# 3.1: Probability Models & Expected Value

**At Grade**Created by: Bruce DeWItt

(Note - Three Separate Videos for this Section)

### Learning Objectives

- Be able to construct a probability model (expected value table) given all possible outcomes and the associated probabilities
- Be able to calculate the expected value for a situation given a probability model
- Be able to calculate missing values in a probability model given information about the expected value in a situation.

Suppose you walk into a casino. You will see all sorts of games varying from blackjack and poker to slot machines. It would not take long for you to notice that there are some players who are winning some money, sometimes a substantial amount. You might wonder how the casino makes money when they are clearly giving some money away.

Casinos have a clear understanding of expected value. The **expected value** for a situation can be thought of as the average result over the long run. In other words, it can be thought of the expected winnings or average payout for a game of chance. Consider the thinking of the owner of a casino. While there are some people who win a little, and occasionally a few people who win a lot, most people end up losing some money at the casino. The casino actually expects some people to occasionally win big. In fact it makes for great advertisement! As long as the mathematics show that the expected value is in the casino's favor, the casino will continue to make money in the long run. In this section, we will focus on how to calculate the expected value.

As mentioned above, the expected value can be thought of as the average result over the long run. Recall that to find the average value of a series of numbers, we simply add up the numbers and divide by how ever many numbers there are. For example, the average of 3, 4, 5, and 6 is 4.5 because \begin{align*}\frac{3+4+5+6}{4}=4.5\end{align*}*The expected value for a situation does not have to be one of the possible values.*

Use the concept of averages to find the expected value for the example below.

#### Example 1

A game is played in which a coin is flipped one time. If the coin lands on tails, the player wins $5. If the coin lands on heads, the player wins $10. What is the expected value for a player who plays this game one time?

#### Solution

The expected value is $7.50. This is strange because it is actually impossible for a player to win $7.50. They could only win either $5 or $10 but the average win will be $7.50. One way to see this is to actually play the game two times. If the flips come out matching their theoretical probabilities, one of the flips will be heads for $5 and the other will be tails for $10. The player will have won $15 in two games so the average win or expected value would be \begin{align*}\frac{\$5+\$10}{2}=\frac{\$15}{2}=\$7.50\end{align*}

$5+$102=$152=$7.50 .

This method works quite well in simple situations, but it gets more cumbersome as the situations get more complex. Consider the example below.

#### Example 2

Student council is raising money to support a program called "Shoes for the Homeless". A booth was set up in the lunchroom at which students could pledge a donation of $1, $5, or $10 for money towards a large shoe purchase. 125 students pledged money for this fundraiser. Eighty students pledged $1, 25 students pledged $5, and 20 students pledged $10.

a) Build a probability model for this situation.

b) What was the average donation per student?

#### Solution

a) Remember that a probability model needs probabilities, not just counts. For $1 pledges we have \begin{align*}\frac{80}{125}=0.64\end{align*}

80125=0.64 , for $5 pledges we have \begin{align*}\frac{25}{125}=0.2\end{align*}25125=0.2 , and for $10 pledges we have \begin{align*}\frac{20}{125}=0.16\end{align*}20125=0.16 . Notice that 0.64 + 0.2 + 0.16 = 1 or 100%.

b) There were 80 students who pledged $1 each for a total of $80, there were 25 students who pledged $5 each for a total of $125, and there were 20 students who pledged $10 each for a total of $200. All the pledges added together give us \begin{align*}\$80+\$125+\$200=\$405\end{align*}

$80+$125+$200=$405 . We now divide to get \begin{align*}\frac{\$405}{125}=\$3.24\end{align*}$405125=$3.24 per student. The average donation per student was $3.24.

You may have noticed that the values in the probability model in Example 2 can be used to find the average donation as well. \begin{align*}Average Donation=\left (\$1 \right )\left ({0.64} \right )+\left (\$5 \right )\left ({0.2} \right )+\left (\$10 \right )\left ({0.16} \right )=\$3.24\end{align*}

#### Example 3

What is the expected value for the total of a roll for two 6-sided dice?

#### Solution

We will address this two ways. The first method will be done by using averaging and the second method will be done by using the expected value formula.

Begin by building the sample space for the sum of two dice. As in section 1.1, we get the dice chart shown below. Notice that there are exactly 36 equally likely spaces on the grid. So instead of playing just one time, suppose we play 36 times. If everything matches the theoretical probabilities, each of these outcomes would happen exactly one time. Add the values for each of the 36 spaces and divide by 36. For simplicity, we will add diagonally to get \begin{align*}\frac{2+3+3+4+4+4+...+10+10+10+11+11+12}{36}=\frac{252}{36}=7\end{align*}

2+3+3+4+4+4+...+10+10+10+11+11+1236=25236=7 . The expected value is 7 which means that 7 is the average value of a roll of two 6-sided dice.

