5.2: Expected Value
- Page ID
- 91509
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Students will be able to:
- Identify valid probability distributions.
- Compute the expected value from a probability distribution or an experiment.
- Understand the implications of the Gamblers' Fallacy and how it related to streaks.
Probability Distributions
A probability distribution (probability space) is a sample space paired with the probabilities for each outcome in the sample space. If we toss a fair coin and see which side lands up, there are two outcomes, heads and tails. Since the coin is fair these are equally likely outcomes and have the same probabilities. The probability distribution would be P(heads) = 1/2 and P(tails) = 1/2. This is often written in table form:
|
Outcome |
Heads |
Tails |
|---|---|---|
|
Probability |
1/2 |
1/2 |
|
A probability distribution for an experiment is a list of all the possible outcomes and their corresponding probabilities. |
In probability problems when we roll two dice, it is helpful to think of the dice as being different colors. Let’s assume that one die is red and the other die is green. We consider getting a three on the red die and a five on the green die different than getting a five on the red die and a three on the green die. In other words, when we list the outcomes the order matters. The possible outcomes of rolling two dice and looking at the sum are given in Table \(\PageIndex{2}\).
|
1+1 = 2 |
1+2 = 3 |
1+3 = 4 |
1+4 = 5 |
1+5 = 6 |
1+6 = 7 |
|
2+1 = 3 |
2+2 = 4 |
2+3 = 5 |
2+4 = 6 |
2+5 = 7 |
2+6 = 8 |
|
3+1 = 4 |
3+2 = 5 |
3+3 = 6 |
3+4 = 7 |
3+5 = 8 |
3+6 = 9 |
|
4+1 = 5 |
4+2 = 6 |
4+3 = 7 |
4+4 = 8 |
4+5 = 9 |
4+6 = 10 |
|
5+1 = 6 |
5+2 = 7 |
5+3 = 8 |
5+4 = 9 |
5+5 = 10 |
5+6 = 11 |
|
6+1 = 7 |
6+2 = 8 |
6+3 = 9 |
6+4 = 10 |
6+5 = 11 |
6+6 = 12 |
|
Sum |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
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Probability |
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Reduced Probability |
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Are the following valid probability distributions?
|
Outcome |
A |
B |
C |
D |
E |
|---|---|---|---|---|---|
|
Probability |
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|
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This is a valid probability distribution. All the probabilities are between zero and one inclusive and the sum of the probabilities is 1.00.
|
Outcome |
A |
B |
C |
D |
E |
F |
|---|---|---|---|---|---|---|
|
Probability |
0.45 |
0.80 |
-0.20 |
-0.35 |
0.10 |
0.20 |
This is not a valid probability distribution. The sum of the probabilities is 1.00, but some of the probabilities are not between zero and one, inclusive.
|
Outcome |
A |
B |
C |
D |
|---|---|---|---|---|
|
Probability |
0.30 |
0.20 |
0.40 |
0.25 |
This is not a valid probability distribution. All of the probabilities are between zero and one, inclusive, but the sum of the probabilities is 1.15 not 1.00.
Would you buy a lottery ticket with the numbers 1, 2, 3, 4, 5? Do you think that a winning ticket with five consecutive numbers is less likely than a winning ticket with the numbers 2, 14, 18, 23 and 32? If you are playing a slot machine in Las Vegas and you have lost the last 10 times, do you keep playing the same machine because you are “due for a win?” Have you ever wondered how a casino can afford to offer meals and rooms at such cheap rates? Should you play a game of chance at a carnival? How much should an organization charge for raffle tickets for their next fund raiser? All of these questions can be answered using probabilities.
Suppose the there are \(n\) outcomes for a given experiment where the values of the outcomes are given by \(x\) where \(x\) can take on the \(n\) values \(x_{1}, x_{2}, x_{3}, \ldots, x_{n}\). If the probability that each of these values occurs is \(p_{1}, p_{2}, p_{3}, \ldots, p_{n}\), respectively, then the expected value of the outcomes is
\[E(x)=x_{1} p_{1}+x_{2} p_{2}+x_{3} p_{3}+\ldots+x_{n} p_{n} \label{expectedvalue}\]
The expected value is the average gain or loss of an event if the experiment is repeated many times.
We can use the probability distribution table to compute the expected value by multiplying each outcome by the probability of that outcome, then adding up the products.
