The concept of expected values is best illustrated by games. Let’s ‘play’ some in the next two examples.

Example 21.1 – Coin Flip

Setup: This first game is rather simple, and not very exciting, but it lays the foundation for the second, more complicated game.

• In this game, I flip a coin.
• If I flip a Head, I’ll pay you a dollar.
• If I flip a Tail, you’ll pay me a dollar.

Question: Can either one of us EXPECT to earn anything in the long run?

Solution: Let us first of all define the term, “in the long run”:

• It simply means if we play the game many, many times.
• I guess most people will agree that the answer is NO.
• Half the time, I win a dollar.
• Half the time, I lose a dollar.

In terms of mathematics, we can state the above sentence this way:

$$\frac{1}{2} (+\$1)+ \frac{1}{2} (-\$1)=\$0$$

Let us denote a more complex game in the next example. This time, we will use a six-sided die and win certain amounts of money depending on the number we roll.

Example 21.2.1 – Die Roll

Setup: In this game, I will roll a fair regular 6 sided die.

• You pay me $2 to play each round.
• Then I roll the die.
• If I roll a 6, I will pay you $5.50.
• If I roll a 5, I will pay you $3.40.
• If I roll a 4, I will pay you $2.50.
• If I roll a 3, 2, or 1, I will pay you nothing.

Question: Would you like to play this game with me? Think about it for a moment.

You Try: Let us now analyze this game from your point of view.

• If needed, we can calculate the standard deviation of a discrete general probability distribution.
• This is not needed if we have the original data that we used to create the probability distribution
• We could just use a standard deviation calculation on the original data instead if it is available
• The standard deviation for discrete general probability distributions is:
  $$\sigma(X) =\sqrt{\Sigma [(x_i-\mu)^2 \cdot p(x_i)]}$$

Example 21.2.2

Setup: Let us re-visit the previous example of the die rolling game.

Question: What is the standard deviation of the winnings for this game?

Solution: Let us use the formula to solve for the standard deviation:

$\begin{align*} \sigma(X) &=\sqrt{(3.50-(-0.10))^2 \cdot \frac{1}{6} + (1.40-(-0.10))^2 \cdot \frac{1}{6}+(0.50-(-0.10))^2 \cdot \frac{1}{6}+(-2.00-(-0.10))^2 \cdot \frac{3}{6}} \\ \\ &= \sqrt{(3.60)^2 \cdot \frac{1}{6} + (1.50)^2 \cdot \frac{1}{6}+(0.60)^2 \cdot \frac{1}{6}+(-1.90)^2 \cdot \frac{3}{6}} \\ \\ &= \sqrt{12.96 \cdot \frac{1}{6} + 2.25 \cdot \frac{1}{6}+0.36 \cdot \frac{1}{6}+3.61 \cdot \frac{3}{6}} \\ \\ &= \sqrt{2.16 + 0.375 + 0.06+ 1.805} \\ \\ &= \sqrt{4.4} \\ \\ &= 2.09762 \end{align*}$

Let us finish this section with an applied example. This is from a forestry course. We used expected values to calculate expected numbers of defective boards.

Example 21.3.2

Setup: What if, instead, you had the following:

• ¹⁄₅ of the boards have a defect
• ¹⁄₂ of the boards with a defect have rot
• ¹⁄₄ of the boards with rot are also bent

Question: Calculate the expected number of boards for each scenario below if there are 3,000 boars in the yards.

You Try:

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