Poker probability (Omaha)

In poker, the probability of many events can be determined by direct calculation. This article discusses how to compute the probabilities for many commonly occurring events in the game of Omaha hold 'em and provides some probabilities and oddsref_label|odds|note 1|^ for specific situations. In most cases, the probabilities and odds are approximations due to rounding.

When calculating probabilities for a card game such as Omaha, there are two basic approaches.

# Determine the number of outcomes that satisfy the condition being evaluated and divide this by the total number of possible outcomes.
# Use conditional probabilities, or in more complex situations, a decision graph.

Often, the key to determining probability is selecting the best approach for a given problem. This article uses both of these approaches, but relies primarily on enumeration.

Starting hands

The probability of being dealt various starting hands can be explicitly calculated. In Omaha, a player is dealt four down (or "hole") cards. The first card can be any one of 52 playing cards in the deck; the second card can be any one of the 51 remaining cards; the third and fourth any of the 50 and 49 remaining cards, respectively. There are 4! = 24 ways (4! is read "four factorial") to order the four cards (ABCD, ABDC, ACBD, ACDB, ...) which gives 52 × 51 × 50 × 49 ÷ 24 = 270,725 possible starting hand combinations. Alternatively, the number of possible starting hands is represented as the binomial coefficient

:$\{52\; choose\; 4\}\; =\; 270,725$

which is the number of possible combinations of choosing 4 cards from a deck of 52 playing cards.

The 270,725 starting hands can be reduced for purposes of determining the probability of starting hands for Omaha—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 cards in the hand share the same suit. Of the 270,725 combinations, there are 16,432 distinct starting hands grouped into 16 "shapes". Throughout this article, hand shape is indicated with the ranks denoted using uppercase letters and suits denoted using lower case letters. For example, the hand shape XaXbYaYc is any hand containing two pair (XX and YY) that share one suit (a), but not the other suits (b and c). The 16 hand shapes can be organized into the following five hand types based on the ranks of the cards.

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Starting hands for straights

In addition to the rank type and suit type of a starting hand, each starting hand also has a "sequence type" that is useful for estimating the possibility of improving to a straight or straight flush. The sequence type is based on the sequential proximity of the ranks in the hand—the number of different ranks in the hand that can be combined to fill a straight on the board. The ace is a special case in the sequence type because it can be either high or low (i.e. can make a straight with A-2-3-4-5 or 10-J-Q-K-A), and is also both the high and low card in the rank sequence from which straights are formed: A-2-3-4-5-6-7-8-9-10-J-Q-K-A.

The sequence type of the hand is only relevant in determining the probability of making a straight or straight flush. In order to make a straight, exactly three community cards must be combined with exactly two cards from the starting hand. Thus the "sequence shape" of a hand is the number of different combinations of three cards that can make a straight when combined with two cards from the hand. There are 20 different sequence shapes ranging from hands like 2-2-8-K that can't make a straight (0 straight combinations) to hands like 8-9-10-J that can make a straight with 20 different combinations of three ranks (5-6-7, 6-7-8, 6-7-9, 6-7-10, 7-8-9, 7-8-10, 7-8-J, 7-9-10, 7-9-J, 7-10-J, 8-9-Q, 8-10-Q, 8-J-Q, 9-10-Q, 9-J-Q, 9-Q-K, 10-J-Q, 10-Q-K, J-Q-K, Q-K-A). The 20 sequence shapes can be organized by the number of ranks in the starting hand. This is similar to the rank type of the hand, the only difference being that both the rank types two pair (XXYY) and three of a kind (XXXY) have two ranks. The following table shows the four sequence types based on the number of distinct ranks in the starting hand.

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Low starting hands

Omaha Hi-Low is a high-low split variant where the best qualifying low hand, if any, splits the pot with the high hand. Different cards can be used to form the high and low hands, each using two cards from the player's hand and three from the board, and a single player can win both the high and low pots. In Omaha/8, the most common form played in American casinos, a qualifying low hand is 8-high or lower (8-7-6-5-4 or lower). A less common variant of Omaha Hi-Lo uses a qualifying low hand of 9-high or lower (9-8-7-6-5 or lower).

