# Poker probability (Texas hold 'em)

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In poker, the probability of many events can be determined by direct calculation. This article discusses computing probabilities for many commonly occurring events in the game of Texas hold 'em and provides some probabilities and odds for specific situations. In most cases, the probabilities and odds are approximations due to rounding.

When calculating probabilities for a card game such as Texas Hold 'em, there are two basic approaches. The first approach is to determine the number of outcomes that satisfy the condition being evaluated and divide this by the total number of possible outcomes. For example, there are six outcomes (ignoring order) for being dealt a pair of aces in Hold 'em: A A, A A, A A, A A, A A and A A. There are 52 ways to pick the first card and 51 ways to pick the second card and two ways to order the two cards yielding $\tfrac{52 \times 51}{2} = 1326$ possible outcomes when being dealt two cards (also ignoring order). This gives a probability of being dealt two aces of $\begin{matrix} \frac{6}{1326} = \frac{1}{221} \end{matrix}$.

The second approach is to use conditional probabilities, or in more complex situations, a decision tree. There are 4 ways to be dealt an ace out of 52 choices for the first card resulting in a probability of $\begin{matrix} \frac{4}{52} = \frac{1}{13} \end{matrix}.$ There are 3 ways of getting dealt an ace out of 51 choices on the second card after being dealt an ace on the first card for a probability of $\begin{matrix} \frac{3}{51} = \frac{1}{17} \end{matrix}.$ This value is the conditional probability that second card dealt is an ace given the condition that the first card dealt is an ace. The joint probability of being dealt two aces is the product of the two probabilities: $\begin{matrix} \frac{1}{13} \times \frac{1}{17} = \frac{1}{221} \end{matrix}.$ This article uses both of these approaches.

The odds presented in this article use the notation x : 1 which translates to x to 1 odds against the event happening. The odds are calculated from the probability p of the event happening using the formula: odds = [(1 − p) ÷ p] : 1, or odds = [(1 ÷ p) − 1] : 1. Another way of expressing the odds x : 1 is to state that there is a 1 in x+1 chance of the event occurring or the probability of the event occurring is 1/(x+1). So for example, the odds of a roll of a fair six-sided dice coming up three is 5 : 1 against because there are 5 chances for a number other than three and 1 chance for a three; alternatively, this could be described as a 1 in 6 chance or $\begin{matrix}\frac{1}{6}\end{matrix}$ probability of a three being rolled because the three is 1 of 6 equally-likely possible outcomes.

## Starting hands

### Single hand

The probability of being dealt various starting hands can be explicitly calculated. In Texas Hold 'em, a player is dealt two down (or hole or pocket) cards. The first card can be any one of 52 playing cards in the deck and the second card can be any one of the 51 remaining cards. This gives 52 × 51 ÷ 2 = 1,326 possible starting hand combinations. Since the order in which you receive the first two cards is not significant, the 2,652 permutations are divided by the 2 ways of ordering two cards. Alternatively, the number of possible starting hands is represented as the binomial coefficient

${52 \choose 2} = 1,326$

which is the number of possible combinations of choosing 2 cards from a deck of 52 playing cards.

The 1,326 starting hands can be reduced for purposes of determining the probability of starting hands for Hold 'em—since suits have no relative value in poker, many of these hands are identical in value before the flop. The only factors determining the strength of a starting hand are the ranks of the cards and whether the cards share the same suit. Of the 1,326 combinations, there are 169 distinct starting hands grouped into three shapes: 13 pocket pairs (paired hole cards), 13 × 12 ÷ 2 = 78 suited hands and 78 unsuited hands; 13 + 78 + 78 = 169. The relative probability of being dealt a hand of each given shape is different. The following shows the probabilities and odds of being dealt each type of starting hand.

Hand shapeNumber
of hands
Suit combinations
for each hand
CombinationsDealt specific handDealt any hand
ProbabilityOddsProbabilityOdds
Pocket pair13${4 \choose 2} = 6$13 × 6 = 78$\begin{matrix} \frac{6}{1326} \approx 0.00452 \end{matrix}$220 : 1$\begin{matrix} \frac{78}{1326}=\frac{3}{51} \approx 0.0588 \end{matrix}$16 : 1
Suited cards78${4 \choose 1} = 4$78 × 4 = 312$\begin{matrix} \frac{4}{1326} \approx 0.00302 \end{matrix}$331 : 1$\begin{matrix} \frac{312}{1326}=\frac{12}{51} \approx 0.2353 \end{matrix}$3.25 : 1
Unsuited cards non paired78${4 \choose 1}{3 \choose 1} = 12$78 × 12 = 936$\begin{matrix} \frac{12}{1326} \approx 0.00905 \end{matrix}$110 : 1$\begin{matrix} \frac{936}{1326}=\frac{36}{51} \approx 0.7059 \end{matrix}$0.417 : 1

Here are the probabilities and odds of being dealt various other types of starting hands.

