19.7: Expected Value

Learning Objectives

• Calculate and interpret the expected value of an experiment.
• Use expected value to analyze real-world applications.

The casino game roulette offers dozens of different bets. These bets vary in their probabilities of winning and also in their payouts. In general, the lower the probability of winning a bet, the higher the potential payout. With so many options, is there one bet that’s “smarter” than the rest? What’s the best play to make at a roulette table? In this section, we’ll develop the tools needed to answer these questions.

Expected Value

Many experiments have numerical values associated with their outcomes. Some are easy to define; for example, if you roll two dice, the sum of the numbers shown is a natural outcome to consider. In some card games, cards have different point values; for instance, in certain forms of 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 typically associated with an amount of money won or lost.

These outcome values are used to calculate 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. 

That definition may seem a little vague - how many is “a large number”? In practice, it depends on the experiment. The number must be large enough that every outcome would be expected to appear at least a few times. For example, if we roll a standard six-sided die and record the number shown, a few dozen repetitions should be enough for the mean to be representative. Since the probability of each outcome is 1/6, we would expect to see each number appear about eight times over 48 rolls.

However, consider the Powerball lottery, where the probability of winning the jackpot is about 1 in 292,000,000. In that case, we would need several billion repetitions to ensure that every outcome appears at least a few times. Fortunately, we can calculate the theoretical expected value before running the experiment even once.

FORMULA: Expected Value

If \(x_i\)​ represents the value associated with an outcome of an experiment, then the expected value of the experiment is:

\[ x_1p(x_1)+x_2p(x_2)+x_3p(x_3)+⋯+x_np(x_n) =  \sum_{i=1}^{n} x_i P(x_i) \]

where \(P(x_i)\) is the probability of the outcome with value \(x_i\), \(n\) is the number of outcomes in the sample space, and \(\sum\) denotes the sum - that is, we add the results of the formula over all possible outcomes. 

Example \(\PageIndex{1}\): Finding Expected Values

Find the expected values of the following experiments:

1. Roll a standard six-sided die and note the number shown.
2. Roll two standard six-sided dice and note the sum of the numbers shown.
3. Draw a card from a well-shuffled standard deck and note its rummy value (15 for aces; 10 for tens, jacks, queens, and kings; and 5 for all other cards).

Solution

1. Step 1: Let’s start by creating a table for this experiment. To find the expected value, we need to compute \(x_i\times P(i)\) for each possible outcome in the table below.

   Value \(x_i\)	Probability, \(P(i)\)	\(x_i\times P(i)\)
   1	\(\displaystyle \frac{1}{6}\)	\(\displaystyle\frac{1}{6}\)
   2	\(\displaystyle \frac{1}{6}\)	\(\displaystyle\frac{2}{6}=\frac{1}{3}\)
   3	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{3}{6}=\frac{1}{2}\)
   4	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{4}{6}=\frac{2}{3}\)
   5	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{5}{6}\)
   6	\(\displaystyle \frac{1}{6}\)	\(\displaystyle \frac{6}{6}=1\)

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

2. Step 1: Let’s use Figure \(\PageIndex{1}\) to create a table for this experiment. To find the expected value, we need to compute  \(x_i \times P(i)\) for each possible outcome as shown in the table below.

   

   Figure \(\PageIndex{1}\)

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

   Step 2: We add all of the values in that last column: \(\displaystyle \frac{1}{18}+\frac{1}{6}+\frac{1}{3}+\frac{5}{9}+\frac{5}{6}+\frac{7}{6}+\frac{10}{9}+1+\frac{5}{6}+\frac{11}{18}+\frac{1}{3}=7\). So, the expected value is 7.

3. Step 1: Let’s make a table for this experiment. There are three events we care about, so we’ll include those events in the table and add columns for their probabilities, their values, and the product of each value and its probability.

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

   Step 2: We’ll find the sum of the last column: \(\displaystyle \frac{15}{13}+\frac{40}{13}+\frac{40}{13}=\frac{95}{13} \approx 7.3\). Thus, the expected Rummy value of a randomly selected card is about 7.3.

Try It \(\PageIndex{1}\)

1. Find the expected value of the number shown when you roll a special six-sided die with faces {1, 1, 2, 3, 5, 8}.
2. Find the expected value of the number of heads showing if you flip a coin three times.
3. You are about to play a game, where you flip a coin three times. If all three flips result in heads, you win $20. If you get two heads, you win $10. If you flip one or no heads, you win nothing. What is the expected value of your winnings?

