- Odds Of Getting Pocket Aces
- Odds Of Getting Pocket Aces In Texas Holdem
- Odds Of Getting Back To Back Pocket Aces
- Odds Of Getting Pocket Aces In Texas Holdem
- Odds Of Poker Hand
We have a 1 in (13. 17) chance of being dealt Pocket Aces, or 1 in 221. So, we have a less than 0.5% chance of being dealt pocket Aces. Please note: the number of players at the table is irrelevant in determining your chances of being dealt pocket Aces.
- To put this in perspective, if you’re playing poker at your local casino and are dealt 30 hands per hour, you can expect to receive pocket Aces an average of once every 7.5 hours. The odds of receiving any of the thirteen possible pocket pairs (twos up to Aces) is: (13/221) = (1/17) ≈ 5.9%.
- If you're a regular hold'em player, you're probably well acquainted with probabilities related to starting hands, and therefore know the chances of getting dealt pocket aces (1 in 221), a pocket.
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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
| 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% |
| 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% |
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
| 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-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% |
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.
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
| 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.
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.
The main underpinning of poker is math – it is essential. For every decision you make, while factors such as psychology have a part to play, math is the key element.
In this lesson we’re going to give an overview of probability and how it relates to poker. This will include the probability of being dealt certain hands and how often they’re likely to win. We’ll also cover how to calculating your odds and outs, in addition to introducing you to the concept of pot odds. And finally we’ll take a look at how an understanding of the math will help you to remain emotional stable at the poker table and why you should focus on decisions, not results.
What is Probability?
Probability is the branch of mathematics that deals with the likelihood that one outcome or another will occur. For instance, a coin flip has two possible outcomes: heads or tails. The probability that a flipped coin will land heads is 50% (one outcome out of the two); the same goes for tails.
Probability and Cards
When dealing with a deck of cards the number of possible outcomes is clearly much greater than the coin example. Each poker deck has fifty-two cards, each designated by one of four suits (clubs, diamonds, hearts and spades) and one of thirteen ranks (the numbers two through ten, Jack, Queen, King, and Ace). Therefore, the odds of getting any Ace as your first card are 1 in 13 (7.7%), while the odds of getting any spade as your first card are 1 in 4 (25%).
Unlike coins, cards are said to have “memory”: every card dealt changes the makeup of the deck. For example, if you receive an Ace as your first card, only three other Aces are left among the remaining fifty-one cards. Therefore, the odds of receiving another Ace are 3 in 51 (5.9%), much less than the odds were before you received the first Ace.
Want to see how poker math intertwines with psychology and strategy to give you a MASSIVE EDGE at the tables? Check out CORE and learn poker in the quickest and most systematic way:
Pre-flop Probabilities: Pocket Pairs
In order to find the odds of getting dealt a pair of Aces, we multiply the probabilities of receiving each card:
(4/52) x (3/51) = (12/2652) = (1/221) ≈ 0.45%.
To put this in perspective, if you’re playing poker at your local casino and are dealt 30 hands per hour, you can expect to receive pocket Aces an average of once every 7.5 hours.
Odds Of Getting Pocket Aces
The odds of receiving any of the thirteen possible pocket pairs (twos up to Aces) is:
(13/221) = (1/17) ≈ 5.9%.
In contrast, you can expect to receive any pocket pair once every 35 minutes on average.
Pre-Flop Probabilities: Hand vs. Hand
Players don’t play poker in a vacuum; each player’s hand must measure up against his opponent’s, especially if a player goes all-in before the flop.
Here are some sample probabilities for most pre-flop situations:
Post-Flop Probabilities: Improving Your Hand
Now let’s look at the chances of certain events occurring when playing certain starting hands. The following table lists some interesting and valuable hold’em math:
Many beginners to poker overvalue certain starting hands, such as suited cards. As you can see, suited cards don’t make flushes very often. Likewise, pairs only make a set on the flop 12% of the time, which is why small pairs are not always profitable.
Odds Of Getting Pocket Aces In Texas Holdem
PDF Chart
We have created a poker math and probability PDF chart (link opens in a new window) which lists a variety of probabilities and odds for many of the common events in Texas hold ‘em. This chart includes the two tables above in addition to various starting hand probabilities and common pre-flop match-ups. You’ll need to have Adobe Acrobat installed to be able to view the chart, but this is freely installed on most computers by default. We recommend you print the chart and use it as a source of reference.
Odds and Outs
If you do see a flop, you will also need to know what the odds are of either you or your opponent improving a hand. In poker terminology, an “out” is any card that will improve a player’s hand after the flop.
One common occurrence is when a player holds two suited cards and two cards of the same suit appear on the flop. The player has four cards to a flush and needs one of the remaining nine cards of that suit to complete the hand. In the case of a “four-flush”, the player has nine “outs” to make his flush.
A useful shortcut to calculating the odds of completing a hand from a number of outs is the “rule of four and two”. The player counts the number of cards that will improve his hand, and then multiplies that number by four to calculate his probability of catching that card on either the turn or the river. If the player misses his draw on the turn, he multiplies his outs by two to find his probability of filling his hand on the river.
In the example of the four-flush, the player’s probability of filling the flush is approximately 36% after the flop (9 outs x 4) and 18% after the turn (9 outs x 2).
Pot Odds
Another important concept in calculating odds and probabilities is pot odds. Pot odds are the proportion of the next bet in relation to the size of the pot.
For instance, if the pot is $90 and the player must call a $10 bet to continue playing the hand, he is getting 9 to 1 (90 to 10) pot odds. If he calls, the new pot is now $100 and his $10 call makes up 10% of the new pot.
Experienced players compare the pot odds to the odds of improving their hand. If the pot odds are higher than the odds of improving the hand, the expert player will call the bet; if not, the player will fold. This calculation ties into the concept of expected value, which we will explore in a later lesson.
Bad Beats
A “bad beat” happens when a player completes a hand that started out with a very low probability of success. Experts in probability understand the idea that, just because an event is highly unlikely, the low likelihood does not make it completely impossible.
A measure of a player’s experience and maturity is how he handles bad beats. In fact, many experienced poker players subscribe to the idea that bad beats are the reason that many inferior players stay in the game. Bad poker players often mistake their good fortune for skill and continue to make the same mistakes, which the more capable players use against them.
Decisions, Not Results
One of the most important reasons that novice players should understand how probability functions at the poker table is so that they can make the best decisions during a hand. While fluctuations in probability (luck) will happen from hand to hand, the best poker players understand that skill, discipline and patience are the keys to success at the tables.
A big part of strong decision making is understanding how often you should be betting, raising, and applying pressure.
The good news is that there is a simple system, with powerful shortcuts & rules, that you can begin using this week. Rooted in GTO, but simplified so that you can implement it at the tables, The One Percent gives you the ultimate gameplan.
This 7+ hour course gives you applicable rules for continuation betting, barreling, raising, and easy ratios so that you ALWAYS have the right number of bluffing combos. Take the guesswork out of your strategy, and begin playing like the top-1%.
Conclusion
A strong knowledge of poker math and probabilities will help you adjust your strategies and tactics during the game, as well as giving you reasonable expectations of potential outcomes and the emotional stability to keep playing intelligent, aggressive poker.
Remember that the foundation upon which to build an imposing knowledge of hold’em starts and ends with the math. I’ll end this lesson by simply saying…. the math is essential.
Odds Of Getting Back To Back Pocket Aces
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By Gerald Hanks
Odds Of Getting Pocket Aces In Texas Holdem
Gerald Hanks is from Houston Texas, and has been playing poker since 2002. He has played cash games and no-limit hold’em tournaments at live venues all over the United States.