Poker probability

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In poker, the probability of each type of 5-card hand can be computed by calculating the proportion of hands of that type among all possible hands.

1 Frequency of 5-card poker hands
- 1.1 Derivation of frequencies of 5-card poker hands
2 Frequency of 7-card poker hands
- 2.1 Derivation of frequencies of 7-card poker hands
3 Frequency of 5-card lowball poker hands
- 3.1 Derivation of frequencies for 5-card lowball hands
4 Frequency of 7-card lowball poker hands
- 4.1 Derivation of frequencies for 7-card lowball hands
5 See also
6 External links

[edit] Frequency of 5-card poker hands

The following enumerates the frequency of each hand, given all combinations of 5 cards randomly drawn from a full deck of 52, without wild cards. The probability is calculated based on $\begin{matrix} {52 \choose 5} = 2,598,960 \end{matrix}$ , the total number of 5-card combinations. Here, the probability is the frequency of the hand divided by the total number of 5-card hands, and the odds are defined by (1/p) − 1 : 1, where p is the probability. (The frequencies given are exact; the probabilities and odds are approximate.)

Hand	Frequency	Probability	Cumulative	Odds
Straight flush	40	0.00154%	0.00154%	64,973 : 1
Four of a kind	624	0.0240%	0.0256%	4,164 : 1
Full house	3,744	0.144%	0.170%	693 : 1
Flush	5,108	0.197%	0.367%	508 : 1
Straight	10,200	0.392%	0.76%	254 : 1
Three of a kind	54,912	2.11%	2.87%	46.3 : 1
Two pair	123,552	4.75%	7.62%	20.0 : 1
One pair	1,098,240	42.3%	49.9%	1.37 : 1
No pair	1,302,540	50.1%	100%	0.995 : 1
Total	2,598,960	100%	100%	0 : 1

The royal flush is included as a straight flush above. The royal flush can be formed 4 ways (one for each suit), giving it a probability of 0.000154% and odds of 649,739 : 1.

When ace-low straights and ace-low straight flushes are not counted, the probabilities of each are reduced: straights and straight flushes each become 9/10 as common as they otherwise would be. The 4 missed straight flushes become flushes and the 1,020 missed straights become no pair.

Note that since suits have no relative value in poker, two hands can be considered identical if one hand can be transformed into the other by swapping suits. For example, the hand 3♣ 7♣ 8♣ Q♠ A♠ is identical to 3♦ 7♦ 8♦ Q♥ A♥ because replacing all of the clubs in the first hand with diamonds and all of the spades with hearts produces the second hand. So eliminating identical hands that ignore relative suit values, there are only 134,459 distinct hands.

The number of distinct poker hands is even smaller. For example, 3♣ 7♣ 8♣ Q♠ A♠ and 3♦ 7♣ 8♦ Q♥ A♥ are not identical hands when just ignoring suit assignments because one hand has three suits, while the other hand has only two—that difference could affect the relative value of each hand when there are more cards to come. However, even through the hands are not identical from that perspective, they still form equivalent poker hands because each hand is an A-Q-8-7-3 high card hand. There are 7,462 distinct poker hands.

[edit] Derivation of frequencies of 5-card poker hands

The following computations show how the above frequencies for 5-card poker hands were determined. To understand these derivations, the reader should be familiar with the basic properties of the binomial coefficients and their interpretation as the number of ways of choosing elements from a given set. See also: sample space and event (probability theory).

Straight flush — Each straight flush is uniquely determined by its highest ranking card; and these ranks go from 5 (A-2-3-4-5) up to A (T-J-Q-K-A) in each of the 4 suits. Thus, the total number of straight flushes is:

${10 \choose 1}{4 \choose 1} = 40$

Four of a kind — Any one of the thirteen ranks can form the four of a kind, leaving 52 − 4 = 48 possibilities for the final card. Thus, the total number of four-of-a-kinds is:

${13 \choose 1}{48 \choose 1} = 624$

Full house — The full house comprises a triple (three of a kind) and a pair. The triple can be any one of the thirteen ranks, and three of the four cards of this rank are chosen. The pair can be any one of the remaining twelve ranks, and two of the four cards of the rank are chosen. Thus, the total number of full houses is:

${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2} = 3,744$

Flush — The flush contains any five of the thirteen ranks, all of which belong to one of the four suits, minus the 40 straight flushes. Thus, the total number of flushes is:

${13 \choose 5}{4 \choose 1} - 40 = 5,108$

Straight — The straight consists of any one of the ten possible sequences of five consecutive cards, from 5-4-3-2-A to A-K-Q-J-T. Each of these five cards can have any one of the four suits. Finally, as with the flush, the 40 straight flushes must be excluded, giving:

${10 \choose 1}{4 \choose 1}^5 - 40 = 10,200$

Three of a kind — Any of the thirteen ranks can form the three of a kind, which can contain any three of the four suits. The other cards can have any two of the remaining twelve ranks, and each can have any one of the four suits. Thus, the total number of three-of-a-kinds is:

${13 \choose 1}{4 \choose 3}{12 \choose 2}{4 \choose 1}^2 = 54,912$

Two pair — The pairs can have any two of the thirteen ranks, and each pair can have two of the four suits. The final card can have any one of the eleven remaining ranks, and any suit. Thus, the total number of two-pairs is:

${13 \choose 2}{4 \choose 2}^2{11 \choose 1}{4 \choose 1} = 123,552$

Pair — The pair can have any one of the thirteen ranks, and any two of the four suits. The remaining three cards can have any three of the remaining twelve ranks, and each can have any of the four suits. Thus, the total number of pair hands is:

${13 \choose 1}{4 \choose 2}{12 \choose 3}{4 \choose 1}^3 = 1,098,240$

No pair — A no-pair hand contains five of the thirteen ranks, discounting the ten possible straights, and each card can have any of the four suits, discounting the four possible flushes. Alternatively, a no-pair hand is any hand that does not fall into one of the above categories; that is, any way to choose five out of 52 cards, discounting all of the above hands. Thus, the total number of no-pair hands is:

$\left[{13 \choose 5} - 10\right](4^5 - 4) = {52 \choose 5} - 1,296,420 = 1,302,540$

[edit] Frequency of 7-card poker hands

In some popular variations of poker, a player uses the best five-card poker hand out of seven cards. The frequencies are calculated in a manner similar to that shown for 5-card hands, except additional complications arise due to the extra two cards in the 7-card poker hand. The total number of distinct 7-card hands is $\begin{matrix} {52 \choose 7} = 133,784,560 \end{matrix}$ . It is notable that the probability of a no-pair hand is less than the probability of a one-pair or two-pair hand. (The frequencies given are exact; the probabilities and odds are approximate.)

Hand	Frequency	Probability	Cumulative	Odds
Straight flush	41,584	0.0311%	0.0311%	3,216 : 1
Four of a kind	224,848	0.168%	0.199%	594 : 1
Full house	3,473,184	2.60%	2.80%	37.5 : 1
Flush	4,047,644	3.03%	5.82%	32.1 : 1
Straight	6,180,020	4.62%	10.4%	20.6 : 1
Three of a kind	6,461,620	4.83%	15.3%	19.7 : 1
Two pair	31,433,400	23.5%	38.8%	3.26 : 1
One pair	58,627,800	43.8%	82.6%	1.28 : 1
No pair	23,294,460	17.4%	100%	4.74 : 1
Total	133,784,560	100%	100%	0 : 1

Since only five cards are used to form a poker hand, the number of distinct poker hands with 7 cards is the same as it is with 5 cards. However, since suits have no relative value in poker, two hands can be considered identical if one hand can be transformed into the other by swapping suits. Eliminating identical hands that ignore relative suit values leaves 6,009,159 distinct hands.

[edit] Derivation of frequencies of 7-card poker hands

See "7-Card Poker Hands" by Brian Alspach for the article on which this explanation is based.

The following computations show how the above frequencies for 7-card poker hands were determined. To understand these derivations, the reader should be familiar with the basic properties of the binomial coefficients and their interpretation as the number of ways of choosing elements from a given set. See also: sample space and event (probability theory).