Suits and cards higher than the maximum qualifying low hand do not factor into low hands and neither do straights and flushes. Based on the ranks of cards, low starting hands in Omaha Hi-Lo are grouped into 12 different low-hand shapes, seven of which have the possibility of making a qualifying low hand. The low hand shapes can be organized by the number of distinct low card ranks in the hand: 0 or 1 low ranks (no low hand possible), 2 low ranks, 3 low ranks and 4 low ranks. The number of distinct low hands depends on the low-hand qualifier.

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Calculating the probability of three low ranks on the board for the turn and river is slightly more complicated because there are multiple possibilities for the fourth card. By the turn a qualifying low is possible with either four low ranks, three low ranks and a pair, or three low ranks and a high card. The probability $P\_t$ of making at least three cards to a qualifying low hand by the turn is the sum of the probabilities for each of these configurations. Each probability is calculated by dividing the number of combinations that satisfy the conditions by the possible boards on the turn.

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Notice that while three of a kind is a 60% favorite to be the nuts after the flop, it's less than 2% to still be on top at the river—although the three of a kind has a good chance of improving to a full house or four of a kind, if it doesn't improve, chances are the nut hand at the river is a straight, flush or straight flush. At the river, having the nuts be four of a kind is more likely (45.2%) than all of the hands ranked below four of a kind combined (44.8%). Also, despite the rarity of straight flushes at showdown, 10% of the boards will have one as the nut hand by the river.

Straight flush

A straight flush is possible whenever the board contains at least three cards of the same suit where the ranks of the suited cards can create a straight with the addition of exactly two ranks. For three ranks, the two lower ranks must be chosen from the up to four next lower ranks, counting the rank of ace as low when trying to make the straight A-2-3-4-5. There are 10 possible straights (Ace high to 5 high). A straight is also possible when the high rank is 4 combined with two of the three lower ranks A to 3, or when the three ranks are 3-2-A, which gives ways. This gives

:$64\; imes\; \{4\; choose\; 1\}\; =\; 256$

combinations for a straight flush after the flop. With four or five cards of different ranks, the determination of the number of "rank sets" that yield a straight with the addition of exactly two cards is more involved because any enumeration must eliminate rank sets that are counted more than once, but it turns out that there are 432 such rank sets with four ranks and 1,208 with five ranks. [cite web
year=2003
author=Brian Alspach
url=http://www.math.sfu.ca/~alspach/comp39
title=Rank Sets and Straights
accessdate = 2006-10-30 ] At the turn there are two ways to make a straight flush—there can either be four cards of the same suit with a rank set that allows a straight, or three cards of the same suit that allow a straight combined with any of the 39 cards with a different suit. This gives

:$432\; imes\; \{4\; choose\; 1\}\; +\; 64\; imes\; \{4\; choose\; 1\}\{39\; choose\; 1\}\; =\; 11,712$

combinations after the turn. At the river a straight flush is possible with a suited rank set of either five cards, four cards combined with one of the 39 cards of another suit, or three cards combined with two of the remaining 39 cards, giving

:$1,208\; imes\; \{4\; choose\; 1\}\; +\; 432\; imes\; \{4\; choose\; 1\}\{39\; choose\; 1\}\; +\; 64\; imes\; \{4\; choose\; 1\}\{39\; choose\; 2\}\; =\; 261,920$

combinations.

Four of a kind

Four of a kind is the nuts whenever there is a pair or three of a kind on the board and no possibility for a straight flush. After the flop, three of a kind is possible by choosing one of the 13 ranks and three of the four cards in that rank; a pair is possible by choosing one of the 13 ranks and two of the four cards in that rank combined with a card in one of the other 12 ranks in any of the four suits. So on the flop there are

:$\{13\; choose\; 1\}\{4\; choose\; 3\}\; +\; \{13\; choose\; 1\}\{4\; choose\; 2\}\{12\; choose\; 1\}\{4\; choose\; 1\}\; =\; 3,796$

boards with a pair or three of a kind. At the turn there can either be three of a kind and another rank, two pair, or one pair and two other ranks. In the case with one pair, any straight flushes made possible by the three different ranks must be subtracted. At the turn, the number of possible straight flushes with a pair on the board is one of the 64 rank sets with three cards that can make a straight in one of the four suits combined with a card that pairs one of the three cards to the straight flush, which is . So there are

