### Probability

#### Probability Basics

#### Probability Problems

#### Poker Probability

- Probability in stud poker
- Probability of straight
- Probability of flush
- Cards of equal rank
- Probability of no pair

#### Random Variables

#### Discrete Distributions

#### Continuous Distributions

### Probability: Table of Contents

#### Probability Basics

#### Probability Problems

#### Poker Probability

- Probability in stud poker
- Probability of straight
- Probability of flush
- Cards of equal rank
- Probability of no pair

#### Random Variables

#### Discrete Distributions

#### Continuous Distributions

# How to Compute the Probability of Equal-Rank Cards in Stud Poker

In this lesson, we explain how to compute the probability of being dealt two or more cards of equal rank in stud poker. (For a brief description of stud poker, click here.)

## Hands With Equal-Rank Cards

In stud poker, there are five types of hands that include two or more cards of equal rank.

**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♥.**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♥.

In this lesson, we will compute probabilities for each of these hands.

## How to Compute Poker Probabilities

In a previous lesson, we explained how to compute probability for any type of poker hand. For convenience, here is a brief review:

- Count the number of possible five-card hands that can be dealt from a standard deck of 52 cards
- Count the number of ways that a particular type of poker hand can occur
- The probability of being dealt any particular type of hand is equal to the number of ways it can occur divided by the total number of possible five-card hands.

So, how do we count the number of ways that different types of poker hands can occur? We recognize that every poker hand consists of five cards, and the order in which cards are arranged does not matter. 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

_{n}C_{r} = n(n - 1)(n
- 2) . . . (n - r + 1)/r! = n! / r!(n - r)!

In summary, we use the combination formula to count (a) the number of possible five-card hands and (b) the number of ways a particular type of hand can be dealt. To find probability, we divide the latter by the former.

## Probability of Four of a Kind

Let's execute the analytical plan described above to find the probability of four of a kind.

- First, we count the number of five-card hands that can be dealt from a standard deck of 52 cards. This is a combination problem.
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:

Hence, there are 2,598,960 distinct poker hands._{52}C_{5}= 52! / 5!(52 - 5)! = 52! / 5!47! = 2,598,960 - Next, we count the number of ways that five cards can be dealt to produce four of a kind. It requires three independent choices to produce four of a kind:
- Choose the rank of the card that appears four times 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 four of a kind, we choose 1 rank from a set of 13 ranks. The number of ways to do this is
_{13}C_{1}. - Choose one rank for the fifth card. There are 12 remaining ranks, from which we choose one.
The number of ways to do this is
_{12}C_{1}. - Choose a suit for the fifth card. There are four suits, from which we choose one.
The number of ways to do this is
_{4}C_{1}.

The number of ways to produce four of a kind (Num

_{4}) is equal to the product of the number of ways to make each independent choice. Therefore,Num

Conclusion: There are 624 different ways to deal a poker hand that can be classified as four of a kind._{4}=_{13}C_{1}*_{12}C_{1}*_{4}C_{1}= 13 * 12 * 4 = 624 - Choose the rank of the card that appears four times 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 four of a kind, we choose 1 rank from a set of 13 ranks. The number of ways to do this is
- Finally, we compute the probability. There are 2,598,960 unique poker hands. Of those, 624 are four of a kind. Therefore, the probability
of being dealt four of a kind (P
_{4}) is:P

_{4}= 624 / 2,598,960 = 0.0002400960384

The probability of being dealt four of a kind is 0.0002400960384. On average, four of a kind is dealt one time in every 4,165 deals.

## Probability of a Full House

We follow a similar process to find the probability of a full house.

- First, count the number of five-card hands that can be dealt from a standard deck of 52 cards. We did this in the previous section, and found that there are 2,598,960 distinct poker hands.
- Next, count the number of ways that five cards can be dealt to produce a full house. It requires 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
_{13}C_{2}. - Choose one rank for the three-card combination. There are 2 ranks in a full house, from which we choose one.
The number of ways to do this is
_{2}C_{1}. - Choose suits for the three-card combination. There are four suits, from which we choose three.
The number of ways to do this is
_{4}C_{3}. - Choose suits for the two-card combination. There are four suits, from which we choose two.
The number of ways to do this is
_{4}C_{2}.

The number of ways to produce full house (Num

_{fh}) is equal to the product of the number of ways to make each independent choice. Therefore,Num

_{fh}=_{13}C_{2}*_{2}C_{1}*_{4}C_{3}*_{4}C_{2}Num

_{fh}= 78 * 2 * 4 * 6 = 3,744 - 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
- Finally, compute the probability. There are 2,598,960 unique poker hands. Of those, 3,744 are examples of a full house. Therefore, the probability
of being dealt a full house (P
_{fh}) is:P

_{fh}= 3,744 / 2,598,960 = 0.00144057623

Based on these results, we can project that a full house will be dealt, on average, approximately one time in every 694 deals.

## Probability of Three of a Kind

We use the same general approach to find the probability of three of a kind.

