There are 52 cards in a deck, 13 of each suit, and 4 of each rank with 1326 poker hands in total. To simplify things just focus on memorizing all of the potential combos to start: 16 possible hand combinations of every unpaired hand; 12 combinations of every unpaired offsuit hand; 4 combinations of each suited hand; 6 possible combinations of. Any five card poker hand — The total number of five card hands that can be drawn from a deck of cards is found using a combination selecting five cards, in any order where n refers to the number of items that can be selected and r to the sample size; the '!' Is the factorial operator. The is the best possible hand you can get in standard five-card Poker is called a royal. There are 1,326 possible combinations of cards from a standard deck but there are only 169 non-equivalent starting hands in poker. This number is made up of 13 pocket pairs, 78 suited hands and 78. However, let’s look at these hands by comparing the total combinations for each hand: AA = 6 combinations (21.5%) KK = 6 combinations (21.5%) AK = 16 combinations (57%) There are more AK hands in a range of AA, KK, AK than there are AA and KK hands combined. So out of 28 possible combinations made up from AA, KK and AK, 16 of them come from AK.
- How Many Possible Poker Hands
- Possible Poker Hand Combinations
- Total Possible Poker Hands Signals
- Total Possible Poker Hands Game
Sanderson M. Smith
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In many forms of poker, one is dealt 5 cards from astandard deck of 52 cards. The number of different 5 -card pokerhands is
A wonderful exercise involves having students verify probabilitiesthat appear in books relating to gambling. For instance, inProbabilities in Everyday Life, by John D. McGervey, one findsmany interesting tables containing probabilities for poker and othergames of chance.
How Many Possible Poker Hands
This article and the tables below assume the reader is familiarwith the names for various poker hands. In the NUMBER OF WAYS columnof TABLE 2 are the numbers as they appear on page 132 in McGervey'sbook. I have done computations to verify McGervey's figures. Thiscould be an excellent exercise for students who are studyingprobability.
There are 13 denominations (A,K,Q,J,10,9,8,7,6,5,4,3,2) in thedeck. One can think of J as 11, Q as 12, and K as 13. Since an acecan be 'high' or 'low', it can be thought of as 14 or 1. With this inmind, there are 10 five-card sequences of consecutive dominations.These are displayed in TABLE 1.
TABLE 1The following table displays computations to verify McGervey'snumbers. There are, of course , many other possible poker handcombinations. Those in the table are specifically listed inMcGervey's book. The computations I have indicated in the table doyield values that are in agreement with those that appear in thebook.
Possible Poker Hand Combinations
N = NUMBER OF WAYS listed by McGervey | |||
Straight flush | There are four suits (spades, hearts, diamond, clubs). Using TABLE 1,4(10) = 40. | ||
Four of a kind | (13C1)(48C1) = 624. Choose 1 of 13 denominations to get four cards and combine with 1 card from the remaining 48. | ||
Full house | (13C1)(4C3)(12C1)(4C2) = 3,744. Choose 1 denominaiton, pick 3 of 4 from it, choose a second denomination, pick 2 of 4 from it. | ||
Flush | (4C1)(13C5) = 5,148. Choose 1 suit, then choose 5 of the 13 cards in the suit. This figure includes all flushes. McGervey's figure does not include straight flushes (listed above). Note that 5,148 - 40 = 5,108. | ||
Straight | (4C1)5(10) = 45(10) = 10,240 Using TABLE 1, there are 10 possible sequences. Each denomination card can be 1 of 4 in the denomination. This figure includes all straights. McGervey's figure does not include straight flushes (listed above). Note that 10,240 - 40 = 10,200. | ||
Three of a kind | (13C1)(4C3)(48C2) = 58,656. Choose 1 of 13 denominations, pick 3 of the four cards from it, then combine with 2 of the remaining 48 cards. This figure includes all full houses. McGervey's figure does not include full houses (listed above). Note that 54,912 - 3,744 = 54,912. | ||
Exactly one pair, with the pair being aces. | (4C2)(48C1)(44C1)(40C1)/3! = 84,480. Choose 2 of the four aces, pick 1 card from remaining 48 (and remove from consider other cards in that denomination), choose 1 card from remaining 44 (and remove other cards from that denomination), then chose 1 card from the remaining 40. The division by 3! = 6 is necessary to remove duplication in the choice of the last 3 cards. For instance, the process would allow for KQJ, but also KJQ, QKJ, QJK, JQK, and JKQ. These are the same sets of three cards, just chosen in a different order. | ||
Two pairs, with the pairs being 3's and 2's. | McGervey's figure excludes a full house with 3's and 2's. (4C2)(4C1)(44C1) = 1,584. Choose 2 of the 4 threes, 2 of the 4 twos, and one card from the 44 cards that are not 2's or 3's. |
'I must complain the cards are ill shuffled 'til Ihave a good hand.'
-Swift, Thoughts on Various Subjects
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