:$\{13\; choose\; 1\}\{4\; choose\; 3\}\{12\; choose\; 1\}\{4\; choose\; 1\}\; +\; \{13\; choose\; 2\}\{4\; choose\; 2\}^2\; +\; left\; [\; \{13\; choose\; 2\}\{4\; choose\; 2\}\{12\; choose\; 2\}\{4\; choose\; 1\}^2\; -\; 2,304\; ight\; ]\; =\; 85,368$

combinations that make four of a kind possible at the turn. On the river, four of a kind can be made when there is either a full house on the board, three of a kind and two other ranks, two pair and one other rank, or a pair and three other ranks. With three of a kind or two pair, any straight flushes made possible by the three different ranks must be subtracted and with a pair, any straight flushes made possible by the four different ranks are subtracted. For three of a kind, choose one of the three cards for the straight flush and then choose 2 of the 3 remaining cards of that rank to make three of a kind for possible straight flushes. With two pair, choose two of the three cards that make a possible straight flush and then choose one of the three remaining cards for each rank to make one of straight flushes. With a pair there are three cases where a straight flush is possible that have a total of 97,536 combinations:
* the three non-pair ranks have the same suit as one of the cards in the pair and the four suited cards form one of the 432 rank sets that allows a straight (example 3♠ 9♠ J♠ Q♠ J♦), so choosing one of the four suited ranks and one of the three remaining cards of that rank gives possible straight flushes;
* two of the non-pair ranks have the same suit as one of the cards in the pair and the three suited cards form one of the 64 rank sets that allows a straight (example 8♠ 9♠ J♠ J♥ K♦), so choosing one of the three suited ranks and one of the three remaining cards of that rank to combine with a card from one of the 10 remaining ranks in one of the three remaining suits gives possible straight flushes;
* the three non-pair ranks share a suit that is different than either of the suits in the pair and the three suited cards form one of the 64 rank sets that allows a straight (example 4♠ 6♠ 7♠ J♥ J♦), so choose one of the 10 remaining ranks not used by the suited cards and choose two of the three cards of that rank that have a different suit to give possible straight flushes.

Altogether there are

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Making a straight flush

:"See the section "Starting hands for straight flushes" for a description of straight flush starting hand types."

The probability of making a straight flush depends primarily on the number of different sets of three cards that can fill a straight flush in the hand. For convenience, the term "straight flush sequence" means a three-card set that can make a straight flush when combined with the starting hand. A secondary factor to the number of straight flush sequences, although much less significant, is the amount of overlap (shared cards) in the straight flush sequences—the more overlap, the lower the probability for a straight flush on the turn and river. More overlap reduces the probability because some of the board combinations make more than one straight flush and are thus counted multiple times.

For example, the hand 5♥ 6♥ 7♥ A♣ makes a straight flush with either of the two straight flush sequences {2♥ 3♥ 4♥} or {8♥ 9♥ 10♥}, which have no overlap. The hand 5♥ 8♥ A♣ A♠ makes a straight flush with either of the two straight flush sequences {4♥ 6♥ 7♥} or {6♥ 7♥ 9♥}, which have an overlap of two cards—the 6♥ and 7♥. On the turn, the board 4♥ 6♥ 7♥ 9♥ satisfies both sets and is only counted once, making the frequency of boards that make a straight flush on the turn for 5♥ 8♥ A♣ A♠ one less than for 5♥ 6♥ 7♥ A♣. On the river, any of the 44 remaining cards combined with 4♥ 6♥ 7♥ 9♥ makes two straight flushes, making the frequency of boards 44 less.

To make a straight flush on the flop, the three cards on the board must exactly match one of the straight flush sequences for the hand. If $s$ is the number of straight flush sequences for a hand, then the frequency $F\_f$ of boards that make a straight flush on the flop is

:$F\_f\; =\; \{s\; choose\; 1\}$

On the turn, one of the $s$ straight flush sequences can be combined with any of the remaining 45 cards. Enumerating the frequencies this way ends up counting any board that can form two different straight flushes twice. On the turn, for every pair of three card sets that share two cards (for example {A♠ 3♠ 4♠} and {3♠ 4♠ 6♠}) there is exactly one board that makes two straight flushes ({A♠ 3♠ 4♠ 6♠} in the example). Where $n\_\{42\}$ is the number of boards containing four cards that make exactly two straight flushes, then the frequency $F\_t$ of boards that make a straight flush on the turn is

