2.5 Expected Value

Robert Foth

2.5 Expected Value

2.5: Expected Value

Learning Objectives

Upon completion of this section, you should be able to

Compute expected values

Expected Value

Expected value is perhaps the most useful probability concept we will cover in this chapter. It provides a method to calculate the average outcome of an uncertain event if it were to be repeated many times. It has many applications, from insurance policies to making financial decisions, and it’s one thing that the casinos and government agencies that run gambling operations and lotteries hope most people never learn about.

Example 1

Roulette wheel In the casino game roulette, a wheel with 38 spaces (18 red, 18 black, and 2 green) is spun. In one possible bet, the player bets $1 on a single number. If that number is spun on the wheel, then they receive $36 (their original $1 + $35). Otherwise, they lose their $1. On average, how much money should a player expect to win or lose if they play this game repeatedly?

Solution

Suppose you bet $1 on each of the 38 spaces on the wheel, for a total of $38 bet.

After the wheel is spun just one of those 38 numbers is going to win, so you are paid $36 on that wining number. While you won on that one number, overall you’ve lost $2 total as bet an initial $38. On a per-space basis, you have “won” $- \frac{$ 2}{$ 38} \approx - $ 0.053$ .

In other words, on average you lose 5.3 cents per space you bet on.

Video Solution Example 1 (2 mins and 13 seconds – CC) Another example starts after 2:42 mark on the video.

In the above example we call this average gain or loss the expected value of playing roulette. Notice that no one ever loses exactly 5.3 cents: most people (in fact, about 37 out of every 38) lose $1 and a very few people (about 1 person out of every 38) gain $35 (the $36 they win minus the $1 they spent to play the game).

There is another way to compute expected value without imagining what would happen if we play every possible space. There are 38 possible outcomes when the wheel spins, so the probability of winning is $\frac{1}{38}$ . The complement, the probability of losing, is $\frac{37}{38}$ .

Summarizing these along with the values, we get this table:

Outcome	Probability of outcome
$35	$\frac{1}{38}$
-$1	$\frac{37}{38}$

Notice that if we multiply each outcome by its corresponding probability we get $$ 35 \cdot \frac{1}{38} = 0.9211$ and $- $ 1 \cdot \frac{37}{38} = - 0.9737$ , and if we add these numbers we get $0.9211 + (- 0.9737) \approx - 0.053$ , which is the expected value we computed above.

The video below summarizes the results from above.
Introduction to Expected Values (10 mins 23 secs)

Expected Value

Expected Value is the average gain or loss of an event if the procedure is repeated many times. It is often referred to as the “long-term” average or mean for a probability experiment.

We can compute the expected value by multiplying each outcome in the sample space by the probability of that outcome, then adding up those products.

If $S = {x_{1}, x_{2}, \dots x_{k}}$ , then

$Expected Value = x_{1} \cdot P (x_{1}) + x_{2} \cdot P (x_{2}) + \dots x_{k} \cdot P (x_{k})$

Example 2

Bob wants to know if you would play his game. It costs $2 to play and you do not get the $2 back if you win or lose. The game consists of selecting a “winning” colored marble from the bag.

–There is 1 green marble and if you select one you win $30.

–There are 4 red marbles and if you select one you win $3.

–There are 8 white marbles and if you select one you win $1.

–There are 7 black marbles and if you select one you win $0.

Is it in your favor to play this game? Answer the question using the idea of Expected Value (calculate the expected value and decide for yourself).

Solution

Start with looking at the outcomes of the game overall (the net profit). There are four things that can happen and each will have thir own net profit and probabilities associated with them. Create a table showing the net profits and probability.

If we draw a green marble it states we win $30, but the actual net profit is $$ 30 - $ 2 = $ 28$ . We can do that for each of the other three outcomes of the experiment and put that in a table (Selecting red has a net profit of $1, white has a net profit of $-1, and black has a net profit of $-2)

Marble Color Selected

Green

Red

White

Black

Net Profit (Outcomes)

28

1

-1

-2

Probability

Next let us add in the probability for each of those outcomes. We will assume each marble is equal likely to be selected. The probability will be the number of marbles of the given color over the total number of marbles. Enter these values into the table.

Net Profit (Outcomes)	28	1	-1	-2
Probability	$\frac{1}{20}$	$\frac{4}{20}$	$\frac{8}{20}$	$\frac{7}{20}$

To find the expected value we need to multiply each outcome times their respective probability, so add one more row on the table that shows the producs. We will then add all those products on the last row to calculate the expected value.