Straight flush — Each straight flush is uniquely determined by its highest ranking card; these ranks go from 5 (A-2-3-4-5) up to A (10-J-Q-K-A) in each of the 4 suits. For any particular suit where the straight flush is ace-high, the extra 2 cards may be chosen from the remaining 47 cards. In the 9 remaining cases when the straight flush is not ace-high, the extra 2 cards may be chosen from the remaining 47 cards, minus the card in that suit directly above the high-card (which would change the rank of the hand). Thus, the total number of straight flushes is:

${4 \choose 1}\left[{47 \choose 2} + {9 \choose 1}{46 \choose 2}\right] = 41,584$

Four of a kind — Any 1 of the 13 ranks can form the four of a kind. The extra 3 cards may be chosen from the remaining 48 cards. Thus, the total number of four of a kinds is:

${13 \choose 1}{48 \choose 3} = 224,848$

Full house — With 7 cards, a full house may be constructed in 1 of 3 ways:

1 triple, 1 pair and 2 kickers

The triple may be 1 of 13 ranks, and by definition 3 of the 4 of that rank are chosen. The pair may be 1 of the remaining 12 ranks, and (again, by definition) 2 of the 4 of that rank are chosen. The ranks of the 2 kickers are chosen from the remaining 11 ranks, and 1 of the 4 of each rank are chosen. Thus, the total number of full houses in this form is:

${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}{11 \choose 2}{4 \choose 1}^2 = 3,294,720$

1 triple and 2 pairs

The triple is chosen the same way as before, the ranks of the two pairs are chosen from the remaining 12 ranks, and the 2 of the 4 of each rank are chosen as usual. Thus, the total number of full houses in this form is:

${13 \choose 1}{4 \choose 3}{12 \choose 2}{4 \choose 2}^2 = 123,552$

2 triples and 1 kicker

The ranks of both triples are chosen from the 13, then the rank of the kicker is chosen from the remaining 11 ranks. Thus, the total number of full houses in this form is:

${13 \choose 2}{4 \choose 3}^2{11 \choose 1}{4 \choose 1} = 9,152$

Thus, the total number of full houses is:

$3,294,720 + 123,552 + 9,152 = 3,473,184\,$

Flush — A flush may be formed with either 5, 6 or 7 cards in any of the 4 suits. The flush cards are chosen from the 13 in that suit, and the extra cards (if any) are chosen from the other 3 suits. The number of straight flushes must then be subtracted from the total. Thus, the total number of flushes is:

${4 \choose 1}\left[{13 \choose 5}{39 \choose 2} + {13 \choose 6}{39 \choose 1} + {13 \choose 7}\right] - 41,584 = 4,047,644$

Straight — Significantly more complications arise when working out the frequencies for a 7-card straight due to the possibility of a straight and a flush (though not necessarily a straight flush) being formed simultaniously, and the fact that pairs and triples of cards can appear. Therefore, the calculations must be broken down into several separate sections:

7 distinct ranks

In this type of straight, all 7 cards are of unique ranks (ie. no pairs occur). First, ignoring suits, the total number of possible sets (combinations) that form a straight with 7 distinct ranks is found. As with straight flushes, a straight is defined by its high card. With an ace-high straight, the ranks of the 2 extra cards may be chosen from any of the remaining 8 ranks, while with the 9 other possible straights, any of the ranks but the rank directly above the high card may be chosen. Thus, the total number of sets of ranks is:

${8 \choose 2} + {9 \choose 1}{7 \choose 2} = 217$

Next, the total number of possible sets of suits, for any of the sets of ranks, is found. Given that each card is of a distinct rank, the total number of sets of suits is:

${4 \choose 1}^7 = 16,384\,$

However, the instances where a flush is formed must be subtracted from the total; there are 3 ways of achieving this: There is 1 case per suit where all 7 are of the same suit. If 6 of the 7 are in the same suit, then the remaining card is chosen from the remaining 3 suits. If 5 of the 7 in the same suit, then 2 independent choices are made for each of the extra cards. Thus, the total number of cases where a flush is formed with 7 distinct ranks is:

