Re: Inclusion/Exclusion Question
 

[ QUOTE ]
4th term: 4 players hold pairs. Possibilities are xxxxyyyy, xxxxyyzz, and wwxxyyzz (now things start getting a bit hairy)

[/ QUOTE ]

Yes it does. This is where I'm focusing. If the sum of the combinations (66) can be expressed as favorable combinations of the 44 cards 4,6,8... (permutations with restrictions rather than combinations) it would be a lot easier to apply. Back to studying matching problems ...

Re: Inclusion/Exclusion Question
 

[ QUOTE ]
Say you hold 3-3...what is the probability that at least one of your opponents holds a pair higher than 3-3.

Can someone rattle this out using inclusion/exclusion? I'm curious how this applies to this situation.

I'd imagine the first term is C(9,1) * [11 * C(4,2)]/C(50,2)

[/ QUOTE ]

Computing 9 terms of inclusion-exclusion is often complicated, and virtually always unnecessary and ridiculous since you are trying to compute terms with a minuscule contribution. In most poker problems, you can get to within 0.1% in 2 or 3 terms. This problem requires 4 terms for that accuracy since there are so many pairs. Since the terms get smaller, your result will be accurate to within less than the magnitude of the last change. In this case, there is a pattern which would allow you to produce all 9 terms, preferably by writing a computer script to do it. This can be done recursively. It is important that you understand the pattern.

The nth term is multiplied by C(9,n) to denote n opponents having a pair. The denominator of the nth term is C(50,2)/C(48,2)/C(46,2)... (n terms). For the 1st term, the player has a choice of 11*6 pairs. For the 2nd term, the first player has a choice of 11*6 pairs, leaving the second player with a choice of 10*6 + 1 pairs. For the 3rd term, the first player has a choice of 11*6, leaving the second player with a choice of 10*6 + 1, and if he chooses one of the 10*6, the 3rd player is left with a choice of 9*6 + 2, otherwise he is left with a choice of 10*6. The terms get more complicated after that, and you can write out a tree-like structure to keep track of it. Here are the first 4 terms:

9*11*6 / C(50,2) -

C(9,2)*11*6*(10*6+1) / C(50,2)/C(48,2) +

C(9,3)*11*6*[10*6*(9*6+2) + 1*10*6] / C(50,2)/C(48,2)/C(46,2) -

C(9,4)*11*6*{10*6*[9*6*(8*6 + 3) + 2*(9*6 + 1)] + 1*10*6*(9*6+1)} / C(50,2)/C(48,2)/C(46,2)/C(44,2)

=~ 39.2%

Re: Inclusion/Exclusion Question
 

I had a feeling it was going to tail off to the point of irrelevance. Thanks much for posting this...I was very curious how you'd approach it.