Net Profit (Outcomes)

28

1

-1

-2

Probability

\frac{1}{20}

\frac{4}{20}

\frac{8}{20}

\frac{7}{20}

Outcome·Probabilty

28 \cdot \frac{1}{20}

1 \cdot \frac{4}{20}

- 1 \cdot \frac{8}{20}

- 2 \cdot \frac{7}{20}

Now add the products together to get the expected value

$\begin{array}{l} Expected Value = x_{1} \cdot P (x_{1}) + x_{2} \cdot P (x_{2}) + \dots x_{k} \cdot P (x_{k}) \\ Expected Value = (28) \cdot \frac{1}{20} + (1) \cdot \frac{4}{20} + (- 1) \cdot \frac{8}{20} + (- 2) \cdot \frac{7}{20} \\ Expected Value = \frac{28}{20} + \frac{4}{20} - \frac{8}{20} - \frac{14}{20} \\ Expected Value = \frac{28 + 4 - 8 - 14}{20} \\ Expected Value = \frac{10}{20} \\ Expected Value = $ 0.50 \end{array}$

The expected value or long run average is $0.50 for the game. This means if you were play this game from Bob many many times and average your winnings and losses that you expect to earn $0.50 per game played. It is definitely in your favor to play.

Video Solution for Example 2 (1 min and 44 seconds – CC)

Try it Now 1

You purchase a raffle ticket to help out a charity. The raffle ticket costs $5. The charity is selling 2000 tickets. One of them will be drawn and the person holding the ticket will be given a prize worth $4000. Compute the expected value for this raffle.

Hint 1 (click to Show/Hide)

In order to find the expected value we need all of the outcomes and associated probabilities. Think of the outcomes as either the ticket holder wins $3995 (the prize minus the cost of the ticket) or the ticket holder loses $5. What is the probability of each of those outcomes? Once we have these outcomes and probabilities use the definition of expected value.

Expected Value is the average gain or loss of an event if the procedure is repeated many times.

We can compute the expected value by multiplying each outcome by the probability of that outcome, then adding up the products.

Answer (click to Show/Hide)

Start by looking for the probability of winning $3995 and losing $5. Then calculate the expected value where the outcomes were $3995 and -$5.

$\begin{array}{l} P (win $ 3995) = \frac{1}{2000} \\ P (lose $ 5) = \frac{1999}{2000} \end{array}$

$($ 3, 995) \cdot \frac{1}{2000} + (- $ 5) \cdot \frac{1999}{2000} \approx - $ 3.00$

The conclusion is that on average a player expects to lose $3 for each raffle ticket they purchase (or that on average $3 is being donated to charity for each raffle ticket that is purchased).

Example 3

In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. If they match 5 numbers, then win $1,000. It costs $1 to buy a ticket. Find the expected value.

Solution

Earlier, we calculated the probability of matching all 6 numbers and the probability of matching 5 numbers:

Probability for matching all 6 numbers,
$P (matching 6) = \frac{6 C 6}{48 C 6} = \frac{1}{2271512} \approx 0 .0000000815$
Probability for matching 5 numbers,
$P (matching 5) = \frac{(6 C 5) (42 C 1)}{48 C 6} = \frac{252}{2271512} \approx 0 .0000205$
Probability for not matching 5 or 6 numbers can be found using the complement rule,
$P (Not matching 5 or 6) = 1 - P (matching 5 or 6)$

Now since matching 5 or matching 6 are mutually exclusive we can add their probabilities together to find the probability of matching 5 or 6.

$\begin{array}{l} P (Not matching 5 or 6) = 1 - (P (matching 5) + P (matching 6)) \\ P (Not matching 5 or 6) = 1 - (\frac{1}{2271512} + \frac{252}{2271512}) \\ P (Not matching 5 or 6) = 1 - (\frac{253}{2271512}) \\ P (Not matching 5 or 6) = \frac{12271259}{2271512} \end{array}$

Our probabilities and outcome values are:

Outcome	Probability of outcome
$999,999	$\frac{1}{12271512}$
$999	$\frac{252}{12271512}$
-$1	$1 - \frac{253}{12271512} = \frac{12271259}{12271512}$

The expected value, then is:

$($ 999, 999) \cdot \frac{1}{12271512} + ($ 999) \cdot \frac{252}{12271512} + (- $ 1) \cdot \frac{12271259}{12271512} \approx - $ 0.898$

On average, one can expect to lose about 90 cents on a lottery ticket. Of course, most players will lose $1.