HandProbabilityOdds
AKs (or any specific suited cards)0.00302331 : 1
AA (or any specific pair)0.00452220 : 1
AKs, KQs, QJs, or JTs (suited cards)0.012181.9 : 1
AK (or any specific non-pair incl. suited)0.012181.9 : 1
AA, KK, or QQ0.013672.7 : 1
AA, KK, QQ or JJ0.018154.3 : 1
Suited cards, jack or better0.018154.3 : 1
AA, KK, QQ, JJ, or TT0.022643.2 : 1
Suited cards, 10 or better0.030232.2 : 1
Suited connectors0.039224.5 : 1
Connected cards, 10 or better0.048319.7 : 1
Any 2 cards with rank at least queen0.049819.1 : 1
Any 2 cards with rank at least jack0.090510.1 : 1
Any 2 cards with rank at least 100.1435.98 : 1
Connected cards (cards of consecutive rank)0.1575.38 : 1
Any 2 cards with rank at least 90.2083.81 : 1
Not connected nor suited, at least one 2-90.5340.873 : 1

### Starting hands heads up

For any given starting hand, there are 50 × 49 ÷ 2 = 1,225 hands that an opponent can have before the flop. (After the flop, the number of possible hands an opponent can have is reduced by the three community cards revealed on the flop to 47 × 46 ÷ 2 = 1,081 hands.) Therefore, there are

${52 \choose 2}{50 \choose 2} \div 2 = 812,175$

possible head-to-head match ups in Hold 'em. (The total number of match ups is divided by the two ways that two hands can be distributed between two players to give the number of unique match ups.) However, since there are only 169 distinct starting hands, there are 169 × 1,225 = 207,025 distinct head-to-head match ups.[Note 1]

It is useful to know how two starting hands compete against each other heads up before the flop. In other words, we assume that neither hand will fold, and we will see a showdown. This situation occurs quite often in no limit and tournament play. Also, studying these odds helps to demonstrate the concept of hand domination, which is important in all community card games.

This problem is considerably more complicated than determining the frequency of dealt hands. To see why, note that given both hands, there are 48 remaining unseen cards. Out of these 48 cards, we can choose any 5 to make a board. Thus, there are

${48 \choose 5} = 1,712,304$

possible boards that may fall. In addition to determining the precise number of boards that give a win to each player, we also must take into account boards which split the pot, and split the number of these boards between the players.

The problem is trivial for computers to solve by brute force search; there are many software programs available that will compute the odds in seconds. A somewhat less trivial exercise is an exhaustive analysis of all of the head-to-head match ups in Texas Hold 'em, which requires evaluating each possible board for each distinct head-to-head match up, or 1,712,304 × 207,025 = 354,489,735,600 (≈354 billion) results.[Note 1]

#### Head-to-head starting hand matchups

When comparing two starting hands, the head-to-head probability describes the likelihood of one hand beating the other after all of the cards have come out. Head-to-head probabilities vary slightly for each particular distinct starting hand matchup, but the approximate average probabilities, as given by Dan Harrington in Harrington on Hold'em [p. 125], are summarized in the following table.

Favorite-to-underdog matchupProbabilityOdds for
Pair vs. 2 undercards0.831 : 4.9
Pair vs. lower pair0.821 : 4.5
Pair vs. 1 overcard, 1 undercard0.711 : 2.5
2 overcards vs. 2 undercards0.631 : 1.7
Pair vs. 2 overcards0.551 : 1.2

These odds are general approximations only derived from averaging all of the hand matchups in each category. The actual head-to-head probabilities for any two starting hands vary depending on a number of factors, including:

• Suited or unsuited starting hands;
• Shared suits between starting hands;
• Connectedness of non-pair starting hands;
• Proximity of card ranks between the starting hands (lowering straight potential);
• Proximity of card ranks toward A or 2 (lowering straight potential);
• Possibility of split pot.

For example, A A vs. K Q is 87.65% to win (0.49% to split), but A A vs. 7 6 is 76.81% to win (0.32% to split).

The mathematics for computing all of the possible matchups is simple. However, the computation is tedious to carry out by hand. A computer program can perform a brute force evaluation of the 1,712,304 possible boards for any given pair of starting hands in seconds.

### Starting hands against multiple opponents

When facing two opponents, for any given starting hand the number of possible combinations of hands the opponents can have is

${50 \choose 2}{48 \choose 2} = 1,381,800$

hands. For calculating probabilities we can ignore the distinction between the two opponents holding A J and 8 8 and the opponents holding 8 8 and A J. The number of ways that hands can be distributed between $n$ opponents is $n!$ (the factorial of n). So the number of unique hand combinations $H$ against two opponents is

$H = {50 \choose 2}{48 \choose 2} \div 2! = 690,900$

and against three opponents is

$H = {50 \choose 2}{48 \choose 2}{46 \choose 2} \div 3! = 238,360,500$

and against $n$ opponents is

$H = \prod_{k=1}^n {52 - 2k \choose 2} \div n!,$ or alternatively $H = {50 \choose 2n} \times (2n-1)!!,$

where $(2n-1)!!$ is the number of ways to distribute $2n$ cards between $n$ hands of two cards each.[Note 2] [!! is the double factorial operator: (2n-1)!! is not ((2n-1)!)!.] The following table shows the number of hand combinations for up to nine opponents.

