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?

Solution

Suppose you bet $1 on each of the 38 spaces on the wheel, for a total of $38 bet.

After the wheel is spun just one of those 38 numbers is going to win, so you are paid $36 on that wining number. While you won on that one number, overall you’ve lost $2 total as bet an initial $38. On a per-space basis, you have “won” $-\frac{\$2}{\$38}\approx -\$0.053$.

In other words, on average you lose 5.3 cents per space you bet on.

Video Solution Example 1 (2 mins and 13 seconds – CC) Another example starts after 2:42 mark on the video.

Bob wants to know if you would play his game. It costs $2 to play and you do not get the $2 back if you win or lose. The game consists of selecting a “winning” colored marble from the bag.

–There is 1 green marble and if you select one you win $30.

–There are 4 red marbles and if you select one you win $3.

–There are 8 white marbles and if you select one you win $1.

–There are 7 black marbles and if you select one you win $0.

Is it in your favor to play this game? Answer the question using the idea of Expected Value (calculate the expected value and decide for yourself).

Solution

Start with looking at the outcomes of the game overall (the net profit). There are four things that can happen and each will have thir own net profit and probabilities associated with them. Create a table showing the net profits and probability.

If we draw a green marble it states we win $30, but the actual net profit is $\$30-\$2=\$28$. We can do that for each of the other three outcomes of the experiment and put that in a table (Selecting red has a net profit of $1, white has a net profit of $-1, and black has a net profit of $-2)

Next let us add in the probability for each of those outcomes. We will assume each marble is equal likely to be selected. The probability will be the number of marbles of the given color over the total number of marbles. Enter these values into the table.

To find the expected value we need to multiply each outcome times their respective probability, so add one more row on the table that shows the producs. We will then add all those products on the last row to calculate the expected value.

Now add the products together to get the expected value

$$\begin{array}{l}\text{Expected Value}=x_1\cdot P\left(x_1\right)+x_2\cdot P\left(x_2\right)+\cdots x_k\cdot P\left(x_k\right)\\ \text{Expected Value}=\left(28\right)\cdot \frac{1}{20}+\left(1\right)\cdot \frac{4}{20}+\left(-1\right)\cdot \frac{8}{20}+\left(-2\right)\cdot \frac{7}{20}\\ \text{Expected Value}=\frac{28}{20}+\frac{4}{20}-\frac{8}{20}-\frac{14}{20}\\ \text{Expected Value}=\frac{28+4-8-14}{20}\\ \text{Expected Value}=\frac{10}{20}\\ \text{Expected Value}=\$0.50\end{array}$$

The expected value or long run average is $0.50 for the game. This means if you were play this game from Bob many many times and average your winnings and losses that you expect to earn $0.50 per game played. It is definitely in your favor to play.

Video Solution for Example 2 (1 min and 44 seconds – CC)

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.

Solution

Earlier, we calculated the probability of matching all 6 numbers and the probability of matching 5 numbers:

• Probability for matching all 6 numbers,
  $$P\left(\text{matching 6}\right)=\frac{6C6}{48C6}=\frac{1}{2271512}\approx \text{0}\text{.0000000815}$$
• Probability for matching 5 numbers,
  $$P\left(\text{matching 5}\right)=\frac{\left(6C5\right)\left(42C1\right)}{48C6}=\frac{252}{2271512}\approx \text{0}\text{.0000205}$$
• Probability for not matching 5 or 6 numbers can be found using the complement rule,

  $$P\left(\text{Not matching 5 or 6}\right)=1-P\left(\text{matching 5 or 6}\right)$$

  Now since matching 5 or matching 6 are mutually exclusive we can add their probabilities together to find the probability of matching 5 or 6.

  $$\begin{array}{l}P\left(\text{Not matching 5 or 6}\right)=1-\left(P\left(\text{matching 5}\right)+P\left(\text{matching 6}\right)\right)\\ P\left(\text{Not matching 5 or 6}\right)=1-\left(\frac{1}{2271512}+\frac{252}{2271512}\right)\\ P\left(\text{Not matching 5 or 6}\right)=1-\left(\frac{253}{2271512}\right)\\ P\left(\text{Not matching 5 or 6}\right)=\frac{12271259}{2271512}\end{array}$$

Our probabilities and outcome values are:

The expected value, then is:

$$\left(\$999,999\right)\cdot \frac{1}{12271512}+\left(\$999\right)\cdot \frac{252}{12271512}+\left(-\$1\right)\cdot \frac{12271259}{12271512}\approx -\$0.898$$

On average, one can expect to lose about 90 cents on a lottery ticket. Of course, most players will lose $1.

Video Solution Example 3 (2 mins and 15 seconds – CC)

The pick 5 lottery works by selecting 5 numbers from 1 to 35 (no repetition). It costs $1.

–Pick all 5 win Jackpot (estimated at $50000)

–Pick 4 win $500

–Pick 3 win $5

What is the expected winnings of a ticket?

Solution

Start by listing the outomes for the lottery in terms of the net profit.

To find the probability of each we will need to use both the combination formulas and the multiplication principle for the numerator term where we find the number of ways to select how many numbers match the winning numbers of interest times the number of ways for the remaining non winning numbers.

The expected value will be the sum of each outcomes times their probability:

$$\begin{array}{l}\text{EV}=\left(49999\right)\cdot \left(\frac{1}{324,632}\right)+\left(499\right)\cdot \left(\frac{150}{324,632}\right)+\left(4\right)\cdot \left(\frac{4350}{324,632}\right)+\left(-1\right)\cdot \left(\frac{320131}{324,632}\right)\\ \text{EV}=-\frac{177882}{324632}\approx -0.55\end{array}$$

The result shows the player should expect that on average over a large number of trials of buying the lottery ticket to lose approximately $0.55 per ticket.

Video Solution Example 4 (14 mins and 58 seconds – CC)

An insurance company determined that a 40-year-old man in the U.S. has a 0.242% risk of dying during the next year. The insurance company charges $275 for a 1 year life-insurance policy that pays $100,000 death benefit. What is the expected value for the person buying the insurance?

Solution

The outcomes in this situation is the net profit for the person buying the life insurance. One outcome is that the person is still alive (and probably greatful for that), but is out the $275 used to purchase the life insurance. The other outcome means the person didn’t survive, but the death benefit is paid out to the beneficiary of the policy. This payout net profit would be $\$100,000-\$275=\$99,725$. Below in the table we have the outcomes and the associated probabilities.

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

The person buying the insurance expects to lose overall $33 for each such policy when we look at a long run average. The flip side is more important to why insurance companies can stay in business. They expect to see many of these policies over the year. Each policy give an averge return of $33 for them.

Not surprisingly, the expected value is negative; the insurance company can only afford to offer policies if they, on average, make money on each policy. They can afford to pay out the occasional benefit because they offer enough policies that those benefit payouts are balanced by the rest of the insured people.

For people buying the insurance, there is a negative expected value, but there is a security that comes from insurance that is worth that cost.

Video Solution Example 5 (3 mins and 27 seconds – CC)