Let's now do this problem by building a probability model and using the expected value formula, \begin{align*}EV=\left ({value\; 1}\right )\left ( {prob. \;1}\right )+\left ({value\; 2}\right )\left ({prob.\; 2}\right )+\left ({value \;3}\right )\left ({prob. \;3}\right )+...\end{align*}

EV=(value1)(prob.1)+(value2)(prob.2)+(value3)(prob.3)+...

The probability model for this situation is given below. Does this make sense and did you verify that the total of the probabilities in the table add up to 1?

Using our expected value formula we have

\begin{align*}EV=\left(2\right)\left(\frac{1}{36}\right )+\left (3\right )\left (\frac{2}{36}\right )+\left (4\right )\left (\frac{3}{36}\right )+\left (5\right )\left (\frac{4}{36}\right )+\left (6\right )\left (\frac{5}{36}\right )+\left (7\right )\left (\frac{6}{36}\right )+\left (8\right )\left (\frac{5}{36}\right )+\left (9\right )\left (\frac{4}{36}\right )+\left ({10}\right )\left (\frac{3}{36}\right )+\left ({11}\right )\left (\frac{2}{36}\right )+\left ({12}\right )\left (\frac{1}{36}\right )=7\end{align*}The expected value for the total when two dice are rolled is 7.EV=(2)(136)+(3)(236)+(4)(336)+(5)(436)+(6)(536)+(7)(636)+(8)(536)+(9)(436)+(10)(336)+(11)(236)+(12)(136)=7 .

#### Example 4

A carnival game is being played that has several prizes that a player can win. Table 3.1 on the following page shows the probability model for this game.

a) Find the missing value.

b) Calculate the expected value and explain what it means.

Value | $30 | $20 | $10 | $1 |

Probability | 0.01 | 0.03 | ??? | 0.9 |

#### Solution

a) The probabilities in a probability model must add up to 1. We recognize that 0.01+0.03+???+0.9=1 must be true. The missing value must be 0.06.

b) \begin{align*}EV=\left (\$30\right )\left ( {0.01}\right )+\left ( \$20 \right )\left ( {0.03}\right )+\left ( \$10\right )\left ( {0.06}\right )+\left ( \$1\right )\left ( {0.9}\right )=\$0.30+\$0.60+\$0.60+\$0.90=\$2.40\end{align*}

EV=($30)(0.01)+($20)(0.03)+($10)(0.06)+($1)(0.9)=$0.30+$0.60+$0.60+$0.90=$2.40 . Our expected value is $2.40. In other words, if this game were played many times, the average payout would be $2.40. Note that the expected value of $2.40 is not a possible prize that a player can win.

#### Example 5

Suppose a casino game has an expected payout of $1 every time it is played. A player is paid nothing 45% of the time, they are paid one dollar 35% of the time, and they are paid three dollars 15% of the time. There is one more payout amount in this game.

a) Build a probability model for this situation. Be sure to calculate the percent of time the remaining payout occurred.

b) How much should this payout be so that the expected value is $1?

#### Solution

a) Start by noticing we have used 45%+35%+15%=95% of all outcomes. This means that the remaining outcome must be 5%. This allows us to build a probability model that is mostly complete.

Probability Model Amount $0 $1 $3 ??? Probability 0.45 0.35 0.15 0.05

Our calculation is now based upon the expected value formula.

We will use 'x' to represent the missing amount.

\begin{align*}\left(\$0\right)\left({0.45}\right)+\left(\$1\right)\left({0.35}\right)+\left(\$3\right)\left({0.15}\right)+\left(x\right)\left({0.05}\right)=\$1\end{align*}

($0)(0.45)+($1)(0.35)+($3)(0.15)+(x)(0.05)=$1 .

\begin{align*}\$0+\$0.35+\$0.45+\left(x\right)\left({0.05}\right)=\$1\end{align*}

$0+$0.35+$0.45+(x)(0.05)=$1

\begin{align*}\$0.80+\left(x\right)\left(0.05\right)=\$1\end{align*}

$0.80+(x)(0.05)=$1 .

\begin{align*}\left(x\right)\left(0.05\right)=\$0.20\end{align*}

(x)(0.05)=$0.20

x=$4. The missing value is $4.

#### Example 6

A carnival game has prizes and probabilities as shown in the table below. How much should the game cost if the owner of the game wants to average a $2 profit per player?

#### Solution

First calculate the expected value to get \begin{align*}EV=\$3\times0.65+\$5\times0.30+\$20\times0.05=\$4.45\end{align*}

EV=$3×0.65+$5×0.30+$20×0.05=$4.45 . This means that the average player will be paid $4.45 when they play. Therefore, the owner should charge $2 more than this or $6.45.

### Problem Set 3.1

#### Exercises

1) The table below represents the number of vehicles and the associated probability of having that number of vehicles in an individual household. What is the expected number of vehicles in a typical household?