Video Solution Example 3 (2 mins and 15 seconds – CC)

Example 4

The pick 5 lottery works by selecting 5 numbers from 1 to 35 (no repetition). It costs $1.

–Pick all 5 win Jackpot (estimated at $50000)

–Pick 4 win $500

–Pick 3 win $5

What is the expected winnings of a ticket?

Solution

Start by listing the outomes for the lottery in terms of the net profit.

Outcome	Probability of outcome
$49,999
$499
$4
-$1

To find the probability of each we will need to use both the combination formulas and the multiplication principle for the numerator term where we find the number of ways to select how many numbers match the winning numbers of interest times the number of ways for the remaining non winning numbers.

Outcome	Probability of outcome
$49,999	$\frac{5 C 5}{35 C 5} = \frac{1}{324, 632}$
$499	$\frac{(5 C 4) (30 C 1)}{35 C 5} = \frac{150}{324, 632}$
$4	$\frac{(5 C 3) (30 C 2)}{35 C 5} = \frac{4350}{324, 632}$
-$1	$1 - (\frac{1}{324, 632} + \frac{150}{324, 632} + \frac{4350}{324, 632}) = \frac{320131}{324, 632}$

The expected value will be the sum of each outcomes times their probability:

$\begin{array}{l} EV = (49999) \cdot (\frac{1}{324, 632}) + (499) \cdot (\frac{150}{324, 632}) + (4) \cdot (\frac{4350}{324, 632}) + (- 1) \cdot (\frac{320131}{324, 632}) \\ EV = - \frac{177882}{324632} \approx - 0.55 \end{array}$

The result shows the player should expect that on average over a large number of trials of buying the lottery ticket to lose approximately $0.55 per ticket.

Video Solution Example 4 (14 mins and 58 seconds – CC)

In general, if the expected value of a game is negative, it is not a good idea to play, since on average you will lose money. It would be better to play a game with a positive expected value (good luck trying to find one!), although keep in mind that even if the average winnings are positive it could be the case that most people lose money and one very fortunate individual wins a great deal of money. If the expected value of a game is 0, we call it a fair game, since neither side has an advantage.

Try it Now 2

A friend offers to play a game, in which you roll 3 standard 6-sided dice. If all the dice roll different values, you give him $1. If any two dice match values, you get $2. What is the expected value of this game? Would you play?

Hint 1 (click to Show/Hide)

The outcomes for the experiment is lose $1 or win $2. To find the expected value you will need to find the probability associated with each outcome. If we roll three dice there are 6*6*6=216 different outcomes of the three dice rolls according to the fundamental counting principle. Which probability is easier to find? If we find that probability can we find the probability of the other outcome with a simple formula?

Hint 2 (click to Show/Hide)

Try finding the probability that none are the same and use the complement rule to find the probability that at least two are equal.

Answer (click to Show/Hide)

To find the probability that none are the same we take the number of ways no dice outcome in three rolls is the same over the total number of ways we can have three dice roll outcomes. We already saw the second portion that the number of ways to roll three dice is to have a total of 216 different outcomes in Hint 1.

We can treat the question of getting three different dice outcomes as a permutation problem or a fundamental counting problem. Think of the first dice outcome as having six choices, then the second dice outcome if not being equal to the first will only have five choices, and the third outcome not equaling with the first or second will only have four choices. This gives us:

$n (all three dice outcome different) = 6 \cdot 5 \cdot 4 = 120$

Now find the probability of each outcome.

$\begin{array}{l} P (Lose $1: All different) = \frac{n (all three different)}{n (roll three dice)} \\ P (Lose $1: All different) = \frac{120}{216} = \frac{5}{9} \end{array}$

If we do not win then the only other outcome is lose in this situation – these two events are complements of each other, so to find the probability of winning we use the complement rule.

$\begin{array}{l} P (Win $2: at least two the same) = 1 - P (Lose $1: All different) \\ P (Win $2: at least two the same) = 1 - \frac{5}{9} \\ P (Win $2: at least two the same) = \frac{4}{9} \end{array}$

Now we have all the information needed to find the expected value.

