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/...;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.