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Expected Value

Calculate the expected value of an experiment.

Learning Objectives

After completing this section, you should be able to:

Calculate the expected value of an experiment.
Interpret the expected value of an experiment.
Use expected value to analyze applications.

Expected Value

Many experiments have numbers associated with their outcomes. Some are easy to define; if you roll 2 dice, the sum of the numbers showing is a good example. In some card games, cards have different point values associated with them; for example, in some forms of the game rummy, aces are worth 15 points; 10s, jacks, queens, and kings are worth 10; and all other cards are worth 5. The outcomes of casino and lottery games are all associated with an amount of money won or lost. These outcome values are used to find the expected value of an experiment: the mean of the values associated with the outcomes that we would observe over a large number of repetitions of the experiment. (See Conditional Probability and the Multiplication Rule for more on means.)

That definition is a little vague; How many is “a large number?” In practice, it depends on the experiment; the number has to be large enough that every outcome would be expected to appear at least a few times. For example, if we’re talking about rolling a standard 6-sided die and we note the number showing, a few dozen replications should be enough that the mean would be representative. Since the probability of each outcome is $\frac{1}{6}$, we would expect to see each outcome about 8 times over the course of 48 replications. However, if we’re talking about the Powerball lottery, where the probability of winning the jackpot is about $\frac{1}{292,000,000}$, we would need several billion replications to ensure that every outcome appears a few times. Luckily, we can find the theoretical expected value before we even run the experiment the first time.

Let’s make note of some things we can learn from . First, as Exercises 1 and 3 demonstrate, the expected value of an experiment might not be a value that could come up in the experiment. Remember that the expected value is interpreted as a mean, and the mean of a collection of numbers doesn’t have to actually be one of those numbers.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Interpreting Expected Values

As we noted, the expected value of an experiment is the mean of the values we would observe if we repeated the experiment a large number of times. (This interpretation is due to an important theorem in the theory of probability called the Law of Large Numbers.) Let’s use that to interpret the results of the previous example.

Interpreting Expected Values

Try it.

Interpret the expected values of the following experiments.

Roll a standard 6-sided die and note the number showing.
Roll 2 standard 6-sided dice and note the sum of the numbers showing.
Draw a card from a well-shuffled standard deck of cards and note its Rummy value (15 for aces; 10 for tens, jacks, queens, and kings; 5 for everything else).

Solution

If you roll a standard 6-sided die many times, the mean of the numbers you roll will be around 3.5.
If you roll a pair of standard 6-sided dice many times, the mean of the sums of the numbers you roll will be about 7.
If you draw a card from a well-shuffled deck many times, the mean of the Rummy values of the cards would be around 7.3.

Using Expected Value

Now that we know how to find and interpret expected values, we can turn our attention to using them. Suppose someone offers to play a game with you. If you roll a die and get a 6, you get $10. However, if you get a 5 or below, you lose $1. Is this a game you’d want to play? Let’s look at the expected value: The probability of winning is $\frac{1}{6}$ and the probability of losing is $\frac{5}{6}$, so the expected value is $\$10\times \frac{1}{6}+(-\$1)\times \frac{5}{6}=\frac{5}{6}\approx \$0.83$. That means, on average, you’ll come out ahead by about 83 cents every time you play this game. It’s a great deal! On the other hand, if the winnings for rolling a 6 drop to $3, the expected value becomes $\$3\times \frac{1}{6}+(-\$1)\times \frac{5}{6}=-\frac{1}{3}\approx -\$0.33$, meaning you should expect to lose about 33 cents on average for every time you play. Playing that game is not a good idea! In general, this is how casinos and lottery corporations make money: Every game has a negative expected value for the player.

Expected Value

Expected Values in Football

In the 21st century, data analytics tools have revolutionized the way sports are coached and played. One tool in particular is used in football at crucial moments in the game. When a team faces a fourth down (the last possession in a series of four possessions, a fairly common occurrence), the coach faces a decision: Run one play to try to gain a certain number of yards, or kick the ball away to the other team. Here’s the interesting part of the decision: If the team “goes for it” and runs the play and they are successful, then they keep possession of the ball and can continue in their quest to score more points. If they are unsuccessful, then they lose possession of the ball, giving the other team an opportunity to score points. If, instead, the team punts, or kicks the ball away, then the other team gets possession of the ball, but in a worse position for them than if the original team goes for it and fails. To analyze this situation, data analysts have generated empirical probabilities for every fourth down situation, and computed the expected value (in terms of points) for each decision. Coaches frequently use those calculations when they decide which option to take!

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

The expected value of an experiment is the sum of the products of the numerical outcomes of an experiment with their corresponding probabilities.
The expected value of an experiment is the most likely value of the average of a large number of replications of the experiment.