Expected value is:

$($ 2) P (Win $2) + (- $ 1) P (Lose $1) = (2) (\frac{4}{9}) + (- 1) (\frac{5}{9}) = \frac{3}{9} \approx $ 0.33$

This positive value tell us that it is in your favor (not your friends) to play this game. On average you would win approximately $0.33 per play.

Expected value also has applications outside of gambling. Expected value is very common in making insurance decisions.

Example 5

An insurance company determined that a 40-year-old man in the U.S. has a 0.242% risk of dying during the next year. The insurance company charges $275 for a 1 year life-insurance policy that pays $100,000 death benefit. What is the expected value for the person buying the insurance?

Solution

The outcomes in this situation is the net profit for the person buying the life insurance. One outcome is that the person is still alive (and probably greatful for that), but is out the $275 used to purchase the life insurance. The other outcome means the person didn’t survive, but the death benefit is paid out to the beneficiary of the policy. This payout net profit would be $$ 100, 000 - $ 275 = $ 99, 725$ . Below in the table we have the outcomes and the associated probabilities.

Outcome

Probability of outcome

$99,725

0.00242

-$275

1 - 0.00242 = 0.99758

The expected value is $($ 99, 725) (0.00242) + (- $ 275) (0.99758) = - $ 33$ .

The person buying the insurance expects to lose overall $33 for each such policy when we look at a long run average. The flip side is more important to why insurance companies can stay in business. They expect to see many of these policies over the year. Each policy give an averge return of $33 for them.

Not surprisingly, the expected value is negative; the insurance company can only afford to offer policies if they, on average, make money on each policy. They can afford to pay out the occasional benefit because they offer enough policies that those benefit payouts are balanced by the rest of the insured people.

For people buying the insurance, there is a negative expected value, but there is a security that comes from insurance that is worth that cost.

Video Solution Example 5 (3 mins and 27 seconds – CC)

Exercises

A bag contains 3 gold marbles, 6 silver marbles, and 28 black marbles. Someone offers to play this game: You randomly select on marble from the bag. If it is gold, you win $3. If it is silver, you win $2. If it is black, you lose $1. What is your expected value if you play this game?

Answer (click to Show/Hide)

To find the expected value, we first need to determine the probability of each outcome. Then, we multiply the value (amount won or lost) of each outcome by its probability and sum the results.

Step 1: Find the total number of outcomes and the probability of each outcome.
The total number of marbles in the bag is the total number of possible outcomes.
$3 gold + 6 silver + 28 black = 37 total marbles$
The probabilities are:

$P (Gold) = \frac{3}{37}$

$P (Silver) = \frac{6}{37}$

$P (Black) = \frac{28}{37}$

Step 2: Set up an expected value table.
We can organize the information in a table with outcomes as columns. Remember that a loss is represented by a negative value.

Gold Silver Black

Value (x) $3 $2 -$1

Probability P(x) $\frac{3}{37}$ $\frac{6}{37}$ $\frac{28}{37}$

Step 3: Calculate the expected value.
The expected value is the sum of each outcome’s value multiplied by its probability.
$Expected Value = (3) \times (\frac{3}{37}) + (2) \times (\frac{6}{37}) + (- 1) \times (\frac{28}{37})$

$= \frac{9}{37} + \frac{12}{37} - \frac{28}{37} = \frac{9 + 12 - 28}{37} = \frac{- 7}{37}$
As a decimal, this is approximately -0.189.

Answer: The expected value is -7/37, or about -$0.19. This means that if you were to play this game many times, you would expect to lose an average of about 19 cents per game.
A friend devises a game that is played by rolling a single six-sided die once. If you roll a 6, he pays you $3; if you roll a 5, he pays you nothing; if you roll a number less than 5, you pay him $1. Compute the expected value for this game. Should you play this game?

Answer (click to Show/Hide)

To find the expected value, we will first determine the value and probability of each possible outcome of the die roll. Then, we will multiply the value of each outcome by its probability and sum the results.

Step 1: Find the probability of each outcome.
The sample space for a single die roll is {1, 2, 3, 4, 5, 6}.

Roll a 6: The value is +$3. The probability is 1/6.

Roll a 5: The value is +$0. The probability is 1/6.

Roll less than 5: This includes rolling a 1, 2, 3, or 4. The value is -$1. There are 4 favorable outcomes, so the probability is 4/6 or 2/3.