Answer

1.  \(~\displaystyle \frac{10}{3}\)
2.  \(~\displaystyle \frac{3}{2}\)
3.  \(~\displaystyle \frac{25}{4} = \$6.25\)

Example \(\PageIndex{2}\)

In a town, 10% of the families have three children, 60% have two children, 20% have one child, and 10% have no children. What is the expected number of children per family?

Solution

Expected Value \(= x_1p(x_1)+x_2p(x_2)+x_3p(x_3)+x_4p(x_4)=3(0.10)+2(0.60)+1(0.20)+0(0.10)=1.7.\) So, on average, there are 1.7 children per family.

Let’s take note of a few things we can learn from Example \(\PageIndex{1}\). First, as Questions 1 and 3 demonstrate, the expected value of an experiment might not be a value that could actually occur in the experiment. Remember that the expected value is interpreted as a mean, and the mean of a set of numbers doesn’t necessarily have to be one of those numbers.

Second, looking at Question 1, the expected value (=3.5) was simply the mean of the numbers on the faces of the die: \(\displaystyle \frac{1+2+3+4+5+6}{6}=3.5\). This is no accident. If we break that fraction apart using the addition in the numerator, we get: \(\displaystyle \frac{1}{6}+\frac{2}{6}+\frac{3}{6}+\frac{4}{6}+\frac{5}{6}+\frac{6}{6}\), which can be rewritten as: \(\displaystyle 1\times \frac{1}{6}+2\times\frac{1}{6}+3\times \frac{1}{6}+4\times\frac{1}{6}+5\times\frac{1}{6}+6\times\frac{1}{6}\).

That’s exactly the computation we used to find the expected value. In fact, expected values can always be interpreted as a special kind of mean called a weighted mean, where the weights are the probabilities associated with each value. When the probabilities are all equal, the weighted mean is the same as the regular (arithmetic) mean. 'Weighted mean' will be discussed in the Statistics chapter later.

Example \(\PageIndex{3}\)

To sell an average house, a real estate broker spends $1,200 on advertising expenses. If the house sells within three months, the broker earns $8,000. Otherwise, the broker loses the listing. If there is a 40% chance that the house will sell within three months, what is the expected payoff for the real estate broker?

Solution

The broker makes $8,000 with a probability of 0.40, but he loses $1,200 whether the house sells or not.

payoff (= expected value) \(=\$8,000\times 0.40−\$1,200=\$2,000.\)

Alternatively, the broker makes $(8,000 \(-\) 1,200) with a probability of 0.40, but loses $1,200 with a probability of 0.60.

Therefore, payoff (= expected value) \(=\$6,800\times 0.40−\$1,200\times 0.60= \$2,000.\)

Try It \(\PageIndex{2}\)

In a town, attendance at a football game depends on the weather. On a sunny day, attendance is 60,000; on a cold day, it is 40,000; and on a stormy day, it is 30,000. For the upcoming football season, the weatherman has predicted that 30% of the days will be sunny, 50% will be cold, and 20% will be stormy. What is the expected attendance for a single game?

Answer

44,000

Try It \(\PageIndex{3}\)

A lottery consists of choosing six numbers from a total of 51. A person who matches all six numbers wins $2 million. If a lottery ticket costs $1, what is the expected payoff?

Answer

−$0.89. This means that every time a person spends $1 to buy a ticket, they should expect to lose 89 cents.

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 comes from an important theorem in probability theory called the Law of Large Numbers.) Let’s use this understanding to interpret the results of the previous example.

Example \(\PageIndex{4}\): Interpreting Expected Values

Interpret the expected values of the following experiments.

1. Roll a standard six-sided die and note the number shown.
2. Roll two standard six-sided dice and note the sum of the numbers shown.
3. Draw a card from a well-shuffled standard deck and note its Rummy value (15 for aces; 10 for tens, jacks, queens, and kings; and 5 for all other cards).

Solution

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

Try It \(\PageIndex{4}\)

1. Interpret the expected value of the number shown when you roll a special six-sided die with faces \(\{1, 1, 2, 3, 5, 8\}\).
2. Interpret the expected value of the number of heads shown when you flip a coin three times.
3. You are about to play a game where you flip a coin three times. If all three flips result in heads, you win $20. If you get two heads, you win $10. If you get one or zero heads, you win nothing. Interpret the expected value of your winnings.

