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Texas Hold ’em - Probability - Pairs

With a 52-card deck, what are the odds of drawing a pair of Jacks?

Rick from Gardnerville, USA

Assuming you draw five cards, and count all hands with exactly two jacks, then the probability would be combin(4,2)*combin(48,3)/combin(52,5) = 6*17296/2598960 = 3.99%.

Thank goodness I just discovered your great site. I have been trying to solve the following, and keep getting different answers. If I am dealt a pocket pair (in Hold'em) what are my chances of getting either three of a kind, or four of a kind on the flop (next three cards).

Elliot from Harwich, Massachusetts

For probability questions, I like to take the number of combinations the event you're interested can happen divided by the total number of combinations. First review the combin function in my probabilities in poker section. The number of ways to get a four of a kind is simply the number of singletons in the deck, or 48. The number of ways to get a three of a kind (not including a full house) is the product of the number of ways to get the third card, 2, and the number of ways to get two other singletons, 2*combin(12,2)*42 = 2,112. The total number of ways the cards can come up in the flop are combin(50,3)=19,600. So, the probability of a four of a kind is 48/19600=0.0024, and the probability of a three of a kind is 2,112/19,600=0.1078.

In hold em poker, what are the odds of being dealt pocket aces? And what are the odds of being dealt pocket aces twice in a row?

Adam from Redding, USA

There are 52*51/2 = 1326 ways to arrange 2 cards out of 52. There are 4*3/2=6 ways to arrange 2 aces out of 4. So the answer is 6/1326 = 1/221. The probability of this happening twice in a row is (1/221)2 = 1 in 48,841.

If ten people are each dealt two cards from a single deck what is the probability that two players will get a pair of aces?

"Anonymous" .

First, there are 10*9/2=45 ways you can choose 2 players out of 10. The probability of two specific players getting four aces is 1/combin(52,4)=1/270725. So the probability of any two players getting a pair of aces is 45/270725=0.0001662.

In a 10-handed game of Texas Hold 'em, and the flop is three different ranks, what is the probability that three players have a set?

"Anonymous" .

For those unfamiliar with the terminology, each player gets two cards to himself and the three flop cards are shared among all players. So this is the same as asking if you dealt three community cards, all of different ranks, and ten 2-card hands, what is the probability three of the 2-card hands would be pairs that match one of the three community cards.

The probability player 1 has a set is 3*combin(3,2)/combin(49,2). Then the probability player 2 has a set is 2*combin(3,2)/combin(47,2). Then the probability player 3 has a set is combin(3,2)/combin(45,2). However, any three players can the three sets, not necessarily the first three. There are combin(10,3) ways to choose the 3 players out of 10 that have sets. So the answer is combin(10,3)*(3*combin(3,2)/combin(49,2))*(2*combin(3,2)/combin(47,2))*(combin(3,2)/combin(45,2)) = 0.00000154464 = 1 in 64,740.

Hi - Thank you for your web site. I'm wondering if you can tell me what are the odds if you are dealt Q-Q that any of the remaining 8 people at the table would be dealt A-A, A-K, K-K, or A-Q? Thank you!

"Anonymous" .

For any given person the probability of having AA is combin(4,2)/combin(50,2) = 6/1,225 = 0.0049 because there are 6 ways to pick 2 aces out of 4, and 1225 ways to pick any 2 cards out of the 50 left in the deck. The probability is the same for a pair of kings. For A-K the probability is 4*4/1,225=0.0131, because there are 4 ways to get an ace and 4 ways to get a king. For A-Q the probability is 4*2/1225=0.0065, because there are 4 aces but only 2 queens left in the deck. So the probability any given player will have one of these hands is (6+6+16+8)/1225 = 0.0294. Now the next step is clearly not perfect because if one player doesn't have one of these hands the odds the next player does is a little bit higher. Forgetting this for the sake of simplicity the probability no player has one of these hands is (1-0.0294)8 = 78.77%. So the probability at least one player has one of these hands is 21.23%.

In Texas hold em, if two players are dealt a pocket pair pre-flop, what are the odds of each of these players flopping a set(three of a kind)?

