# How many hand combinations are there in poker?

**Poker odds**

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**Five Card Draw, Texas Hold'em and Omaha Hold'em**

This website is not intended to encourage people to play poker. Rather, using the example of poker, it should be shown how the probabilities for certain card combinations (categories) are calculated in card games. It turns out that even calculating the probability of reaching a certain category straight away can be very complicated. That is why 5 of the 22 calculations listed below were carried out exclusively with a computer program.

In poker, mostly n = 52 cards with k = 4 different colors are used. In the now rare Five Card Draw, each player's hand is formed directly from 5 cards. In Texas Hold'em, on the other hand, a player can put together a hand of 5 cards from 7 cards (5 open and 2 hidden). And in Omaha Hold'em a player can even put together a hand of 5 cards from 9 cards (5 face-up and 4 face-down), whereby exactly 2 of these must come from the 4 face-down cards. In order to calculate the probabilities W to achieve the listed categories right away, it makes sense to first determine the most favorable possibilities (combinations) and then divide them by the total number of combinations. The total number of combinations is calculated as in the lottery 6 out of 49. Only here, not 6 out of 49 balls, but 5, 7 or 9 out of 52 cards are drawn. The five card draw therefore results in a total of (^{52}_{5}) = 2,598,960 combinations, in Texas Hold'em it is (^{52}_{7}) = 133,784,560 combinations and in Omaha Hold'em (^{52}_{9}) · (^{9}_{4}) = 3,679,075,400 · 126 = 463,563,500,400 combinations due to the additional restriction for the hidden cards. The following probabilities then result from this: **Royal flush** (Straight in a suit with Ace as the highest card)

Probability (Five Card Draw):

W = k / (^{n}_{5})

= 4 / (^{52}_{5})

= 4 / 2.598.960 = 1 : 649.740 = 0,000154%

There is a royal flush for each of the 4 colors. So there are 4 favorable combinations for a royal flush.

Probability (Texas Hold'em):

W = k (^{n - 5}_{2}) / (^{n}_{7})

= 4 · (^{47}_{2}) / (^{52}_{7})

= 4324 / 133.784.560 = 1 : 30.940 = 0,003232%

There are 4 royal flushes. For this you need 5 cards each. For the 7 - 5 = 2 remaining cards, 52 - 5 = 47 cards are possible. Therefore there is

(^{47}_{2}) = 1081 possibilities. In total there are 4 · 1081 = 4324 favorable combinations of a total of (^{52}_{7}) Combinations.

Probability (Omaha Hold'em):

W = k (^{n - 5}_{2}) · (^{5}_{2}) · (^{4}_{2}) / ((^{n}_{9}) · (^{9}_{4}))

= 4 · (^{47}_{4}) · (^{5}_{2}) · (^{4}_{2}) / ((^{52}_{9}) · (^{9}_{4}))

= 42.807.600 / 463.563.500.400 = 1 : 10.829 = 0,009234%

There are 4 royal flushes. There are 47 cards left for the 4 remaining cards. Therefore there is (^{47}_{4}) = 178,365 possibilities.

Additionally there are (^{5}_{2}) · (^{4}_{2}) = 60 combinations that among the 4 face down cards there are 2 royal flush cards and 2 other cards.

In total there are 4 178,365 60 = 42,807,600 favorable combinations.**Straight flush** (Street in one color)

Probability (Five Card Draw):

W = (n / k - 4) k / (^{n}_{5})

= 9 · 4 / (^{52}_{5})

= 36 / 2.598.960 = 1 : 72.193,33 = 0,001385%

There are 10 streets for each of the 4 colors. Because of the 4 royal flushes there are 9 straight flushes left for each of the 4 suits.

This results in 4 · 9 = 36 favorable combinations.

Probability (Texas Hold'em):

W = (n / k - 4) k (^{n - 6}_{2}) / (^{n}_{7})

= 9 · 4 · (^{46}_{2}) / (^{52}_{7})

= 37.260 / 133.784.560 = 1 : 3590,568 = 0,027851%

There are 9 straight flushes for each of the 4 suits. For the other two cards there are 47 cards left.

The card of the same suit as the straight flush, the value of which is exactly 1 higher than the highest value of any of the straight flush cards,

is not possible, because otherwise a higher straight flush or even a royal flush would result. So there are 46 cards to choose from.

Therefore, for the two remaining cards (^{46}_{2}) = 1035 possibilities.

In total there are 9 · 4 · 1035 = 37,200 favorable combinations here.

