Inclusion-exclusion principle question

[BugsBunny]

OK using inclusion exclusion I want to figure the chances, at a full ring, that at least one opponent has AA or KK if I hold QQ.

I figure that there should be 4 terms since at most 4 people can hold AA and KK. The first term is simple:

(9 * 6 * 2)/C(50,2) = 0.0881632653

And this is close to the correct answer. But if I want to get the eaxct answer I need the next 3 terms.
So how do I figure the number of combinations for 2 sets of pairs? I'm having a brain freeze at the moment.

The second term will be - x/C(50,4) * C(9,2) the third will be + x/C(50,6) * C(9,3) and the 4th will be - C(9,4)/C(50,8) but I'm having trouble figuring how to calculate x for the 2nd and 3rd terms.

I have 6 possible ways to make each pair and 2 sets of pairs. For 2 people it can run AA AA or KK KK or AA KK, but the AA KK combination can have multiple ways of being made. Is it just 6*6 for the 2 person setup (meaning 38 total combinations after adding on the all A or all K combinations) and 1*6*2 = 12 for the 3 person numerator or am I missing something?

If I'm right than the final formula becomes:
(9 * 6 * 2)/C(50,2) - 38/C(50,4) * C(9,2) + 12/C(50,6) * C(9,3) - C(9,4)/C(50,8) =
.0881632653 - .0059400781589 + .000063433328928 - .00000023468990618 = 0.0822863858
or ~ 8.23%

Is this correct?

[aloiz]

I think you need to seperate the two cases for two people.

So second term would be something like C(9,2)* 2 * 1/C(50,4) + C(9,2) * C(2,1) * 6/C(50,2) * 6/C(48,2)
First half is when we have either AA AA or KK KK, and the second half is when we have AA and KK.

third term (this is the easiest way for me to write it out): C(9,3) * 6 * 6/C(50,2) * 1/C(48,2) * 6/C(46,2)

aloiz

[BugsBunny]

You're right. Knew I was missing something, just didn't feel right.

That makes the final answer:
.0881632653 - .002188449848 + .000012686665786 - .00000023468990618 = 0.0859872674
or ~8.60%