OpponentsNumber of possible hand combinations
11,225
2690,900
3238,360,500
456,372,258,250
5≈9.7073 × 1012 (more than 9 trillion)
6≈1.2620 × 1015 (more than 1 quadrillion)
7≈1.2674 × 1017 (more than 126 quadrillion)
8≈9.9804 × 1018 (almost 10 quintillion)
9≈6.2211 × 1020 (more than 622 quintillion)

An exhaustive analysis of all of the match ups in Texas Hold 'em of a player against nine opponents requires evaluating each possible board for each distinct starting hand against each possible combination of hands held by nine opponents, which is

$169 \times {50 \choose 18} \times 17!! \times {32 \choose 5} \approx 2.117 \times 10^{28}$ (more than 21 octillion).

If you were able to evaluate one trillion (1012) combinations every second, it would take over 670 million years to evaluate all of the hand/board combinations. While it is possible to significantly reduce the total number of combinations by pruning combinations with identical properties, the total number of situations is still well beyond the number that can be evaluated by brute force. For this reason, most software programs compute probabilities and expected values for Hold 'em poker hands against multiple opponents by simulating the play of thousands or even millions of hands to determine statistical probabilities.

### Dominated hands

When evaluating a hand before the flop, it is useful to have some idea of how likely the hand is dominated. A dominated hand is a hand that is beaten by another hand (the dominant hand) and is unlikely to win against it. Often the dominated hand has only a single card rank that can improve the dominated hand to beat the dominant hand (not counting straights and flushes). For example, KJ is dominated by KQ—both hands share the king, and the queen kicker is beating the jack kicker. Barring a straight or flush, the KJ will need a jack on the board to improve against the KQ (and would still be losing if a queen appears on the board along with the jack). A pocket pair is dominated by a pocket pair of higher rank.

#### Pocket pairs

Barring a straight or flush, a pocket pair needs to make three of a kind to beat a higher pocket pair. See the section "After the flop" for the odds of a pocket pair improving to three of a kind.

To calculate the probability that another player has a higher pocket pair, first consider the case against a single opponent. The probability that a single opponent has a higher pair can be stated as the probability that the first card dealt to the opponent is a higher rank than the pocket pair and the second card is the same rank as the first. Where $r$ is the rank of the pocket pair (assigning values from 2–10 and J–A = 11–14), there are (14 − r) × 4 cards of higher rank. Subtracting the two cards for the pocket pair leaves 50 cards in the deck. After the first card is dealt to the player there are 49 cards left, 3 of which are the same rank as the first. So the probability $P$ of a single opponent being dealt a higher pocket pair is

\begin{align} P & = \frac{(14 - r) \times 4}{50} \times \frac{3}{49}\\ & = \frac{84 - 6r}{1225}.\\ \end{align}

The following approach extends this equation to calculate the probability that one or more other players has a higher pocket pair.

1. Multiply the base probability for a single player for a given rank of pocket pairs by the number of opponents in the hand;
2. Subtract the adjusted probability that more than one opponent has a higher pocket pair. (This is necessary because this probability effectively gets added to the calculation multiple times when multiplying the single player result.)

Where $n$ is the number of other players still in the hand and $P_{ma}$ is the adjusted probability that multiple opponents have higher pocket pairs, then the probability that at least one of them has a higher pocket pair is

$P = \left(\frac{84 - 6r}{1225}\right) \times n - P_{ma}= \frac{n(84-6r)-1,225P_{ma}}{1,225}.$

The calculation for $P_{ma}$ depends on the rank of the player's pocket pair, but can be generalized as

$P_{ma} = P_2 + 2P_3 + \cdots + (n-1)P_n,$

where $P_2$ is the probability that exactly two players have a higher pair, $P_3$ is the probability that exactly three players have a higher pair, etc. As a practical matter, even with pocket 2s against 9 opponents, $P_4 < 0.0015$ and $P_5 < 0.00009$, so just calculating $P_2$ and $P_3$ gives an adequately precise result.

The following table shows the probability that before the flop another player has a larger pocket pair when there are one to nine other players in the hand.

Probability of facing a
larger pair when holding
Against 1Against 2Against 3Against 4Against 5Against 6Against 7Against 8Against 9
KK0.00490.00980.01470.01960.02440.02930.03420.03910.0439
QQ0.00980.01950.02920.03880.04840.05790.06730.07660.0859
JJ0.01470.02920.04360.05770.07170.08560.09920.11270.1259
TT0.01960.03890.05780.07640.09460.11240.12990.14700.1637
990.02450.04840.07180.09460.11680.13840.15930.17950.1990
880.02940.05800.08570.11250.13840.16340.18730.21010.2318
770.03430.06740.09940.13010.15950.18740.21380.23870.2619
660.03920.07690.11300.14730.17990.21040.23890.26510.2890
550.04410.08620.12630.16420.19960.23240.26230.28920.3129
440.04900.09560.13950.18060.21860.25320.28410.31090.3334
330.05390.10480.15260.19670.23700.27290.30400.33000.3503
220.05880.11410.16540.21240.25460.29140.32220.34640.3633

The following table gives the probability that a hand is facing two or more larger pairs before the flop. From the previous equations, the probability $P_m$ is computed as

