Number Of Distinct Poker Hands
An Introduction to Thermal Physics Daniel V. Schroeder Problem 2-4 Calculate the number of different 5-card poker hands selected from a standard deck of 52 c. The number of distinct 5-card poker hands that are possible from 7 cards is 4,824. Perhaps surprisingly, this is less than the number of 5-card poker hands from 5 cards because some 5-card hands are impossible with 7 cards (e.g. Derivation of frequencies of 7-card poker hands Edit. There are 52c5 = 2,598,960 ways to choose 5 cards from a 52 card deck. However, since suits are interchangeable in poker, many of these are equivalent - the hand 2H 2C 3H 3S 4D is equivalent to 2D 2S 3D 3C 4H - simply swap the suits around. According to wikipedia, there are 134,459 distinct 5 card hands once you account for possible suit.
This post works with 5-card Poker hands drawn from a standard deck of 52 cards. The discussion is mostly mathematical, using the Poker hands to illustrate counting techniques and calculation of probabilities
The number of different 5 -card poker hands is. 52 C 5 = 2,598,960. A wonderful exercise involves having students verify probabilities that appear in books relating to gambling. For instance, in Probabilities in Everyday Life, by John D. McGervey, one finds many interesting tables containing probabilities for poker and other games of chance. I was doing number counting problem and wanted to check if my result was correct. Problem description: From a standard deck of cards(52 cards, 4 suits, 13 numbers in each suit) there are 5 cards drawn. Calculate number of hands where there are atleast two cards with the same number and atleast two cards with different number. The fifth card can.
Working with poker hands is an excellent way to illustrate the counting techniques covered previously in this blog – multiplication principle, permutation and combination (also covered here). There are 2,598,960 many possible 5-card Poker hands. Thus the probability of obtaining any one specific hand is 1 in 2,598,960 (roughly 1 in 2.6 million). The probability of obtaining a given type of hands (e.g. three of a kind) is the number of possible hands for that type over 2,598,960. Thus this is primarily a counting exercise.
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Preliminary Calculation
Usually the order in which the cards are dealt is not important (except in the case of stud poker). Thus the following three examples point to the same poker hand. The only difference is the order in which the cards are dealt.
These are the same hand. Order is not important.
The number of possible 5-card poker hands would then be the same as the number of 5-element subsets of 52 objects. The following is the total number of 5-card poker hands drawn from a standard deck of 52 cards.
The notation is called the binomial coefficient and is pronounced “n choose r”, which is identical to the number of -element subsets of a set with objects. Other notations for are , and . Many calculators have a function for . Of course the calculation can also be done by definition by first calculating factorials.
Thus the probability of obtaining a specific hand (say, 2, 6, 10, K, A, all diamond) would be 1 in 2,598,960. If 5 cards are randomly drawn, what is the probability of getting a 5-card hand consisting of all diamond cards? It is
This is definitely a very rare event (less than 0.05% chance of happening). The numerator 1,287 is the number of hands consisting of all diamond cards, which is obtained by the following calculation.
The reasoning for the above calculation is that to draw a 5-card hand consisting of all diamond, we are drawing 5 cards from the 13 diamond cards and drawing zero cards from the other 39 cards. Since (there is only one way to draw nothing), is the number of hands with all diamonds.
If 5 cards are randomly drawn, what is the probability of getting a 5-card hand consisting of cards in one suit? The probability of getting all 5 cards in another suit (say heart) would also be 1287/2598960. So we have the following derivation.
Thus getting a hand with all cards in one suit is 4 times more likely than getting one with all diamond, but is still a rare event (with about a 0.2% chance of happening). Some of the higher ranked poker hands are in one suit but with additional strict requirements. They will be further discussed below.
Another example. What is the probability of obtaining a hand that has 3 diamonds and 2 hearts? The answer is 22308/2598960 = 0.008583433. The number of “3 diamond, 2 heart” hands is calculated as follows:
One theme that emerges is that the multiplication principle is behind the numerator of a poker hand probability. For example, we can think of the process to get a 5-card hand with 3 diamonds and 2 hearts in three steps. The first is to draw 3 cards from the 13 diamond cards, the second is to draw 2 cards from the 13 heart cards, and the third is to draw zero from the remaining 26 cards. The third step can be omitted since the number of ways of choosing zero is 1. In any case, the number of possible ways to carry out that 2-step (or 3-step) process is to multiply all the possibilities together.
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Number Of Distinct Poker Hands
The Poker Hands
Here’s a ranking chart of the Poker hands.