Step 2: Set up an expected value table.
We can organize this information in a table with the outcomes as columns.

Roll a 6 Roll a 5 Roll less than 5

Value (x) $3 $0 -$1

Probability P(x) $\frac{1}{6}$ $\frac{1}{6}$ $\frac{4}{6}$

Step 3: Calculate the expected value.
The expected value is the sum of each outcome’s value multiplied by its probability.
$Expected Value = (3) \times (\frac{1}{6}) + (0) \times (\frac{1}{6}) + (- 1) \times (\frac{4}{6})$
$= \frac{3}{6} + 0 - \frac{4}{6} = \frac{3 - 4}{6} = \frac{- 1}{6}$
As a decimal, this is approximately -0.167.

Answer: The expected value is -1/6, or about -$0.17. Because the expected value is negative, you should not play this game if you want to win money over the long term. On average, you will lose about 17 cents each time you play.
In a lottery game, a player picks six numbers from 1 to 23. If the player matches all six numbers, they win 30,000 dollars. Otherwise, they lose $1. Find the expected value of this game.

Answer (click to Show/Hide)

To find the expected value, we need to find the value and probability of each outcome (winning or losing). Since the order of the numbers does not matter, we use combinations to find the total number of possible outcomes.

Step 1: Find the total number of outcomes and the probability of each outcome.
The total number of ways to choose 6 numbers from 23 is C(23, 6).
$C (23, 6) = \frac{23!}{6! (23 - 6)!} = \frac{23!}{6! 17!} = 100,947$
There are 100,947 possible combinations.

Winning: There is only 1 winning combination. The probability is $P (Win) = \frac{1}{100,947}$ .

Losing: The rest of the combinations are losing ones. The probability is $P (Lose) = \frac{100,946}{100,947}$ .

Step 2: Set up an expected value table.
We can organize this information in a table.

Win Lose

Value (x) $30,000 -$1

Probability P(x) $\frac{1}{100,947}$ $\frac{100,946}{100,947}$

Step 3: Calculate the expected value.
The expected value is the sum of each outcome’s value multiplied by its probability.

$\begin{aligned} Expected Value & = (30000) \times (\frac{1}{100947}) + (- 1) \times (\frac{100946}{100947}) \\ = \frac{30000}{100947} - \frac{100946}{100947} \\ = \frac{30000 - 100946}{100947} \\ = \frac{- 70946}{100947} \end{aligned}$

As a decimal, this is approximately -0.7028.

Answer: The expected value is approximately -$0.70. This means that if you play this game repeatedly, you can expect to lose an average of about 70 cents for every ticket you buy.
A game is played by picking two cards from a standard 52 deck of cards. If they are the same value, then you win $5, otherwise you lose $1. What is the expected value of this game?

Answer (click to Show/Hide)

To find the expected value, we need the value and probability of each outcome (winning or losing).

Step 1: Find the probability of winning.
A “win” occurs if the second card drawn has the same value as the first. The first card can be any card, so after the first card is drawn, there are 51 cards remaining in the deck. Of those 51 cards, exactly 3 of them have the same value as the first card. Therefore, the probability of the second card matching the first is 3 out of 51.

$P (Win) = \frac{3}{51} = \frac{1}{17}$

Step 2: Find the probability of losing.
The event “losing” is the complement of “winning.”
$P (Lose) = 1 - P (Win) = 1 - \frac{1}{17} = \frac{16}{17}$

Step 3: Set up an expected value table.

Win Lose

Value (x) $5 -$1

Probability P(x) $\frac{1}{17}$ $\frac{16}{17}$

Step 4: Calculate the expected value.
$\begin{aligned} Expected Value & = (5) \times (\frac{1}{17}) + (- 1) \times (\frac{16}{17}) \\ = \frac{5}{17} - \frac{16}{17} \\ = \frac{- 11}{17} \end{aligned}$
As a decimal, this is approximately -0.647.

Answer: The expected value is -11/17, or about -$0.65. This means that if you were to play this game many times, you would expect to lose an average of about 65 cents per game.
A company estimates that 0.7% of their products will fail after the original warranty period but within 2 years of the purchase, with a replacement cost of $350. If they offer a 2 year extended warranty for $48, what is the company’s expected value of each warranty sold?
Answer (click to Show/Hide)
To find the company’s expected value, we need to consider the value and probability of each possible outcome from the company’s perspective.