Bob from Cincinnati

Let’s assume you have a pair of aces. Before considering that the other player has another pair the probability of flopping a three of a kind is the [nc(one ace)*nc(two ranks out of 12)*nc(one suit out of 4)2 + nc(any other three of a kind)]/nc(any three cards), where nc(x) = number of combinations of x. This equals [2*combin(12,2)*42+12*combin(4,3)]/combin(50,3) = (2112+48)/19600 = 11.020%. Now lets assume the other player has any other pair, but not the same as yours. Then probability becomes [2*(combin(11,2)*42 + 11*2*4 + 11*combin(4,3)]/combin(48,3) = 11.4477%.

What are the chances in a heads up game of Texas Hold’em that each player gets KK. Then on the very next hand both players get KK. We can’t even get a close estimate. If you can figure it out please respond, thanks.

"Anonymous" .

The probability for any given hand is (combin(4,2)/combin(52,2))*(1/combin(50,2)) = 1/270725. So, the probability of this happening twice in a row is 1 in 270,7252 = 1 in 73,292,025,625.

What are the odds of a pair showing on the board on the flop in Holdem? ie., A A 10 or 5 Q 5, etc.

Stephen from Addison

13*12*combin(4,2)*4/combin(52,3) = 3744/22100 = 16.941%.

Thanks for the help your site has given. You’ve probably saved me thousands. I was playing in a NL Texas Hold 'em tournament on-line recently, and was dealt pocket kings (at a 10-man table) only to be dominated by pocket aces. I'd like to know the probability, given the condition that you have a pair, of at least one other player at the 10-man table having a higher pair than yours (in other words, having a "dominated pair"). Thanks again!

Dan from Cairo, Egypt

The following table shows estimated probabilities that a pair will be beaten by at least one higher pair according to the number of players (including yourself). These probabilities are not exact because the hands are not independent. However to find the exact probabilities would get complicated and I think these are pretty close. My formula is 1-(1-r*combin(4,2)/combin(50,2))(n-1), where r=number of higher ranks than your pair, and n = total number of players. The table shows the probability of another player having a pair of aces, when you have a pair of kings, in a 10-player game, to be 4.323%.

Probability Pair Beaten by Higher Pair

Pair 2 Pl. 3 Pl. 4 Pl. 5 Pl. 6 Pl. 7 Pl. 8 Pl. 9 Pl. 10 Pl.
KK 0.49% 0.977% 1.462% 1.945% 2.425% 2.903% 3.379% 3.852% 4.323%
QQ 0.98% 1.95% 2.91% 3.861% 4.803% 5.735% 6.659% 7.573% 8.479%
JJ 1.469% 2.917% 4.344% 5.749% 7.134% 8.499% 9.843% 11.168% 12.473%
TT 1.959% 3.88% 5.763% 7.609% 9.42% 11.194% 12.934% 14.64% 16.312%
99 2.449% 4.838% 7.168% 9.442% 11.66% 13.823% 15.934% 17.992% 20.001%
88 2.939% 5.791% 8.56% 11.247% 13.855% 16.387% 18.844% 21.229% 23.544%
77 3.429% 6.74% 9.937% 13.025% 16.007% 18.887% 21.668% 24.353% 26.947%
66 3.918% 7.683% 11.301% 14.776% 18.115% 21.324% 24.407% 27.369% 30.215%
55 4.408% 8.622% 12.65% 16.501% 20.181% 23.7% 27.063% 30.279% 33.352%
44 4.898% 9.556% 13.986% 18.199% 22.205% 26.016% 29.64% 33.086% 36.363%
33 5.388% 10.485% 15.308% 19.871% 24.188% 28.273% 32.137% 35.794% 39.253%
22 5.878% 11.41% 16.617% 21.517% 26.13% 30.472% 34.559% 38.405% 42.025%

In a three-handed game of Hold 'em, what are the chances of A-A vs. K-K vs. Q-Q?