${4 \choose 1}\left[1 + {7 \choose 6}{3 \choose 1} + {7 \choose 5}{3 \choose 1}^2\right] = 844$

Thus, the total number of sets of suits which produce a straight, but not a flush is:

$16,384 - 844 = 15,541\,$

And as each set of suits occurs for each set of ranks, the total number of straights with 7 distinct ranks is:

$217 \cdot 15,541 = 3,372,180\,$

6 distinct ranks

A straight can also be formed with only 6 distinct ranks (ie. the hand contains 1 pair). In this case, one of the extra cards will have the same rank as one of the cards forming the straight, therefore only one extra rank need be chosen. Thus, the total number of sets of ranks is:

${8 \choose 1} + {9 \choose 1}{7 \choose 1} = 71$

The way to proceed now is to calculate the total number of ways to form a pair, and then calculate the total number of ways to form a straight, but not a flush (given that the pair has already been chosen). The pair can be 1 of the 6 previously chosen ranks, and 2 of the 4 of that rank form the pair. Thus, the total number of ways for form a pair is:

${6 \choose 1}{4 \choose 2} = 36$

The total number of sets of suits for the remaining 5 cards can be calculated in the same way as for 7 cards:

${4 \choose 1}^5 = 1,024\,$

As with 7 distinct ranks, the instances where a flush is formed must be subtracted from the total. The remaining 5 cards can be chosen in two different manners in order to form a flush: either they are all of the same suit, or 4 of them are in the same suit as either of the two paired cards. If all 5 are in the same suit, 1 of the 4 suits is chosen. If 4 of the 5 are in the same suit, 1 of the 2 suits forming the pair is chosen, and the suit of the extra card is chosen from the remaining 3 suits. Thus the total number of ways to form a flush is:

${4 \choose 1} + {5 \choose 4}{2 \choose 1}{3 \choose 1} = 34$

Thus, the total number of sets of suits which produce a straight, but not a flush is:

$1,024 - 34 = 990\,$

Thus the total number of straights with 6 distinct ranks equals the total number of sets of ranks, multiplied by the total number of ways to form the pair, multiplied by the total number of ways to form a straight:

$71 \cdot 36 \cdot 990 = 2,530,440\,$

5 distinct ranks with a triple

There are two ways to form a straight with 5 distinct ranks. The first is using 3 cards of the same rank, and 4 of separate ranks. There are only 10 sets of ranks in this case, as there are no extra ranks to be chosen. The triple can be 1 of the 5 ranks, and 3 of the 4 of that rank make up the triple. Thus, the number of ways to choose the triple is:

${5 \choose 1}{4 \choose 3} = 20$

The total number of sets of suits for the remaining 4 cards is

44

and the only ways to form a flush are if all 4 cards are of the same suit as 1 of the 3 suits forming the triple. Thus, the total number of straights form a straight, but not a flush is:

${4 \choose 1}^4 - {3 \choose 1} = 253\,$

Thus the total number of straights with 5 distinct ranks and a triple is:

$10 \cdot 20 \cdot 253 = 50,600\,$

5 distinct ranks with 2 pairs

The second way to form a straight with 5 distinct ranks is to have 2 pairs and 3 other cards of separate ranks. As before, there are 10 different sets of ranks, however, calculating the number of ways that a flush is formed is complicated, due to the fact that the two pairs can consist of either 2, 3 or 4 suits. Firstly, the ranks for the two pairs are chosen from the 5 available. Thus, the number of ways to choose the ranks for the two pairs is:

${5 \choose 2} = 10$

Then the cards are chosen for each of the pairs. Thus, the number of ways to choose the suits for the pairs is:

${4 \choose 2}^2 = 36$

6 of these ways, the pairs consist of 2 suits, 24 of these ways the pairs consist of 3 suits, and the remaining 6 of these ways they consist of 4 suits. Note that the total number of sets of suits for the remaining 3 cards is