Answer

1. The expected value of the number shown when you roll this die is approximately 3.33. This means that if you roll the die many times, the average value of the numbers you roll will approach 3.33 over the long run. Even though 3.33 is not one of the actual faces, it represents the long-run mean outcome across many repetitions of the experiment.
2. This means that if you repeat the experiment of flipping a coin three times over and over again, the average number of heads per trial will approach 1.5 in the long run. While you can never actually get exactly 1.5 heads in a single trial, the expected value represents the mean outcome across many repetitions of the experiment.
3. The expected value of your winnings is $6.25. This means that if you played this game many times, you would win an average of $6.25 per game in the long run. You might win $20 in some games, $10 in others, and nothing in many - but over time, your average winnings per game would approach $6.25.

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 win $10. However, if you roll 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 \(\displaystyle\frac{1}{6}\)​, and the probability of losing is \(\displaystyle\frac{5}{6}\). So the expected value is: \(\displaystyle \$10 \times \frac{1}{6} + (-\$1) \times \frac{5}{6} = \frac{5}{6} \approx \$0.83\). 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: \(\displaystyle \$3 \times \frac{1}{6} + (-\$1) \times \frac{5}{6} = -\frac{2}{6} \approx -\$0.33\), meaning you should expect to lose about 33 cents on average 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.

Who Knew?: Expected Values in Football

In the 21st century, data analytics tools have revolutionized the way sports are coached and played. One such tool is used in football at crucial moments during the game. When a team faces a fourth down - the final play in a series of four attempts, which is a fairly common occurrence - the coach must make a decision: run a play to try to gain the required number of yards, or kick the ball to the other team.

Here’s the interesting part of the decision: if the team decides to “go for it” and successfully gains the necessary yards, they retain possession of the ball and can continue their drive to score more points. If they are unsuccessful, they lose possession, giving the opposing team a chance to score. If the team instead chooses to punt, or kick the ball away, the opposing team still gains possession, but typically in a less advantageous field position than if the original team had gone for it and failed.

To analyze this situation, data analysts have generated empirical probabilities for every fourth down scenario and calculated the expected value - in terms of points - for each possible decision. Coaches frequently use these calculations when deciding which option to take.

Example \(\PageIndex{5}\): Using Expected Values

In the casino game Keno, a machine randomly selects 20 numbers between 1 and 80 (inclusive) without replacement. Players try to predict which numbers will be chosen. They don’t usually attempt to guess all 20; instead, they generally try to predict between 1 and 10 of the selected numbers. The amount won depends on how many numbers the player chooses and how many of those guesses are correct.

At one casino:

1. A player can choose just one number. If that number is among the 20 selected, the player wins $2; otherwise, they lose $1. What is the expected value?
2. If a player chooses two numbers and both are correct, they win $14; otherwise, they lose $1. What is the expected value?
3. A player chooses three numbers; If two out of three are correct, they win $1. If all three are correct, they win $42. Otherwise, they lose $1. What is the expected value?
4. Which of these games is best for the player? Which is best for the casino?

Solution

1. There are 20 winning numbers out of 80, so if we try to guess one of them, the probability of guessing correctly is \(\displaystyle \frac{20}{80} = \frac{1}{4}\)​. The probability of losing is then \(\displaystyle 1 - \frac{1}{4} = \frac{3}{4},\) so the expected value is:  \(\displaystyle \$2 \times \frac{1}{4} + (-\$1) \times \frac{3}{4} = -\$0.25\).
2. There are \(_{20}C_{2} = 190\) winning combinations out of \(_{80}C_{2} =3,160\) total ways to choose two numbers. So, the probability of winning is \(\displaystyle \frac{190}{3,160}​,\) and the probability of losing is \(\displaystyle 1 - \frac{190}{3,160} = \frac{2,970}{3,160}\). Thus, the expected value of the game is: \(\displaystyle \$14 \times \frac{190}{3,160} + (-\$1) \times \frac{2,970}{3,160} \approx -\$0.10\).