Chris from Hampton

Let's call the players A, B, and C. The probability A has a pair of aces is combin(4,2)/combin(52,2) = 6/1326. The probability B has a pair of kings is combin(4,2)/combin(50,2) = 6/1225. The probability C has a pair of queens is combin(4,2)/combin(48,2) = 6/1128. However there are 3! = 1*2*3 = 6 ways you can arrange three pairs between three players. So, the answer is 6*(6/1326)*(6/1225)*(6/1128) = 0.000000707321.

I’ve been a huge fan for many years (even before you got interested in poker and sports betting) and looked forward to every Ask The Wizard column. It’s great to see you’re doing them again! My question is this: at my local card room, they offer Aces Cracked, Win A Rack during certain hours. That is, if you have pocket Aces in one of their 3-6 or 4-8 Texas Hold ’Em games and you lose the pot, the casino will give you a rack of chips ($100). I’m trying to figure out how often a)I get pocket Aces b)how often they would lose if I played them aggressively as I’m supposed to and c)whether it’s not better to just check all the way down and hope to lose, as $100 is usually better than what the pot would have been anyway. Any stats you may have at the ready would be wonderful and forever appreciated! Thanks again and keep up enlightening the masses!

Shane from Santa Rosa

Thanks for the kind words. The probability you will get pocket aces in any one hand is 6/1326, or once every 221 hands. According to my 10-player Texas Hold ’em section (/games/texas-hold-em/10players.html) the probability of winning with pocket aces is 31.36%, assuming all players stay in until the end. However that is a big if. If forced to make a guess I’d estimate the probability of winning with aces in a real 10-player game is about 70%. So the probability of getting pocket aces and then losing is 0.3*(1/221) = 0.1357%. So, at $100 per incident that is worth 13.57 cents per hand. Over ten people that costs the poker room $1.36 per hand on average, which cuts into the rake quite a bit. I tend to agree with your strategy of calling, which will keep more players in the hand, and increase your chance of losing.

What are the odds of having pocket aces and pocket kings both being dealt in the same hand?

Jake from Loveland, CO

The odds of a specific player having aces is combin(4,2)/combin(52,2) = 6/1326. The odds of the next player having a pair of kings is combin(4,2)/combin(50,2) = 6/1225. However, in a ten-player game there are 10 possible players who could get the aces, and 9 possible players for the kings. So a strong approximation would be 10*9*(6/1326)*(6/1225) = 0.001995, or 1 in 501. This answer is slightly too high, because it double counts the situation where two players have aces, or two have kings, or both.

Hello, thank you for a very interesting and informative site. I have a question of my own that I hope you can answer for me. As a Texas hold 'em player, I pay special attention to pocket pairs, and have particular interest towards 10-10 or J-J or similar, as on the surface they appear strong but can be beaten easily. My question however, is how do you work out the probability of there being at least one person on your table holding a higher pocket pair to yours?

Andrew from St. Albans

The math of this gets very messy due to the possibility of more than one playing having a higher pair, including the same type of pair. For example if you have pocket kings two players could have pocket aces. However it is easy to show the expected number of players who will beat you. This would be n*r*(6/1225), where n is the number of opponents, and r is the number of higher ranks. The following table shows the average number of players who will have a higher pocket pair according to your pocket pair (left column) by the number of opponents (top row).

Expected Number of Higher Pocket Pairs by Number of Opponents

Pair 1 Opp. 2 Opp. 3 Opp. 4 Opp. 5 Opp. 6 Opp. 7 Opp. 8 Opp. 9 Opp.
2,2 0.0588 0.1176 0.1763 0.2351 0.2939 0.3527 0.4114 0.4702 0.529
3,3 0.0539 0.1078 0.1616 0.2155 0.2694 0.3233 0.3771 0.431 0.4849
4,4 0.049 0.098 0.1469 0.1959 0.2449 0.2939 0.3429 0.3918 0.4408
5,5 0.0441 0.0882 0.1322 0.1763 0.2204 0.2645 0.3086 0.3527 0.3967
6,6 0.0392 0.0784 0.1176 0.1567 0.1959 0.2351 0.2743 0.3135 0.3527
7,7 0.0343 0.0686 0.1029 0.1371 0.1714 0.2057 0.24 0.2743 0.3086
8,8 0.0294 0.0588 0.0882 0.1176 0.1469 0.1763 0.2057 0.2351 0.2645
9,9 0.0245 0.049 0.0735 0.098 0.1224 0.1469 0.1714 0.1959 0.2204
T,T 0.0196 0.0392 0.0588 0.0784 0.098 0.1176 0.1371 0.1567 0.1763
J,J 0.0147 0.0294 0.0441 0.0588 0.0735 0.0882 0.1029 0.1176 0.1322
Q,Q 0.0098 0.0196 0.0294 0.0392 0.049 0.0588 0.0686 0.0784 0.0882
K,K 0.0049 0.0098 0.0147 0.0196 0.0245 0.0294 0.0343 0.0392 0.0441


