[Cobra]

I will give this a shot. I am new to the probability area so hopefully someone else will check me.

1. The probability the flop has a pair in it. The flop must come in the form of (x,x,z), therefore,

13*3*48/c(52,3)=.0847%

Thats 13 ways to choose the first x, 3 ways to choose the second and 48 other cards for the third.

2. Assuming 10 handed and you do not have an x, therefore you have two z's. This leaves 2x's and 45 z's left.

Probability breaking the possible hands into vectors.

(z,z) No one has a set. c(2,0)*c(45,18)/c(47,18)=.376%

(x,z) One person has a set. c(2,1)*c(45,17)/c(47,18)=.483%

(x,x) Two people have a set or someone has 4 of a kind.

c(2,2)*c(45,16)/c(47,18)=.142%

Now these percentages are not that they are still there after the flop, just that they were dealt the hands prior to the flop.

If you go down to five handed the percentages are as follows.

(z,z) No one was dealt a set. c(2,0)*c(45,8)/c(47,8)=.685%

(x,z)=.289%

(x,x)=.026%

So the answer is yes it does depend on how many people are dealt cards.

Cobra