3. Step 1: Let’s start with the big prize. There are \(_{20}C_{3} = 1,140\) ways to correctly guess three winning numbers out of \(_{80}C_{3} = 82,160\) total possible combinations of three numbers. That means the probability of winning the big prize is \(\displaystyle \frac{1,140}{82,160} \approx 0.01388.\)

   Step 2: Let’s find the probability of winning the second prize. The denominator remains the same: 82,160. Now, let’s determine the numerator. To win the second prize, the player must pick two of the 20 winning numbers and one of the 60 losing numbers. We can use the Multiplication Rule for Counting: there are \(\displaystyle _{20}C_{2} = 190\) ways to choose two winning numbers and 60 ways to choose one losing number. So, there are \(190 \times 60 = 11,400\) ways to win the second prize. Therefore, the probability of winning the second prize is: \(\displaystyle \frac{11,400}{82,160} \approx 0.13875\).

   Step 3: Since the overall probability of winning is \(\displaystyle \frac{1,140}{82,160} + \frac{11,400}{82,160} = \frac{12,540}{82,160} \approx 0.15263,\)  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 best bet for the player is the one with the highest expected value, which is guessing two numbers. The best bet for the casino is the one with the lowest expected value for the player, which is guessing one number.

Try It \(\PageIndex{5}\)

The casino game Craps involves rolling two standard six-sided dice. While the main game includes repeated rolls, players can also place bets on the outcomes of single rolls. Find the expected values of the following three $1 bets, then decide which bet is best for the player and which is best for the casino.

1. Players can bet that the next roll of the dice will have a sum of 7. Winners receive $4, and losers lose $1.
2. Players can bet that the next roll will have a sum of 12. Winners receive $30, and losers lose $1.
3. Players can bet that the next roll will be “any craps” (a sum of 2, 3, or 12). Winners receive $7, and losers lose $1.

Answer

1. If the player bets on 7, the expected value is \(-\$0.17\).
2. If the player bets on 12, the expected value is \(-\$0.14\).
3. If the player bets on any craps, the expected value is \(-\$0.11\).

The best bet for the player is any craps; the best bet for the casino is the bet on 7.

Check Your Understanding

1. You are about to roll a 20-sided die with the following distribution of faces: five faces show a 1; six faces show a 3; four faces show a 5; three faces show a 7; two faces show a 9.
   1. What is the expected value of the number showing on the die after it’s rolled?
   2. Interpret your answer.
2. You are about to play a game in which you draw three ping-pong balls without replacement from a barrel containing six green balls and four red balls. If all three of your selections are green, you win $5. If two of the three are green, you win $1. If two or more of your selections are red, you lose $5.
   1. What is the expected value of this game?
   2. Interpret your answer.
   3. Is it advantageous for you to play the game? How do you know?

Display Answers

1. 1. 4.1
   2. If you roll this die many times, the mean of the numbers rolled will be around 4.1.
2. 1. \(-$0.33\)
   2. This means that if you play the game many times, you can expect to lose about 33 cents per play on average.
   3. No; the expected value is negative.

Exercises

1. You roll a standard six-sided die and win points equal to the square of the number shown.
   1. What is the expected value of the number of points you win?
   2. Interpret your answer.
2. In the classic board game The Game of Life, players have the chance to play the market. A spinner with 10 equally likely spaces is spun to select a random number. If the result is 3 or less, the player loses $25,000. If the result is 7 or more, the player wins $50,000. If the result is 4, 5, or 6, the player neither wins nor loses anything.
   1. What is the expected value of playing the market?
   2. Interpret the answer. 
3. The Game of Life players also occasionally have the opportunity to speculate. Players choose any 2 of the 10 numbers on the spinner and then give it a spin. If one of their numbers is selected, they win $140,000; if not, they lose $10,000.
   1. What is the expected value of this speculation?
   2. Interpret your answer.
   3. Which is better for The Game of Life players: playing the market or speculating? How do you know?
4. A charitable organization is selling raffle tickets as a fundraiser. They intend to sell 5,000 tickets at $10 each. One ticket will be randomly selected to win the grand prize of a new car worth $35,000.
   1. What is the expected value of a single ticket?
   2. Interpret your answer.
   3. The organization is worried they won’t be able to sell all the tickets, so they announce that, in addition to the grand prize, they will offer 10 second prizes of $500 in cash. What is the new expected value of a single ticket?
   4. Interpret your answer.
5. In the following exercises, you will randomly select golf balls from a bucket. The bucket contains 4 yellow balls (numbered 1–4) and 6 white balls (numbered 1–6).

