Dual pocket pairs probability

[Wetdog]

I have done a search on this forum and have yet to find thia question. If someone has a link to the answer I'd appreciate it.

It seems that there is a surprising number of times in Texas Holdem that there is another pocket pair at the same time as mine. I saw in another forum that there is a "law" that says that this will often happen(removing the paired cards makes it more likely that it happens with the remaining 50). I don't know how credible this person is but, knowing that there are 8+ x 10^67 hands in TH, it seems rather odd that this occurance is so likely.

Could this be explained to a non-calculus guy in his mid 50's how much more likely another pocket pair could occur in a 10, 9 or 6 handed game?

[BruceZ]

[ QUOTE ]
I have done a search on this forum and have yet to find thia question. If someone has a link to the answer I'd appreciate it.

It seems that there is a surprising number of times in Texas Holdem that there is another pocket pair at the same time as mine. I saw in another forum that there is a "law" that says that this will often happen(removing the paired cards makes it more likely that it happens with the remaining 50). I don't know how credible this person is but, knowing that there are 8+ x 10^67 hands in TH, it seems rather odd that this occurance is so likely.

Could this be explained to a non-calculus guy in his mid 50's how much more likely another pocket pair could occur in a 10, 9 or 6 handed game?

[/ QUOTE ]

The exact answer requires the inclusion-exclusion principle; however, I happen to know from doing a similar problem that approximating the hands to be independent is very accurate in this case. That was shown in this thread, which computes the probability that a pair is out at an 8-handed table from a spectator's point of view. That thread discusses both the inclusion-exclusion solution, and the independence approximation.

For your example at a 10-handed table, if you have a pair, there is 1 more pair remaining of the same rank as your pair, and 12*6 other pairs, for a total of 73 remaining pairs. The probability that a particular opponent has a pair is 73/C(50,2) = 73/1225. The probability that he has a non-pair is 1 - 73/1225 = 1152/1225. Using the approximation of independence, the probability that none of your 9 opponents has a pair is (1152/1225)^9. The probability that at least one opponent has a pair is 1 - (1152/1225)^9 =~ 42.5%.

Notice that if you held a non-pair, then there would only be 3 pairs remaining for each of the two ranks that you hold, plus 6*11 other pairs, for a total of 72 pairs. So there is 1 fewer pair remaining than in the case where you held a pair, so it will be slightly more likely that your opponents will hold a pair when you hold a pair, as you said, but there isn't enough difference that you would notice.

[Wetdog]

Thanks, Bruce. Even an old fossil like me could understand that. Good work.