# Owned | 0 | 1 | 2 | 3 | 4 | 5 |

Probability | 0.02 | 0.26 | 0.37 | 0.19 | 0.12 | 0.04 |

2) A student sells products as part of a fundraiser to raise money for a choir trip to New York. She sold 75 items total which included 50 rolls of cookie dough for $6 each, 15 packages of butter braids at $10 each, and 10 bake-at-home bread packs for $12 each.

a) Find the percent of her sales for each item.

b) Build a probability model for this situation.

c) Find the expected value of a sale for this particular student.

3) The owner of Friendly's Casino decides that she will set up her payouts in their 'Fast Cash' game so that the average gambler neither wins nor loses money. For a gambler who plays this game, the chance of getting paid nothing is 30%, the chance of getting paid $5 is 40%, the chance of getting paid $10 is 25%, and the chance of getting paid $30 is 5%. How much will the owner of Friendly's charge for this game?

4) The owner of Greedy's Casino decides he wants to make an average of $1.50 every time a gambler plays the game called 'Funny Money'. The chance of getting paid $2 is 20%, the chance of getting paid $5 is 40%, the chance of getting paid $10 is 30%, and the chance of getting paid $15 is 10%. What should the owner of Greedy's charge to play this game?

5) In a certain racing video game, players try to go around a track as many times as possible. If a racer completes a lap in time, they continue on to the next lap. If they don't complete a lap in time, their race is complete at the end of the lap they are currently finishing. The probability model below gives the probabilities of the maximum number of laps completed by people who play the video game. What is the expected number of laps completed for each racer?

# of Laps | 1 | 2 | 3 | 4 | 5 |

Probability | 0.29 | 0.38 | 0.17 | 0.11 | 0.05 |

6) In a certain casino game, the average payout (expected value) for a player is $2.53. A partially completed probability model for this game is given below.

Amount Paid | $0 | $1 | $3 | ??? | $21 |

Probability | 0.32 | 0.47 | 0.08 | 0.07 | 0.06 |

a) What is the missing amount?

b) If the casino was going to set a price for this game, do you think they would choose to charge $2 to play or $3 to play? Explain your choice.

c) If the casino was going to set a price for this game, do you think they would choose $3 to play or $6 to play? Explain your choice.

7) What is the average roll for a single 6-sided die?

8) A coin is flipped one time. If it lands on heads, you win $20. If it lands on tails, you win $30. Build a probability model and calculate the expected value for this game by using the expected value formula.

9) Two students are given the partially completed probability model below as part of a project. The teacher tells them that the expected value for this situation is $6.95.

Value | $3 | $6 | $10 | $50 |

Probability | 0.25 | 0.35 | ??? | 0.07 |

a) Assuming that the expected value of $6.95 is correct, what should be the value of the missing probability? Explain why this is impossible.

b) Assuming the expected value of $6.95 is incorrect, what should be the value of the missing probability?

c) The given expected value of $6.95 was incorrect. What is the correct expected value?

10) I want to come up with a game that has 5 prizes. There will be a 20% chance of getting paid $1, a 25% chance of getting paid $3, a 15% chance of getting paid $4, and a 30% chance of getting paid $7.

a) What is the probability of winning the 5th prize?

b) What is the value of the 5th prize if the expected payout for the game is $4.75?

11) For Halloween next year, I have decided that I will distribute an average of 1.6 pieces of candy per child who comes to my door. To help me do this, I have set up a game of chance whereby each trick-or-treater gets to play a game that determines how many pieces of candy they get to pick from my bag. I started building a probability model that shows the probabilities of being able to select 0, 1, 2, 3, or 4 pieces of candy. Unfortunately, I did not have time to finish my table. Use what I have so far to answer the questions.

# of Pieces | 0 | 1 | 2 | 3 | 4 |

Probability | 0.03 | 0.45 | x | y | 0.02 |

a) Give the expected value equation using the variables x and y and the expected value of 1.6. Simplify your equation by combining like terms.

b) Give an equation using the variables x and y that uses the fact that the probabilities in a probability model must add up to one. Clean up your equation by combining like terms.

c) Using your answers from parts a) and b), write a system of equations and solve for the variables x and y.

#### Review Exercises

12) A sample of 325 students were asked which electronic device they use most frequently, their cell phone, a computer (including wireless devices), or a television (including video games). The gender of the student was also recorded and the results are shown in the two-way table below.

a) What is the probability that a randomly selected student was a male?

b) What is the probability that a randomly selected student was most likely to say they used a cell phone most frequently?

c) What is the probability that a student was male given that they indicated they used the television most frequently?

d) What is the probability that a student indicated that they used a computer most frequently given that they were a female?

13) One floor of an office building is being remodeled and redecorated and an employee is responsible for picking out three different styles of chair and two different styles of table for their office furniture. Suppose the furniture store has 10 different chair styles and 4 different table styles that would be appropriate for office furniture. In how many different ways can the employee select the three chair styles and two table styles?

14) Suppose 1 card is drawn randomly from a standard 52 card deck. Find each probability.

a) P(Red Card)

b) P(Spade)

c) P(Face)

d) P(Heart|Red)

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