Step 1: Identify the outcomes, their values, and their probabilities.
There are two possible outcomes for the company when they sell a warranty:

Outcome 1: The product fails.
- Value: The company earns $48 for the warranty but has to pay $350 for the replacement. The net value is $48 - 350 = - 302$ .
- Probability: The probability of failure is given as 0.7%, which is 0.007.
Outcome 2: The product does not fail.
- Value: The company earns $48 from the warranty and has no additional costs. The net value is $48.
- Probability: This is the complement of the product failing. The probability is $1 - 0.007 = 0.993$ .
Step 2: Set up an expected value table.

Product Fails Product Does Not Fail

Value (x) -$302 $48

Probability P(x) 0.007 0.993

Step 3: Calculate the expected value.
The expected value is the sum of each outcome’s value multiplied by its probability.
$\begin{aligned} Expected Value & = (- 302) \times (0.007) + (48) \times (0.993) \\ = - 2.114 + 47.664 \\ = 45.55 \end{aligned}$

Answer: The company’s expected value for each warranty sold is $45.55. On average, the company can expect to profit $45.55 for each extended warranty it sells.
An insurance company estimates the probability of an earthquake in the next year to be 0.0013. The average damage done by an earthquake it estimates to be $60,000. If the company offers earthquake insurance for $100, what is their expected value of the policy?

Answer (click to Show/Hide)

To find the company’s expected value, we need to consider the value and probability of each possible outcome from the company’s perspective.

Step 1: Identify the outcomes, their values, and their probabilities.
There are two possible outcomes for the company when they sell a policy:

Outcome 1: An earthquake occurs.

Value: The company earns $100 for the policy but has to pay out $60,000 for damages. The net value is $100 - 60000 = - 59900$ .

Probability: The probability of an earthquake is given as 0.0013.

Outcome 2: No earthquake occurs.

Value: The company earns $100 from the policy and has no additional costs. The net value is $100.

Probability: This is the complement of an earthquake occurring. The probability is $1 - 0.0013 = 0.9987$ .

Step 2: Set up an expected value table.

Earthquake Occurs No Earthquake Occurs

Value (x) -$59,900 $100

Probability P(x) 0.0013 0.9987

Step 3: Calculate the expected value.
The expected value is the sum of each outcome’s value multiplied by its probability.
$\begin{aligned} Expected Value & = (- 59900) \times (0.0013) + (100) \times (0.9987) \\ = - 77.87 + 99.87 \\ = 22 \end{aligned}$

Answer: The company’s expected value for each policy sold is $22.00. This means that, on average, the company can expect to profit $22 for each earthquake insurance policy it sells.
You roll two dice. If the sum of the faces is eight or more, you win $7. If the sum is 7 you win $1 and if the sum is less than 7 you lose $8. Is this a fair game? Justify your answer using expected values.

Answer (click to Show/Hide)

To determine if the game is fair, we need to compute its expected value. A fair game has an expected value of 0.

Step 1: Find the probability of each outcome.
There are 36 possible outcomes when rolling two six-sided dice. We need to count the number of outcomes for each event.

Sum is 8 or more: The outcomes are (2,6), (3,5), (4,4), (5,3), (6,2), (3,6), (4,5), (5,4), (6,3), (4,6), (5,5), (6,4), (5,6), (6,5), (6,6). There are 15 outcomes.
$P (Sum ≥ 8) = \frac{15}{36}$

Sum is 7: The outcomes are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). There are 6 outcomes.
$P (Sum = 7) = \frac{6}{36}$

Sum is less than 7: The remaining outcomes total $36 - 15 - 6 = 15$ .
$P (Sum < 7) = \frac{15}{36}$

Step 2: Set up an expected value table.

Sum ≥ 8 Sum = 7 Sum < 7

Value (x) $7 $1 -$8

Probability P(x) $\frac{15}{36}$ $\frac{6}{36}$ $\frac{15}{36}$

Step 3: Calculate the expected value.
The expected value is the sum of each outcome’s value multiplied by its probability.
$\begin{aligned} Expected Value & = (7) \times (\frac{15}{36}) + (1) \times (\frac{6}{36}) + (- 8) \times (\frac{15}{36}) \\ = \frac{105}{36} + \frac{6}{36} - \frac{120}{36} \\ = \frac{- 9}{36} = - \frac{1}{4} \end{aligned}$

Answer: The game is not fair as the expected value is -$0.25. This means that if you were to play this game many times, you would expect to lose an average of about 25 cents per game.
You pay $10 to play a game of chance. There are twenty balls in a bag. Seven of them are red, five are green, and the rest are yellow. You are to draw one ball from the bag. You will win $14 if you draw a yellow ball, you will win $12 if you draw a red ball, and you win $1 if you draw a green ball. How much do you expect to win or lose after you have played one hundred games? Use the word “win” or “lose” in your answer.