Probability (Omaha Hold'em):

W = 368,486,160 / 463,563,500,400 = 1: 1258.02 = 0.079490%

Calculation with the help of a computer program**Four of a Kind** (Four cards of the same value)

Probability (Five Card Draw):

W = n / k (^{k}_{4}) · (^{n / k - 1}_{1}) K / (^{n}_{5})

= 13 · (^{4}_{4}) · (^{12}_{1}) · 4 / (^{52}_{5})

= 624 / 2.598.960 = 1 : 4165 = 0,024010%

There are 13 different quadruplets. For the fifth card there are still 12 values left,

each of which can have one of the 4 colors. In total there are 13 · 12 · 4 = 624 favorable combinations here.

Probability (Texas Hold'em):

W = n / k (^{4}_{4}) · (^{n - 4}_{3}) / (^{n}_{7})

= 13 · (^{4}_{4}) · (^{48}_{3}) / (^{52}_{7})

= 224.848 / 133.784.560 = 1 : 595 = 0,168067%

There are 13 different quadruplets. For this you need 4 cards each. For the 7 - 4 = 3 remaining cards, 52 - 4 = 48 cards are possible.

Therefore there is (^{48}_{3}) = 17,296 possibilities. In total, there are 1317,296 = 224,848 favorable combinations here.**Full house** (A triplet and a twin)

Probability (Five Card Draw):

W = n / k (^{k}_{3}) · (N / k - 1) · (^{k}_{2}) / (^{n}_{5})

= 13 · (^{4}_{3}) · 12 · (^{4}_{2}) / (^{52}_{5})

= 3744 / 2.598.960 = 1 : 694,167 = 0,144058%

The triplets can have 13 different values. There are 4 different triplets for each value.

There are then 12 different values left for the twins. For each value there is then (^{4}_{2}) = 6 different twins.

In total there are 13 · 4 · 12 · 6 = 3744 favorable combinations.

Probability (Texas Hold'em):

W = (n / k (^{k}_{3}) · (N / k - 1) · (^{k}_{2}) · (^{n / k - 2}_{2}) · K^{2} + n / k (^{k}_{3}) · (^{n / k - 1}_{2}) · (^{k}_{2}) · (^{k}_{2}) + (^{n / k}_{2}) · (^{k}_{3}) · (^{k}_{3}) · (N / k - 2) · k) / (^{n}_{7})

= (13 · (^{4}_{3}) · 12 · (^{4}_{2}) · (^{11}_{2}) · 4^{2} + 13 · (^{4}_{3}) · (^{12}_{2}) · (^{4}_{2}) · (^{4}_{2}) + (^{13}_{2}) · (^{4}_{3}) · (^{4}_{3}) · 11 · 4) / (^{52}_{7})

= 3.473.184 / 133.784.560 = 1 : 38,5193 = 2,596102%

There are 13 values for the triplets and (^{4}_{3}) = 4 possible color combinations. Then 12 values remain for the twin

(^{4}_{2}) = 6 possible color combinations. For the other two cards there are still (^{11}_{2}) = 55 combinations of values

with 4^{2} = 16 color variations. In addition to a triplet, there can also be two twins.

There is then (^{12}_{2}) = 66 combinations of values. And for every twin are (^{4}_{2}) = 6 color combinations possible.

Finally, two triplets are also possible. Therefore there is (^{13}_{2}) = 78 value combinations.

And 4 color combinations are possible for each treble. For the seventh card there are 11 values with 4 colors each. That’s a total of

13 4 12 6 55 16 + 13 4 66 6 6 + 78 4 4 11 4 = 3,294,720 + 123,552 + 54,912 = 3,473,184 favorable combinations. **Flush** (Five cards of one suit)

Probability (Five Card Draw):

W = (^{n / k}_{5}) K / (^{n}_{5}) - W (Straight Flush) - W (Royal Flush)

= ((^{13}_{5}) · 4 – 36 – 4) / (^{52}_{5})

= 5108 / 2.598.960 = 1 : 508,802 = 0,196540%

for each of the 4 colors there is (^{13}_{5}) = 1287 flushes. Because of the 36 straight flushes and 4 royal flushes there

Therefore there are 4 · 1287 - 40 = 5108 favorable combinations.