   1. If you draw a single ball, what is the expected number of yellow balls selected?

   2. Suppose you draw 2 balls with replacement: 
      a. Give a probability distribution (PDF) table for the possible outcomes for the number of yellow balls selected.
      b. What is the expected number of yellow balls selected?

   3. Suppose you draw 2 balls without replacement: 
      a. Give a PDF table for the possible outcomes for the number of yellow balls selected.
      b. What is the expected number of yellow balls selected?

   4. Suppose you draw 3 balls with replacement:
      a. Give a PDF table for the possible outcomes for the number of yellow balls selected.
      b. What is the expected number of yellow balls selected?

   5. Suppose you draw 3 balls without replacement: 
      a. Give a PDF table for the possible outcomes for the number of yellow balls selected.
      b. What is the expected number of yellow balls selected?

   6. If you draw a single ball, what is the expected value of the number on the ball?

   7. Suppose you draw 2 balls with replacement: 
      a. Give a PDF table for the possible outcomes for the sum of the numbers on the selected balls.
      b. What is the expected sum of the numbers on the balls?

   8. Suppose you draw 2 balls without replacement: 
      a. Give a PDF table for the possible outcomes for the sum of the numbers on the selected balls.
      b. What is the expected sum of the numbers on the balls?

6. The following exercises deal with the game Punch a Bunch, which appears on the TV game show The Price Is Right. In this game, contestants have a chance to punch through up to four paper circles on a board; behind each circle is a card with a dollar amount printed on it. There are 50 of these circles, and the dollar amounts are given in the table below. Contestants are shown their selected dollar amounts one at a time, in the order chosen. After each amount is revealed, the contestant is given the option of taking that amount of money or discarding it in favor of the next one. (You can watch the game being played in the video "Punch a Bunch.") 

   Dollar Amount	$25,000	$10,000	$5,000	$2,500	$1,000	$500	$250	$100
   Frequency	1	2	4	8	10	10	10	5

    Anita is playing Punch a Bunch and gets two punches.

   1. If Anita got $500 on her first punch, what is the expected value of her second punch?
   2. If Anita got $500 on her first punch, should she discard the $500 and take the result of her second punch? How do you know?
   3. If Anita got $1,000 on her first punch, what is the expected value of her second punch?
   4. If Anita got $1,000 on her first punch, should she discard the $1,000 and take the result of her second punch? How do you know?
   5. If Anita got $2,500 on her first punch, what is the expected value of her second punch?
   6. If Anita got $2,500 on her first punch, should she discard the $2,500 and take the result of her second punch? How do you know?
7. The following exercises are about the casino game, Roulette. In this game, the dealer spins a marble around a wheel that contains 38 pockets into which the marble can fall. Each pocket has a number (each whole number from 0 to 36, along with a “double zero”) and a color. The 0 and 00 are both green, while the other 36 numbers are evenly divided between black and red. Players place bets on which number (or group of numbers) they think the marble will land on. The figure shows the layout of the numbers and colors, as well as some of the possible bets. 

   

   Roulette Table (credit: "American Roulette Table Layout" by Film8ker/Wikimedia Commons, Public Domain)

   1. If a player makes a $1 bet on a single number, they win $35 if that number comes up, but lose $1 if it doesn’t. What is the expected value of this bet?
   2. Interpret your answer to the previous question.
   3. Suppose a player makes the $1 bet on a single number in 5 consecutive spins. What is the expected value of this series of bets? (Hint: Use the Binomial Distribution.)
   4. Interpret your answer to the previous question.
   5. If a player makes a $10 bet on the first dozen, they win $20 if one of the numbers 1–12 comes up, but lose $10 otherwise. What is the expected value of this bet?
   6. Interpret your answer to the previous question.
   7. Suppose a player makes the $10 bet on the first dozen in 4 consecutive spins. What is the expected value of that series of bets?
   8. Interpret your answer to the previous question.
   9. If a player makes a $10 basket bet, they win $60 if 0, 00, 1, 2, or 3 comes up, but lose $10 otherwise. What is the expected value of this bet?
   10. Interpret your answer to the previous question.
   11. Which is better for the player: a $10 first dozen bet or a $10 basket bet? How do you know?