Probability of a random draw resulting in two teams playing each other

I really need an equation writer, but I think it is:

6*(29*27*25*23*...*1) - (27*25*23*21*...*1)
-----------------------------------------------------
(31*27*25*... *1)

~ 19.4%

We have some good stats guys on here who I'm sure will double check that.

I'm revising my answer. I think it is:

6*(29*27*25*23*...*1) - 3* (27*25*23*21*...*1)
-----------------------------------------------------
(31*27*25*... *1)

or 19.3%

I ran a 100,000 monte carlo and got 19.0% so I'm in the right ballpark.

so if I put all 32 teams in a hat and drew out matches then approximately one in five draws (19%) I can expect to see a match where two of our four teams are drawn out together in a match up? That seems a little too often given the scenario?

If you do the entire draw (all sixteen games) you will have a match about one in five, yes.

Simple math error here, the equation posted yields 19.0%, not 19.3%.

Here is the entire table for you. The first column is the number of players being paired, the next is the probability of a conflict, the third is the number of conflicted matchups, the fourth is the total number of matchups.

6 100.0% 15 15
8 77.1% 81 105
10 61.9% 585 945
12 51.5% 5355 10395
14 44.1% 59535 135135
16 38.5% 779625 2027025
18 34.1% 11756745 34459425
20 30.7% 200675475 654729075
22 27.8% 3824996175 13749310575
24 25.5% 80531676225 3.16234E+11
26 23.5% 1.85616E+12 7.90585E+12
28 21.8% 4.64864E+13 2.13458E+14
30 20.3% 1.25703E+15 6.19028E+15
32 19.0% 3.65013E+16 1.91899E+17