To get the probability that at least one player will beat you I will make the not entirely correct assumption that the number of players with a higher pocket pair is a Poisson random variable with a mean in the above table. Given that assumption the probability that at least one player will beat you is 1-e, where µ is the mean. For examle if you have pocket queens and there are 9 other players the expected number of players with a higher pocket pair is 0.0882, so the probability of at least one player having a higher pocket pair is 1-e-0.0882 = 8.44%. The table below shows those probabilities.

Probability of Higher Pocket Pair by Number of Opponents — Wizard's Approximation

Pair 1 Opp. 2 Opp. 3 Opp. 4 Opp. 5 Opp. 6 Opp. 7 Opp. 8 Opp. 9 Opp.
2,2 5.71% 11.09% 16.17% 20.95% 25.46% 29.72% 33.73% 37.51% 41.08%
3,3 5.25% 10.22% 14.92% 19.39% 23.62% 27.62% 31.42% 35.02% 38.42%
4,4 4.78% 9.33% 13.67% 17.79% 21.72% 25.46% 29.03% 32.42% 35.65%
5,5 4.31% 8.44% 12.39% 16.17% 19.78% 23.24% 26.55% 29.72% 32.75%
6,6 3.84% 7.54% 11.09% 14.51% 17.79% 20.95% 23.99% 26.91% 29.72%
7,7 3.37% 6.63% 9.77% 12.82% 15.75% 18.59% 21.34% 23.99% 26.55%
8,8 2.9% 5.71% 8.44% 11.09% 13.67% 16.17% 18.59% 20.95% 23.24%
9,9 2.42% 4.78% 7.08% 9.33% 11.52% 13.67% 15.75% 17.79% 19.78%
10,10 1.94% 3.84% 5.71% 7.54% 9.33% 11.09% 12.82% 14.51% 16.17%
J,J 1.46% 2.9% 4.31% 5.71% 7.08% 8.44% 9.77% 11.09% 12.39%
Q,Q 0.97% 1.94% 2.9% 3.84% 4.78% 5.71% 6.63% 7.54% 8.44%
K,K 0.49% 0.97% 1.46% 1.94% 2.42% 2.9% 3.37% 3.84% 4.31%


So my approximation of the probability of at least one higher pocket pair is 1-e-n*r*(6/1225).

P.S. After this column appeared one of my fans, Larry B., wrote a brute force combinatorial program to solve this problems. Here are his results.

Probability of Higher Pocket Pair by Number of Opponents — Larry B.'s Exact Probabilities

Pair 1 Opp. 2 Opp. 3 Opp. 4 Opp. 5 Opp. 6 Opp. 7 Opp. 8 Opp. 9 Opp.
2,2 5.88% 11.41% 16.61% 21.5% 26.1% 30.43% 34.5% 38.33% 41.94%
3,3 5.39% 10.48% 15.3% 19.87% 24.18% 28.26% 32.12% 35.77% 39.22%
4,4 4.9% 9.56% 13.99% 18.2% 22.21% 26.03% 29.66% 33.12% 36.4%
5,5 4.41% 8.62% 12.66% 16.52% 20.21% 23.73% 27.11% 30.35% 33.45%
6,6 3.92% 7.69% 11.31% 14.8% 18.15% 21.38% 24.48% 27.47% 30.34%
7,7 3.43% 6.74% 9.95% 13.05% 16.05% 18.95% 21.76% 24.47% 27.09%
8,8 2.94% 5.8% 8.58% 11.28% 13.91% 16.46% 18.95% 21.36% 23.71%
9,9 2.45% 4.84% 7.19% 9.47% 11.71% 13.9% 16.04% 18.13% 20.17%
T,T 1.96% 3.89% 5.78% 7.64% 9.47% 11.27% 13.04% 14.77% 16.48%
J,J 1.47% 2.92% 4.36% 5.78% 7.18% 8.57% 9.93% 11.29% 12.63%
Q,Q 0.98% 1.95% 2.92% 3.88% 4.84% 5.79% 6.73% 7.67% 8.6%
K,K 0.49% 0.98% 1.47% 1.96% 2.44% 2.93% 3.42% 3.91% 4.39%