Formulas

If $O$ represents an outcome of an experiment and $n(O)$ represents the value of that outcome, then the expected value of the experiment is: \[\sum n(O)\times P(O)\]
where $\Sigma$ stands for the sum, meaning we add up the results of the formula that follows over all possible outcomes.

Projects

The Binomial Distribution is one of many examples of a discrete probability distribution. Other examples include the Geometric, Hypergeometric, Multinomial, Poisson, and Negative Binomial Distributions. Choose one of these distributions, and find out what makes it different from the Binomial Distribution. In what situations can it be applied? How is it used? Once you have an idea of how it’s used, write a series of five questions like the ones in this chapter that can be answered with that distribution, and find the answers.
Binomial is a word that also comes up in algebra; the word describes polynomials with two terms. At first glance, there isn’t much to indicate that these two uses of the word are related, but it turns out there is a connection. Explore the connection between the Binomial Distribution and the algebraic concept of binomial expansion, (the process of multiplying out expressions like ${(x+y)}^{n}$ for a positive whole number $n$). Search for a connection with the mathematical object known as Pascal’s Triangle.
Hazard is a dice game that was mentioned in Chaucer’s Canterbury Tales. It was a popular game of chance played in taverns and coffee houses well into the 18th century; its popularity at the time of the foundation of probability theory means that it was a common example in early texts on finding expected values and probabilities. Find the rules of the game, and get some practice playing it. Then, analyze the choices that the caster gets to make, and decide which is most advantageous, using the language of expected values.

Practice (3)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Find the expected values of the following experiments.

Roll a standard 6-sided die and note the number showing.
Roll two standard 6-sided dice and note the sum of the numbers showing.
Draw a card from a well-shuffled standard deck of cards and note its rummy value (15 for aces; 10 for tens, jacks, queens, and kings; 5 for everything else).

Avslöja svaret

Step 1: Let’s start by writing out the PDF table for this experiment.

Value	Probability
1	$\frac{1}{6}$
2	$\frac{1}{6}$
3	$\frac{1}{6}$
4	$\frac{1}{6}$
5	$\frac{1}{6}$
6	$\frac{1}{6}$

Step 2: To find the expected value, we need to find $n(O)\times P(O)$ for each possible outcome in the table below.

Value	Probability	$n(O)\times P(O)$
1	$\frac{1}{6}$	$1\times \frac{1}{6}=\frac{1}{6}$
2	$\frac{1}{6}$	$2\times \frac{1}{6}=\frac{1}{3}$
3	$\frac{1}{6}$	$3\times \frac{1}{6}=\frac{1}{2}$
4	$\frac{1}{6}$	$4\times \frac{1}{6}=\frac{2}{3}$
5	$\frac{1}{6}$	$5\times \frac{1}{6}=\frac{5}{6}$
6	$\frac{1}{6}$	$6\times \frac{1}{6}=1$

Step 3: We add all of the values in that last column: $\frac{1}{6}+\frac{1}{3}+\frac{1}{2}+\frac{2}{3}+\frac{5}{6}+1=\frac{7}{2}=3.5$. So, the expected value of a single roll of a die is 3.5.

Back in , we made this table of all of the equally likely outcomes ():

Step 1: Let’s use to create the PDF for this experiment, as shown in the following table:

Value	Probability
2	$\frac{1}{36}$
3	$\frac{1}{18}$
4	$\frac{1}{12}$
5	$\frac{1}{9}$
6	$\frac{5}{36}$
7	$\frac{1}{6}$
8	$\frac{5}{36}$
9	$\frac{1}{9}$
10	$\frac{1}{12}$
11	$\frac{1}{18}$
12	$\frac{1}{36}$

Step 2: We can multiply each row to find $n(O)\times P(O)$as shown in the following table:

Value	Probability	$n(O)\times P(O)$
2	$\frac{1}{36}$	$\frac{1}{18}$
3	$\frac{1}{18}$	$\frac{1}{6}$
4	$\frac{1}{12}$	$\frac{1}{3}$
5	$\frac{1}{9}$	$\frac{5}{9}$
6	$\frac{5}{36}$	$\frac{5}{6}$
7	$\frac{1}{6}$	$\frac{7}{6}$
8	$\frac{5}{36}$	$\frac{10}{9}$
9	$\frac{1}{9}$	$1$
10	$\frac{1}{12}$	$\frac{5}{6}$
11	$\frac{1}{18}$	$\frac{11}{18}$
12	$\frac{1}{36}$	$\frac{1}{3}$

Step 3: We can add the last column to get the expected value: \[\frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}=7\]So, the expected value is 7.