$P_m = P_2 + P_3 + \cdots + P_n.$
Probability of facing multiple
larger pairs when holding
Against 2Against 3Against 4Against 5Against 6Against 7Against 8Against 9
KK< 0.000010.000010.000030.000040.000070.000090.000120.00016
QQ0.000060.000180.000370.000610.000910.001280.001710.00220
JJ0.000170.000510.001020.001710.002570.003600.004820.00621
TT0.000330.000990.002000.003350.005040.007090.009500.01226
990.000540.001640.003300.005530.008360.011770.015800.02045
880.000810.002440.004930.008280.012530.017690.023780.03084
770.001120.003410.006890.011600.017580.024870.033510.04353
660.001490.004540.009180.015500.023530.033350.045030.05861
550.001910.005830.011820.019980.030400.043180.058400.07619
440.002390.007280.014800.025060.038210.054380.073710.09635
330.002910.008900.018120.030750.046980.066990.090990.11919
220.003490.010680.021800.037060.056730.081070.110340.14484

From a practical perspective, however, the odds of out drawing a single pocket pair or multiple pocket pairs are not much different. In both cases the large majority of winning hands require one of the remaining two cards needed to make three of a kind. The real difference against multiple overpairs becomes the increased probability that one of the overpairs will also make three of a kind.

#### Hands with one ace

When holding a single ace (referred to as Ax), it is useful to know how likely it is that another player has a better ace—an ace with a higher second card, since a weaker ace is dominated by a better ace. The probability that a single opponent has a better ace is the probability that he has either AA or Ax where x is a rank other than ace that is higher than the player's second card. When holding Ax, the probability that a chosen single player has AA is $\begin{matrix} \frac{3}{50} \times \frac{2}{49} \approx 0.00245 \end{matrix}$. In the case of a table with $n$ opponents, the probability of one of them holding AA is $(1-(1-0.00245)^n)$. If the player is holding Ax against 9 opponents, there is a probability of approximately 0.0218 that one opponent has AA.

Where $x$ is the rank 2–K of the second card (assigning values from 2–10 and J–K = 11–13) the probability that a single opponent has a better ace is calculated by the formula

\begin{align} P & = \left(\frac{3}{50} \times \frac{2}{49}\right) + \left(\frac{3}{50} \times \frac{(13 - x) \times 4}{49} \times 2\right)\\ & = \frac{3}{1225} + \frac{12 \times (13 - x)}{1225}\\ & = \frac{159 - 12x}{1225}.\\ \end{align}

The probability $\begin{matrix} \frac{3}{50} \times \frac{(13 - x) \times 4}{49} \end{matrix}$ of a player having Ay, where y is a rank such that x < y <= K, is multiplied by the two ways to order the cards A and y in the hand.

The following table shows the probability that before the flop another player has an ace with a larger kicker in the hand.

Probability of facing an ace
with larger kicker when holding
Against 1Against 2Against 3Against 4Against 5Against 6Against 7Against 8Against 9
AK0.002450.004890.007330.009760.012190.014600.017020.019420.02183
AQ0.012240.024340.036290.048090.059740.071260.082630.093860.10496
AJ0.022040.043600.064680.085290.105450.125170.144450.163310.18175
AT0.031840.062660.092500.121390.149370.176450.202670.228050.25263
A90.041630.081530.119770.156420.191540.225200.257450.288370.31799
A80.051430.100210.146490.190380.232020.271520.308980.344520.37823
A70.061220.118700.172660.223310.270860.315500.357410.396750.43369
A60.071020.137000.198290.255230.308120.357260.402910.445310.48471
A50.080820.155100.223380.286150.343840.396870.445610.490410.53160
A40.090610.173010.247950.316090.378060.434420.485670.532270.57465
A30.100410.190730.271990.345090.410850.470000.523220.571090.61416
A20.110200.208260.295520.373150.442230.503700.558400.607060.65037

## The flop

The value of a starting hand can change dramatically after the flop. Regardless of initial strength, any hand can flop the nuts—for example, if the flop comes with three 2s, any hand holding the fourth 2 has the nuts. Conversely, the flop can undermine the perceived strength of any hand—a player holding A A would not be happy to see 8 9 10 on the flop because of the straight and flush possibilities.

There are

${50 \choose 3} = 19,600$

possible flops for any given starting hand. By the turn the total number of combinations has increased to

${50 \choose 4} = 230,300$

and on the river there are

${50 \choose 5} = 2,118,760$

possible boards to go with the hand.

The following are some general probabilities about what can occur on the board. These assume a "random" starting hand for the player.

Board consisting ofMaking on flopMaking by turnMaking by river
Prob.OddsProb.OddsProb.Odds
Three or more of same suit (other suit can have two)0.0517718.3 : 10.175374.70 : 10.371071.69 : 1
Four or more of same suit0.0105693.7 : 10.0449021.3 : 1
Rainbow flop (all different suits)0.397651.51 : 10.105508.48 : 1
Three cards of consecutive rank (but not four consecutive)0.0347527.8 : 10.105448.48 : 10.199104.02 : 1
Four cards to a straight (but not five)0.0104095.1 : 10.0376325.6 : 1
Three or more cards of consecutive rank and same suit0.00217459 : 10.00675147 : 10.0130575.6 : 1
Three of a kind (but not a full house or four of a kind)0.00235424 : 10.00922107 : 10.0211346.3 : 1
A pair (but not two pair or three or four of a kind)0.169414.90 : 10.304252.29 : 10.422571.37 : 1
Two pair (but not a full house)0.0103795.4 : 10.0475420.0 : 1

One can see from the table above that more than 60% of the flops will have at least two of the same suit.