Later Stephen Z. suggested a simple approximation. Take the number of higher pairs, multiply by the number of other players, and divide by 2. That is the percentage probability that there will be at least one higher pair. For example, with a pair of jacks in a 10 player game the probability of a higher pocket pair is 3*9/2 = 13.5%. Using that formula you get the following for all situations.

Probability of Higher Pocket Pair by Number of Opponents — Stephen Z. Approximation

Pair 1 Opp. 2 Opp. 3 Opp. 4 Opp. 5 Opp. 6 Opp. 7 Opp. 8 Opp. 9 Opp.
2,2 6% 12% 18% 24% 30% 36% 42% 48% 54%
3,3 5.5% 11% 16.5% 22% 27.5% 33% 38.5% 44% 49.5%
4,4 5% 10% 15% 20% 25% 30% 35% 40% 45%
5,5 4.5% 9% 13.5% 18% 22.5% 27% 31.5% 36% 40.5%
6,6 4% 8% 12% 16% 20% 24% 28% 32% 36%
7,7 3.5% 7% 10.5% 14% 17.5% 21% 24.5% 28% 31.5%
8,8 3% 6% 9% 12% 15% 18% 21% 24% 27%
9,9 2.5% 5% 7.5% 10% 12.5% 15% 17.5% 20% 22.5%
T,T 2% 4% 6% 8% 10% 12% 14% 16% 18%
J,J 1.5% 3% 4.5% 6% 7.5% 9% 10.5% 12% 13.5%
Q,Q 1% 2% 3% 4% 5% 6% 7% 8% 9%
K,K 0.5% 1% 1.5% 2% 2.5% 3% 3.5% 4% 4.5%

I have looked all over the net about the probability of at least getting a pair by the river card in hold’em if you have two different cards dealt to you. I have tried to work it out using a probability tree but my answer seems too high. Also on the net I have read different answers some suggesting it is around 1/3 or 2/5 or 1/2. What is the probability of at least pairing and is it possible to work this out using a probability tree? Your help will be much appreciated thanks.

Nathan S. from New Plymouth

For those not familiar with the hold 'em terminology, you are asking for the probability of at least a pair in six cards, given that the first two are (the hole cards) of different ranks. I hope you'll forgive me if I just do the probability of getting exactly a pair, including hands that also form a straight or flush.

The number of ways to pair one of your hole cards is six (2 hole cards * 3 suits remaining). The other three cards must all be of different ranks from the 11 left. There are combin(11,3)=165 ways to choose 3 ranks out of 11. For each of these there are four suits to choose from. So the number of ways to pair one of your hole cards is 6*165*43=63,360.

Now let's look at the number of ways to get a pair outside of the two hole cards. There are 11 ranks to choose from for the pair. Once the pair is chosen there are combin(4,2)=6 ways to choose 2 suits out of 4. For the other two cards there are combin(10,2)=45 ways to choose 2 ranks out of the 10 fully intact ranks left. For both of those ranks there are 4 possible suits. So the total combinations for a pair, not including the hole cards, is 11*6*45*42=47,520.

The total number of ways to choose 4 cards out of the 50 left in the deck is combin(50,4)=230,300. So the probability of getting exactly a pair in six cards is (63,360+47,520)/230,300 = 48.15%.

