Texas Hold 'em odds for pairs?

I have a relatively simple question about Texas Hold 'em. What is the probability of two players out of about 6 both being dealt any kind of pocket pair at the same time? And what about the odds of two players being dealt any kind of pair at any time during that hand not just limited to hole cards? All responses would be very appreciated. Thank you.

Let me solve part of the problem. Suppose Art and Bob are playing Texas Hold 'em with any other number of players. What is the probability that both Art and Bob are dealt pocket pairs? (We don't care what the other players have--even if all of them have pocket pairs.)

For Art the probability of a pocket pair is 3/51 = 1/17. Given that Art has a pocket pair, the conditional probability that Bob has the same pocket pair (ignoring suits, of course) is (2/50) * (1/49). The conditional probability that Bob has a particular different pocket pair is (4/50) * (3/49).

The answer to my question is therefore
(1/17) * [(2/50 * (1/49) + 12 * (4/50) * (3/49)] = 73/20875, which is approximately 0.35%.

In your original question with six players, I presume you don't care who gets the pocket pairs, so the probability is much higher, by a factor of about 15. (Hey, if you can say "about six," I am allowed to say "about 15." :-) But this ignores the probabilities of three getting pocket pairs, four getting pocket pairs, etc. You didn't say if you wanted the probability of exactly two getting pockets pairs or the probability of two or more getting pocket pairs.

Getting pairs at any time is more difficult to calculate.

Quote: TD33

I have a relatively simple question about Texas Hold 'em. What is the probability of two players out of about 6 both being dealt any kind of pocket pair at the same time? And what about the odds of two players being dealt any kind of pair at any time during that hand not just limited to hole cards? All responses would be very appreciated. Thank you.

The probability that at least 2 players out of 6 are dealt pairs for their hole cards is exactly

C(6,2)*78*73/C(52,2)/C(50,2) -

2*C(6,3)*78*(72*68 + 1*72)/C(52,2)/C(50,2)/C(48,2) +

3*C(6,4)*78*(72*(66*63 + 2*67) + 1*72*67)/C(52,2)/C(50,2)/C(48,2)/C(46,2) -

4*C(6,5)*78*(72*(66*(60*58 + 3*62) + 2*(66*62 + 1*66)) + 1*72*(66*62 + 1*66))/
C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2) +

5*C(6,6)*78*(72*(66*(60*(54*53 + 4*57) + 3*(60*57 + 2*61)) + 2*(66*(60*57 + 2*61) + 1*66*61)) +
1*72*(66*(60*57 + 2*61) + 1*66*61))/C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)/C(42,2)

≈ 0.0446835057167044

or odds of about 21.4-to-1 by a generalization of the inclusion-exclusion principle.

On the first line, you can see SDSDNSR's answer of 0.35% multiplied by C(6,2) = 15 pairs of players. This counts the cases where exactly 3 players get pairs C(3,2) = 3 times, so the second line computes the probability of 3 specific players getting pairs, multiplies that by C(6,3) = 20 ways to choose 3 players, and then multiplies that by 2 and subtracts from the first line so these are only counted once. But that first line also counted the probabilty of 4 players getting pairs C(4,2) = 6 times, and the second line subtracted them 2*C(4,3) = 8 times, which means we have counted those a total of -2 times. So the 3rd line computes the probability that 4 specific players have pairs, multiplies by C(6,4) = 15 ways to choose the 4 players, then multiplies all that by 3 and adds to the result of the first 2 lines so that we have now counted those only once. The remaining 2 lines handle in the same way the cases of 5 and 6 players getting pairs. Basically we do the same type of over counting at each step that SDSDNSR did when he multiplied by 15, and then we proceed to fix this over counting by doing more over counting of the same type, until we get to the end where it becomes exact since there is only 1 way for 6 players to have pairs. SDSDSNR said it would be about 15 times the value for 2 players, and we knew it would be less. In fact it is about 12.7 times that value.

Note that we alternate subtraction and addition. First we include too many, then we subtract too many, then we add back too many, etc., hence the name inclusion-exclusion. The results become more precise at each step, and you can always bound your answer between the last 2 values computed. We could have quit after the 3rd line and gotten an answer between 4.41% and 4.47%. Doing the whole thing exactly is a bit ridiculous, but I wanted to illustrate that the terms follow an easy pattern for pairs. As we consider each additional player, he can be dealt some number of pairs of which there are 6 of each, and some number (possibly 0) of which there is only 1 of each remaining. So for example on the 3rd line, a player can get one of 68 pairs, which would mean there are 11 pairs of which there are 6, and 2 pairs of which there are 1 each. So we break the 68 into 66+2, and note that if he gets one of the 66, the total number of pairs will decrease by 5 to 68-5=63, whereas if he gets one of the other 2, the total will decrease by 1 to 68-1 = 67. So this factor of 68 will be replaced by (68*63 + 2*67).

Note the leading coefficients of 1,2,3,4,5. Had we been asking about at least 1 player with a pair, those would have all been 1. Had we been asking about at least 3 players with pairs, they would have been 1,3,6,10. In general, for at least N, they come from the Nth diagonal of Pascal's triangle. We can also ask about exactly 1 pair, exactly 2 pairs, etc. The coefficients for exactly N pairs turn out to be the same as for at least N+1 pairs. I discovered this pattern for myself years ago only to realize much later that I had it on my shelf all along in the classic work by Feller. It's in a chapter titled "Combination of Events".

I did a simulation to confirm this calculation since people aren't always convinced by a calculation that's as long as your arm.


For your second question about at least 2 people making a pair including the board, I'll assume you want all players to stay in to the river, but that isn't realistic. It isn't clear if you want to include pairs only on the board. If we don't count pairs on the board, it's about 90.9%, and if we do count pairs on the board, it's about 98.3%. These are by the simulations below. Since the board normally pairs about half the time, we can see that when the board pairs fewer than 2 players and nobody has a pair in the hole, the board is much more likely than usual to have fewer than 5 different ranks. This makes sense from Bayes' theorem since it is much more likely that no one will pair the board when the board has fewer than 5 ranks.