Pocket pairs in the same hand - Gambling and Probability

m2510k · 07-25-2012, 02:38 AM

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.

statmanhal · 07-25-2012, 08:46 PM

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.

**BruceZ** · 07-25-2012, 09:53 PM

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.

**BruceZ** · 07-25-2012, 11:05 PM

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

statmanhal · 07-25-2012, 11:41 PM

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%

m2510k · 07-27-2012, 08:30 AM

You guys are wizzards, thank you

Sherman · 07-27-2012, 09:21 AM

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.

**BruceZ** · 07-28-2012, 05:39 PM

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.