[Cobra]

When I read this I looked at it a little differently than LetYouDown. I am going to answer this question that at a six person table what are the odds that on the next deal you get a flop of XXY and any two people have an XY or XZ.

Without consideration of hold cards the flop will come XXY:
=13*(4c2)*48/(52c3) = 3744/22100 = 16.94%.

Now that the flop has come XXY we will determine the odds that two people were dealt XY or XZ. In order to do this you use inclusion exclusion. In this case there can be no more than two people who are dealt an XY or XZ therefore there are only two terms in the inclusion exculusion exercise. We are interested in the second term which gives the probability that two people are dealt that above hands.

= 4412*(6c2)/(49c2)/(47c2) = 5.21%

So the probability of this event occuring on the next deal is.

5.2%*16.9% = .882% or 1 in 113.3 hands.

Assuming you are a bystander and you see a flop of AAK the probability that two people were dealt an ace is 5.21%.

Cobra