43

. When the pairs consist of 2 suits, a flush will be formed when the remaining 3 cards are all in either of those two suits. There are 2 ways of this happening which must be subtracted from the total. When there are 3 suits, a flush will be formed when the remaining 3 cards are all in the suit of the 2 cards of matching suit in the pairs. There is 1 way of this happening. When there are 4 suits there are no ways of making a flush. Thus, the total number of sets of suits that do not form a flush is:

$6 \cdot \left[64 - {2 \choose 1}\right] + 24 \cdot (64 - 1) + 6 \cdot 64 = 2,268\,$

Thus, the total number of straights with 5 distinct ranks and 2 pairs is:

$10 \cdot 10 \cdot 2,268 = 226,800\,$

Thus, the total number of straights is:

$3,372,180 + 2,530,440 + 50,600 + 226,800 = 6,180,020\,$

Three of a kind — A three of a kind must consist of 5 of the 13 ranks, but the 10 combinations that form straights must be subtracted, giving the total number of sets of ranks as:

${13 \choose 5} - 10 = 1,277$

The rank of the triple is chosen from the 5 available, and 3 of the 4 of that rank are chosen. Thus, the total number of ways of choosing the triple is:

${5 \choose 1}{4 \choose 3} = 20$

There are

44

ways to choose the suits of the remaining 4 cards, minus the ways in which all 4 match one of the 3 suits in the triple (making a flush):

${4 \choose 1}^4 - {3 \choose 1} = 253$

Thus, the total number of three of a kinds is:

$1,277 \cdot 20 \cdot 253 = 6,461,620\,$

Two pair — A two pair can be formed in 2 ways:

3 pairs with 1 kicker

The 4 ranks are chosen, then 3 of the 4 are chosen for the 3 pairs, and 2 of the 4 of each rank are chosen for each pair. The kicker is then chosen from the 4 cards in the remaining rank. Thus, the total number of 3 pairs with 1 kicker is:

${13 \choose 4}{4 \choose 3}{4 \choose 2}^3{4 \choose 1} = 2,471,040$

2 pairs with 3 kickers

A two pair hand must consist of 5 of the 13 ranks, but the 10 combinations that form straights must be subtracted. 2 of the ranks are chosen for the pairs and as with the calculations for straights with 5 ranks and two pairs, there are 2,268 sets of suits that do not form flushes. Thus, the total number of 2 pairs with 3 kickers is:

$\left[{13 \choose 5} - 10\right]{5 \choose 2} \cdot 2,268 = 28,962,360$

Thus, the total number of two pairs is:

$2,471,040 + 28,962,360 = 31,433,400\,$

Pair — A pair hand must consist of 6 of the 13 ranks, but the combinations that form straights must be subtracted. There are 9 ways to form a 6-card straight (6- to ace-high). With 5-card straights, when the straight is either 5- or ace-high, the remaining card may be selected from any of the 8 other ranks, minus the rank at the open end of the straight (6 and 9 respectively). In any of the other 8 situations, the remaining card may be selected from any of the other 8 ranks, minus the two ranks at either end of the straight. Thus, the total number of sets of ranks that do not form straights is:

${13 \choose 6} - 9 - \left[2 \cdot {7 \choose 1} + 8 \cdot {6 \choose 1}\right] = 1,645$

There are

45

ways of choosing the ranks of the kickers, and as with the calculations for straights with 6 distinct suits, there are 34 sets of suits that form flushes, therefore the total number of sets of suits that do not form flushes is:

${4 \choose 1}^5 - 34 = 990$

There are 6 different ranks to choose for the pair and the pair can be formed from 2 of the 4 cards in that rank, therefor the number of ways to choose the pair is:

${6 \choose 1}{4 \choose 2} = 36$

Thus, the total number of pair hands is:

$1645 \cdot 990 \cdot 36 = 58,627,800$

No pair — The 7 ranks are chosen, but the combinations that form straights must be subtracted. There are 8 ways to form a 7-card straight (7- to ace-high). With 6-card straights, as with 5-card straights in the pair hand calculations, any of the remaining ranks minus 1 may be chosen for the highest and lowest straight (6 ranks), while in the other cases, any remaining rank minus 2 may be chosen (5 ranks). With 5-card straights, the calculations are the same as with pairs, but 2 cards must be chosen rather than 1. Thus, the total number of sets of ranks that do not form straights is:

${13 \choose 6} - 8 - \left[2 \cdot {6 \choose 1} + 7 \cdot {5 \choose 1}\right] - \left[2 \cdot {7 \choose 2} + 8 \cdot {6 \choose 2}\right] = 1,499$

There are

47

ways of choosing the suits of the cards, and as with the calculations for straights with 7 distinct suits, there are 844 sets of suits that form flushes, therefore the total number of sets of suits that do not form flushes is:

${4 \choose 1}^7 - 844 = 15,540$

Thus, the total number of no pair hands is:

$1,499 \cdot 15,540 = 23,294,460\,$

[edit] Frequency of 5-card lowball poker hands

See Rank of hands (poker)#Low-poker ranking for a more complete discussion of lowball poker hands.

Some variants of poker, called lowball, use a low hand to determine the winning hand. In most variants of lowball, the ace is counted as the lowest card and straights and flushes don't count against a low hand, so the lowest hand is the five-high hand A-2-3-4-5, also called a wheel. The probability is calculated based on $\begin{matrix} {52 \choose 5} = 2,598,960 \end{matrix}$ , the total number of 5-card combinations. (The frequencies given are exact; the probabilities and odds are approximate.)

Hand	Distinct hands	Frequency	Probability	Cumulative	Odds
5-high	1	1,024	0.0394%	0.0394%	2,537.05 : 1
6-high	5	5,120	0.197%	0.236%	506.61 : 1
7-high	15	15,360	0.591%	0.827%	168.20 : 1
8-high	35	35,840	1.38%	2.21%	71.52 : 1
9-high	70	71,680	2.76%	4.96%	35.26 : 1
10-high	126	129,024	4.96%	9.93%	19.14 : 1
Jack-high	210	215,040	8.27%	18.2%	11.09 : 1
Queen-high	330	337,920	13.0%	31.2%	6.69 : 1
King-high	495	506,880	19.5%	50.7%	4.13 : 1
Total	1,287	1,317,888	50.7%	50.7%	0.97 : 1

As can be seen from the table, just over half the time a player gets a hand that has no pairs or three- or four-of-a-kind.

If aces are not low, simply rotate the hand descriptions so that 6-high replaces 5-high for the best hand and ace-high replaces king-high as the worst hand.

[edit] Derivation of frequencies for 5-card lowball hands

The following computations show how the above frequencies for 5-card lowball poker hands were determined. To understand these derivations, the reader should be familiar with the basic properties of the binomial coefficients and their interpretation as the number of ways of choosing elements from a given set. See also: sample space and event (probability theory).

The probability for any specific low hand with 5 distinct ranks (i.e. no paired cards) is the same. The frequency of a 5-high hand or any a specific low hand is calculated by making 5 independent choices for the suit for each rank, which is:

${4 \choose 1}^5 = 1,024$

There is one way to choose the ranks for a five-high hand:

${5 \choose 1} = 1$

To determine the number of distinct six-high hands, once the six is chosen, the other 4 ranks are chosen from the 5 ranks A to 5, which is:

${5 \choose 4} = 5$

This can be generalized for any non-paired low hand. Where $r$ is the highest rank in the hand (numbering jack–king as 11–13), the number of distinct low hands $D$ is:

$D = {r-1 \choose 4}$

and the frequency of low hands that are $r$ -high is $1,024 \cdot D$ .

Derivation for lowball hands without straights and flushes:

In the case where straights and flushes count against a low hand, the frequency of a specific hand must subtract the 4 combinations of suits that yeild a flush, and the calculation for the number of distinct hands must subtract the $r - 4$ combinations of ranks that yeild a straight. This gives the following frequency for low hands of rank $r$ that do not include a straight or a flush:

$\left[ {4 \choose 1}^5 - 4 \right] \cdot \left[ {r-1 \choose 4} - (r - 4) \right]$

[edit] Frequency of 7-card lowball poker hands

See Rank of hands (poker)#Low-poker ranking for a more complete discussion of lowball poker hands.