Answer (click to Show/Hide)

First, we need to find the expected value for a single game. Then, we can multiply that result by 100 to find the expected outcome after one hundred games.

Step 1: Find the probability of each outcome.
There are 20 balls in total.

Red balls: 7, so $P (Red) = \frac{7}{20}$

Green balls: 5, so $P (Green) = \frac{5}{20}$

Yellow balls: $20 - 7 - 5 = 8$ , so $P (Yellow) = \frac{8}{20}$

Step 2: Determine the net value of each outcome.
We must subtract the $10 cost to play from the winnings for each outcome.

Yellow: $14 - 10 = 4$

Red: $12 - 10 = 2$

Green: $1 - 10 = - 9$

Step 3: Set up an expected value table.

Yellow Red Green

Value (x) $4 $2 -$9

Probability P(x) $\frac{8}{20}$ $\frac{7}{20}$ $\frac{5}{20}$

Step 4: Calculate the expected value for one game.
$\begin{aligned} Expected Value & = (4) \times (\frac{8}{20}) + (2) \times (\frac{7}{20}) + (- 9) \times (\frac{5}{20}) \\ = \frac{32}{20} + \frac{14}{20} - \frac{45}{20} \\ = \frac{32 + 14 - 45}{20} \\ = \frac{1}{20} = 0.05 \end{aligned}$
The expected value of a single game is $0.05.

Step 5: Calculate the expected outcome for 100 games.
$Total Expected Value = 0.05 \times 100 = 5$

Answer: You expect to win $5 after playing one hundred games.
In a gambling game a person is paid $3 if they draw a Jack or a Queen and $5 if they draw a King or an Ace from an ordinary deck of 52 playing cards. If they draw any other card, they lose. How much should they pay to play if the game is to be fair.

Answer (click to Show/Hide)

For a game to be fair, the expected value must be $0. This means the cost to play must be equal to the expected winnings per game. First, we will calculate the expected winnings.

Step 1: Find the probability of each outcome.
There are 52 cards in a standard deck.

Win $3 (Jack or Queen): There are 4 Jacks and 4 Queens, for a total of 8 cards.
$P (Win $3) = \frac{8}{52}$

Win $5 (King or Ace): There are 4 Kings and 4 Aces, for a total of 8 cards.
$P (Win $5) = \frac{8}{52}$

Lose (Win $0): The remaining cards are $52 - 8 - 8 = 36$ .
$P (Win $0) = \frac{36}{52}$

Step 2: Set up an expected value table.

Jack or Queen King or Ace Other Card

Winnings (x) $3 $5 $0

Probability P(x) $\frac{8}{52}$ $\frac{8}{52}$ $\frac{36}{52}$

Step 3: Calculate the expected winnings.
$\begin{aligned} Expected Winnings & = (3) \times (\frac{8}{52}) + (5) \times (\frac{8}{52}) + (0) \times (\frac{36}{52}) \\ = \frac{24}{52} + \frac{40}{52} + 0 \\ = \frac{64}{52} = \frac{16}{13} \\ \approx 1.23 \end{aligned}$

Answer: The expected winnings are approximately $1.23. Therefore, to make the game fair, they should pay about $1.23 to play.

Attributions

This page contains modified content from David Lippman, “Math In Society, 2nd Edition.” Licensed under CC BY-SA 4.0.
The photo of roulette wheel is from http://www.flickr.com/photos/stoneflower/. Licensed under CC-BY-SA 4.0.
This page contains content by Robert Foth, Math Faculty, Pima Community College, 2021.
Portions of the exercise solutions in this answer key were generated with the assistance of Gemini, a large language model from Google.

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	Product Fails	Product Does Not Fail
Value (x)	-$302	$48
Probability P(x)	0.007	0.993

	Earthquake Occurs	No Earthquake Occurs
Value (x)	-$59,900	$100
Probability P(x)	0.0013	0.9987

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