Probability (Texas Hold'em):

W = ((^{n / k}_{5}) K (^{n - 13}_{2}) + (^{n / k}_{6}) K (n - 13) + (^{n / k}_{7}) K) / (^{n}_{7}) - W (Straight Flush) - W (Royal Flush)

= ((^{13}_{5}) · 4 · (^{39}_{2}) + (^{13}_{6}) · 4 · 39 + (^{13}_{7}) · 4 – 36 · (^{46}_{2}) – 4 · (^{47}_{2})) / (^{52}_{7})

= 4.047.644 / 133.784.560 = 1 : 33,0525 = 3,025494%

for each of the 4 colors there is (^{13}_{5}) = 1287 possibilities for 5 cards of the same suit, (^{13}_{6}) = 1716 possibilities

for 6 cards of the same suit and (^{13}_{7}) = 1716 possibilities for 7 cards of the same suit. For the remaining two, one and zero cards

there are 39 cards left to choose from and there are (^{39}_{2}) = 741, (^{39}_{1}) = 39 and (^{39}_{0}) = 1 possibility.

Together this is initially 4 (1287 741 + 1716 39 + 1716 1) = 4,089,228 combinations.

If you consider the 4324 favorable combinations for a royal flush and the 37,260 favorable combinations

subtracts for a straight flush, this leaves a total of 4,047,644 favorable combinations. **Straight** (Five cards in a row)

Probability (Five Card Draw):

W = (n / k - 3) * k^{5} / (^{n}_{5}) - W (Straight Flush) - W (Royal Flush)

= (10 · 4^{5} – 36 – 4) / (^{52}_{5})

= 10.200 / 2.598.960 = 1 : 254,8 = 0,392465%

There are a total of 10 different streets in 4^{5} = 1024 color variations. Then when one of the

10 · 1024 = 10,240 possibilities the 36 straight flushes and the 4 royal flushes subtract, 10,200 favorable combinations remain.

Probability (Texas Hold'em):

W = 6,180,020 / 133,784,560 = 1: 21.6479 = 4.619382%

Calculation with the help of a computer program**Three of a Kind** (Three cards of the same value)

Probability (Five Card Draw):

W = n / k (^{k}_{3}) · (^{n / k - 1}_{2}) · K^{2} / (^{n}_{5})

= 13 · (^{4}_{3}) · (^{12}_{2}) · 4^{2} / (^{52}_{5})

= 54.912 / 2.598.960 = 1 : 47,3295 = 2,112845%

The triplets can have 13 different values. There are triplets in 4 different colors for each value.

There are then 12 different values left for the other two cards. Therefore there is (^{12}_{2}) = 66 combinations of values.

And for every combination of values there are 4^{2} = 16 color variations.

In total there are 13 · 4 · 66 · 16 = 54,912 favorable combinations.

Probability (Texas Hold'em):

W = 6,461,620 / 133,784,560 = 1: 20.7045 = 4.829870%

Calculation with the help of a computer program**Two Pairs** (Two twins)

Probability (Five Card Draw):

W = (^{n / k}_{2}) · (^{k}_{2}) · (^{k}_{2}) · (^{n / 4 - 2}_{1}) · 4 / (^{n}_{5})

= (^{13}_{2}) · (^{4}_{2}) · (^{4}_{2}) · (^{11}_{1}) · 4 / (^{52}_{5})

= 123.552 / 2.598.960 = 1 : 21,0354 = 4,753902%

The two twins can each have one of 13 different values. Therefore there is

(^{13}_{2}) = 78 possibilities. For each of the two values there are twins in (^{4}_{2}) = 6 different color combinations.

For the fifth card there are still 11 values left, each of which can have one of the 4 colors.

In total there are 78 6 6 11 4 = 123,552 favorable combinations.

Probability (Texas Hold'em):

W = 31,433,400 / 133,784,560 = 1: 4.25613 = 23.495536%

Calculation with the help of a computer program**One pair** (One twin, two cards of the same value)

Probability (Five Card Draw):

W = n / k (^{k}_{2}) · (^{n / k - 1}_{3}) · 4^{3} / (^{n}_{5})

= 13 · (^{4}_{2}) · (^{12}_{3}) · 4^{3} / (^{52}_{5})

= 1.098.240 / 2.598.960 = 1 : 2,36648 = 42,256903%

The twin can have one of 13 different values. For every value there are twins in

(^{4}_{2}) = 6 different color combinations. There are then 12 values left for the three remaining cards.

This results in (^{12}_{3}) = 220 value combinations, each of which is 4^{3} = Can have 64 color variations.

In total there are 13 6 220 64 = 1,098,240 favorable combinations.

