Odds Of Pocket Aces In Texas Holdem

• What Are The Chances Of Pocket Aces
• What Are The Odds Of Winning With Pocket Aces
• Calculating Odds In Texas Holdem
• What Are The Odds Of Pocket Aces

Pocket Aces or Kings: 110 to 1 (1%) From the poker hand odds preflop above, you have a 15% chance of being dealt an ace in Texas Hold em poker. There are other Texas Hold’em poker odds that come with different betting rounds like “on the flop” and “after plop.”. Pocket Aces (or AA) is the most powerful starting hand of all 169 hands in Texas hold'em. This hand consists of two aces. What are the odds to be dealt pocket aces? The odds are 1/221.

Let's start with some rather simple but quite important odds: being dealt aces. There are 1,326 different hole-card combinations in Texas Hold'em poker and 6 of them are aces. The odds of being dealt any specific pocket pair, such as aces, is 220-to-1 but the odds of being dealt a pair of aces and then someone at the same table being dealt pocket kings is slightly less as the two aces have been removed from the 52-card deck. This means that the odds of someone being dealt a pair of kings when you have aces is 205-to-1. How is a Texas Hold’em poker odds calculator useful? A poker odds calculator shows you the exact odds of your hand winning in any scenario. For example, you can give yourself pocket Aces, opponent 1 pocket Kings, and opponent 2 pocket Queens. The poker odds software will then calculate how often each hand wins.

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%.

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.

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.

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.

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.

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%.

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%.

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.

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

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

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.

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.

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.

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

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

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

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

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%.

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.

What Are The Chances Of Pocket Aces

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.

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

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.

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%.

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.

What Are The Odds Of Winning With Pocket Aces

Want to find out the probabilities of being dealt an ace before the flop in Texas Hold'em? Use the tables below to help you find the different aces odds based on the situation and the number of players at the table.

Aces probabilities odds charts.

Calculating Odds In Texas Holdem

Probability of being dealt an ace in any Texas Hold'em game is 15%.

How to use aces odds.

These tables are not going to be of much use to you whilst you play, but it is useful information on the probabilities of hands with aces turning up at your table.

The most useful information from these aces odds tables is that the likelihood of a player being dealt an ace increases with the number of players at the table.

The likelihood of another player holding an ace if you hold an ace yourself table (top right) is interesting. There is a fairly high probability that another player at a 9-seater table holds an ace if you do as well. This means that we should be cautious when dealt weak aces and should fold them in early position. There is a chance that we were the only player at the table holding an ace, but we are saving ourselves more money in the long run for when we come up against another player with a better ace than ours.

Playing aces in Texas Hold'em.

An ace is the highest ranking card in Texas Hold'em and can be played either high or low. This means that if you are dealt an ace preflop, there is a fair chance that you hold one of the strongest hands at the table. If you are all-in with an opponent and they do not have a pocket pair or a better ace, you are the favorite to win the hand.

However, you should be careful to not become overly attached to the hand when you are dealt an Ace, because there is a fair chance that you will lose a sizeable proportion of your stack if you get into a hand with another player that holds a better ace than you (e.g. A9 vs AQ).

Be careful to not overvalue hands with aces, as they can get you in to a lot of trouble if you do not have a good kicker.

For strategy guides on how to play hands with aces, take a look at the rag aces and pocket aces articles.

Go back to the poker odds charts.

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What Are The Odds Of Pocket Aces

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