3.3: Expected Value
 Page ID
 107214
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left#1\right}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section we look at expectation of a result that is determined by chance. Although the outcomes of an experiment is random and cannot be predicted on any one trial, we need a way to describe what should happen on average over the long run.
Expected value is perhaps the most useful probability concept we will discuss. It has many applications, from insurance policies to making financial decisions. It's one thing that the casinos and government agencies that run gambling operations and lotteries hope most people never learn about.
Let's begin with a very simple example of a game that Rena and Larson are playing. Rena rolls a regular sixsided die. If the die lands on an even number, Rena gets $1 from Larson. However, if the die lands on a odd number, Rena must give $1 to Larson. Either player could win $1 on each roll of the die. But, who would win money in the long run if Rena and Larson play this game many times?
Intuitively and based on our understanding of probabilities on a die, we expect Rena to win $1 about half the time and Larson to win $1 about half the time. We would expect Rena to break even, winning as much as she loses. Mathematically, we can determine Rena's expected gain or loss using the following procedure:
\(\begin{aligned} \text{Rena's expected gain or loss} &= \Bigg({\text{amount Rena wins} \times P(\text{Rena wins})}\Bigg) + \Bigg({\text{amount Rena loses} \times P(\text{Rena loses})}\Bigg) \\ &= \left($1 \times \dfrac{1}{2}\right) +\left($1 \times \dfrac{1}{2}\right) \\ &= $0.50 + \left($0.50\right) \\ &= $0 \end{aligned}\)
Rena has an expected gain or loss of $0 as we predicted. The expected gain or loss is called the expected value and indicates that Rena should expect to break even if she plays the game many times. If Rena's expected value had been positive, she would have gained money after playing many times. If Rena's expected value had been negative, she would have lost money after playing many times. However, it is important to understand that this does not tell Rena what to expect on any one game  just the longterm expectation.
As in the procedure above, we compute the expected value by multiplying the amount gained/lost for each outcome by the probability of that outcome. Then, we add the products.
Expected Value \((EV)\) is the average gain or loss if an experiment or procedure with a numerical outcome is repeated many times.
\(EV =X_{1} \cdot P_{1}+X_{2} \cdot P_{2}+X_{3} \cdot P_{3}+\ldots+X_{n} \cdot P_{n}\)
where each \(X\) is the net amount gained or lost on each outcome of the experiment and \(P\) is the probability of that outcome.
A company plans to invest in a particular construction project. There is a 35% chance that it will lose $30,000, a 40% chance that it will break even, and a 25% chance that it will make a profit of $55,000. How much can the company expect to make or lose on this project?
Solution
It is helpful to make a table that summarizes all the information before finding the expected value. Represent the the loss outcome with a negative number, the break even outcome with 0, and the profit outcome with a positive number.
Outcome (Amount of Profit or Loss) 
\($30,000\)  \($0\)  \($55,000\) 

Probability  \(35\% = 0.35\)  \(40\% = 0.40\)  \(25\% = 0.25\) 
Now use the formula for the expected value:
\(\begin{aligned}
EV &=$30,000\left(0.35\right)+ \$0\left(0.40\right)+ $55,000\left(0.25\right)\\
&= $3,250
\end{aligned}\)
The company has an expectation, or average expected gain, of $3,250. If this company makes investments in projects like this one with these particular probabilities and amounts, in the long run it would make an average profit of $3,250 per project. However, there is still a 40% chance of not making any money and a 35% chance of losing $35,000.
Not all problems involving expectation are about games or money. The outcome can be anything that is numerical as the following problem shows.
The organizers at a balloon festival know that if it’s windy in town, only 9,000 people will show up for the festival. But if it’s not windy, 30,000 people will show up. The probability of wind is 45%. What is the expectation for the number of people who attend?
Solution
There are two outcomes in this scenario. It is windy and it is not windy. The numerical values for those outcomes are 9,000 people and 30,000 people respectively. The probability of a windy day is \(45\%\) so the probability it is not windy is \(100\%  45\% = 55\%\). Organizing these values into a table, we have
Outcome (attendance) 
Windy \(9,000\) 
Not Windy \(30,000\) 

Probability  \(45\% = 0.45\)  \(55\% = 0.55\) 
Now use the formula for the expected value:
\(\begin{aligned}
EV &=9,000\left(0.45\right)+ 30,000\left(0.55\right)\\
&= 20,550
\end{aligned}\)
The organizers have an expectation that 20,550 people will attend this festival based on these attendance estimates and probabilities of wind.
Try it Now 1
A hot dog vendor knows that there is a 20% chance of selling a hot dog with no toppings, a 50% chance of a hot dog with 1 topping, and a 30% chance of selling a hot dog with 2 toppings. What is the expected number of toppings sold per hot dog?
 Answer

1.1 toppings
Expectation is very important in analyzing games of chance. While the expected value won't tell you how much you will win or lose on any one play of the game, it will give you a sense of what should happen in the long run if you play the game many times. If the expected value is positive, then you more likely have an advantage overall. If the expected value is negative, then you more likely have an disadvantage overall. If the expected value of a game is 0, then we call it a fair game.
A game that has an expected value of zero is called a fair game.
In the following example, we will analyze a game to determine whether it is fair in terms of winnings.
A carnival game consists of drawing one ball from a box containing two yellow balls, five red balls, and eight white balls. If the ball is yellow, you win $5. If the ball is red, you win $2. If the ball is white, you lose $3.00.
Is this a fair game? If you play the game many times, how much would you expect to gain or lose per game?
Solution
It is helpful to make a table that summarizes all the information before finding the expected value.
Outcome (amount won or lost) 
Yellow \($5\) 
Red \($2\) 
White \($3\) 

