Hand Combinations in Poker. The term “combination” (or “combo” for short) refers to the different ways we can make a specific type of poker hand. Example: We are dealt AK preflop in Hold’em. How many different combinations of AK are there? If we were so inclined, we could list every possible way of making AK. With such a hand one can have reasonable results, especially if you get a three-of-a-kind with flop. This hand is worse than it seems and raising will be improper. 6: 10.8%: Raise. A hand full of high cards must raise to reduce opponents. Side cards are lousy but two kings are enough for the start. Probabilities of poker combinations In this part of the material we will tell you about the mathematical probabilities for forming various poker combinations. As you can see royal flush is the rarest and the strongest poker combination. The probability of collecting royal flush in poker is 1 to 649 740.

• Poker Hand Combinations Probability Calculator
• Poker Hand Combinations Probability Chart
• Poker Hand Combinations Probability Formula
• Poker Hand Combinations Probability Table

While fluctuations in probability (luck) will happen from hand to hand, the best poker players understand that skill, discipline and patience are the keys to success at the tables. A big part of strong decision making is understanding how often you should be betting, raising, and applying pressure. Of the 1,326 combinations, there are 169 distinct starting hands grouped into three shapes: 13 pocket pairs (paired hole cards), 13 × 12 ÷ 2 = 78 suited hands and 78 unsuited hands; 13 + 78 + 78 = 169. The relative probability of being dealt a hand of each given shape is different.

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.

Poker Hand Combinations Probability Calculator

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.

We can use permutations and combinations to help us answer more complex probability questions

Example 1

A 4 digit PIN is selected. What is the probability that there are no repeated digits?

There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 × 10 × 10 × 10 = 10
⁴ = 10000 total possible PINs.

To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 × 9 × 8 × 7, or notice that this is the same as the permutation
₁₀P₄ = 5040.

The probability of no repeated digits is the number of 4 digit PINs with no repeated digits divided by the total number of 4 digit PINs. This probability is

[latex]displaystylefrac{{{}_{{10}}{P}_{{4}}}}{{{10}^{{4}}}}=frac{{5040}}{{10000}}={0.504}[/latex]

Example 2

In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. In this lottery, the order the numbers are drawn in doesn’t matter. Compute the probability that you win the million-dollar prize if you purchase a single lottery ticket.

In order to compute the probability, we need to count the total number of ways six numbers can be drawn, and the number of ways the six numbers on the player’s ticket could match the six numbers drawn from the machine. Since there is no stipulation that the numbers be in any particular order, the number of possible outcomes of the lottery drawing is
₄₈C₆ = 12,271,512. Of these possible outcomes, only one would match all six numbers on the player’s ticket, so the probability of winning the grand prize is:

[latex]displaystylefrac{{{}_{{6}}{C}_{{6}}}}{{{}_{{48}}{C}_{{6}}}}=frac{{1}}{{12271512}}approx={0.0000000815}[/latex]

Example 3

In the state lottery from the previous example, if five of the six numbers drawn match the numbers that a player has chosen, the player wins a second prize of $1,000. Compute the probability that you win the second prize if you purchase a single lottery ticket.

As above, the number of possible outcomes of the lottery drawing is
₄₈C₆ = 12,271,512. In order to win the second prize, five of the six numbers on the ticket must match five of the six winning numbers; in other words, we must have chosen five of the six winning numbers and one of the 42 losing numbers. The number of ways to choose 5 out of the 6 winning numbers is given by ₆C₅ = 6 and the number of ways to choose 1 out of the 42 losing numbers is given by ₄₂C₁ = 42. Thus the number of favorable outcomes is then given by the Basic Counting Rule: ₆C₅ × ₄₂C₁ = 6 × 42 = 252. So the probability of winning the second prize is

[latex]displaystylefrac{{{left({}_{{6}}{C}_{{5}}right)}{left({}_{{42}}{C}_{{1}}right)}}}{{{}_{{48}}{C}_{{6}}}}=frac{{252}}{{12271512}}approx{0.0000205}[/latex]

Try it Now 1

A multiple-choice question on an economics quiz contains 10 questions with five possible answers each. Compute the probability of randomly guessing the answers and getting exactly 9 questions correct.

