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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. 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.

Hand	Frequency	Probability	Odds
Straight flush	40	.0000154	64,973 : 1
Four of a kind	624	.000240	4,164 : 1
Full house	3,744	.00144	693 : 1
Flush	5,108	.00197	508 : 1
Straight	10,200	.00392	254 : 1
Three of a kind	54,912	.0211	46.3 : 1
Two pair	123,552	.0475	20.0 : 1
One pair	1,098,240	.423	1.37 : 1
No pair	1,302,540	.501	0.995 : 1
Total	2,598,960	1.00	0 : 1

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

• Four of a kind -- First, we choose one of the 13 ranks for the 4 of a kind; then there are 52 − 4 = 48 cards remaining from which to choose the final card. Thus, the total number of four of a kinds is

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

• Full house -- First, we choose one of the 13 ranks and 3 of the 4 suits for the 3 of a kind; then we choose one of the remaining 12 ranks and 2 of the 4 suits for the pair. 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 -- First, we choose one of four suits; then we choose 5 of the 13 possible ranks. Finally, we must subtract the 40 straight flushes, since these are ranked as straight flushes, not flushes. Thus, the total number of flushes is

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

• Straight -- First, we choose the highest ranking card; there are 10 of these, from 5 (A-2-3-4-5) to A (T-J-Q-K-A). Then we choose one of four suits for each of the 5 cards. Finally, we must subtract the 40 straight flushes, since these are ranked as straight flushes, not straights. Thus, the total number of straights is

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

• Three of a kind -- First, we choose one rank out of 13 for the 3 of a kind; then we choose 3 out of 4 suits for the 3 of a kind. Then we choose 2 distinct ranks out of the remaining 12 for the other 2 cards, as well as suits for each of those cards. 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 -- First, we choose 2 of the 13 ranks for the 2 pairs; then we choose 2 out of 4 suits for each of those 2 pairs. The final card can be any one of the 44 remaining cards not comprising the ranks of the 2 pairs. Thus, the total number of two pairs is

<math>{13 \choose 2}{4 \choose 2}^2{44 \choose 1} = 123,552<math>

• One pair -- First, we choose one of the 13 ranks for the pair; then we choose 2 out of 4 suits for that pair. For the other 3 cards, we choose 3 ranks out of the remaining 12 and one of 4 suits for each of the 3 cards. Thus, the total number of one pairs is

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

• No pair -- We can choose 5 out of 13 ranks, discounting the 10 possible straights. Then we choose one of 4 suits for each of the 5 cards, discounting the 4 possible flushes. Alternatively, since we are looking for any hand which does not fall into one of the above categories, we can take the total number of 5 card hands and subtract the sum of the above. 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>