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 07-25-2012, 02:38 AM #1 enthusiast   Join Date: May 2010 Posts: 77 Pocket pairs in the same hand You're playing 6-max NLH and are dealt a pocket pair. What are the odds that one of the other five players has a pocket pair as well? And another one, let's say you are dealt 77. What are the odds someone has a higher/lower pocket pair? How do you calculate this? Thanks.
 07-25-2012, 08:46 PM #2 veteran   Join Date: Jan 2009 Posts: 2,336 Re: Pocket pairs in the same hand Working out probabilities like this with more than one opponent is often difficult. To overcome this I wrote a simulation program that looks at these types of problems. With you holding a pair and with 5 other opponents, the chance at least one player has a pair is 26.4%. With a pair of sevens, the chance at least one opponent has a higher pair is 16.1% The chance for a lower pair is 12.1%. These results were based on 5 million simulations. I would be interested to see an analytical solution or another simulation result to verify these. Although I have made hundreds of checks on the program, an undiscovered bug is always possible.
07-25-2012, 09:53 PM   #3
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,899
Re: Pocket pairs in the same hand

Quote:
 Originally Posted by statmanhal Working out probabilities like this with more than one opponent is often difficult. To overcome this I wrote a simulation program that looks at these types of problems. With you holding a pair and with 5 other opponents, the chance at least one player has a pair is 26.4%. With a pair of sevens, the chance at least one opponent has a higher pair is 16.1% The chance for a lower pair is 12.1%. These results were based on 5 million simulations. I would be interested to see an analytical solution or another simulation result to verify these. Although I have made hundreds of checks on the program, an undiscovered bug is always possible.
For pairs, the independence approximation is so good that it makes exact calculations with the inclusion-exclusion principle or simulated results unnecessary for all practical purposes. The approximate results are generally within a tenth of a percent.

By this approximation, when you hold 77, the probability that one of your 5 opponents holds a pair is

1 - [1 - 73/C(50,2)]5

=~ 26.45%

Inclusion-exclusion gives 26.41%.

For a higher pair it is:

1 - [1 - 42/C(50,2)]5

=~ 16.00%

Inclusion-exclusion gives 16.05%.

For a lower pair it is:

1 - [1 - 30/C(50,2)]5

=~ 11.66%

Inclusion-exclusion gives 11.71%. Since this is the only one that differs from yours, here are the first 3 terms of inclusion-exclusion which is good to much better than 0.01%:

5*30/C(50,2) -
C(5,2)*30*25/C(50,2)/C(48,2) +
C(5,3)*30*(24*20 + 1*24)/C(50,2)/C(48,2)/C(46,2)

=~ 11.71%

You should check that last one or run it some more. You should be closer than 12.1% after 5 million.

Last edited by BruceZ; 07-25-2012 at 10:38 PM.

 07-25-2012, 11:05 PM #4 Carpal \'Tunnel     Join Date: Sep 2002 Posts: 8,899 Re: Pocket pairs in the same hand I see what you did. 12.1% includes another pair of sevens, so it isn't for a lower pair. I wrote my own quick simulation and got a result consistent with the 11.71% for a lower pair. Code: ```deck = c(rep(1:13,2),1:6,8:13,1:6,8:13) sims = 0 count = 0 while (1) { deal = sample(deck, 10, F) j = 1 pair = FALSE while (j <= 9 & !pair) { if ( (deal[j] == deal[j+1]) & (deal[j] < 7) & (deal[j] > 1) ) { count = count + 1 pair = TRUE } j = j + 2 } sims = sims + 1 } p = count/sims error = 3.29*sqrt(p*(1-p)/sims) p p - error p + error sims``` Output: > p [1] 0.1170673 > p - error [1] 0.1168082 > p + error [1] 0.1173264 > sims [1] 16663222 Last edited by BruceZ; 07-25-2012 at 11:47 PM.
07-25-2012, 11:41 PM   #5
veteran

Join Date: Jan 2009
Posts: 2,336
Re: Pocket pairs in the same hand

Quote:
 Originally Posted by BruceZ I see what you did. 12.1% includes another pair of sevens, so it isn't for a lower pair. I wrote my own quick simulation and got a result consistent with the 11.7% for a lower pair.
Yes - one thing my program is very bad at doing is telling the dummy inputter, me, that he entered the wrong value. When I entered 66- (for pairs lower than 7), I got 11.7%

 07-27-2012, 08:30 AM #6 enthusiast   Join Date: May 2010 Posts: 77 Re: Pocket pairs in the same hand You guys are wizzards, thank you
 07-27-2012, 09:21 AM #7 Carpal \'Tunnel     Join Date: Jun 2005 Location: Psychology Department Posts: 7,426 Re: Pocket pairs in the same hand Bruce, the 3.29 is for a 99.9% confidence interval? <- note the question mark. edit. I'm 99% confident that is correct, I just wanted you to confirm.
07-28-2012, 05:39 PM   #8
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 8,899
Re: Pocket pairs in the same hand

Quote:
 Originally Posted by Sherman Bruce, the 3.29 is for a 99.9% confidence interval? <- note the question mark. edit. I'm 99% confident that is correct, I just wanted you to confirm.
Yes.

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