would be

52/52 * 48/51 * 3/50 * 3/49 = 1 in 289.23611

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I think this should be (52/52 * 3/51) * (48/50 * 3/49) =

p(player A receives pocket pair) * p(player B is dealt a first card that does not match player A's pair) * p(player B pairs this first card) =~ 0.003457

then, of 48C3 possible flops, 4 will make quads and a full house.

4/(48C3) = ~0.000231267

therefore, the probabilty of this happening at a 2 person table is 0.003457 * 0.00023126 = ~ 8e-7 or roughly 1 in 1.25 million

I am anxious to see how to extend this result to a 6 person table.

I thought of using a binomial distribution here- however I am fairly certain it does not consider the times when 3 or more people have been dealt pocket pairs- and is therefore not applicable?!?

Many thanks to Bruce for all his help.