- First, count the number of five-card hands that can be dealt from a standard deck of 52 cards. We did this previously, and found that there are 2,598,960 distinct poker hands.
- Next, count the number of ways that five cards can be dealt to produce three of a kind. It requires five independent choices to produce three of a kind:
- Choose the rank for cards of matching rank. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace.
For three of a kind, we choose 1 rank from a set of 13 ranks. The number of ways to do this is
_{13}C_{1}. - Choose the rank of non-matching cards. There are 12 remaining ranks, from which we choose two.
The number of ways to do this is
_{12}C_{2}. - Choose suits for the three-card combination. There are four suits, from which we choose three.
The number of ways to do this is
_{4}C_{3}. - Choose a suit for one of the non-matching cards. There are four suits, from which we choose one.
The number of ways to do this is
_{4}C_{1}. - Choose a suit for the other non-matching card. There are four suits, from which we choose one.
The number of ways to do this is
_{4}C_{1}.

The number of ways to produce three of a kind (Num

_{3}) is equal to the product of the number of ways to make each independent choice. Therefore,Num

_{3}=_{13}C_{1}*_{12}C_{2}*_{4}C_{3}*_{4}C_{1}*_{4}C_{1}Num

_{3}= 13 * 66 * 4 * 4 * 4 = 54,912 - Choose the rank for cards of matching rank. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace.
For three of a kind, we choose 1 rank from a set of 13 ranks. The number of ways to do this is
- Finally, compute the probability. There are 2,598,960 unique poker hands. Of those, 54,912 are three of a kind. Therefore, the probability
of being dealt three of a kind (P
_{3}) is:P

_{3}= 54,912 / 2,598,960 = 0.021128451138

In stud poker, players get three of a kind about one time in every 47 deals.

## Probability of Two Pair

To find the probability for two pair, we execute the same analytical plan that we've used to compute the other probabilities.

- First, count the number of five-card hands that can be dealt from a standard deck of 52 cards. We did this previously, and found that there are 2,598,960 distinct poker hands.
- Next, count the number of ways that five cards can be dealt to produce two pair. It requires five independent choices to produce two pair:
- Choose the rank for cards of matching rank. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace.
For two pair, we choose 2 ranks from a set of 13 ranks. The number of ways to do this is
_{13}C_{2}. - Choose the rank of the remaining non-matching card. There are 11 remaining ranks, from which we choose one.
The number of ways to do this is
_{11}C_{1}. - Choose suits for the first two-card combination. There are four suits, from which we choose two.
The number of ways to do this is
_{4}C_{2}. - Choose suits for the second two-card combination. There are four suits, from which we choose two.
The number of ways to do this is
_{4}C_{2}. - Choose a suit for the non-matching card. There are four suits, from which we choose one.
The number of ways to do this is
_{4}C_{1}.

The number of ways to produce two pair (Num

_{tp}) is equal to the product of the number of ways to make each independent choice. Therefore,Num

_{tp}=_{13}C_{2}*_{11}C_{1}*_{4}C_{2}*_{4}C_{2}*_{4}C_{1}Num

_{tp}= 78 * 11 * 6 * 6 * 4 = 123,552 - Choose the rank for cards of matching rank. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace.
For two pair, we choose 2 ranks from a set of 13 ranks. The number of ways to do this is
- Finally, compute the probability. There are 2,598,960 unique poker hands. Of those, 123,552 are two pair. Therefore, the probability
of being dealt two pair (P
_{tp}) is:P

_{tp}= 123,552 / 2,598,960 = 0.04753901561

On average, players get two pair about one time in every 21 deals.

## Probability of One Pair

To find the probability for one pair, we can use the same, general approach that we've used to compute each of the other probabilities for this lesson.

- First, count the number of five-card hands that can be dealt from a standard deck of 52 cards. We did this previously, and found that there are 2,598,960 distinct poker hands.
- Next, count the number of ways that five cards can be dealt to produce one pair. It requires six independent choices to produce one pair:
- Choose the rank for cards of matching rank. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace.
For one pair, we choose 1 rank from a set of 13 ranks. The number of ways to do this is
_{13}C_{1}. - Choose a rank for each of the remaining non-matching cards. There are 12 remaining ranks and three non-matching cards,
so we choose three ranks from the remaining 12.
The number of ways to do this is
_{12}C_{3}. - Choose suits for the cards of matching rank. There are four suits, from which we choose two.
The number of ways to do this is
_{4}C_{2}. - Choose a suit for the first non-matching rank. There are four suits, from which we choose one.
The number of ways to do this is
_{4}C_{1}. - Choose a suit for the second non-matching rank. There are four suits, from which we choose one.
The number of ways to do this is
_{4}C_{1}. - Choose a suit for the third non-matching rank. There are four suits, from which we choose one.
The number of ways to do this is
_{4}C_{1}.

The number of ways to produce one pair (Num

_{op}) is equal to the product of the number of ways to make each independent choice. Therefore,Num

_{op}=_{13}C_{1}*_{12}C_{3}*_{4}C_{2}*_{4}C_{1}*_{4}C_{1}*_{4}C_{1}Num

_{op}= 13 * 220 * 6 * 4 * 4 * 4 = 1,098,240 - Choose the rank for cards of matching rank. A playing card can have a rank of 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king, or ace.
For one pair, we choose 1 rank from a set of 13 ranks. The number of ways to do this is
- Finally, compute the probability. There are 2,598,960 unique poker hands. Of those, 1,098,240 are one pair. Therefore, the probability
of being dealt one pair (P
_{op}) is:P

_{op}= 1,098,240 / 2,598,960 = 0.4225690276

In stud poker, on any given hand, there is about a 42% chance that a player will be dealt one pair.

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