\setcounter{ExampleCounter}{1}
An investor is considering a \$10,000 investment in a start-up company. She estimates that she has a probability 0.25 of a \$20,000 loss, probability 0.20 of a \$10,000 profit, probability 0.15 of a \$50,000 profit, and probability 0.40 of breaking even (a profit of \$0). Would you advise her to make the investment?
To answer a question like this, we need to find the \textbf{expected value} of this random variable.\\
\begin{proc}{Expected Value}
The \textbf{mean} or \textbf{expected value} of a random variable is the long-term average result if the experiment is repeated many times.\\
To find the expected value, multiply each value by its probability and add them up:
\[\mu_X = E(X) = \sum x \cdot P(x)\]
\end{proc}
In the example with the investor, we have the following probability distribution, where $X$ represents the earnings on this investment:
\begin{center}
\begin{tabular}{c | c}
$x$ & $P(x)$\\
\hline
\$0 & 0.40\\
\$10,000 & 0.20\\
\$50,000 & 0.15\\
-\$20,000 & 0.25
\end{tabular}
\end{center}
The expected value is
\[E(X) = (\$0)(0.40) + (\$10,000)(0.20) + (\$50,000)(0.15) + (-\$20,000)(0.25) = \$4500\]
Since the expected value is positive, that means she can expect to make a profit.\\
\paragraph{Note:} It's impossible for her to actually make \$4500, so this expected value isn't what we expect her to make. Instead, this means that if a bunch of investors made this investment, the majority of them would make money, and their average profits would be \$4500.
\vfill
\pagebreak
\begin{example}{Expected Value}
Find the mean of the random variable with the following probability distribution.
\begin{center}
\begin{tabular}{c | c}
$x$ & $P(x)$\\
\hline
0 & 0.03125\\
1 & 0.15625\\
2 & 0.31250\\
3 & 0.31250\\
4 & 0.15625\\
5 & 0.03125
\end{tabular}
\end{center}
\begin{align*}
\mu_X = E(X) &= 0(0.03125) + 1(0.15625) + 2(0.31250) + 3(0.31250)\\ &+ 4(0.15625) + 5(0.03125)\\
&= 2.5
\end{align*}
\end{example}
\begin{example}{Circuit Board Defects}
The following table presents the probability distribution of the number of defects $X$ in a randomly chosen printed circuit board.
\begin{center}
\begin{tabular}{c | c c c c}
$x$ & 0 & 1 & 2 & 3\\
\hline
$P(x)$ & 0.5 & 0.3 & 0.1 & 0.1
\end{tabular}
\end{center}
Compute the mean $\mu_X$.\\
\[\mu_X = 0(0.5) + 1(0.3) + 2(0.1) + 3(0.1) = 0.8\]
This means that we expect the average number of defects in all these circuit boards to be less than 1.
\end{example}
\vfill
\pagebreak
\begin{example}{Lottery}
In the NY State Numbers Lottery, you pay \$1 and pick a number from 000 to 999. If your number comes up, you win \$500, which is a profit of \$499. If you lose, you lose \$1. What is your expected value?\\
The probability distribution below describes your possible winnings.
\begin{center}
\begin{tabular}{c | c}
$x$ & $P(x)$\\
\hline
\$499 & 0.001\\
-\$1 & 0.999
\end{tabular}
\end{center}
Therefore, the expected value is \[E(X) = \$499(0.001) - \$1(0.999) = -\$0.50\]
If you played this lottery many times, you would lose an average of fifty cents per play.
\end{example}
When casinos design games, you better believe the player's expected value is negative. This is why ``the house always wins'' in the long run, even if a few players win money now and again.
\begin{example}{Craps}
In the game of craps, two dice are rolled, and people bet on the outcome. For example, you can bet \$1 that the dice will total 7, and if you win, your profit is \$4. What is your expected value?
