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Using Combinations to Calculate Probabilities

By Jim Frost 6 Comments

Combinations in probability theory and other areas of mathematics refer to a sequence of outcomes where the order does not matter. For example, when you’re ordering a pizza, it doesn’t matter whether you order it with ham, mushrooms, and olives or olives, mushrooms, and ham. You’re getting the same pizza!

Photograph showing the Powerball lottery drawing machine with the numbered balls.When calculating probabilities, you often need to calculate the number of possible combinations.

In this post, I’ll show you how to calculate the number of combinations with and without repetition and teach you the standard notation for combinations. I’ll also work through example problems so you can see the process in action. You’ll learn how to calculate the probability of winning the jackpot in the Powerball lottery!

Comparing Combinations to Permutations

In probability theory, combinations and permutations are distinct concepts.

  • Combinations: The order of outcomes does not matter.
  • Permutations: The order of outcomes does matter.

Let’s use three letters to illustrate both concepts: C, J, M.

Suppose we’re using these letters to signify people on a team, such as Carl, Jack, and Mary. As a combination of team members, C, J, M is equivalent to M, J, C. Either way, you have the same people on the team. Consequently, the order of the letters does not matter because you have the same team. These three letters form one combination.

However, if you’re using these letters for a short passcode, then the specific arrangement of letters matters. If the correct code is CJM, then MJC will not work. CJM is one possible permutation of these three letters. Using just these three letters, you can create 33 = 27 possible permutations.

When you have at least two permutations, the number of permutations is greater than the number of combinations. Learn more about the differences between permutation vs combination.

In this post, I’m working only with combinations. To learn about permutations, read my post about Using Permutations to Calculate Probabilities.

Combinations with Repetition

When the outcomes in a combination can repeat, you have repetition. With our pizza example, if you can choose a topping multiple times, such as double or triple ham, we have combinations with repetition. It’s also known as combinations with replacement.

To calculate the number of combinations with repetitions, use the following equation.

Equation for combinations with repetition.

Where:

  • n = the number of options.
  • r = the size of each combination.

The exclamation mark (!) represents a factorial. In general, n! equals the product of all numbers up to n. For example, 3! = 3 * 2 * 1 = 6. The exception is 0! = 1, which simplifies equations.

Photograph of a pizza.Let’s work through an example of the number of combinations with repetition. Imagine we own a pizza restaurant and we have six toppings from which customers can choose. We have a special offer for pizzas with three toppings. How many potential combinations fall under our special? Because we’re allowing repetition, customers can get triple bacon pizzas if they want to risk a hospital visit!

To solve these types of problems, you need to determine the n and r for the equation. The number of possible outcomes equals the number of choices for toppings, which is six. The size of each combination is three because that’s what we’re allowing for our special. So, n = 6 and r = 3.

Using the standard notation for combinations of nCr, we’d write this as 6C3.

Inputting those numbers into the equation:

Example calculations for combinations with repetition.

6C3 = 56.

There are 56 combinations of three toppings when you have six topping choices and allow repeats.

Combinations without Repetition

When the outcomes cannot repeat, you are working with combinations without repetition. These calculations use the following equation for permutations without repetition as a starting point.

Equation for permutations without repetition.

In permutations, the order does matter. However, the order does not matter for combinations without repetition, which produces a lower number than the permutations. In the following equation, the first portion captures the number of permutations where the order matters. The second portion reduces that number because the order does not matter for combinations. The product of the two produces the final equation.

Equation for combinations without repetitions.

In the simplified form:

Simplified equation for combinations without repetition.

Where:

  • n = the number of possible outcomes at the start.
  • r = the size of each combination.

For example, imagine a process that draws five people to be on a team out of a pool of 30. The order in which the process draws team members does not matter. Suppose a team comprises A, B, C, D, E. That is equivalent to a team where the same people were drawn but in a different order. Additionally, because the process can select a person only once, there is no repetition.

In this example, n = 30 because we’re drawing from a pool of 30 candidates. R = 5 because the team can have only five members. Inputting those numbers into the equation:

Example calculation for combinations without repetition.

Using the standard notation: 30C5 = 142,506.

There are 142,506 combinations of five person teams when you draw from a pool of 30 individuals.

You can also use Pascal’s triangle to find the number of combinations without repetition.

