Probability Of Pocket Pair

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Understanding Probability In Texas Holdem Is An Essential Aspect Of Profitable Play
– Pre-Flop Probability Guide

Probability Of Being Dealt Pocket Pair
Probability Of Flopping A Set With Pocket Pair
Probability Of Pocket Pair

The probability of getting pocket aces in any one hand if 6/1326. When playing against nine players, the probability of winning with pocket aces is 31.36% – assuming all players stay until the end. Poker aces win 85% of the time against an opponent, although this varies depending on the other hands around the table and the number of opponents. Pocket Pairs: I. Poker Digest Vol. 1, December 28-January 10, 2002. Every now and then in hold'em, a table is treated to two players who've been dealt the same pocket pair. Determining the probability of this happening is an easy application of inclusion-exclusion.

Ever wondered how often you will be dealt Aces? Or what the chances of facing an over-pair are when you hold Jack-Jack? This section of the site will give you the probability of certain hands before the flop (this page) and in later pages the chances of certain flops (for example one suit flop probability) and the chances of dominating hands being out there to your ace-x hand (where x is one of several small to medium cards).

We start by looking at how the cards in the deck can be dealt based on a random distribution of the 1326 ways of 2 cards falling and how often you’ll expect to be dealt certain hands. Then we take suitedness into account – since there are less ways of being dealt 2 suited cards. Later articles in this series will continue with the essential poker probability and card distribution stats you need.

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Firstly, where did the number 1326 come from? Well with 52 cards in the deck your first card is 1/52 this is then multiplied by 1/51 and the total divided by 2 (since it does not matter what order your cards are dealt in) giving 1326 combinations of all cards in any suit.

Of course suitedness is not often important, especially for low cards. The next question is then – what is the number of unique starting hands in Texas Holdem, not counting suits? The answer here is 169 unique hands.

Pre-Flop Poker Probability - 169 Distinct Starting Hands

Now we are getting somewhere. Next we can use the numbers above to work out what the probability of being dealt AA, KK (or in fact any pair) is. This is a case of taking the total number of possible hands and then seeing how many of these are your pair. We will take a pair of Kings as an example. Of the 1326 possible combinations there are 6 ways of being dealt this hand pre-flop. The 6 combinations possible are Kh-Kd, Kh-Kc, Kh-Ks, Kd-Kc, Kd-Ks and Kc-Ks.

6 / 1326 = 0.00435 or 221-to-1

So the chances of being dealt any specific pair are 221:1 against, in fact with 13 possible pairs the chances of being dealt any pair go up to 16-to-1 (there are 78 pair combinations from 1326 total).

Next we can look at unpaired hands, a specific example is the number of ways of being dealt Ace-king pre-flop. Here we have more possible combinations, since there are 8 cards that can be dealt first and then 3 remaining cards to make this hand (we will ignore suitedness for the moment). This gives 16 ways in which A-K can be dealt out of the 1326 combinations – a probability of 0.0121% or approximately 82-to-1. In fact this is the same for any unpaired hand when you ignore the suits.

Pre-Flop Poker Probability - Probability Of Hands Pre-Flop Chart

The reference table below gives probabilities of being dealt specific hands pre-flop: The next article in this series will look at the chances of being dealt hands at the same time as one or more opponent is dealt a higher hand – for example AA vs KK and AK vs QQ.

Pre-flop Hand	Odds
AA	0.045	Same for any pocket pair
AK (any suits)	0.012	Any 2 cards not inc. suits
AK (suited)	0.003
Pair 10-10 or better	0.023	10-10, JJ, QQ, KK or AA
AK, AQ or AJ	0.036	Suits not considered
2 Suited Cards 10+	0.03
Any Suited Connector	0.039	23 suited or better
2 Cards Jack +	0.09	Suits not considered

Finally we can look at how combinations of pre-flop hands work, asking the question what is the probability of being dealt a playable hand for each position. We will use arbitary early, middle and late position combinations here to demonstrate – the actually hands you play is up to your personal style!

Position	Combinations
Early Position	AA to JJ, AKo, AQs	46	29:1
Early Mid Position	AA to 99, AQo+, AJs	18:1
Mid Position	77+, A10o+ A8s+ KQ+	140	9,5:1
Late Position	55+, A7+, QJo+, 78s+	5,5:1

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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?

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

Probability Of Being Dealt Pocket Pair

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^-µ, 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%

Congratulations on a great site. I totally understand your anger over the spread of 6 to 5 Blackjack payouts but am very curious as to why Americans seem to accept 00 Roulette without any argument. This Roulette is almost criminal and should be ranked along side Keno & Slots.

Thanks. You make a valid point. The house edge in 6 to 5 blackjack is 1.44% under the usual rules, while double zero roulette is 5.26%. That is 3.7 times as bad. However I have learned through the years that it is almost hopeless getting players to leave a game they like, regardless of how bad the house edge is. So the best I can do is advise them how to play their game of choice. For blackjack players there is still no shortage of 3 to 2 games out there. Playing 6 to 5 is giving the casino an extra 0.8% advantage for no reason at all. I also stress the importance of looking for single-zero roulette if you are a roulette player. So I see no inconsistency.

