# The probabilities of poker hands

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

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.

$\displaystyle \binom{52}{5}=\frac{52!}{5! \ 47!}=2,598,960$

The notation $\binom{n}{r}$ is called the binomial coefficient and is pronounced “n choose r”, which is identical to the number of $r$-element subsets of a set with $n$ objects. Other notations for $\binom{n}{r}$ are $_nC_r$, $C_{n,r}$ and $C(n,r)$. Many calculators have a function for $_nC_r$. 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

$\displaystyle P(\text{all diamond})=\frac{1287}{2598960}=0.000495198=0.0495198 \%$

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.

$\displaystyle \binom{13}{5} \times \binom{39}{0}=\frac{13!}{5! \ 8!} \times 1=1,287$

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 $\binom{39}{0}=1$ (there is only one way to draw nothing), $\binom{13}{5}=1287$ 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.

\displaystyle \begin{aligned} P(\text{one suit})&=P(\text{all diamond})+P(\text{all heart})+P(\text{all club})+P(\text{all spade}) \\&=\frac{1287}{2598960}+\frac{1287}{2598960}+\frac{1287}{2598960}+\frac{1287}{2598960} \\&=\frac{5148}{2598960} \\&=0.001980792 \\&=0.198 \% \end{aligned}

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:

$\displaystyle \binom{13}{3} \times \binom{13}{2} \times \binom{26}{0}=\frac{13!}{3! \ 11!} \times \frac{13!}{2! \ 13!}=286 \times 78=22308$

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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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 $\binom{4}{3} \times \binom{4}{2}$ = 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

$\displaystyle 13 \times 12 \times \binom{4}{3} \times \binom{4}{2}= 13 \times 12 \times 4 \times 6 =3744$

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 $\binom{13}{2}$ = 78?

Flush
There are $\binom{13}{5}$ = 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 $10 \times 4^5= 10240$. 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.

$\displaystyle \binom{13}{1} \times \binom{4}{3} \times \binom{12}{2} \times \binom{4}{1} \times \binom{4}{1}=13 \times 4 \times 66 \times 4 \times 4 = 54912$

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
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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$\copyright$ 2017 – Dan Ma