The chart lists the rankings with an example for each ranking. The examples are a good reminder of the definitions. The highest ranking of them all is the royal flush, which consists of 5 consecutive cards in one suit with the highest card being Ace. There is only one such hand in each suit. Thus the chance for getting a royal flush is 4 in 2,598,960.
Royal flush is a specific example of a straight flush, which consists of 5 consecutive cards in one suit. There are 10 such hands in one suit. So there are 40 hands for straight flush in total. A flush is a hand with 5 cards in the same suit but not in consecutive order (or not in sequence). Thus the requirement for flush is considerably more relaxed than a straight flush. A straight is like a straight flush in that the 5 cards are in sequence but the 5 cards in a straight are not of the same suit. For a more in depth discussion on Poker hands, see the Wikipedia entry on Poker hands.
The counting for some of these hands is done in the next section. The definition of the hands can be inferred from the above chart. For the sake of completeness, the following table lists out the definition.
Definitions of Poker Hands
Poker Hand | Definition | |
---|---|---|
1 | Royal Flush | A, K, Q, J, 10, all in the same suit |
2 | Straight Flush | Five consecutive cards, |
all in the same suit | ||
3 | Four of a Kind | Four cards of the same rank, |
one card of another rank | ||
4 | Full House | Three of a kind with a pair |
5 | Flush | Five cards of the same suit, |
not in consecutive order | ||
6 | Straight | Five consecutive cards, |
not of the same suit | ||
7 | Three of a Kind | Three cards of the same rank, |
2 cards of two other ranks | ||
8 | Two Pair | Two cards of the same rank, |
two cards of another rank, | ||
one card of a third rank | ||
9 | One Pair | Three cards of the same rank, |
3 cards of three other ranks | ||
10 | High Card | If no one has any of the above hands, |
the player with the highest card wins |
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Counting Poker Hands
Straight Flush
Counting from A-K-Q-J-10, K-Q-J-10-9, Q-J-10-9-8, …, 6-5-4-3-2 to 5-4-3-2-A, there are 10 hands that are in sequence in a given suit. So there are 40 straight flush hands all together.
Four of a Kind
There is only one way to have a four of a kind for a given rank. The fifth card can be any one of the remaining 48 cards. Thus there are 48 possibilities of a four of a kind in one rank. Thus there are 13 x 48 = 624 many four of a kind in total.
Full House
Let’s fix two ranks, say 2 and 8. How many ways can we have three of 2 and two of 8? We are choosing 3 cards out of the four 2’s and choosing 2 cards out of the four 8’s. That would be = 4 x 6 = 24. But the two ranks can be other ranks too. How many ways can we pick two ranks out of 13? That would be 13 x 12 = 156. So the total number of possibilities for Full House is
Note that the multiplication principle is at work here. When we pick two ranks, the number of ways is 13 x 12 = 156. Why did we not use = 78?
Flush
There are = 1,287 possible hands with all cards in the same suit. Recall that there are only 10 straight flush on a given suit. Thus of all the 5-card hands with all cards in a given suit, there are 1,287-10 = 1,277 hands that are not straight flush. Thus the total number of flush hands is 4 x 1277 = 5,108.
Straight
There are 10 five-consecutive sequences in 13 cards (as shown in the explanation for straight flush in this section). In each such sequence, there are 4 choices for each card (one for each suit). Thus the number of 5-card hands with 5 cards in sequence is . Then we need to subtract the number of straight flushes (40) from this number. Thus the number of straight is 10240 – 10 = 10,200.
Three of a Kind
There are 13 ranks (from A, K, …, to 2). We choose one of them to have 3 cards in that rank and two other ranks to have one card in each of those ranks. The following derivation reflects all the choosing in this process.
Two Pair and One Pair
These two are left as exercises.
High Card
The count is the complement that makes up 2,598,960.
The following table gives the counts of all the poker hands. The probability is the fraction of the 2,598,960 hands that meet the requirement of the type of hands in question. Note that royal flush is not listed. This is because it is included in the count for straight flush. Royal flush is omitted so that he counts add up to 2,598,960.
Probabilities of Poker Hands
Poker Hand | Count | Probability | |
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2 | Straight Flush | 40 | 0.0000154 |
3 | Four of a Kind | 624 | 0.0002401 |
4 | Full House | 3,744 | 0.0014406 |
5 | Flush | 5,108 | 0.0019654 |
6 | Straight | 10,200 | 0.0039246 |
7 | Three of a Kind | 54,912 | 0.0211285 |
8 | Two Pair | 123,552 | 0.0475390 |
9 | One Pair | 1,098,240 | 0.4225690 |
10 | High Card | 1,302,540 | 0.5011774 |
Total | 2,598,960 | 1.0000000 |
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2017 – Dan Ma
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.
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.
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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