Is it required to flip over your cards if you call a player going all in? I’ve seen this done on television numerous times, but can I just wait until then end of the hand?

According to various sources it is required to turn your cards over in a tournament game, but optional in a cash game.

I’m coming out to Vegas next month during March Madness. I’ll probably be bringing a couple thousand dollars with me, some of my own, and some for some friends I’ll be placing bets for on the NCAA games. I’m not sure I’m comfortable carrying that amount of cash around. Would I be able to get a casino marker for a few thousand dollars or are they typically only issued for ’whales’, and can you draw on it for non-table gaming such as the sportsbook? Are there any costs or pitfalls associated with getting a marker (other than having your personal banking information in another database). Thanks for the great site and your time.

Gamblers in your range certainly do use markers. You should try to establish credit with the casino before you go. Alternatively you can wire the casino money, that way you won’t have to go through a credit check. Either way, do so at least a week in advance. The use of markers and wire is very commonplace in the casino and from what I hear the process usually goes very smoothly.

I’ve heard that casinos are looking at using RFID in chips to speed up counting, reduce errors, and defeat fakes. Is the use of this new technology expected to eliminate the ability to use those inadvertently pocketed chips from one casino at another one? Thanks for the site and your time.

For those who may not know, RFID stands for Radio Frequency Identification. I’m not an expert on this topic but it is my understanding they will be used to track player betting patterns, which will help for both comping and catching card counters. However counterfeit chips seems to be a growing problem and that may be another benefit. Currently casinos like it when you leave with chips and never cash them. That is why they create so many chips for special occasions, hoping chip collectors will hoard them. Again, I’m not an expert, but I don’t think it would be cost effective to create these chips if the expense were more than the face value. So I think you’ll be safe pocketing chips.

I have been with my boyfriend for 2years and 8mths we are doing ok, we have talked about our futures and how we want to spend them together, but latley he has been acting different and today i asked him a question about a friend of mine she said that he talked to her well we always tell each other when we talk to the opposite sex but for some reason i guess he didnt and when i asked him about it he just said he didnt talk to her and he got really mad and defensive whats going on with him?

He is probably sick of this policy of reporting to you whenever he talks to someone of the opposite sex. If you can’t trust each other to have an innocent conversation with someone of the opposite sex then the relationship is doomed. So I don’t blame him for getting mad and defensive. I think you are making a small problem into a big one over this. My advice is drop it, and drop this ridiculous confession rule while you’re at it.

In the NBA, there are three divisions per conference, and 8 teams per conference make the playoffs. The top 3 seeds in each conference are the respective division winners, and the #4 through #8 seeds are the non-division winners with the best records. This year, 2 teams from the same division in the Western Conference, the Spurs and Mavericks, have the top record in the West. If this keeps up, it means that the 2nd best team in the Western Conference will end up with the #4 seed and will have to face the best team in the 2nd round if they both win. Many people are pointing to this as a problem with the system, while the NBA considers it an anomaly. In an effort to explain that its not an anomaly because it could happen frequently, an analyst from ESPN recently made the following statement in a blog: 'There are 15 teams in each conference, and five teams in each of the three divisions. That means that there's a 4-in-14 chance that the team with the second-best record will be from the same division as the team with the best record.' Is he correct that there is a 4/14 chance of it happening in a particular conference? How would you figure this out? If he is correct, than it would happen in at least 1 of the conferences 57% of the time, right?

Yes, he is right. There are combin(15,2)=105 ways to choose the two best teams out of 15. There are 3*combin(5,3)=30 ways to choose them from the save division. So the probability the two best teams are from the same division is 30/105 = 4/14. The probability of this happening in at least one conference is 1-(10/14)² = 48.98%.

A reader wrote in to express his comments about the assumptions I made in my answer. Here is a link to his commentary.

In blackjack appendix 10 you explain how CSM machines actually decrease the houses edge, stating that because the way people play basic there is a tendency for more '10' cards to be drawn. Just playing basic, not counting cards at all, would this not mean that the closer you get to the end of the deck/shoe the worse your odds would be since there there would be a tendency for the 10s to have already been extracted thus leaving more smaller cards to the end?

Probability Of Flopping A Set With Pocket Pair

There is no particular time to expect the small cards. The last hand in a cut card has just about the same odds as the overall shoe. However if the dealer deals out much more than the average number of hands in a cut card game the last hands tend to be very bad for the player. This is because in the early hands the players and dealer didn’t hit much, which in turn is because lots of large cards came out, leaving more small cards for later in the shoe. So if you notice that you have already passed the average number of hands and the cut card is still a ways off then the deck is probably small card rich and it would be a good time to drop your bet or take a walk. However with other players jumping in and out of the game, and inconsistent cut card placement the practicality of this strategy is very small.

I recently broke up with my boyfriend of 5 years as I found out he was cheating on me with a 'friend.' After we broke up, I found out that I was pregnant. Because of his new 'friend' he wants nothing to do with our child and refuses to acknowledge my being pregnant. I would like our child to know his family. Is it okay to tell his family that I am pregnant?