Last night I played a hand where three players all hit sets on the flop. Luckily for me I had AA against QQ and 22. What is the probability that three players hit sets on the flop? Cheers

Gareth H. from Auckland, NZ

The probability of three different ranks in the flop is combin(13,3)×43/combin(52,3) = 0.828235. There are combin(10,3)=120 ways you can choose three players out of ten. Of the three, the probability the first will have a set is 3×combin(3,2)/combin(49,2) = 0.007653061. The probability the second will have a set is 2×combin(3,2)/combin(47,2) = 0.005550416. The probability the third will have a set is combin(3,2)/combin(45,2) = 0.003030303. Take the product of all this and the probability is 0.828235 × 120 × 0.007653061 × 0.005550416 × 0.003030303 = 0.00001279, or 1 in 78,166.

In 55,088 hands of poker I had a pair going into the flop 2,787 times. Of those 2,787 I hit a set 273 times. How does that square with expectations?

Linus from Alingsås, Sweden

For readers who may not know, a "set" is a three of a kind after the flop, including a pocket pair. The probability of not making a set is (48+combin(48,3))/combin(50,3) = 17,344/19600 = 88.49%. So the probability of making a set is 11.51%. In 2,787 pairs you should have made a set 320.8 times. So you are 47.8 sets under expectations. The variance is n × p × (1-p), where n = number of hands, and p = probability of making the set. In this case the variance is 2,787 × .1176 × .8824 = 283.86. The standard deviation is the square root of that, or 16.85. So you are 47.8/16.85 = 2.84 standard deviations south of expectations. The probability of luck this bad or worse can be found in any Standard Normal table, or in Excel as norsdist(-2.84) = 0.002256, or 1 in 443.

I feel I was cheated in a poker game. According to my math, AA vs. KK will happen once every 45,000 hands heads up, but it happened to me 3 times in 400 hands. Is this unlikely enough to suspect something?

Rafael

The probability of being on the losing end of KK vs. AA is (combin(4,2)/combin(52,2)) × (combin(4,2)/combin(50,2)) = 0.000022162, for each opponent at the table. That is once every 45,121 hands, so your math was right. The expected number of times that would happen in 400 hands is 400 × 0.000022162 = 0.008865084, per opponent. The following table shows the probability of 3 or more instances of having KK against AA in 400 hands, by the number of opponents.

3+ KK vs AA probability in 400 hands

Opponents Probability Inverse
1 0.0000001145 1 in 8,734,376
2 0.0000009133 1 in 1,094,949
3 0.0000030658 1 in 326,182
4 0.0000072234 1 in 138,438
5 0.0000140202 1 in 71,325
6 0.0000240728 1 in 41,541
7 0.000037981 1 in 26,329
8 0.0000563277 1 in 17,753
9 0.0000796798 1 in 12,550

So, yes, I would say this looks fishy. The fewer the players, the more fishy it looks. I would be interested to know where this game was.

Great Site!! If I have pocket Queens, what is the chance that an Ace or King will come by the river? A simple fundamental question, but one that will help me tremendously.

Ed Miller from Banning CA

Thanks. There are 50 cards left in the deck, and 42 of them are not aces or kings. The probability of not seeing any aces or kings in five community cards is combin(42,5)/combin(50,5) = 850,668/2,118,760=40.15%. So, the probability of seeing at least one ace or king is 100% - 40.15% = 59.85%.

This happened to me this week, and am eagerly curious as to the statistic. Over two nights, I held Pocket Aces 3 times in total, and all 3 times I had them there was another player on the table of 10 players also with Pocket Aces. I have not been able to find the probability of this happening anywhere and I hope you can shed some light on this. On a full table of 10 players, what is the chance of this happening?

Rob T. from Hong Kong

The probability of a specific other player having pocket aces, given that you do, is (2/50)×(1/49) = 1 in 1,225. Given 9 other players, the probability is 9 times that, or 1 in 136. This might seem like an abuse of taking the sum of probabilities. However, it is okay if only one player can get the two aces. To answer your question, the probability that another player had pockets aces three out of the three times you had pockets aces is (9×(2/50)×(1/49))3 = 1 in 2,521,626.