The probability distribution below describes your possible winnings.
\begin{center}
\begin{tabular}{c | c}
$x$ & $P(x)$\\
\hline
& \\
\$4 & $\dfrac{6}{36}$\\
& \\
-\$1 & $\dfrac{30}{36}$
\end{tabular}
\end{center}
Therefore, the expected value is \[E(X) = \$4\left(\dfrac{1}{6}\right) - \$1\left(\dfrac{5}{6}\right) = -\$0.1667\]
\end{example}
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\pagebreak
\begin{example}{Carnival Game}
Consider the following game: you flip a fair coin. If you get heads on the flip, you win \$200 and the game is over. If you get tails on the flip, you get to flip the coin a second time; if you get heads on the second flip, you win \$40 and the game is over. If you get tails on the second flip, you win nothing and the game is over.
\begin{comment} %%For the student version
\begin{enumerate}[(a)]
\item Fill in the following probability distribution with the possible winnings and their associated probabilities.
\begin{center}
\begin{tabular}{c| c}
&\\
$x$ & \hspace*{4in} \text{}\\
&\\
\hline
&\\
$P(x)$ & \\
&\\
\end{tabular}
\end{center}
\vspace{0.25in}
\item What is the expected value for this game?
\vspace{0.75in}
\item What is the probability that you win at most \$40 when you play this game once?
\end{enumerate}
\end{comment}
\begin{enumerate}[(a)]
\item Fill in the following probability distribution with the possible winnings and their associated probabilities.
\begin{center}
\begin{tabular}{c| c c c}
$x$ & \$200 & \$40 & \$0\\
\hline
$P(x)$ & 0.5 & 0.25 & 0.25\\
\end{tabular}
\end{center}
\item What is the expected value for this game?
\[E(X) = \$200(0.5) + \$40(0.25) + \$0(0.25) = \$110\]
\item What is the probability that you win at most \$40 when you play this game once?
\[P(\$0) + P(\$40) = 0.25 + 0.25 = 0.5\]
\end{enumerate}
\end{example}
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\begin{example}{Draw Four Cards}
You are playing a game in which you draw four cards from a standard deck of 52 cards, and the cards are replaced in the deck after each draw. You guess the suit of each card before it is drawn; you pay \$1 to play, and if you guess the correct suit each time, you get your money back plus \$275. Will you make money playing this game in the long run?\\
Since the card is replaced each time, the trials are independent, so the probability of guessing the correct suit four times in a row is
\[P(win) = \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} = \dfrac{1}{256}\]
\pagebreak
Therefore, the probability distribution looks like
\begin{center}
\begin{tabular}{c | c}
$x$ & $P(x)$\\
\hline
& \\
\$275 & $\dfrac{1}{256}$\\
& \\
-\$1 & $\dfrac{255}{256}$
\end{tabular}
\end{center}
The expected value of this game is
\[E(X) = \$275\left(\dfrac{1}{256}\right) - \$1\left(\dfrac{255}{256}\right) = \$0.078.\]
You can expect to make an average of 8 cents every time you play, so yes, this is a profitable game, but the profit margin is so small that it probably isn't worth it.
\end{example}
\begin{example}{Biased Coin}
Suppose you play a game with a biased coin, where the probability of heads is 2/3 and the probability of tails is 1/3. You toss the coin once; if your toss is heads, you pay \$6, and if your toss is tails, you win \$10. If you play this game many times, will you come out ahead?\\
The probability distribution looks like
\begin{center}
\begin{tabular}{c | c}
$x$ & $P(x)$\\
\hline
& \\
\$10 & $\dfrac{1}{3}$\\
& \\
-\$6 & $\dfrac{2}{3}$
\end{tabular}
\end{center}
Therefore, the expected value of this game is
\[E(X) = \$10\left(\dfrac{1}{3}\right) - \$6\left(\dfrac{2}{3}\right) = -\$0.6667,\]
so you could expect to lose an average of 67 cents per play.
\end{example}
\vfill
\pagebreak