The formula for combinations without repetition is a crucial part of the binomial distribution. To learn how, read my post, The Binomial Distribution.

Using Combinations to Calculate Probabilities

In the previous examples, we calculated the number of combinations for different scenarios. However, we didn’t use them to calculate probabilities. In this context, a probability is the number of combinations considered to be an event divided by the total number of combinations.

When you’re given a probability problem using combinations, you need to determine several things to solve the problem.

  1. Set up a ratio to determine the probability.
  2. Determine whether the numerator and denominator require combinations, permutations, or a mix? For this post, we’ll stick with combinations.
  3. Are these combinations with or without repetitions?
  4. Both types of combinations require you to identify the n and r to enter into the equations.

Let’s work through an example probability calculation using combinations!

Problem: What is the probability of winning the basic game and the jackpot in the Powerball lottery?

Related post: Probability Fundamentals

Probability of winning the basic game in the Powerball lottery

The classic example of probability using combinations without repetition is a lottery where machines randomly choose balls with numbers from a pool of balls. The order of the numbers does not matter. You just need to match the numbers. Because each ball can be drawn only once, it is without repetition.

For this example, we’ll look at the Powerball lottery, which is a lottery offered in 45 states in the United States. In the basic game, there are 69 white balls in a machine that randomly selects five of the balls. The following ratio defines the probability of winning:

Probability = {\displaystyle \frac {{\text{Winning combination}}}{{\text{Total number of combinations}}}}

The numerator is simple. There is one winning combination without repetition. You could use the equation to calculate 5C5, which equals the number of combinations where you have five numbers in the winning combination, and there are five winning numbers from which to choose. But it makes sense intuitively. There’s just one winning combination.

For the denominator, you need to calculate 69C5, which equals the number of combinations when you draw five numbers from a total of 69 numbers. Let’s enter these numbers into the equation:

Example calculations for winning the Powerball basic game using combinations.

69C5 = 11,238,513

When you draw five numbers out of 69 without repetition, there are 11,238,513 combinations.

Using the standard notation, the probability of winning the basic game in Powerball is the following:

Probability calculations for winning the Powerball basic game using combinations.

You have a 1 in 11,238,513 chance of winning the basic game in the Powerball lottery.

The probability of winning the jackpot in Powerball

To win the jackpot in the Powerball, you need to win the basic game and match one red Powerball number. The Powerballs are 26 red balls in a separate machine that randomly draws one of them. Let’s calculate the probability of winning the jackpot!

For the numerator, there is again one winning combination for the Powerball. For the denominator, we need to calculate the number of combinations for 26C1 = 26. That’s not surprising. There are 26 combinations when you choose one number from 26 options! Thus, the chance of matching the Powerball is the following:

Probability for matching the single Powerball.

To include this in the probability calculations for the basic game, we need to use the multiplication rule for independent events and multiply the two ratios.

Using the standard notation for combinations, the probability of winning the jackpot in Powerball is the following:

Probability calculations for winning the jackpot in the Powerball lottery using combinations.

You have a 1 in 292,201,338 chance of winning the jackpot in the Powerball lottery.

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Filed Under: Probability Tagged With: analysis example, choosing analysis, conceptual

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Comments

  1. George says

    June 11, 2021 at 12:25 am

    Minor detail in light of your wonderful work!

    Reply
    • Jim Frost says

      June 11, 2021 at 12:44 am

      Thanks, George!

      Reply
  2. George says

    June 10, 2021 at 6:23 am

    Hi Jim,

    But shouldn’t the permutation with replacement be n^r (i.e., 3^3=27)? How do you get 9??

    Best,
    George

    Reply
    • Jim Frost says

      June 10, 2021 at 3:14 pm

      Hi George,

      Thanks, that’s a great catch! You’re absolutely right. I’ve changed the text. Apparently, my brain slipped a gear! 🙂

      Reply
  3. Stalin Amirtharaj K says

    May 10, 2021 at 6:55 am

    We can create only 6 permutations out of three letters(CJM in this example) right? whereas it is mentioned as 9 permutations in this article

    Reply
    • Jim Frost says

      May 10, 2021 at 12:03 pm

      Hi, it’s 9 with repeats, and 6 without.

      Reply

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