Expected Value and Finding the Average Outcome

Expected value is easiest to think of as the average outcome. In statistics and probability analysis, the expected value is calculated by multiplying each of the possible outcomes by the likelihood that each outcome will occur and then summing all of those values. By calculating expected values, we can find the scenario most likely to give the desired outcome.

The Formula for Expected Value (EV) Is:
\(
E V=\sum P\left(X_i\right) \times X_i
\)
where:
\(X\) is a random variable
\(P(X)\) is the probability of the random variable
Thus, the \(\mathrm{EV}\) of a random variable \(\mathrm{X}\) is taken as each value of the random variable multiplied by its probability, and each of those products is summed.

Example 1: What is the expected value of one roll of a standard six-faced die?

Solution:

Once you roll the die, it has an equal one-sixth chance of landing on one, two, three, four, five, or six. Given this information, the calculation is straightforward:

\(
\begin{aligned}
\left(\frac{1}{6} \times 1\right) & +\left(\frac{1}{6} \times 2\right)+\left(\frac{1}{6} \times 3\right) \\
& +\left(\frac{1}{6} \times 4\right)+\left(\frac{1}{6} \times 5\right)+\left(\frac{1}{6} \times 6\right)=3.5
\end{aligned}
\)
If you were to roll a six-sided die an infinite amount of times, you see the average value equals \(3.5\).

Example 2: A men’s soccer team plays soccer zero, one, or two days a week. The probability that they play zero days is 0.2, the probability that they play one day is 0.5, and the probability that they play two days is 0.3. Find the long-term average or expected value of the number of days per week the men’s soccer team plays soccer.

Solution:

To do the problem, first, let the random variable \(X=\) the number of days the men’s soccer team plays soccer per week. \(X\) takes on the values \(0,1,2\). Construct a table adding a column \(x \cdot P(x)\). In this column, you will multiply each \(x\) value by its probability.
\(
\begin{array}{lll}
\boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) & \boldsymbol{x}.\boldsymbol{P}(\boldsymbol{x}) \\
0 & 0.2 & (0)(0.2)=0 \\
1 & 0.5 & (1)(0.5)=0.5 \\
2 & 0.3 & (2)(0.3)=0.6
\end{array}
\)
\(
\text { Expected value: }(0)(0.2)+(1)(0.5)+(2)(0.3)=0+0.5+0.6=1.1
\)

The expected value is 1.1. The men’s soccer team would, on average, expect to play soccer 1.1 days per week. 

Example 3: A hospital researcher is interested in the number of times the average post-op patient will ring the nurse during a 12-hour shift. For a random sample of 50 patients, the following information was obtained. What is the expected value?

Solution:

\(
\begin{array}{ll}
x & P(x) \\
0 & P(x=0)=\frac{4}{50} \\
1 & P(x=1)=\frac{8}{50} \\
2 & P(x=2)=\frac{16}{50} \\
3 & P(x=3)=\frac{14}{50} \\
4 & P(x=4)=\frac{6}{50} \\
5 & P(x=5)=\frac{2}{50}
\end{array}
\)
The expected value is \(2.24\),
\((0) \frac{4}{50}+(1) \frac{4}{50}+(2) \frac{16}{50}+(3) \frac{14}{50}+(4) \frac{6}{50}+(5) \frac{2}{50}=0+\frac{8}{50}+\frac{32}{50}+\frac{42}{50}+\frac{24}{50}+\frac{10}{50}=\frac{116}{50}=2.32\)

Example 4: You are playing a game of chance in which four cards are drawn from a standard deck of 52 cards. You guess the suit of each card before it is drawn. The cards are replaced in the deck on each draw. You pay \(\$1\) to play. If you guess the right suit every time, you get your money back and \(\$256\). What is your expected profit of playing the game over the long term?

Solution:

Let \(X=\) the amount of money you profit. The \(x\)-values are \(-\$1\) and \(\$256\).
The probability of guessing the right suit each time is
\(
\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)=\frac{1}{256}=0.0039
\)
The probability of losing is
\(
\begin{aligned}
& 1-\frac{1}{256}=\frac{255}{256}=0.9961 \\
& (0.0039) 256+(0.9961)(-1)=0.9984+(-0.9961)=0.0023 \text { or } 0.23 \text { cents. }
\end{aligned}
\)