### Flopping overcards when holding a pocket pair

It is also useful to look at the chances different starting hands have of either improving on the flop, or of weakening on the flop. When holding a pocket pair, cards of higher rank than the pair weaken the hand because of the potential that such a card has paired a card in an opponent's hand. The hand gets worse the more such cards there are on the board and the more opponents that are in the hand because the probability that one of the overcards has paired a hole card increases. To calculate the probability of no overcard, take the total number of outcomes without an overcard divided by the total number of outcomes.

Where $x$ is the rank 3–K of the pocket pair (assigning values from 3–10 and J–K = 11–13), then the number of overcards is $\begin{matrix}(14 - x) \times 4\end{matrix}$ and the number of cards of rank $x$ or less is $\begin{matrix}50 - (14 - x) \times 4 = 4x - 6\end{matrix}$. The number of outcomes without an overcard is the number of combinations that can be formed with the remaining cards, so the probability $P$ of no overcard on the flop is

$P = {(4x - 6) \choose 3} \div {50 \choose 3},$

and on the turn and river are

$P = {(4x - 6) \choose 4} \div {50 \choose 4}$ and $P = {(4x - 6) \choose 5} \div {50 \choose 5},$ respectively.

The following table gives the probability that no overcards will come on the flop, turn and river, for each of the pocket pairs from 3 to K.

Holding pocket pairNo overcard on flopNo overcard by turnNo overcard by river
Prob.OddsProb.OddsProb.Odds
KK0.77450.29 : 10.70860.41 : 10.64700.55 : 1
QQ0.58570.71 : 10.48601.06 : 10.40151.49 : 1
JJ0.43041.32 : 10.32052.12 : 10.23693.22 : 1
TT0.30532.28 : 10.20143.97 : 10.13136.61 : 1
990.20713.83 : 10.11907.40 : 10.067313.87 : 1
880.13276.54 : 10.064914.40 : 10.031031.21 : 1
770.078611.73 : 10.031830.48 : 10.012479.46 : 1
660.041623.02 : 10.013374.26 : 10.0040246.29 : 1
550.018652.85 : 10.0043229.07 : 10.00091,057.32 : 1
440.0061162.33 : 10.00091,095.67 : 10.00018,406.78 : 1
330.0010979.00 : 10.000115,352.33 : 10.0000353,125.67 : 1

Notice that there is a better than 35% probability that an ace will come by the river if holding pocket kings, and with pocket queens, the odds are slightly in favor of an ace or a king coming by the turn, and a full 60% in favor of an overcard to the queen by the river. With pocket jacks, there's only a 43% chance that an overcard will not come on the flop and it is better than 3 : 1 that an overcard will come by the river.

Notice, though, that those probabilities would be lower if we consider that at least one opponent happens to hold one of those overcards.

## After the flop – outs

During play—that is, from the flop and onwards—drawing probabilities come down to a question of outs. All situations which have the same number of outs have the same probability of improving to a winning hand over any unimproved hand held by an opponent. For example, an inside straight draw (e.g. 3-4-6-7 missing the 5 for a straight), and a full house draw (e.g. 6-6-K-K drawing for one of the pairs to become three-of-a-kind) are equivalent. Each can be satisfied by four cards—four 5s in the first case, and the other two 6s and other two kings in the second.

The probabilities of drawing these outs are easily calculated. At the flop there remain 47 unseen cards, so the probability is (outs ÷ 47). At the turn there are 46 unseen cards so the probability is (outs ÷ 46). The cumulative probability of making a hand on either the turn or river can be determined as the complement of the odds of not making the hand on the turn and not on the river. The probability of not drawing an out is (47 − outs) ÷ 47 on the turn and (46 − outs) ÷ 46 on the river; taking the complement of these conditional probabilities gives the probability of drawing the out by the river which is calculated by the formula

$P = 1 - \left(\frac{47 - outs}{47} \times \frac{46 - outs}{46}\right) = \frac{93outs-outs^2}{2,162}.$

For reference, the probability and odds for some of the more common numbers of outs are given here.