In some variants of poker a player uses the best five-card low hand made from seven cards. In most variants of lowball, the ace is counted as the lowest card and straights and flushes don't count against a low hand, so the lowest hand is the five-high hand A-2-3-4-5, also called a wheel. The probability is calculated based on $\begin{matrix} {52 \choose 7} = 133,784,560 \end{matrix}$ , the total number of 7-card combinations. (The frequencies given are exact; the probabilities and odds are approximate.)

Hand	Frequency	Probability	Cumulative	Odds
5-high	781,824	0.584%	0.584%	170.12 : 1
6-high	3,151,360	2.36%	2.94%	41.45 : 1
7-high	7,426,560	5.55%	8.49%	17.01 : 1
8-high	13,171,200	9.85%	18.3%	9.16 : 1
9-high	19,174,400	14.3%	32.7%	5.98 : 1
10-high	23,675,904	17.7%	50.4%	4.65 : 1
Jack-high	24,837,120	18.6%	68.9%	4.39 : 1
Queen-high	21,457,920	16.0%	85.0%	5.23 : 1
King-high	13,939,200	10.4%	95.4%	8.60 : 1
Total	127,615,488	95.4%	95.4%	0.05 : 1

As can be seen from the table, 95.4% of the time a player can make a 5-card low hand that has no pairs or three- or four-of-a-kind.

If aces are not low, simply rotate the hand descriptions so that 6-high replaces 5-high for the best hand and ace-high replaces king-high as the worst hand.

[edit] Derivation of frequencies for 7-card lowball hands

The following computations show how the above frequencies for 7-card lowball poker hands were determined. To understand these derivations, the reader should be familiar with the basic properties of the binomial coefficients and their interpretation as the number of ways of choosing elements from a given set. See also: sample space and event (probability theory).

To make a low hand of a specific rank four ranks are chosen that are lower than the high rank. Where $r$ is the highest rank in the hand (numbering jack–king as 11–13), the number of sets of 5 ranks that can make a low hand is:

${r-1 \choose 4}$

There are then three different ways to choose the remaining two cards that are not used in the low hand. Each of these cases must be considered separately:

7 distinct ranks

In this type of hand the two additional ranks are chosen from the ranks higher than $r$ , so this type of hand can only occur when there are at least two ranks greater than $r$ —that is, jack-high or better hands. The suits can be assigned by making 7 independent choices for the suit for each rank, so the number of ways to make a low hand with two distinct higher ranks is:

${13-r \choose 2}{4 \choose 1}^7 = {13-r \choose 2} \cdot 16,384$

6 distinct ranks

In this type of hand there are 6 distinct ranks and one pair. The additional rank is chosen from the ranks higher than $r$ , so this type of hand can only occur when there is at least one rank greater than $r$ —that is, queen-high or better hands. One of the 6 ranks is chosen for the pair and two of the four cards in that rank are chosen. The suits for the remaining 5 ranks are assigned by making 5 independent choices for each rank, so the number of ways to make a low hand with one higher ranks and a pair is:

${13-r \choose 1}{6 \choose 1}{4 \choose 2}{4 \choose 1}^5 = {13-r \choose 1} \cdot 36,864$

5 distinct ranks

There are two ways to choose 5 distinct ranks for seven cards. Either two pair and three unpaired ranks or three of a kind and four unpaired ranks.

Two pair

In this type of hand there are 5 distinct ranks and two pair. Two of the 5 ranks are chosen for the pairs and two of the four cards in each rank are chosen. The suits for the remaining 3 ranks are assigned by making 3 independent choices for each rank, so the number of ways to make a low hand with two pair is:

${5 \choose 2}{4 \choose 2}^2{4 \choose 1}^3 = 23,040$

Three of a kind

In this type of hand there are 5 distinct ranks and three of a kind. One of the 5 ranks is chosen for the three of a kind and three of the four cards in the rank are chosen. The suits for the remaining 4 ranks are assigned by making 4 independent choices for each rank, so the number of ways to make a low hand with three of a kind is:

${5 \choose 1}{4 \choose 3}{4 \choose 1}^4 = 5,120$