Inclusion/Exclusion Question

[LetYouDown]

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)

[SheetWise]

44 * (3/49) = 2.69

Can't be that simple. I'll look up "using inclusion/exclusion"

[LetYouDown]

http://archiveserver.twoplustwo.com/showthreaded.php?Cat=&Board=&Number=417383&page=&v iew=&sb=5&o=&fpart=

I believe there should be 9 terms to get an exact answer.

[SheetWise]

I just read the post by BruceZ (sorry if I got that wrong), it's interesting. I'm comparing it to a formula I use .... it seems we're getting the same place by different routes .... I'm going to try to assimilate the differences and grasp his method ...

[aloiz]

Think this is exact against three opponents but no guarantees.

1st term: C(3,1) * 11 * 6 / C(50,2)
-
2nd term: ( C(3,2) * C(11,2) * 6^2 * 2 / (50!/(2*2*46!)) + C(3,2) * 11 / C(50,4) )
+
3rd term: (C(11,3) * 6^3 * 3 * 2/ (50!/(2*2*2*44!)) + C(3,2) * C(11,2) * C(4,2) / (50!/(4!*2*44!))

1st term is pretty straight forward. The 2nd term I split into to halves. The first case when the two chosen opponents have different pairs, and the second when they have the same. The third term is also split into two cases; when all three opponents have different pp, and when a chosen two share a pair.

aloiz

[SheetWise]

I'm following the thought - I just saw the post
by aloiz -- I think there's a cleaner way.
I need to check this, work is interrupting my play ...

C(9,1) * C(66,1) / C(50,2) -
C(9,2) * C(66,2) / C(50,4) +
C(9,3) * C(66,3) / C(50,6) -
C(9,4) * C(66,4) / C(50,8) +
C(9,5) * C(66,5) / C(50,10) -
C(9,6) * C(66,6) / C(50,12) +
C(9,7) * C(66,7) / C(50,14) -
C(9,8) * C(66,8) / C(50,16) +
C(9,9) * C(66,9) / C(50,18) = .29054

SheetWise

I noticed the new font doesn't hold left margin spacing -- is there a way to set font?

[SheetWise]

Sheetwise you dummy -- that won't work. Try again after you get home.

[SheetWise]

It would be interesting to restrict the positioning of the cards (44,42,40,...) in order to determine the number of remaining pair combinations to make this format work.

[LetYouDown]

Bump. Any more thoughts? I didn't have any time this weekend to look at this question further. I'm hoping to have some thoughts to post later today.

[BugsBunny]

For a 10 player table:

1st term:
C(9,1) * [11 * C(4,2)]/C(50,2)

- 2nd term: 2 players hold pairs. The combinations can be in the format xxxx (same pair held by 2 players) or xxyy (2 players hold 2 different pairs)

C(9,2) * {[C(4,4) * 11] + [C(4,2) * C(4,2) * 11 * 10]/2}/C(50,4)

+ 3rd term: 3 players hold pairs. We have xxxxyy and xxyyzz as possibilities.

C(9,3) *{[C(4,4) * 11 * (C(4,2) * 10] + [C(4,2) * C(4,2) * 11 * 10 * 9 ]/3}/C(50,6)

- 4th term: 4 players hold pairs. Possibilities are xxxxyyyy, xxxxyyzz, and wwxxyyzz (now things start getting a bit hairy)
C(9,4) * {[C(4,4) * C(4,4) * 11 * 10]/2 + [C(4,4) * 11] * [C(4,2) * C(4,2) * 10 * 9]/2 + [C(4,2) * C(4,2) * C(4,2) * C(4,2) * 11 * 10 * 9 * 8]/4}/C(50,8)

I'm pretty sure that's right, as far as it goes. I'll leave the remainder of the terms for someone else.

[SheetWise]

[ 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 ...

[BruceZ]

[ 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%

[LetYouDown]

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.