Poker Hand Combinations Probability Chart

Example 4

Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace.

In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands,
₅₂C₅. This number will go in the denominator of our probability formula, since it is the number of possible outcomes.

For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will be
₄C₁ ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be ₄₈C₄ ways to select the four non-Aces. Now we use the Basic Counting Rule to calculate that there will be ₄C₁ × ₄₈C₄ ways to choose one ace and four non-Aces.

Putting this all together, we have

[latex]displaystyle{P}{left(text{one Ace}right)}=frac{{{left({}_{{4}}{C}_{{1}}right)}{left({}_{{48}}{C}_{{4}}right)}}}{{{}_{{52}}{C}_{{5}}}}=frac{{778320}}{{2598960}}approx{0.299}[/latex]

Example 5

Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces.

The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same:

It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.

Try it Now 2

Compute the probability of randomly drawing five cards from a deck of cards and getting three Aces and two Kings.

Birthday Problem

Let’s take a pause to consider a famous problem in probability theory:

Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?

Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let’s see if we should listen to our intuition. Let’s start with a simpler problem, however.

Example 6

Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people?

There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves “What is the alternative to having at least one shared birthday?” In this case, the alternative is that there are
no shared birthdays. In other words, the alternative to “at least one” is having none. In other words, since this is a complementary event,

P(at least one) = 1 – P(none)

We will start, then, by computing the probability that there is no shared birthday. Let’s imagine that you are one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday? There are 365 days in the year (let’s ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364/365. Now we move to the third person. What is the probability that this third person does not have the same birthday as either you or the second person? There are 363 days that will not duplicate your birthday or the second person’s, so the probability that the third person does not share a birthday with the first two is 363/365.

We want the second person not to share a birthday with you
and the third person not to share a birthday with the first two people, so we use the multiplication rule:

[latex]displaystyle{P}{left(text{no shared birthday}right)}=frac{{365}}{{365}}cdotfrac{{364}}{{365}}cdotfrac{{363}}{{365}}approx{0.9918}[/latex]

and then subtract from 1 to get

P(shared birthday) = 1 – P(no shared birthday) = 1 – 0.9918 = 0.0082.

This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small. Let’s make our group a bit bigger.

Example 7

Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people?

Continuing the pattern of the previous example, the answer should be

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{365}}{{365}}cdotfrac{{364}}{{365}}cdotfrac{{363}}{{365}}cdotfrac{{362}}{{365}}cdotfrac{{361}}{{365}}approx{0.0271}[/latex]

Note that we could rewrite this more compactly as

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{5}}}}{{365}^{{5}}}approx{0.0271}[/latex]

which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.

Example 8

Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people?

Poker Hand Combinations Probability Formula

Here we can calculate

Poker Hand Combinations Probability Table

[latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{30}}}}{{365}^{{30}}}approx{0.706}[/latex]

which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday!

If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn’t studied probability!) You wouldn’t be guaranteed to win, but you should win more than half the time.

This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts. If you still don’t believe the math, you can carry out a simulation. Just so you won’t have to go around rounding up groups of 30 people, someone has kindly developed a Java applet so that you can conduct a computer simulation. Go to this web page:
http://www-stat.stanford.edu/~susan/surprise/Birthday.html, and once the applet has loaded, select 30 birthdays and then keep clicking Start and Reset. If you keep track of the number of times that there is a repeated birthday, you should get a repeated birthday about 7 out of every 10 times you run the simulation.

Try it Now 3

Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?

1. [latex]displaystyle{P}{left({9} text{ answers correct}right)}=frac{9cdot4}{(5^{10})}approx0.0000037[/latex] chance
2. [latex]displaystyle{P}{left(text{three Aces and two Kings}right)}=frac{{{left({}_{{4}}{C}_{{3}}right)}{left({}_{{4}}{C}_{{2}}right)}}}{{{}_{{52}}{C}_{{5}}}}=frac{{24}}{{2598960}}approx{0.0000092}[/latex]
3. [latex]displaystyle{P}{left(text{shared birthday}right)}={1}-frac{{{}_{{365}}{P}_{{10}}}}{{365}^{{10}}}approx{0.117}[/latex]

David Lippman, Math in Society, “Probability,” licensed under a CC BY-SA 3.0 license.