Poker probability

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.

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 2,598,960, 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	Odds against
Straight flush	40	.00154 %	64,973 : 1
Four of a kind	624	.0240 %	4,164 : 1
Full house	3,744	.144 %	693 : 1
Flush	5,108	.197 %	508 : 1
Straight	10,200	.392 %	254 : 1
Three of a kind	54,912	2.11 %	46.3 : 1
Two pair	123,552	4.75 %	20.0 : 1
One pair	1,098,240	42.3 %	1.37 : 1
No pair	1,302,540	50.1 %	0.995 : 1
Total	2,598,960	100 %	0 : 1

The royal flush is included as a straight flush above. By itself, the royal flush can be formed 4 ways (one for each suit), giving it a probability of .00000154 and odds of 649,739 : 1.

When ace-low straights and straight flushes are not counted, the probabilities of each are reduced: straights and straight flushes become 9/10 as common as they otherwise would be.

Derivation

The following computations show how the above frequencies 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:

<math>{4 \choose 1}{10 \choose 1} = 40<math>

• 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:

<math>{13 \choose 1}{48 \choose 1} = 624<math>

• 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 any three of the four suits. The triple can be any one of the remaining twelve ranks, and any two of the four suits. Thus, the total number of full houses is:

<math>{13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2} = 3,744<math>

• 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:

<math>{4 \choose 1}{13 \choose 5} - 40 = 5,108<math>

• 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:

<math>{10 \choose 1}{4 \choose 1}^5 - 40 = 10,200<math>

• 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:

<math>{13 \choose 1}{4 \choose 3}{12 \choose 2}{4 \choose 1}^2 = 54,912<math>

• 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:

<math>{13 \choose 2}{4 \choose 2}^2{11 \choose 1}{4 \choose 1} = 123,552<math>

• 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:

<math>{13 \choose 1}{4 \choose 2}{12 \choose 3}{4 \choose 1}^3 = 1,098,240<math>

• 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 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:

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

External links

• MathWorld: Poker (http://mathworld.wolfram.com/Poker.html)
• Probability and Poker (http://www.indepthinfo.com/probability-poker/index.shtml)
• Brian Alspach's mathematics and poker page (http://www.math.sfu.ca/~alspach)
• Brian Alspach Poker Math articles (http://www.math.sfu.ca/~alspach/pokerdigest.html)
• Brian Alspach Poker Math computations (http://www.math.sfu.ca/~alspach/computations.html)