Odds you have only pair

[Schizo]

Anyone know what percent of the time you are going to have the only pair HU? Against 2 people? Against 3?

Assume random cards of course.

[Cobra]

Using inclusion/exclusion I came up with the following probabilities. The following is a list of probabilities that no one has a pocket pair given that you do.

1 opponent = 94.0%
2 opponents = 88.4%
3 opponents = 83.2%

If you know excel I came up with a formula for the probability that given you have a pair that someone has a pair higher than your.

I did figure out an excel computation for at least one person holding a higher pocket pair than you have. I used inclusion-exclusion with three terms to get an approximation.

N = Number of opponents
X = Number of pair above yours(You have JJ, X = 3)

First term - 6X/C(50,2)
Second term - (36*Power(X,2)-30x)/c(50,2)/c(48/2)
Third term -

(216*Power(X,3)-540*Power(X,2)+324x)/c(50,2)/c(48,2)/c(46,2)

Take the first term and multiply by # opponent combo's, etc.

Term 1 * c(N,1) - Term 2 * c(N,2) + Term 3 * c(N,3)

This will give you the proboability that at least one person will hold a pocket pair higher than you with N number of opponents.

Cobra

[MickeyHoldem]

[ QUOTE ]
Using inclusion/exclusion I came up with the following probabilities. The following is a list of probabilities that no one has a pocket pair given that you do.

1 opponent = 94.0%
2 opponents = 88.4%
3 opponents = 83.2%

[/ QUOTE ]

for 1 opponent = 1152/1225 = 0.940408163
for 2 opponents = 10184/11515 = 0.884411637

just wondering if these are the same results you get... hard to tell with the decimals.

[ QUOTE ]
I did figure out an excel computation for at least one person holding a higher pocket pair than you have. I used inclusion-exclusion with three terms to get an approximation.

N = Number of opponents
X = Number of pair above yours(You have JJ, X = 3)

First term - 6X/C(50,2)
Second term - (36*Power(X,2)-30x)/c(50,2)/c(48/2)
Third term -

(216*Power(X,3)-540*Power(X,2)+324x)/c(50,2)/c(48,2)/c(46,2)

Take the first term and multiply by # opponent combo's, etc.

Term 1 * c(N,1) - Term 2 * c(N,2) + Term 3 * c(N,3)

[/ QUOTE ]
For n=9 and x=2...pocket Qs... your result is 0.085992939

actual results for QQ = 53494035690665815245 / 622114224555778329375 = 0.0859874820076061

There is a little difference here... and it grows as x increases...

x=3 gives 0.126311616 by your method compared to 0.127125134

[Cobra]

Yes Mickeyholdem I got the same exact answers as you did, in this example I was able to determine the exact number of terms that were required, three:

1 opponent = 94.040816%
2 opponents = 88.441164%
3 opponents = 83.179097%

For my equation on overpairs I only figured out three terms due to the complexity of the computations. If I was able to figure out more terms I believe our answers would be closer. Maybe I will try to figure the fourth term and see if I can get a little closer to your answer. Now that I read your post closer we should have gotten exactly the same answer for QQ because there is only two terms in that example. I will take a look at my equation and see if I made a mistake.

Cobra

[Cobra]

Mickeyholdem

If you could run your program for KK with 9 opponents and compare it to mine, then tell me if they different, this will tell me which term to look at.

Cobra

[Cobra]

Mickeyholdem

I just used my formula for QQ and came up with

.0859748155 this differs from what you said my formula would produce. I have double checked it by doing the inclusion/exclusion again and get the same answer.

Term one = 12/1225 = .0097959184

Term two =2*6*(1+6)=84/(50c2)/(48c2)=.0000607903

term one *9=.088163265
term two *36=.00218845

Prob of one or more = .088163265-.00218845= .0859748155

Cobra

[MickeyHoldem]

OK... I don't think your inclusion exclusion formula is handling all the possible combinations of pairs correctly... there should be additional terms for 3 and 4 pairs against you!

see this post (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=probability&Number=1806545 &Forum=,f11,&Words=face%20pairs&Searchpage=1&Limit =25&Main=1802299&Search=true&where=bodysub&Name=&d aterange=1&newerval=&newertype=&olderval=1&olderty pe=w&bodyprev=#Post1806545)
--Mike