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Poker probabilityIn 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.
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).
\n\n| Hand | \nFrequency | \nProbability | \nOdds | \n \n\n| Straight flush | \n40 | \n.0000154 | \n64,973 : 1 | \n \n\n| Four of a kind | \n624 | \n.000240 | \n4,164 : 1 | \n \n\n| Full house | \n3,744 | \n.00144 | \n693 : 1 | \n \n\n| Flush | \n5,108 | \n.00197 | \n508 : 1 | \n \n\n| Straight | \n10,200 | \n.00392 | \n254 : 1 | \n \n\n| Three of a kind | \n54,912 | \n.0211 | \n46.3 : 1 | \n \n\n| Two pair | \n123,552 | \n.0475 | \n20.0 : 1 | \n \n\n| One pair | \n1,098,240 | \n.423 | \n1.366 | \n \n\n| No pair | \n1,302,540 | \n.501 | \n0.995 : 1 | \n \nTotal | \n2,598,960 | \n1.00 | \n0 : 1 | \n
Derivation
The following computations show how the above frequencies were determined.
- 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
- 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
- Full house -- First, we choose one of the 13 ranks and one of the 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
- 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
- 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
- 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
- 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
- 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
- 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 no pair is 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
External links\n* MathWorld: Poker |
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