Step 1: Let’s make a PDF table for this experiment. There are 3 events that we care about, so let’s use those events in the table below:
Event Probability
{A} $\frac{1}{13}$
{10, J, Q, K} $\frac{4}{13}$
{2, 3, 4, 5, 6, 7, 8, 9} $\frac{8}{13}$

Step 2: Let’s add a column to the following table for the values of each event:

Event Probability Value
{A} $\frac{1}{13}$ 15
{10, J, Q, K} $\frac{4}{13}$ 10
{2, 3, 4, 5, 6, 7, 8, 9} $\frac{8}{13}$ 5

Step 3: We’ll add the column for the product of the values and probabilities to the table below:

Event Probability Value $n(O)\times P(O)$
{A} $\frac{1}{13}$ 15 $\frac{15}{13}$
{10, J, Q, K} $\frac{4}{13}$ 10 $\frac{40}{13}$
{2, 3, 4, 5, 6, 7, 8, 9} $\frac{8}{13}$ 5 $\frac{40}{13}$

Step 4: We’ll find the sum of the last column: $\frac{15}{13}+\frac{40}{13}+\frac{40}{13}=\frac{95}{13}\approx 7.3$. Thus, the expected Rummy value of a randomly selected card is about 7.3.

Interpret the expected values of the following experiments.
1. Roll a standard 6-sided die and note the number showing.
2. Roll 2 standard 6-sided dice and note the sum of the numbers showing.
3. Draw a card from a well-shuffled standard deck of cards and note its Rummy value (15 for aces; 10 for tens, jacks, queens, and kings; 5 for everything else).
Avslöja svaret
1. If you roll a standard 6-sided die many times, the mean of the numbers you roll will be around 3.5.
2. If you roll a pair of standard 6-sided dice many times, the mean of the sums of the numbers you roll will be about 7.
3. If you draw a card from a well-shuffled deck many times, the mean of the Rummy values of the cards would be around 7.3.
In the casino game keno, a machine chooses at random 20 numbers between 1 and 80 (inclusive) without replacement. Players try to predict which numbers will be chosen. Players don’t try to guess all 20, though; generally, they’ll try to predict between 1 and 10 of the chosen numbers. The amount won depends on the number of guesses they made and the number of guesses that were correct.
1. At one casino, a player can try to guess just 1 number. If that number is among the 20 selected, the player wins $2; otherwise, the player loses $1. What is the expected value?
2. At the same casino, if a player makes 2 guesses and they’re both correct, the player wins $14; otherwise, the player loses $1. What is the expected value?
3. Players can also make 3 guesses. If 2 of the 3 guesses are correct, the player wins $1. If all 3 guesses are correct, the player wins $42. Otherwise, the player loses $1. What is the expected value?
4. Which of these games is the best for the player? Which is the best for the casino?
Avslöja svaret
1. There are 20 winning numbers out of 80, so if we try to guess one of them, the probability of guessing correctly is $\frac{20}{80}=\frac{1}{4}$. The probability of losing is then $\frac{3}{4}$, and so the expected value is $\$2\times \frac{1}{4}+(-\$1)\times \frac{3}{4}=-\$0.25$.
2. There are ${}_{20}{C}_{2}=190$ winning choices out of ${}_{80}{C}_{2}=3,160$ total ways to choose 2 numbers. So, the probability of winning is $\frac{190}{3,160}$ and the probability of losing is $\frac{3,160-190}{3,160}=\frac{2,970}{3,160}$. So, the expected value of the game is $\$14\times \frac{190}{3,160}+(-\$1)\times \frac{2970}{3160}\approx -\$0.10$.
3. Step 1: Let’s start with the big prize. There are ${}_{20}{C}_{3}=1,140$ ways to correctly guess 3 winning numbers out of ${}_{80}{C}_{3}=82,160$ ways to guess three numbers total. That means the probability of winning the big prize is $\frac{1,140}{82,160}\approx 0.01388$.
  Step 2: Let’s find the probability of the second prize. The denominator is the same: 82,160. Let’s figure out the numerator. To win the second prize, the player must pick 2 of the 20 winning numbers and one of the 60 losing numbers. The number of ways to do that can be found using the Multiplication Rule for Counting: there are ${}_{20}{C}_{2}=190$ ways to pick 2 winning numbers and 60 ways to pick 1 losing number, so there are $190\times 60=11,400$ ways to win the second prize. So, the probability of winning that second prize is $\frac{11,400}{82,160}\approx 0.13875$.
  
