Online Poker

Poker probability

In poker, the probability of each type of 5 poker card hand can be computed by calculating the proportion of poker hands of that type among all possible poker hands or not.

Frequency of 5 poker card hands

The following enumerates the frequency of each hand, given all combinations of 5 poker cards randomly drawn from a full poker deck of 52, without wild poker cards. The probability is calculated based on 2,598,960, the total number of 5 poker card combinations. Here, the probability is the frequency of the poker hand divided by the total number of 5 poker 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
Royal flush	4	.00000154	649,739 : 1
Straight flush	36	.0000139	72,192 : 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

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).

Royal flush -- Each suit has just one possible royal flush, containing the T-J-Q-K-A of that suit. Thus, the total number of straight flushes is

${4 \choose 1} = 4$

Straight flush -- Each straight flush is uniquely determined by its highest ranking poker card; and these poker card ranks go from 5 (A-2-3-4-5) up to K (9-T-J-Q-K) in each of the 4 suits. (The ace is omitted because that would form a royal flush instead). Thus, the total number of straight flushes is

${4 \choose 1}{9 \choose 1} = 36$

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

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

Full house -- First, choose one of the 13 poker card ranks and 3 of the 4 suits for the 3 of a kind; then choose one of the remaining 12 poker card ranks and 2 of the 4 suits for the pair. Thus, the total number of full houses is

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

Flush -- First, choose one of four suits; then choose 5 of the 13 possible poker card ranks. Finally, subtract the 40 straight flushes, since these are ranked as straight flushes, not flushes. Thus, the total number of flushes is

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

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

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

Three of a kind -- First, choose one rank out of 13 for the 3 of a kind; then choose 3 out of 4 suits for the 3 of a kind. Then choose 2 distinct poker card ranks out of the remaining 12 for the other 2 poker cards, as well as suits for each of those poker cards. 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 -- First, choose 2 of the 13 poker card ranks for the 2 pairs; then choose 2 out of 4 suits for each of those 2 pairs. The final poker card can be any one of the 44 remaining poker cards whose poker card ranks are different from those of the 2 pairs. Thus, the total number of two-pair hands is

${13 \choose 2}{4 \choose 2}^2{44 \choose 1} = 123,552$

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

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

No pair -- Choose 5 out of 13 poker card ranks, discounting the 10 possible straights. Then choose one of 4 suits for each of the 5 poker 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-poker card hands and subtract from that the sum of all the above hands. Thus, the total number of no pair hands is

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

External links

MathWorld: Poker
Probability and Poker
Brian Alspach's mathematics and poker page
Brian Alspach Poker Math articles
Brian Alspach Poker Math computations