Example drawing toOutsMake on turnMake on riverMake on turn or river
Prob.OddsProb.OddsProb.Odds
Inside straight flush; Four of a kind10.021346.0 : 10.021745.0 : 10.042622.5 : 1
Open-ended straight flush; Three of a kind20.042622.5 : 10.043522.0 : 10.084210.9 : 1
High pair30.063814.7 : 10.065214.3 : 10.12497.01 : 1
Inside straight; Full house40.085110.8 : 10.087010.5 : 10.16475.07 : 1
Three of a kind or two pair50.10648.40 : 10.10878.20 : 10.20353.91 : 1
Either pair60.12776.83 : 10.13046.67 : 10.24143.14 : 1
Full house or four of a kind;[A]
Inside straight or high pair
70.14895.71 : 10.15225.57 : 10.27842.59 : 1
Open-ended straight80.17024.88 : 10.17394.75 : 10.31452.18 : 1
Flush90.19154.22 : 10.19574.11 : 10.34971.86 : 1
Inside straight or pair100.21283.70 : 10.21743.60 : 10.38391.60 : 1
Open-ended straight or high pair110.23403.27 : 10.23913.18 : 10.41721.40 : 1
Inside straight or flush; Flush or high pair120.25532.92 : 10.26092.83 : 10.44961.22 : 1
130.27662.62 : 10.28262.54 : 10.48101.08 : 1
Open-ended straight or pair140.29792.36 : 10.30432.29 : 10.51160.955 : 1
Open-ended straight or flush; Flush or pair;
Inside straight, flush or high pair
150.31912.13 : 10.32612.07 : 10.54120.848 : 1
160.34041.94 : 10.34781.88 : 10.56980.755 : 1
170.36171.76 : 10.36961.71 : 10.59760.673 : 1
Inside straight or flush or pair;
Open-ended straight, flush or high pair
180.38301.61 : 10.39131.56 : 10.62440.601 : 1
190.40431.47 : 10.41301.42 : 10.65030.538 : 1
200.42551.35 : 10.43481.30 : 10.67530.481 : 1
Open-ended straight, flush or pair210.44681.24 : 10.45651.19 : 10.69940.430 : 1
1. ^ When drawing to a full house or four of a kind with a pocket pair that has hit a set (three of a kind) on the flop, there are 6 outs to get a full house by pairing the board and one out to make four of a kind. This means that if the turn does not pair the board or make four of a kind, there will be 3 additional outs on the river, for a total of 10, to pair the turn card and make a full house. This makes the probability of drawing to a full house or four of a kind on the turn or river 0.334 and the odds are 1.99 : 1. This makes drawing to a full house or four of a kind by the river about 8½ outs.

If a player doesn't fold before the river, a hand with at least 14 outs after the flop has a better than 50% chance to catch one of its outs on either the turn or the river. With 20 or more outs, a hand is a better than 2 : 1 favorite to catch at least one out in the two remaining cards.

See the article on pot odds for examples of how these probabilities might be used in gameplay decisions.

### Estimating probability of drawing outs - The rule of four and two

Many poker players do not have the mathematical ability to calculate odds in the middle of a poker hand. One solution is to just memorize the odds of drawing outs at the river and turn since these odds are needed frequently for making decisions. Another solution some players use is an easily calculated approximation of the probability for drawing outs, commonly referred to as the "Rule of Four and Two". With two cards to come, the percent chance of hitting x outs is about (x × 4)%. This approximation gives roughly accurate probabilities up to about 12 outs after the flop, with an absolute average error of 0.9%, a maximum absolute error of 3%, a relative average error of 3.5% and a maximum relative error of 6.8%. With one card to come, the percent chance of hitting x is about (x × 2)%. This approximation has a constant relative error of an 8% underestimation, which produces a linearly increasing absolute error of about 1% for each 6 outs.

A slightly more complicated, but significantly more accurate approximation of drawing outs after the flop is to use (x × 4)% for up to 9 outs and (x × 3 + 9)% for 10 or more outs. This approximation has a maximum absolute error of less than 1% for 1 to 19 outs and maximum relative error of less than 5% for 2 to 23 outs. A more accurate approximation for the probability of drawing outs after the turn is (x × 2 + (x × 2) ÷ 10)%. This is easily done by first multiplying x by 2, then rounding the result to the nearest multiple of ten and adding the 10's digit to the first result. For example, 12 outs would be 12 × 2 = 24, 24 rounds to 20, so the approximation is 24 + 2 = 26%. This approximation has a maximum absolute error of less than 0.9% for 1 to 19 outs and a maximum relative error of 3.5% for more than 3 outs. The following shows the approximations and their absolute and relative errors for both methods of approximation.

