Poker Hands Probability Explained
Becoming a strong poker player starts with knowing the hands and odds of getting those hands. In this way, players can better predict the outcome of a poker game and make better bets. Sure, you could sit down and memorize all of the hands and statistics, but players can really perfect their poker skills through experience. Example 1: Using Pot Odds Preflop. NOTE: We’ll be going over cash game examples, but this same process works for tournament hands as well. Suppose you’re playing a $0.50/$1.00, 6-handed online cash game. Poker hands from highest to lowest 1. Royal flush A, K, Q, J, 10, all the same suit. How to Play Poker Poker Rules Poker Hand Rankings Poker Tools Hold'em Poker Odds Calculator Omaha Poker. The tables below show the probabilities of being dealt various poker hands with different wild card specifications. Each poker hand consists of dealing 5 random cards. While the results on the main Poker Probabilities page can be calculated via direct combinatorics, the introduction of wild cards greatly complicates the combinatoric calculations.
Sanderson M. Smith
Home About Sanderson Smith Writings and Reflections Algebra 2 AP Statistics Statistics/Finance Forum
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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Copyright © 2003-2009 Sanderson Smith
Brian Alspach
18 January 2000
Abstract:
One of the most popular poker games is 7-card stud. The way hands areranked is to choose the highest ranked 5-card hand contained amongst the7 cards. People frequently encounter difficulty in counting 7-card handsbecause a given set of 7 cards may contain several different types of5-card hands. This means duplicate counting can be troublesome as canomission of certain hands. The types of 5-card poker hands in decreasingrank are
- straight flush
- 4-of-a-kind
- full house
- flush
- straight
- 3-of-a-kind
- two pairs
- a pair
- high card
The total number of 7-card poker hands is .
We shall count straight flushes using the largest card in the straightflush. This enables us to pick up 6- and 7-card straight flushes. Whenthe largest card in the straight flush is an ace, then the 2 other cardsmay be any 2 of the 47 remaining cards. This gives us straight flushes in which the largest card is an ace.
If the largest card is any of the remaining 36 possible largest cards ina straight flush, then we may choose any 2 cards other than theimmediate successor card of the particular suit. This gives usstraight flushes of the second type, and41,584 straight flushes altogether.
In forming a 4-of-a-kind hand, there are 13 choices for the rank ofthe quads, 1 choice for the 4 cards of the given rank, and choices for the remaining 3 cards. This implies there are 4-of-a-kind hands.
There are 3 ways to get a full house and we count them separately. Oneway of obtaining a full house is for the 6-card hand to contain 2 setsof triples and a singleton. There are ways tochoose the 2 ranks, 4 ways to choose each of the triples, and 44 ways tochoose the singleton. This gives us fullhouses of this type. A second way of getting a full houseis for the 7-card hand to contain a triple and 2 pairs. There are 13ways to choose the rank of the triple, ways tochoose the ranks of the pairs, 4 ways to choose the triple of the givenrank, and 6 ways to choose the pairs of each of the given ranks. Thisproduces full house of the secondkind. The third way to get a full house is for the 7-card hand tocontain a triple, a pair and 2 singletons of distinct ranks. There are13 choices for the rank of the triple, 12 choices for the rank of thepair, choices for the ranks of the singletons,4 choices for the triple, 6 choices for the pair, and 4 choices for eachof the singletons. We obtain full houses of the last kind. Adding the 3 numbers gives us3,473,184 full houses.
To count the number of flushes, we first obtain some useful informationon sets of ranks. The number of ways of choosing 7 distinct ranks from13 is .We want to remove the sets of rankswhich include 5 consecutive ranks (that is, we are removing straightpossibilities). There are 8 rank sets of the form .Another form to eliminate is ,where y is neither x-1 nor x+6. If x is ace or 9, thereare 6 choices for y. If x is any of the other 7 possibilities, thereare 5 possibilities for y. This produces sets with 6 consecutive ranks. Finally, the remaining form to eliminateis ,where neither y nor z is allowed totake on the values x-1 or x+5. If x is either ace or 10, theny,z are being chosen from a 7-subset. If x is any of the other 8possible values, then y,z are being chosen from a 6-set. Hence, thenumber of rank sets being excluded in this case is .In total, we remove 217 sets of ranks ending upwith 1,499 sets of 7 ranks which do not include 5 consecutive ranks.Thus, there are flushes having all 7 cards in thesame suit.
