Poker Hand Combinations Probability

Brian Alspach

The answer is 52.51, or 2,652. Carry this out to three-card poker: 52.51.50=132,600. With four cards, you could see 6,497,400 potential hands. Finally, we get to five-card poker.

18 January 2000

1. Remember, don't get confused with the method used in applying combinatorics in the overall game of poker. What is being worked out here is the probability of starting hands being dealt to you, nothing else. As far as applicability to real poker, and the methods used, there are 16 combinations of AK off-suit, but only four combinations of AK suited.
2. 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 be confident the numbers are correct. Here is a table summarizing the number of 7-card poker hands. The probability is the probability of having the hand dealt to you when dealt 7 cards.
3. In 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.

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.

The main underpinning of poker is math – it is essential. For every decision you make, while factors such as psychology have a part to play, math is the key element.

In this lesson we’re going to give an overview of probability and how it relates to poker. This will include the probability of being dealt certain hands and how often they’re likely to win. We’ll also cover how to calculating your odds and outs, in addition to introducing you to the concept of pot odds. And finally we’ll take a look at how an understanding of the math will help you to remain emotional stable at the poker table and why you should focus on decisions, not results.

What is Probability?

Probability is the branch of mathematics that deals with the likelihood that one outcome or another will occur. For instance, a coin flip has two possible outcomes: heads or tails. The probability that a flipped coin will land heads is 50% (one outcome out of the two); the same goes for tails.

Probability and Cards

When dealing with a deck of cards the number of possible outcomes is clearly much greater than the coin example. Each poker deck has fifty-two cards, each designated by one of four suits (clubs, diamonds, hearts and spades) and one of thirteen ranks (the numbers two through ten, Jack, Queen, King, and Ace). Therefore, the odds of getting any Ace as your first card are 1 in 13 (7.7%), while the odds of getting any spade as your first card are 1 in 4 (25%).

Unlike coins, cards are said to have “memory”: every card dealt changes the makeup of the deck. For example, if you receive an Ace as your first card, only three other Aces are left among the remaining fifty-one cards. Therefore, the odds of receiving another Ace are 3 in 51 (5.9%), much less than the odds were before you received the first Ace.

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Pre-flop Probabilities: Pocket Pairs

In order to find the odds of getting dealt a pair of Aces, we multiply the probabilities of receiving each card:

(4/52) x (3/51) = (12/2652) = (1/221) ≈ 0.45%.

To put this in perspective, if you’re playing poker at your local casino and are dealt 30 hands per hour, you can expect to receive pocket Aces an average of once every 7.5 hours.

The odds of receiving any of the thirteen possible pocket pairs (twos up to Aces) is:

(13/221) = (1/17) ≈ 5.9%.

In contrast, you can expect to receive any pocket pair once every 35 minutes on average.

Pre-Flop Probabilities: Hand vs. Hand

Players don’t play poker in a vacuum; each player’s hand must measure up against his opponent’s, especially if a player goes all-in before the flop.

Here are some sample probabilities for most pre-flop situations:

Post-Flop Probabilities: Improving Your Hand

Now let’s look at the chances of certain events occurring when playing certain starting hands. The following table lists some interesting and valuable hold’em math:

Many beginners to poker overvalue certain starting hands, such as suited cards. As you can see, suited cards don’t make flushes very often. Likewise, pairs only make a set on the flop 12% of the time, which is why small pairs are not always profitable.

PDF Chart

We have created a poker math and probability PDF chart (link opens in a new window) which lists a variety of probabilities and odds for many of the common events in Texas hold ‘em. This chart includes the two tables above in addition to various starting hand probabilities and common pre-flop match-ups. You’ll need to have Adobe Acrobat installed to be able to view the chart, but this is freely installed on most computers by default. We recommend you print the chart and use it as a source of reference.

Odds and Outs

If you do see a flop, you will also need to know what the odds are of either you or your opponent improving a hand. In poker terminology, an “out” is any card that will improve a player’s hand after the flop.

One common occurrence is when a player holds two suited cards and two cards of the same suit appear on the flop. The player has four cards to a flush and needs one of the remaining nine cards of that suit to complete the hand. In the case of a “four-flush”, the player has nine “outs” to make his flush.

A useful shortcut to calculating the odds of completing a hand from a number of outs is the “rule of four and two”. The player counts the number of cards that will improve his hand, and then multiplies that number by four to calculate his probability of catching that card on either the turn or the river. If the player misses his draw on the turn, he multiplies his outs by two to find his probability of filling his hand on the river.

In the example of the four-flush, the player’s probability of filling the flush is approximately 36% after the flop (9 outs x 4) and 18% after the turn (9 outs x 2).

Pot Odds

Another important concept in calculating odds and probabilities is pot odds. Pot odds are the proportion of the next bet in relation to the size of the pot.

For instance, if the pot is $90 and the player must call a $10 bet to continue playing the hand, he is getting 9 to 1 (90 to 10) pot odds. If he calls, the new pot is now $100 and his $10 call makes up 10% of the new pot.

Experienced players compare the pot odds to the odds of improving their hand. If the pot odds are higher than the odds of improving the hand, the expert player will call the bet; if not, the player will fold. This calculation ties into the concept of expected value, which we will explore in a later lesson.

Poker Hand Combinations Probability Calculator

Bad Beats

A “bad beat” happens when a player completes a hand that started out with a very low probability of success. Experts in probability understand the idea that, just because an event is highly unlikely, the low likelihood does not make it completely impossible.

A measure of a player’s experience and maturity is how he handles bad beats. In fact, many experienced poker players subscribe to the idea that bad beats are the reason that many inferior players stay in the game. Bad poker players often mistake their good fortune for skill and continue to make the same mistakes, which the more capable players use against them.

Decisions, Not Results

One of the most important reasons that novice players should understand how probability functions at the poker table is so that they can make the best decisions during a hand. 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.
The good news is that there is a simple system, with powerful shortcuts & rules, that you can begin using this week. Rooted in GTO, but simplified so that you can implement it at the tables, The One Percent gives you the ultimate gameplan.

Poker Hand Combinations Probability Formula

This 7+ hour course gives you applicable rules for continuation betting, barreling, raising, and easy ratios so that you ALWAYS have the right number of bluffing combos. Take the guesswork out of your strategy, and begin playing like the top-1%.

Conclusion

A strong knowledge of poker math and probabilities will help you adjust your strategies and tactics during the game, as well as giving you reasonable expectations of potential outcomes and the emotional stability to keep playing intelligent, aggressive poker.

Remember that the foundation upon which to build an imposing knowledge of hold’em starts and ends with the math. I’ll end this lesson by simply saying…. the math is essential.

By Gerald Hanks

Gerald Hanks is from Houston Texas, and has been playing poker since 2002. He has played cash games and no-limit hold’em tournaments at live venues all over the United States.

Poker Hand Combinations Probability Chart

Poker Hand Combinations Probability Table