OutsMake on turn or riverMake on river
Actual(x × 4)%(x × 3 + 9)%Actual(x × 2)%(x × 2 + (x × 2) ÷ 10)%
Est.Error % ErrorEst.Error % ErrorEst.Error % ErrorEst.Error % Error
14.2553%4%−0.26%6.00%4%−0.26%6.00%2.1739%2%−0.17%8.00%2%−0.17%8.00%
28.4181%8%−0.42%4.97%8%−0.42%4.97%4.3478%4%−0.35%8.00%4%−0.35%8.00%
312.4884%12%−0.49%3.91%12%−0.49%3.91%6.5217%6%−0.52%8.00%7%+0.48%7.33%
416.4662%16%−0.47%2.83%16%−0.47%2.83%8.6957%8%−0.70%8.00%9%+0.30%3.50%
520.3515%20%−0.35%1.73%20%−0.35%1.73%10.8696%10%−0.87%8.00%11%+0.13%1.20%
624.1443%24%−0.14%0.60%24%−0.14%0.60%13.0435%12%−1.04%8.00%13%−0.04%0.33%
727.8446%28%+0.16%0.56%28%+0.16%0.56%15.2174%14%−1.22%8.00%15%−0.22%1.43%
831.4524%32%+0.55%1.74%32%+0.55%1.74%17.3913%16%−1.39%8.00%18%+0.61%3.50%
934.9676%36%+1.03%2.95%36%+1.03%2.95%19.5652%18%−1.57%8.00%20%+0.43%2.22%
1038.3904%40%+1.61%4.19%39%+0.61%1.59%21.7391%20%−1.74%8.00%22%+0.26%1.20%
1141.7206%44%+2.28%5.46%42%+0.28%0.67%23.9130%22%−1.91%8.00%24%+0.09%0.36%
1244.9584%48%+3.04%6.77%45%+0.04%0.09%26.0870%24%−2.09%8.00%26%−0.09%0.33%
1348.1036%52%+3.90%8.10%48%−0.10%0.22%28.2609%26%−2.26%8.00%29%+0.74%2.62%
1451.1563%56%+4.84%9.47%51%−0.16%0.31%30.4348%28%−2.43%8.00%31%+0.57%1.86%
1554.1166%60%+5.88%10.87%54%−0.12%0.22%32.6087%30%−2.61%8.00%33%+0.39%1.20%
1656.9843%64%+7.02%12.31%57%+0.02%0.03%34.7826%32%−2.78%8.00%35%+0.22%0.62%
1759.7595%68%+8.24%13.79%60%+0.24%0.40%36.9565%34%−2.96%8.00%37%+0.04%0.12%
1862.4422%72%+9.56%15.31%63%+0.56%0.89%39.1304%36%−3.13%8.00%40%+0.87%2.22%
1965.0324%76%+10.97%16.86%66%+0.97%1.49%41.3043%38%−3.30%8.00%42%+0.70%1.68%
2067.5301%80%+12.47%18.47%69%+1.47%2.18%43.4783%40%−3.48%8.00%44%+0.52%1.20%
2169.9352%84%+14.06%20.11%72%+2.06%2.95%45.6522%42%−3.65%8.00%46%+0.35%0.76%
2272.2479%88%+15.75%21.80%75%+2.75%3.81%47.8261%44%−3.83%8.00%48%+0.17%0.36%
2374.4681%92%+17.53%23.54%78%+3.53%4.74%50.0000%46%−4.00%8.00%51%+1.00%2.00%

Either of these approximations is generally accurate enough to aid in most pot odds calculations.

### Runner-runner outs

Some outs for a hand require drawing an out on both the turn and the river—making two consecutive outs is called a runner-runner. Examples would be needing two cards to make a straight, flush, or three or four of a kind. Runner-runner outs can either draw from a common set of outs or from disjoint sets of outs. Two disjoint outs can either be conditional or independent events.

#### Common outs

Drawing to a flush is an example of drawing from a common set of outs. Both the turn and river need to be the same suit, so both outs are coming from a common set of outs—the set of remaining cards of the desired suit. After the flop, if $x$ is the number of common outs, the probability $P$ of drawing runner-runner outs in Texas hold 'em is

$P = \frac{x}{47} \times \frac{x-1}{46}= \frac{x^2-x}{2,162}.$

Since a flush would have 10 outs, the probability of a runner-runner flush draw is $\begin{matrix} \frac{10}{47} \times \frac{9}{46} = \frac{90}{2162} \approx \frac{1}{24} \approx 0.04163 \end{matrix}$. Other examples of runner-runner draws from a common set of outs are drawing to three or four of a kind. When counting outs, it is convenient to convert runner-runner outs to "normal" outs (see "After the flop"). A runner-runner flush draw is about the equivalent of one "normal" out.

The following table shows the probability and odds of making a runner-runner from a common set of outs and the equivalent normal outs.

Likely drawing toCommon outsProbabilityOddsEquivalent outs
Four of a kind (with pair)
Inside-only straight flush
20.000931,080 : 10.02
Three of a kind (with no pair)30.00278359 : 10.07
40.00556179 : 10.13
50.00925107 : 10.22
Two pair or three of a kind (with no pair)60.0138871.1 : 10.33
70.0194350.5 : 10.46
80.0259037.6 : 10.61
90.0333029.0 : 10.78
Flush100.0416323.0 : 10.98

#### Disjoint outs

Two outs are disjoint when there are no common cards between the set of cards needed for the first out and the set of cards needed for the second out. The outs are independent of each other if it does not matter which card comes first, and one card appearing does not affect the probability of the other card appearing except by changing the number of remaining cards; an example is drawing two cards to an inside straight. The outs are conditional on each other if the number of outs available for the second card depends on the first card; an example is drawing two cards to an outside straight.

After the flop, if $x$ is the number of independent outs for one card and $y$ is the number of outs for the second card, then the probability $P$ of making the runner-runner is

$P = \frac{x}{47} \times \frac{y}{46} \times 2 = \frac{xy}{1081}.$

For example, a player holding J Q after the flop 9 5 4 needs a 10 and either a K or 8 on the turn and river to make a straight. There are 4 10s and 8 Ks and 8s, so the probability is $\begin{matrix}\frac{4 \times 8}{1081} \approx 0.0296\end{matrix}$.