Now suppose we have 6 cards in the same suit. Again there are 1,716sets of 6 ranks for these cards in the same suit. We must excludesets of ranks of the form of which thereare 9. We also must exclude sets of ranks of the form ,where y is neither x-1 nor x+5. So if x is aceor 10, y can be any of 7 values; whereas, if x is any of the other8 possible values, y can be any of 6 values. This excludes 14 + 48= 62 more sets. Altogether 71 sets have been excluded leaving 1,645sets of ranks for the 6 suited cards not producing a straight flush.The remaining card may be any of the 39 cards from the other 3 suits.This gives us flushes with 6 suitedcards.
Finally, suppose we have 5 cards in the same suit. The remaining 2cards cannot possibly give us a hand better than a flush so all we needdo here is count flushes with 5 cards in the same suit. There arechoices for 5 ranks in the same suit. We mustremove the 10 sets of ranks producing straight flushes leaving us with1,277 sets of ranks. The remaining 2 cards can be any 2 cards from theother 3 suits so that there are choices for them.Then there are flushes of this lasttype. Adding the numbers of flushes of the 3 types produces 4,047,644flushes.
We saw above that there are 217 sets of 7 distinct ranks which include5 consecutive ranks. For any such set of ranks, each card may be anyof 4 cards except we must remove those which correspond to flushes.There are 4 ways to choose all of them in the same suit. There areways to choose 6 of them in the same suit. For 5of them in the same suit, there are ways to choosewhich 5 will be in the same suit, 4 ways to choose the suit of the 5cards, and 3 independent choices for the suits of each of the 2 remainingcards. This gives choices with 5 in the samesuit. We remove the 844 flushes from the 47 = 16,384 choices of cardsfor the given rank set leaving 15,540 choices which produce straights.We then obtain straights when the 7-cardhand has 7 distinct ranks.
We now move to hands with 6 distinct ranks. One possible form is,where x can be any of 9 ranks. The otherpossible form is ,where y is neither x-1nor x+5. When x is ace or 10, then there are 7 choices for y.When x is between 2 and 9, inclusive, there are 6 choices for y.This implies there are sets of 6 distinctranks corresponding to straights. Note this means there must be a pairin such a hand. We have to ensure we do not count any flushes.
As we just saw, there are 71 choices for the set of 6 ranks. Thereare 6 choices for which rank will have a pair and there are 6 choicesfor a pair of that rank. Each of the remaining 5 cards can be chosenin any of 4 ways. Now we remove flushes. If all 5 cards were chosenin the same suit, we would have a flush so we remove the 4 ways ofchoosing all 5 in the same suit. In addition, we cannot choose 4 ofthem in either suit of the pair. There are 5 ways to choose 4 cardsto be in the same suit, 2 choices for that suit and 3 choices for thesuit of the remaining card. So there are choices which give a flush. This means there are 45 - 34 = 990choices not producing a flush. Hence, there are straights of this form.
We also can have a set of 5 distinct ranks producing a straight whichmeans the corresponding 7-card hand must contain either 2 pairs or3-of-a-kind as well. The set of ranks must have the formand there are 10 such sets. First we supposethe hand also contains 3-of-a-kind. There are 5 choices for the rankof the trips, and 4 choices for trips of that rank. The cards of theremaining 4 ranks each can be chosen in any of 4 ways. This gives44 = 256 choices for the 4 cards. We must remove the 3 choices for whichall 4 cards are in the same suit as one of the cards in the 3-of-a-kind.So we have straights which alsocontain 3-of-a-kind.
Next we suppose the hand also contains 2 pairs. There are choices for the 2 ranks which will be paired. There are 6choices for each of the pairs giving us 36 ways to choose the 2 pairs.We have to break down these 36 ways of getting 2 pairs because differentsuit patterns for the pairs allow different possibilities for flushesupon choosing the remaining 3 cards. Now 6 of the ways of getting the2 pairs have the same suits represented for the 2 pairs, 24 of themhave exactly 1 suit in common between the 2 pairs, and 6 of them haveno suit in common between the 2 pairs.
There are 43 = 64 choices for the suits of the remaining 3 cards.In the case of the 6 ways of getting 2 pairs with the same suits, 2of the 64 choices must be eliminated as they would produce a flush(straight flush actually). In the case of the 24 ways of getting 2pairs with exactly 1 suit in common, only 1 of the 64 choices need beeliminated. When the 2 pairs have no suit in common, all 64 choicesare allowed since a flush is impossible. Altogether we obtain
straights which alsocontain 2 pairs. Adding all the numbers together gives us 6,180,020straights.