Valley View Elementary is trying to raise money to buy tablets for their classrooms. The PTA sells 2000 raffle tickets at $3 each. First prize is a flat-screen TV worth $500. Second prize is an android tablet worth $375. Third prize is an e-reader worth $200. Five $25 gift certificates will also be awarded. What are the expected winnings for a person who buys one ticket?
Solution
We need to write out the probability distribution before we find the expected value.
A total of eight tickets are winners and the other 1992 tickets are losers.
|
Outcome |
Win $500 |
Win $375 |
Win $200 |
Win $25 |
Win $0 |
|---|---|---|---|---|---|
|
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 (Equation \ref{expectedvalue}).
\[\begin{aligned}
E &=\$500\left(\dfrac{1}{2000}\right)+ \$375\left(\dfrac{1}{2000}\right)+ \$200\left(\dfrac{1}{2000}\right)+ \$25\left(\dfrac{1}{400}\right)+ \$0\left(\dfrac{249}{250}\right) \\
&=0.60
\end{aligned}\]
It costs $3 to buy a ticket but we only win an average of $0.60 per ticket. 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 ∙ 2000 = $4800.
In general, if the expected value of a game is negative, it is not a good idea to play the game, 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.
Not surprisingly, the expected value for casino games is negative for the player, which is positive for the casino. It must be positive or they would go out of business. Players just need to keep in mind that when they play a game repeatedly, their expected value is negative. That is fine so long as you enjoy playing the game and think it is worth the cost. But it would be wrong to expect to come out ahead.
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 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).
The probability distribution is
|
Outcome |
$50,000 |
$10,000 |
-$25,000 |
|---|---|---|---|
|
Probability |
0.40 |
0.45 |
0.15 |
\[\begin{aligned}
E &=50,000(0.40)+10,000(0.45)+(-25,000)(0.15) \\
&=\$ 20,750
\end{aligned}\]
The expected profit from the purchase is $20,750.
The cost of a $50,000 life insurance policy is $150 per year for a person who is 21-years old. Assume the probability that a person will die at age 21 is 0.001. What is the company’s expected profit if the company sells 10,000 policies to 21-year olds?
Solution
There are two outcomes. If the person lives the insurance company makes a profit of $150. The probability that the person lives is 1-0.001=0.999. If the person dies the company takes in $150 and pays out $50,000 for a loss of $49,850.
|
Outcome |
$150 |
-$49,850 |
|---|---|---|
|
Probability |
0.999 |
0.001 |
The expected value for one policy is:
\[E(x)=(\$ 150)(0.999)+(-\$ 49,850)(0.001)=\$ 149.80 \nonumber\]
If the company sells 10,000 policies at a profit of $149.80 each, the total expected profit is .
Gamblers’ Fallacy and Streaks
Often times a gambler on a losing streak will keep betting in the belief that his/her luck must soon change. Consider flipping a fair coin. Each toss of the coin is independent of all the other tosses, meaning the past outcomes do not influence the probability of getting a heads or tails on the next coin toss. Assume the coin has landed heads up the last eight times. Some people erroneously believe that the coin is more likely to land tails up on the next toss. In reality, the coin still has a 50% chance of landing tails up. It does not matter what happened the last eight tosses.
The gambler’s fallacy is the mistaken belief that a streak of losses makes a person due for a streak of wins and vice versa.
Toss a fair coin seven times and record which side lands up. For example HHTHTTT would represent getting a head on the first, second, and fourth tosses and a tail on the other tosses. Is a streak of all heads less likely than the other possible outcomes?
Solution
As we will see in Section 3.5, there are 128 possible ways to toss a coin seven times. Some of the possibilities are \(HHTTHHT\), \(HTHTHTH\), \(HHHHTTT\), and \(HTTHTTH\). Because tossing coins are independent events and the coin is fair each of these 128 possibilities has the same probability.
\[P(\mathrm{HHTTHHT})=\dfrac{1}{128} \nonumber\]
\[P(\mathrm{HTHTHTH})=\dfrac{1}{128} \nonumber\]
etc.
This also means that the probability of getting all heads is
\[P(\mathrm{HHHHHHH})=\dfrac{1}{128}. \nonumber\]
Getting a streak of all heads has exactly the same probability as any other possible outcome.