# Dice: Finding Expected Values of Games of Chance

An error occurred trying to load this video.

Try refreshing the page, or contact customer support.

Coming up next: Blackjack: Finding Expected Values of Games of Chance with Cards

### You're on a roll. Keep up the good work!

Replay
Your next lesson will play in 10 seconds
• 0:01 The Game of Dice
• 6:38 Putting It All Together
• 8:18 Rolling Three Dice
• 10:30 Four-Sided Dice
• 12:09 Lesson Summary

Want to watch this again later?

Timeline
Autoplay
Autoplay
Speed

#### Recommended Lessons and Courses for You

Lesson Transcript
Instructor: Rudranath Beharrysingh
This lesson examines the various combinations and probabilities behind rolling dice. We will look at a game of dice and what to expect to win or lose in a game. In addition we will extend these concepts to playing with different sided dice.

## The Game of Dice

Imagine walking into a casino for a game of dice, knowing exactly what to bet and how much you will win or lose. This is called the expected value of the game, and in this lesson we will learn how to calculate it.

In casinos, there is a game called 'craps,' which involves rolling two six-sided dice. The rules are somewhat complicated, but understanding the probabilities of the different combinations of the dice gives a person a solid advantage in the game.

Suppose you play a game where you roll two six-sided die. And, suppose the rules of the game are: you lose \$3 if you get a sum of 2, 4, or 10. You lose \$2 if you get a sum of 7, but win \$1 for anything else. If you continue to play this game, what can you expect to win or lose in the long run?

So you may recall that the expected value, called E of X, of a situation is the sum of each value, called X, times its probability, called P of X. In mathematical notation this is written like this:

For this game, there are three events to consider:

1. getting a sum of 2, 4 or 10
2. getting a sum of 7
3. getting anything else

And each of these occurrences has a value associated with them:

1. losing \$3
2. losing \$2
3. winning \$1, respectively

We can think of the amount won or lost as the values of the discrete random variable X. And so, X is the amount of money won or lost per roll of the dice. Plus, each of these values has a probability associated with it.

So the key to this is to know the probabilities of each event. For this, we need to examine what happens when two dice are rolled. Even though the die may look the same, from a mathematical perspective they are two distinct die. They are independent of each other. In probability, independence means they don't affect each other. So each die is independent of the other because one die doesn't affect the outcome of the other.

To illustrate, let's make one die blue and the other die red.

Each die has six sides, and so there are six possible outcomes for each die when rolled individually. However, when rolled together, there are 6 * 6 = 36 possible outcomes! This may be surprising, but the diagram below illustrates all the outcomes and all the sums when two dice are rolled.

So, for example, you could get a 1 on the red die and a 1 on the blue die, which adds to 2. And, actually, this is the only way to get a sum of 2. And so the probability of getting a sum of 2 when you roll two dice is 1 out of 36, which is about 0.028, or a 2.8% chance!

We can look at the table to find the probability of any of the sums for the two dice. So, for a sum of 4 the possibilities are 1 on red + 3 on blue, 2 on red + 2 on blue, or 3 on red + 1 on blue. The reason we do not count 2 + 2 twice is because 2 on red + 2 on blue is the same as 2 on blue + 2 on red!

Thus, the probability of a sum of 4 is 3 out of 36, which is about 0.083.

And the probability of a sum of 10 when you roll a dice is? You can pause the video here to calculate this from the table.

I hope you calculated 3/36. You can see from the table that there are three ways to get a sum of 10: 4 + 6, 5 + 5 or 6 + 4.

So let's go back to the game. You lose \$3 if you get a sum of 2, 4 or 10. And now we know there is a 1/36 probability of getting a sum of 2, a 3/36 probability of getting a sum of 4 and also a 3/36 probability of getting a sum of 10. Since we are interested in all of these outcomes, we get a total probability of 1/36 + 3/36 + 3/36 = 7/36, or about a 0.194 probability of getting a sum of 2, 4 or a 10 when you roll two dice.

What about getting a sum of 7? What is its probability? You can pause the video here to see if you can calculate it.

So I hope you got 6/36, which is about 17%! So, we see there are 6 ways to get a sum of 7: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, 6 + 1.

There are more ways to get a sum of 7 than any other sum, and it is sometimes called lucky seven! Although in this case, it is unlucky for you since you will lose \$2 if you get it!

So how do you win this game? You need to roll anything else but the above sums of 2, 4, 10 or 7. And what is the probability of doing this? One of the properties of probabilities for a particular situation is that they all must add to 1. More formally: the sum of all the probabilities in a probability distribution add to 1.

A probability distribution lists all the outcomes of a particular situation and all the corresponding probabilities. In this case we are talking about the sum of the numbers on the two six-sided dice.

So, we already know that the probability of getting a sum of 2, 4 or 10 is 7/36 and the probability of getting a sum of 7 is 6/36. So what is the probability of getting some other sum?

Since they must all add to 1, we can just total the probabilities we know and subtract the result from 1! In other words the probability of getting a sum of anything else other than 2, 4, 10 or 7 is: 1 - 7/36 - 6/36 = 23/36, or approximately 0.64 or 64%. So there's a 64% chance you win something in this game, and that something is a dollar!

Here's the table from before with the probabilities:

## Putting It All Together

What can you expect in this game if you play it many times?

To calculate the expected value we multiply the value of each event by its probability and then add the results. So, for the event of getting a sum of 2, 4 or 10, we multiply -3 times 7/36, which equals -21/36. And for the event of getting a sum of 7, we multiply -2 times 6/36, which equals -12/36. And finally, for the event of getting anything else, we multiply +1 times 23/36 which equals 23/36.

Adding them all together, we get an expected value of: -21/36 - 12/36 + 23/36 = -10/36 or -0.28. In other words, if you continued to play this game, your average winnings would be -0.28 per roll, or a loss of 28 cents per roll of the dice!

This is summarized in the table below:

And remember, your loss is the casino's gain. So if 100,000 people played this game over the course of a week, what would the casino make on this one game? Well, they would make 28 cents per person times 100,000 = \$28,000 times the number of rolls each person played! This is not too shabby a return on investment for the casino!

## Rolling Three Dice

We have seen that there are 36 possible outcomes when you roll two six-sided dice. Suppose you play a game using three six-sided dice. How many possible outcomes are there now?

To unlock this lesson you must be a Study.com Member.

### Register to view this lesson

Are you a student or a teacher?

#### See for yourself why 30 million people use Study.com

##### Become a Study.com member and start learning now.
Back
What teachers are saying about Study.com

### Earning College Credit

Did you know… We have over 160 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level.