# How to Compute Probability for Poker Hands

In this lesson, we explain how to compute probabilities for every type of poker hand (e.g., full house, flush, staight) in stud poker.

## Stud Poker in a Nutshell

Five-card stud poker is a classic game of chance, in which each player is dealt five cards. The first card is dealt face down; the remaining cards, face up. And the player with the best five-card hand wins.

The nine types of poker hand are listed below, from strongest to weakest. Stronger hands beat weaker hands. Thus, a straight flush beats four of a kind; four of a kind beats a full house; and so on.

• Straight flush. Five cards of the same suit in sequence, such as 3, 4, 5, 6, 7.
• Four of a kind. Four cards of the same rank, such as 9♠, 9, 9, 9♣, 2
• Full house. Three cards of one rank and two of another rank, such as 9♠, 9, 9, 2♣, 2.
• Flush. Five cards of the same suit with at least one card out of sequence, such as 3, 4, 5, 6, K.
• Straight. Five cards in sequence, with at least two cards of different suits. Ace can be high or low, but not both. Thus, A♠, 2, 3, 4♣, 5 and 10♠, J, Q, K♣, A are valid straights; but Q♠, K, A, 2♣, 3 is not.
• Three of a kind. Three cards of the same rank and two cards of different ranks, such as 9♠, 9, 9, 4♣, Q.
• Two pair. Two cards of equal rank, two other cards of equal but different rank, and another card of different rank, such as A♠, A, 5, 5♣, 7.
• One pair. Two cards of equal rank and three cards of different rank, such as 8♠, 8, 2, J♣, K.
• High card. Five cards of different rank with at least two different suits, such as 3, 5, Q,K♣, A.

## Poker as a Statistical Experiment

We want to compute the probabilities associated with each type of poker hand. That is, we want to compute the probability of being dealt a straight flush, four of a kind, a full house, etc. Where to begin?

First, recognize that the game of five-card stud is actually a statistical experiment. It is a statistical experiment because:

• Every deal can have more than one possible outcome.
• Every possible outcome can be specified in advance.
• The outcome of the experiment (i.e., the particular hand dealt to a player) depends on chance.

In this statistical experiment, the sample space is the set of all possible five-card hands that can be dealt from a standard deck of 52 cards. Each of the possible five-card hands is a sample point in the sample space. And each type of poker hand (i.e., straight flush, full house, etc.) is an event in the sample space.

One more thing: Every sample point is equally likely to occur. This means that every possible five-card hand is equally likely to occur.

## How to Calculate Poker Probability

Because stud poker is a statistical experiment in which every sample point is equally likely to occur, we can compute the probability of any type of poker hand in three steps:

• Count the number of five-card combinations that can be classified as each type of hand. For example, count the number of five-card combinations that can be classified as a straight flush. Count the number that can be classified as four of a kind. Count the number that can be classifed as a full house. And so on.
• Count the number of possible five-card hands that can be dealt from a standard deck of 52 cards
• The probability of being dealt any particular type of hand is equal to the number of five-card combinations of that kind divided by the total number of possible five-card hands.

In short, the formula for computing the probability that a particular type of poker hand will be dealt is:

 Prob = Number of hands of a particular type Number of possible hands

An example will make this clear. Suppose there are X five-card combinations that can be classified as a straight flush. And suppose the total number of possible five-card hands is Y. Then, the probability of being dealt a straight flush is X divided by Y.

## How to Count Poker Hands

So, to compute probabilities, we need to be able to count poker hands. How can we do that?

First, recognize that every poker hand consists of five cards. The order in which cards are arranged does not matter. For example, the following two sets of cards represent the same poker hand:

9♠, 9, 9, 2♣, 2

2♣, 2, 9♠, 9, 9

When you talk about all the possible ways to count a set of objects without regard to order, you are talking about counting combinations. Luckily, we have a formula to do that:

Counting combinations. The number of combinations of n objects taken r at a time is

nCr = n(n - 1)(n - 2) . . . (n - r + 1)/r! = n! / r!(n - r)!