Probability (Texas Hold'em):

W = 58,627,800 / 133,784,560 = 1: 2.28193 = 43.822546%

Calculation with the help of a computer program**Highest card** (None of the previous categories)

Probability (Five Card Draw):

W = ((^{n / k}_{5}) - (n / k - 3)) · (k^{5} - k) / (^{n}_{5})

=((^{13}_{5}) – 10) · (4^{5} – 4) / (^{52}_{5})

= 1.302.540 / 2.598.960 = 1 : 2,36648 = 50,117739%

There are (^{13}_{5}) = 1287 value combinations to choose from 13 different values 5.

Among them are 10 streets that do not count here. There are 1277 combinations of values left.

There are 4 for each of the remaining 1277 value combinations^{5} = 1024 color variations.

These include 4 variations in which all 5 colors are the same. Since these do not count here either, there are 1020 color variations left.

The product of 1020 and 1277 is then 1,302,540 and indicates the number of favorable combinations.

Probability (Texas Hold'em):

W = ((^{n / k}_{7}) – (6 · (^{4}_{0}) + 2 · (^{5}_{0}) + 7 · (^{5}_{1}) + 2 · (^{6}_{1}) + 8 · (^{6}_{2}) + 2 · (^{7}_{2}))) · (K^{7} - k ((^{7}_{5}) · 3^{2} + (^{7}_{6}) 3^{1} + (^{7}_{7}) · 3^{0})) / (^{n}_{7})

= ((^{13}_{7}) – (6 · (^{4}_{0}) + 2 · (^{5}_{0}) + 7 · (^{5}_{1}) + 2 · (^{6}_{1}) + 8 · (^{6}_{2}) + 2 · (^{7}_{2}))) · (4^{7} – 4 · ((^{7}_{5}) · 3^{2} + (^{7}_{6}) 3^{1} + (^{7}_{7}) · 3^{0})) / (^{52}_{7})

= 23.294.460 / 133.784.560 = 1 : 5,74319 = 17,411920%

The calculation strategy is to subtract from the total number of combinations those that belong to the remaining 9 categories.

Since no value may appear more than once in the case of "High Card", it is initially assumed that there are 13 cards with 13 different values.

This already excludes the 5 categories "Four of a Kind", "Full House", "Three of a Kind", "Two Pairs" and "One Pair".

There is then (^{13}_{7}) = 1716 combinations of values, choose 7 cards from the 13 cards.

You now have to subtract the combinations from which you can form any straight ("Royal Flush", "Straight Flush" or "Straight").

Underneath there are 8 streets (consisting of 5 cards), where the two remaining cards (^{6}_{2}) = 15 possibilities.

Below that there are 2 streets (consisting of 5 cards), where the two remaining cards (^{7}_{2}) = 21 possibilities.

Among them there are 7 streets (consisting of 6 maps), where there are 5 possibilities for the rest of the map.

Below that there are 2 streets (consisting of 6 cards) with 6 options for the rest of the card.

Below are 8 streets (consisting of 7 maps).

For example, with the 7 streets, which consist of 6 cards (e.g. "6 7 8 9 10 Jack"), only the 5 options 2, 3, 4, king or ace remain for the 7th card.

With the 2 streets, which consist of 6 cards (namely "2 3 4 5 6 7" and "9 10 Jack Queen King Ace") there are 6 possibilities for the 7th card,

namely 9, 10, Jack, Queen, King or Ace or 2, 3, 4, 5, 6 or 7.

In total you have to subtract 8 · 15 + 2 · 21 + 7 · 5 + 2 · 6 + 8 = 217 combinations with which roads can be built.

There are 4 for each of the remaining 1499 value combinations^{7} = 16,384 color variations, because each of the 7 cards can appear in 4 colors.

From these color variations you now have to subtract the variations that can form a "flush".

If 5 colors are the same there is (^{7}_{5}) = 21 combinations to assign this color to the 7 cards. For the other two cards, 3 remain^{2} = 9 color variations.

If 6 colors are the same there is (^{7}_{6}) = 7 combinations to assign this color to the 7 cards. There are 3 colors left for the rest of the card.

If all 7 colors are the same, there is only one combination.

The same considerations apply to each of the 4 colors. Together this results in (21 9 + 7 3 + 1) 4 = 844 color variations,

with which you can form a "flush" and which must therefore be subtracted from the 16,384 color variations.

There remain 15,540 color variations for each of the 1499 value combinations. So there are a total of 15,5401499 = 23,294,460 favorable combinations.

Copyright © Werner Brefeld, 2009 ... 2021

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