The probability of making a conditional runner-runner depends on the condition. For example, a player holding 9 10 after the flop 8 2 A can make a straight with J Q, 7 J or 6 7. The number of outs for the second card is conditional on the first card—a Q or 6 (8 cards) on the first card leaves only 4 outs (J or 7, respectively) for the second card, while a J or 7 (8 cards) for the first card leaves 8 outs ({Q, 7} or {J, 6}, respectively) for the second card. The probability $P$ of a runner-runner straight for this hand is calculated by the equation

$P = \left(\frac{8}{47} \times \frac{4}{46}\right) + \left(\frac{8}{47} \times \frac{8}{46}\right) = \frac{96}{2162} \approx 0.0444$

The following table shows the probability and odds of making a runner-runner from a disjoint set of outs for common situations and the equivalent normal outs.

Drawing toProbabilityOddsEquivalent outs
Outside straight0.0444021.5 : 11.04
Inside+outside straight0.0296032.8 : 10.70
Inside-only straight0.0148066.6 : 10.35
Outside straight flush0.00278359 : 10.07
Inside+outside straight flush0.00185540 : 10.04

The preceding table assumes the following definitions.

Outside straight and straight flush
Drawing to a sequence of three cards of consecutive rank from 3-4-5 to 10-J-Q where two cards can be added to either end of the sequence to make a straight or straight flush.
Inside+outside straight and straight flush
Drawing to a straight or straight flush where one required rank can be combined with one of two other ranks to make the hand. This includes sequences like 5-7-8 which requires a 6 plus either a 4 or 9 as well as the sequences J-Q-K, which requires a 10 plus either a 9 or A, and 2-3-4 which requires a 5 plus either an A or 6.
Inside-only straight and straight flush
Drawing to a straight or straight flush where there are only two ranks that make the hand. This includes hands such as 5-7-9 which requires a 6 and an 8 as well as A-2-3 which requires a 4 and a 5.

#### Compound outs

The strongest runner-runner probabilities lie with hands that are drawing to multiple hands with different runner-runner combinations. These include hands that can make a straight, flush or straight flush, as well as four of a kind or a full house. Calculating these probabilities requires adding the compound probabilities for the various outs, taking care to account for any shared hands. For example, if $P_s$ is the probability of a runner-runner straight, $P_f$ is the probability of a runner-runner flush, and $P_{sf}$ is the probability of a runner-runner straight flush, then the compound probability $P$ of getting one of these hands is

$P = P_s + P_f - P_{sf}.$

The probability of the straight flush is subtracted from the total because it is already included in both the probability of a straight and the probability of a flush, so it has been added twice and must therefore be subtracted from the compound outs of a straight or flush.

The following table gives the compound probability and odds of making a runner-runner for common situations and the equivalent normal outs.

Drawing toProbabilityOddsEquivalent outs
Flush, outside straight or straight flush0.0832611.0 : 11.98
Flush, inside+outside straight or straight flush0.0693813.4 : 11.65
Flush, inside-only straight or straight flush0.0555017.0 : 11.30

Some hands have even more runner-runner chances to improve. For example, holding the hand J Q after a flop of 10 J 7 there are several runner-runner hands to make at least a straight. The hand can get two cards from the common outs of {J, Q} (5 cards) to make a full house or four of a kind, can get a J (2 cards) plus either a 7 or 10 (6 cards) to make a full house from these independent disjoint outs, and is drawing to the compound outs of a flush, outside straight or straight flush. The hand can also make {7, 7} or {10, 10} (each drawing from 3 common outs) to make a full house, although this will make four of a kind for anyone holding the remaining 7 or 10 or a bigger full house for anyone holding an overpair. Working from the probabilities from the previous tables and equations, the probability $P$ of making one of these runner-runner hands is a compound probability

$P = 0.08326 + 0.00925 + \frac{2 \times 6}{1081} + (0.00278 \times 2) \approx 0.1092$

and odds of 8.16 : 1 for the equivalent of 2.59 normal outs. Almost all of these runner-runners give a winning hand against an opponent who had flopped a straight holding 8, 9,[Note 3] but only some give a winning hand against A 2 (this hand makes bigger flushes when a flush is hit) or against K Q (this hand makes bigger straights when a straight is hit with 8 9). When counting outs, it is necessary to adjust for which outs are likely to give a winning hand—this is where the skill in poker becomes more important than being able to calculate the probabilities.

## See also

Poker topics:

Math and probability topics:

## Notes

1. ^ a b By removing reflection and applying aggressive search tree pruning, it is possible to reduce the number of unique head-to-head hand combinations from 207,025 to 47,008. Reflection eliminates redundant calculations by observing that given hands $h_1$ and $h_2$, if $w_1$ is the probability of $h_1$ beating $h_2$ in a showdown and $s$ is the probability of $h_1$ splitting the pot with $h_2$, then the probability $w_2$ of $h_2$ beating $h_1$ is $w_2 = 1 - (s + w_1)$, thus eliminating the need to evaluate $h_2$ against $h_1$. Pruning is possible, for example, by observing that Q♥ J has the same chance of winning against both 8 7 and 8 7 (but not the same probability as against 8 7 because sharing the heart affects the flush possibilities for each hand).
2. ^ See "Capital Pi notation for multiplication" for a description of the $\prod$ (capital π or pi) symbol.
3. ^ In the example, if the opponent is holding either 8 9 or 8 9, then the opponent wins with a flush if the player makes a straight using two hearts or two diamonds, respectively. If the opponent is holding 8 9, then the opponent wins with a straight flush if the player makes a full house with 10 J.