Each type of poker hand (straight flush, four of a kind, etc.) is defined by a particular arrangement of card ranks and card suits. It is possible to create any type of poker hand through a concise series of independent choices. The number of ways to produce any type of poker hand is equal to the product of the number of ways to make each independent choice.

For example, consider a full house. A full house has three cards of one rank, and two cards of another rank. Each of the three matching cards is a different suit. And each of the two matching cards is a different suit. It requires the following four independent choices to produce a full house:

• Choose the rank of cards in the hand. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace. For a full house, we choose 2 ranks from a set of 13 ranks. The number of ways to do this is 13C2.
• Choose the rank of the three-card set. From the two ranks chosen in Step 1, we choose one rank for the three-card set. The number of ways to do this is 2C1.
• Choose a suit for each card in the three-card set. There are four suits, from which we choose three. The number of ways to do this is 4C3.
• Choose a suit for each card in the two-card set. There are four suits, from which we choose two. The number of ways to do this is 4C2.

The number of ways to produce a full house (Numfull) is equal to the product of the number of ways to make each independent choice. Therefore,

Numfull = 13C2 * 2C1 * 4C3 * 4C2

Numfull = 78 * 2 * 4 * 6 = 3744

Conclusion: There are 3744 different poker hands that fall in the category of full house.

## Warning

The use of combinations to count poker hands is not hard, but it can be tricky. The most common error is overcounting the number of ways you can make a choice.

For example, suppose you want count the number of ways to choose 2 different ranks from a deck of 13 ranks, without regard to order. This is a single-choice combinations problem. The correct way to count the number of options (N) available for a single choice is through a single computation:

N = 13C2 = 78

Some folks try a different approach. First, they choose 1 rank from a deck of 13 ranks. There are 13C1 = 13 ways to make this choice. Next, they choose 1 rank from the remaining 12 ranks. There are 12C1 = 12 ways to make this choice. You might think the number of ways to make both choices would be:

N = 13C1 * 12C1 = 13 * 12 = 156

But this approach is incorrect. The approach fails, because the two choices are not independent. You know the choices are not independent, because the set of ranks available for the second choice depends on the rank chosen in the first choice. Because the choices are not independent, the product of 13C1 and 12C1 counts some card combinations twice. The result is an overcount.

## Test Your Understanding

In this section, three problems illustrate how to count poker hands and how to compute the probability of a poker hand. Specifically, we will compute the probability of being dealt a royal flush.

Problem 1

Five-card stud is a poker game, in which a player is dealt 5 cards from an ordinary deck of 52 playing cards. How many distinct poker hands could be dealt? (Hint: In this problem, the order in which cards are dealt is NOT important; For example, if you are dealt the ace, king, queen, jack, ten of spades, that is the same as being dealt the ten, jack, queen, king, ace of spades.)

Solution: For this problem, it would be impractical to list all of the possible poker hands. However, the number of possible poker hands can be easily calculated using combinations.

The number of combinations is n! / r!(n - r)!. We have 52 cards in the deck so n = 52. And we want to arrange them in unordered groups of 5, so r = 5. Thus, the number of combinations is:

52C5 = 52! / 5!(52 - 5)! = 52! / 5!47! = 2,598,960

Hence, there are 2,598,960 distinct poker hands.

Problem 2

In poker, a royal flush is a straight flush made up of the 10, jack, queen, king, and ace. How many distinct poker hands can be classified as a royal flush?

Solution: The easiest way to solve this problem is to simply list all of possible royal-flush hands; then, count the list. Here's the list:

10, J, Q, K, A

10, J, Q, K, A

10, J, Q, K, A

10, J, Q, K, A

So, there are four poker hands that can be described as a royal flush.

Problem 3

On any single deal, what is the probability that a player will be dealt a royal flush?

Solution: The formula for computing the probability that a particular type of poker hand will be dealt is:

 Prob = Number of hands of a particular type Number of possible hands

From Problem 2, we know that there are four hands that qualify as a royal flush. And, from Problem 1, we know that there are 2,598,960 distinct poker hands. Therefore, the probability of dealing a royal flush is:

Prob = 4 / 2,598,960 = 0.000001539

The probability of being dealt a royal flush is 0.000001539. On average, a royal flush is dealt one time in every 649,740 deals.