Probability  \(\dfrac{2}{15}\)  \(\dfrac{5}{15}=\dfrac{1}{3} \)  \(\dfrac{8}{15}\) 
Now use the formula for the expected value:
\(EV =\$5\left(\dfrac{2}{15}\right)+ \$2\left(\dfrac{1}{3}\right)+ $3\left(\dfrac{8}{15}\right) \approx $0.27\)
Since the expected value is not zero this is not a fair game. The expected value is negative which indicates that a player can expect to lose about 27 cents per game on average when playing this game many times. That means that the carnival will make an average of $0.27 for every game played.
Picking a ball from the bag is a random experiment so we cannot predict what will happen if we play the game once. We can predict what will happen only if we play the game many times.
Try it Now 2
A game consists of rolling a colored die with three green sides, two red sides, and one blue side. If you roll green, you earn 2 points. A roll of a red earns you 1 point. If you roll blue, you lose 8 points. Is this a fair game? If not, how many points can you expect to earn or lose per game?
 Answer

This is a fair game because the expected value is 0 points.
Valley View Elementary is raising money to buy tablets for the classrooms. The PTA sells 2,000 raffle tickets at $3 each.
 First prize is a flatscreen TV worth $500.
 Second prize is an android tablet worth $375.
 Third prize is an ereader worth $200.
 Five $25 gift certificates will also be awarded.
What are the expected winnings for a person who buys one ticket?
Solution
A total of 8 tickets are winners and the other 1,992 tickets are losers. The cost is $3 per ticket so we must reduce all values won by $3 to find the net amount won or lost.
Outcome (net amount won or lost) 
First Prize \($500$3=$497\) 
Second Prize \($375$3=$372\) 
Third Prize \($200$3=$197\) 
Gift Certificates \($25$3=$22\) 
No Prize \(0$3=$3\) 

Probability  \(\dfrac{1}{2000}\)  \(\dfrac{1}{2000}\)  \(\dfrac{1}{2000}\)  \(\dfrac{5}{2000}=\dfrac{1}{400}\)  \(\dfrac{1992}{2000}=\dfrac{249}{250}\) 
Now use the formula for the expected value:
\(\begin{aligned}
EV &=\$497\left(\dfrac{1}{2000}\right)+ \$372\left(\dfrac{1}{2000}\right)+ \$197\left(\dfrac{1}{2000}\right)+ \$22\left(\dfrac{1}{400}\right)+ \$3\left(\dfrac{249}{250}\right) \\
&=2.40 \end{aligned}\)
That means the expected winnings per ticket are $0.60  $3 = $2.40.
We would expect to lose an average of $2.40 for each ticket bought. This means that the school will earn an average of $2.40 for each ticket bought for a profit of \(\$2.40 \cdot 2,000 \text{ tickets } = \$4,800.\)
A real estate investor buys a parcel of land for $150,000. He estimates the probability that he can sell it for $200,000 to be 0.40, the probability that he can sell it for $160,000 to be 0.45 and the probability that he can sell it for $125,000 to be 0.15. What is the expected profit for this purchase?
Solution
Find the net profit for each situation first: $200,000 – $150,000 = $50,000 profit, $160,000  $150,000 = $10,000 profit, and $125,000  $150,000 = $25,000 profit (loss).
Outcome (net profit or loss) 
$50,000  $10,000  $25,000 

Probability  0.40  0.45  0.15 
Now use the formula for the expected value:
\(\begin{aligned}
EV &=50,000(0.40)+10,000(0.45)+(25,000)(0.15) \\
&=$20,750
\end{aligned}\)
The expected profit from the purchase is $20,750.
Expected value is very common in making insurance decisions. Insurance companies take into account various factors to determine probabilities that policy holders will have an accident, fire, or death. They set prices on policies to make ensure the company will remain profitable even though they will occasionally pay out large monetary claims. In other words, even though the insurance companies expect some negative outcomes, it is important to their bottom line that their expected return remain positive.
Let's look at an illustration.
The cost of a $50,000 life insurance policy is $150 per year for a person who is 21years old. Assume the probability that a person will die at age 21 is 0.001. What is the company’s expected profit on a policy of this type? How much can the company expect to earn in profit if it sells 10,000 of these policies to 21year olds?
Solution
There are two outcomes here. If the person lives, the insurance company makes a pure profit of $150. The probability that the person lives is 10.001=0.999. If the person dies, the company takes in $150 and but must also pay out $50,000 for a loss of $49,850.
Outcome (insurance company's net profit or loss) 
Person Lives \($150\) 
Person Dies \($150$50,000=$49,850\) 

Probability  0.999  0.001 
The expected value for one policy is
\(EV = 150(0.999)+($49,850)(0.001)=$100\).
If the company sells 10,000 policies at a profit of $100 each, the total expected profit is \($100(10,000)=$1,000,000\).
Try it Now 3
You take out a fire insurance policy on your home. The yearly premium is $250. In case of fire, the insurance company will pay you $200,000. The probability of a house fire in your area is 0.003%. What is the insurance company's expected value for selling one of these policy?
 Answer

$190