example

In the casino game roulette, a wheel with 38 spaces (18 red, 18 black, and 2 green) is spun. In one possible bet, the player bets $1 on a single number. If that number is spun on the wheel, then they receive $36 (their original $1 + $35). Otherwise, they lose their $1. On average, how much money should a player expect to win or lose if they play this game repeatedly?

Answer: Suppose you bet $1 on each of the 38 spaces on the wheel, for a total of $38 bet. When the winning number is spun, you are paid $36 on that number. While you won on that one number, overall you’ve lost $2. On a per-space basis, you have “won” -$2/$38 ≈ -$0.053. In other words, on average you lose 5.3 cents per space you bet on. We call this average gain or loss the expected value of playing roulette. Notice that no one ever loses exactly 5.3 cents: most people (in fact, about 37 out of every 38) lose $1 and a very few people (about 1 person out of every 38) gain $35 (the $36 they win minus the $1 they spent to play the game). There is another way to compute expected value without imagining what would happen if we play every possible space.  There are 38 possible outcomes when the wheel spins, so the probability of winning is [latex]\frac{1}{38}[/latex]. The complement, the probability of losing, is [latex]\frac{37}{38}[/latex]. Summarizing these along with the values, we get this table:

Outcome	Probability of outcome
$35	[latex]\frac{1}{38}[/latex]
-$1	[latex]\frac{37}{38}[/latex]

Notice that if we multiply each outcome by its corresponding probability we get [latex]\$35\cdot \frac{1}{38}=0.9211[/latex] and [latex]-\$1\cdot \frac{37}{38}=-0.9737[/latex], and if we add these numbers we get 0.9211 + (-0.9737) ≈ -0.053, which is the expected value we computed above.

Example

In a certain state's lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them are drawn at random. If the six numbers drawn match the numbers that a player had chosen, the player wins $1,000,000. If they match 5 numbers, then win $1,000.   It costs $1 to buy a ticket. Find the expected value.

Answer: Earlier, we calculated the probability of matching all 6 numbers and the probability of matching 5 numbers: [latex]\frac{{}_{6}{{C}_{6}}}{{}_{48}{{C}_{6}}}=\frac{1}{12271512}\approx0.0000000815[/latex] for all 6 numbers, [latex]\frac{\left({}_{6}{{C}_{5}}\right)\left({}_{42}{{C}_{1}}\right)}{{}_{48}{{C}_{6}}}=\frac{252}{12271512}\approx0.0000205[/latex] for 5 numbers. Our probabilities and outcome values are:

Outcome	Probability of outcome
$999,999	[latex]\frac{1}{12271512}[/latex]
$999	[latex]\frac{252}{12271512}[/latex]
-$1	[latex]1-\frac{253}{12271512}=\frac{12271259}{12271512}[/latex]

The expected value, then is: [latex-display]\left(\$999,999 \right)\cdot \frac{1}{12271512}+\left( \$999\right)\cdot\frac{252}{12271512}+\left(-\$1\right)\cdot\frac{12271259}{12271512}\approx-\$0.898[/latex-display] On average, one can expect to lose about 90 cents on a lottery ticket. Of course, most players will lose $1.

View more about the expected value examples in the following video. https://youtu.be/pFzgxGVltS8

Example

A 40-year-old man in the U.S. has a 0.242% risk of dying during the next year.[footnote]According to the estimator at http://www.numericalexample.com/index.php?view=article&id=91[/footnote] An insurance company charges $275 for a life-insurance policy that pays a $100,000 death benefit. What is the expected value for the person buying the insurance?

Answer: The probabilities and outcomes are

Outcome	Probability of outcome
$100,000 - $275 = $99,725	0.00242
-$275	1 – 0.00242 = 0.99758

The expected value is ($99,725)(0.00242) + (-$275)(0.99758) = -$33.

The insurance applications of expected value are detailed in the following video. https://youtu.be/Bnai8apt8vw