A hand which is a 3-of-a-kind hand must consist of 5 distinct ranks.There are sets of 5 distinct ranks fromwhich we must remove the 10 sets corresponding to straights. Thisleaves 1,277 sets of 5 ranks qualifying for a 3-of-a-kind hand. Thereare 5 choices for the rank of the triple and 4 choices for the tripleof the chosen rank. The remaining 4 cards can be assigned any of 4suits except not all 4 can be in the same suit as the suit of one ofcards of the triple. Thus, the 4 cards may be assigned suits in 44-3=253 ways. Thus, we obtain 3-of-a-kind hands.
Next we consider two pairs hands. Such a hand may contain either 3pairs plus a singleton, or two pairs plus 3 remaining cards of distinctranks. We evaluate these 2 types of hands separately. If the hand has3 pairs, there are ways to choose the ranks ofthe pairs, 6 ways to choose each of the pairs, and 40 ways to choosethe singleton. This produces 7-card hands with 3 pairs.
The other kind of two pairs hand must consist of 5 distinct ranks andas we saw above, there are 1,277 sets of ranks qualifying for a twopairs hand. There are choices for the two ranksof the pairs and 6 choices for each of the pairs. The remaining cardsof the other 3 ranks may be assigned any of 4 suits, but we must removeassignments which result in flushes. This results in exactly thesame consideration for the overlap of the suits of the two pairs asin the final case for flushes above. We then obtain
hands of two pairs of the second type. Adding the two gives 31,433,4007-card hands with two pairs.
Now we count the number of hands with a pair. Such a hand must have6 distinct ranks. We saw above there are 1,645 sets of 6 ranks whichpreclude straights. There are 6 choices for the rank of the pair and6 choices for the pair of the given rank. The remaining 5 ranks canhave any of 4 suits assigned to them, but again we must remove thosewhich produce a flush. We cannot choose all 5 to be in the same suitfor this results in a flush. This can happen in 4 ways. Also, wecannot choose 4 of them to be in the same suit as the suit of eitherof the cards forming the pair. This can happen in ways. Hence, there are 45-34 = 990 choices for the remaining 4 cards.This gives us hands with a pair.
We could determine the number of high card hands by removing the handswhich have already been counted in one of the previous categories.Instead, let us count them independently and see if the numbers sumto 133,784,560 which will serve as a check on our arithmetic.
A high card hand has 7 distinct ranks, but does not include straights.So we must eliminate sets of ranks which have 5 consecutive ranks.Above we determined there are 1,499 sets of 7 ranks not containing 5consecutive ranks, that is, there are no straights. Now the card ofeach rank may be assigned any of 4 suits giving 47 = 16,384 assignmentsof suits to the ranks. We must eliminate those which resulkt in flushes.There are 4 ways to assign all 7 cards the same suit. There are 7choices for 6 cards to get the same suit, 4 choices of the suit to beassigned to the 6 cards, and 3 choices for the suit of the other card.This gives assignments in which 6 cards end upwith the same suit. Finally, there are choices for5 cards to get the same suit, 4 choices for that suit, and 3 independentchoices for each of the remaining 2 cards. This gives assignments producing 5 cards in the same suit. Altogether wemust remove 4 + 84 + 756 = 844 assignments resulting in flushes. Thus,the number of high card hands is 1,499(16,384 - 844)=23,294,460.
If we sum the preceding numbers, we obtain 133,784,560 and we can beconfident the numbers are correct.
Here is a table summarizing the number of 7-card poker hands. Theprobability is the probability of having the hand dealt to you whendealt 7 cards.
| hand | number | Probability |
| straight flush | 41,584 | .00031 |
| 4-of-a-kind | 224,848 | .0017 |
| full house | 3,473,184 | .026 |
| flush | 4,047,644 | .030 |
| straight | 6,180,020 | .046 |
| 3-of-a-kind | 6,461,620 | .048 |
| two pairs | 31,433,400 | .235 |
| pair | 58,627,800 | .438 |
| high card | 23,294,460 | .174 |
You will observe that you are less likely to be dealt a hand withno pair (or better) than to be dealt a hand with one pair. Thishas caused some people to query the ranking of these two hands.In fact, if you were ranking 7-card hands based on 7 cards, theorder of the last 2 would switch. However, you are basing the rankingon 5 cards so that if you were to rank a high card hand higher than a handwith a single pair, people would choose to ignore the pair in a7-card hand with a single pair and call it a high card hand. Thiswould have the effect of creating the following distortion. Thereare 81,922,260 7-card hands in the last two categories containing5 cards which are high card hands. Of these 81,922,260 hands,58,627,800 also contain 5-card hands which have a pair. Thus, thelatter hands are more special and should be ranked higher (as theyindeed are) but would not be under the scheme being discussed inthis paragraph.
Poker Hands Probability Explained Probability
last updated 18 January 2000