  Step 3: Since the overall probability of winning is $\frac{1,140}{82,160}+\frac{11,400}{82,160}=\frac{12,540}{82,160}\approx 0.15263$, the probability of losing must be $1-0.15263=0.84737$. So, the expected value is $\$42\times 0.01388+\$1\times 0.13875+(-\$1)\times 0.84737\approx -\$0.13$.
4. The bet that’s the best for the player is the one with the highest expected value for the player, which is guessing two numbers. The best one for the casino is the one with the lowest expected value for the player, which is guessing one number.

Symbols used here

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

P(A),\ P(A \mid B)

probability, conditional probability
Chance of A; chance of A given that B happened.

\approx

approximately equal
Equal to the precision shown, not exactly.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

\bar{x},\ \mu

sample mean, population mean
Average of the data; average of the whole population.

\sigma,\ s,\ \sigma^2

standard deviation, sample s.d., variance
Typical distance from the mean; its square.

E[X],\ \operatorname{Var}(X)

expected value, variance
Probability-weighted average of X; its spread.

N(\mu, \sigma^2),\ z

normal distribution, z-score
The bell curve with mean μ and variance σ²; (x − μ)/σ.

How to: Expected Value

Calculate the expected value of an experiment.
Interpret the expected value of an experiment.
Use expected value to analyze applications.
Roll a standard 6-sided die and note the number showing.
Roll two standard 6-sided dice and note the sum of the numbers showing.
Draw a card from a well-shuffled standard deck of cards and note its rummy value (15 for aces; 10 for tens, jacks, queens, and kings; 5 for everything else).
Back in
Roll a standard 6-sided die and note the number showing.

Questions people ask

Mean or median — which should I use?

Median when the data have outliers or a long tail (incomes, house prices); mean when the data are roughly symmetric and you want every value to count. Report both if they disagree — the gap is itself information.

What does a p-value actually say?

The probability of seeing data at least this extreme if the null hypothesis were true. It is not the probability that the null hypothesis is true.

Why divide by n − 1 for the sample variance?

The sample mean sits closer to the sample than the true mean does, so squared deviations from it are slightly too small on average; dividing by n − 1 instead of n corrects the bias.

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Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Mer information Statistics & Probability

Sampling and data Describing data with graphs Mean, median and mode Probability Counting: permutations and combinations Discrete random variables Continuous random variables The normal distribution The central limit theorem Confidence intervals Hypothesis testing Comparing two samples Chi-square tests Linear regression and correlation

Value	Probability
1	\(\frac{1}{6}\)
2	\(\frac{1}{6}\)
3	\(\frac{1}{6}\)
4	\(\frac{1}{6}\)
5	\(\frac{1}{6}\)
6	\(\frac{1}{6}\)

Value	Probability	\(n(O)\times P(O)\)
1	\(\frac{1}{6}\)	\(1\times \frac{1}{6}=\frac{1}{6}\)
2	\(\frac{1}{6}\)	\(2\times \frac{1}{6}=\frac{1}{3}\)
3	\(\frac{1}{6}\)	\(3\times \frac{1}{6}=\frac{1}{2}\)
4	\(\frac{1}{6}\)	\(4\times \frac{1}{6}=\frac{2}{3}\)
5	\(\frac{1}{6}\)	\(5\times \frac{1}{6}=\frac{5}{6}\)
6	\(\frac{1}{6}\)	\(6\times \frac{1}{6}=1\)

Value	Probability
2	\(\frac{1}{36}\)
3	\(\frac{1}{18}\)
4	\(\frac{1}{12}\)
5	\(\frac{1}{9}\)
6	\(\frac{5}{36}\)
7	\(\frac{1}{6}\)
8	\(\frac{5}{36}\)
9	\(\frac{1}{9}\)
10	\(\frac{1}{12}\)
11	\(\frac{1}{18}\)
12	\(\frac{1}{36}\)

Value	Probability	\(n(O)\times P(O)\)
2	\(\frac{1}{36}\)	\(\frac{1}{18}\)
3	\(\frac{1}{18}\)	\(\frac{1}{6}\)
4	\(\frac{1}{12}\)	\(\frac{1}{3}\)
5	\(\frac{1}{9}\)	\(\frac{5}{9}\)
6	\(\frac{5}{36}\)	\(\frac{5}{6}\)
7	\(\frac{1}{6}\)	\(\frac{7}{6}\)
8	\(\frac{5}{36}\)	\(\frac{10}{9}\)
9	\(\frac{1}{9}\)	\(1\)
10	\(\frac{1}{12}\)	\(\frac{5}{6}\)
11	\(\frac{1}{18}\)	\(\frac{11}{18}\)
12	\(\frac{1}{36}\)	\(\frac{1}{3}\)

Event	Probability
{A}	\(\frac{1}{13}\)
{10, J, Q, K}	\(\frac{4}{13}\)
{2, 3, 4, 5, 6, 7, 8, 9}	\(\frac{8}{13}\)