:$F\_t\; =\; \{s\; choose\; 1\}\{45\; choose\; 1\}\; -\; n\_\{42\}$

On the river, one of the $s$ straight flush sequences can be combined with any two of the remaining 45 cards. Now all boards that make exactly two straight flushes are counted twice, and all boards the make exactly three straight flushes are counted three time. Where $n\_\{52\}$ is the number of boards containing five cards that make exactly two straight flushes and $n\_\{53\}$ is the number of boards containing five cards that make exactly three straight flushes, then the frequency $F\_r$ of boards that make a straight flush on the river is

:$F\_r\; =\; \{s\; choose\; 1\}\{45\; choose\; 2\}\; -\; n\_\{52\}\; -\; 2n\_\{53\}$

The probabilities of making a straight flush are the same for any two starting hands that can make a straight flush with exactly two straight flush sequences that contain no overlap. So A♥ 2♥ 10♥ A♣ (which makes a straight flush with either {3♥ 4♥ 5♥} or {J♥ Q♥ K♥}) and 5♥ 6♥ 7♥ A♣ have the same probabilities on the flop, turn and river. Likewise, 2♥ 3♥ A♣ A♠ (which makes a straight flush with either {A♥ 4♥ 5♥} or {4♥ 5♥ 6♥}) and 5♥ 8♥ A♣ A♠ each make a straight flush with one of two straight flush sequences, with two cards overlapping.

A complete straight flush hand pattern is then the number of straight flush sequences for the hand combined with the overlaps between all of the straight flush sequences. The following rules can be used to derive a notation for describing complete straight flush hand patterns:

#Label each element in each straight flush sequence according to the number of straight flush sequences that contain it.
#Sort the labelled elements of each straight flush sequence to derive the labeled straight flush sequence. Arbitrarily, choose to sort the elements in decending order.
#Sort the labelled straight flush sequences. Arbitrarily, choose to sort the straight flush sequences in decending order.

Each element can be either the low, middle, or high rank of a straight flush sequence. Using numbers to label the straight flush sequence elements, each element in a straight flush sequence is assigned a label from 1–3 depending on whether it appears in 1, 2 or 3 straight flush sequences. For example, the two straight flush sequences for a hand {2♥ 3♥ 4♥} and {8♥ 9♥ 10♥} are labelled as 111 and 111 since each element appears in only a single straight flush sequence, and the straight flush hand pattern is 111+111. The two straight flush sequences {4♥ 6♥ 7♥} and {6♥ 7♥ 9♥} are labelled as 221 and 221, and the straight flush hand pattern is 221+221.

To determine the probability of making a straight flush from any starting hand, first identify all of the straight flush hand patterns, and then determine the probabilities for each hand pattern. It turns out that there are 32 hand patterns possible using a single suit to make the straight flush, with either 2, 3, or 4 cards from the suit being used to make straight flushes. The following table shows each of the single-suit straight flush hand patterns, listed in order of probability of making a straight flush on the river, from highest to lowest probability.

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Of the 32 hand patterns that can make a straight flush in one suit, four of the hand patterns (111, 211+211, 322+321+321, and 332+332+321+321) use exactly two cards in the suit. For hands that can make a straight flush in two suits, each of these hand patterns can be used by one of the two suits. This gives different combinations of single suit hand patterns for making a straight flush in one of two suits. There is no overlap in the straight flush sequences between suits and it is not possible to make a straight flush in more than one suit. Therefore, where $P\_1$ and $P\_2$ are the probabilities of making a flush in each of the two suits, then the probability $P$ of making a straight flush in either suit is $P\; =\; P\_1\; +\; P\_2$. The following table gives the double-suit straight flush hand patterns, listed in order of probability of making a straight flush on the river, from highest to lowest probability.

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See also

Derivations and probability tables for Omaha:
* Probability of making the nut low hand in Omaha hold 'em
* Probability derivations for making rank-based hands in Omaha hold 'em

Poker topics:
* Omaha hold 'em
* Poker probability
* Poker strategy
* Pot odds

Math and probability topics:
* Probability
* Odds
* Sample space
* Event (probability theory)
* Binomial coefficient
* Combination
* Permutation
* Combinatorial game theory
* Game complexity
* Set theory

Notes

# 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 role of a fair six-sided die 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 probability of a three being rolled because the three is 1 of 6 equally-likely possible outcomes.

References

External links

* [http://www.math.sfu.ca/~alspach/computations.html Poker Computations] by Brian Alspach
* [http://www.cardplayer.com/poker_odds/omaha Omaha odds calculator] on www.cardplayer.com