## Section4.3Expected Value

Expected value is one of the useful probability concept we will discuss. It has many applications, from insurance policies to making financial decisions, and it's one thing that the casinos and government agencies that run gambling operations and lotteries may hope most people never learn about.

### Subsection4.3.1Expected Value

The expected value is the average gain or loss of an event if the procedure is repeated many times. To help get a better understanding of what expected value is and how it is used, consider the following scenario:

You are commissioned to design a game for a local carnival. Your proposed game will have players roll a six-sided die. If it comes up 6, they win $10. If not, they get to roll again. If they get a 6 on the second roll, then they win 3. If they do not get a 6 on the second roll, they lose. With the game design complete, you now need to decide how much the carnival game owner should charge players in order to make a profit over the long run. To make a profit, the carnival needs to know how much they will pay in winnings, on average, over the long run and charge more than that. In other words, they must charge more than the expected value of the game. One way to organize the outcomes and probabilities is with a probability model. A probability model or probability distribution is a table listing the possible outcomes and their corresponding probabilities. The outcomes will be the amounts a player can win, and we will calculate the probabilities using what we have learned about theoretical probability. As we have seen with complements, probabilities in a probability distribution must add to 1, so that is important to check. Here is the probability model for the carnival game: Outcome ($ won) Rolling Event Probability
$10 Roll a 6 on the first roll $$P(\text{roll a 6})=\frac{1}{6}$$$3 Roll not a 6 on the first roll
and a 6 on the second roll
$$P(\text{roll a 6 then roll a 6})=\frac{5}{6}\cdot\frac{1}{6}=\frac{5}{36}$$
$0 Roll not a 6 on the first roll and not a 6 on the second roll $$P(\text{roll a 6 then not a 6})=\frac{5}{6}\cdot\frac{5}{6}=\frac{25}{36}$$ Think of the expected value as a weighted average. We could take the average of$10, $3, and$0, but they are not all equally likely. It is much more likely to win $0 than to win$10. So, to find the average, we multiply each outcome by the chance it will happen and add the products together.

###### Expected Value.

Multiply each outcome by its probability and add up the products

In this case we have:

\begin{gather*} \text{Expected winnings}=\$10(\frac{1}{6})+\$3(\frac{5}{36})+\$0(\frac{25}{36})=\$2.08 \end{gather*}

This tells us that over the long run, players can expect to win $2.08 per game. This also means that the carnival owner will be paying out an average of$2.08 per game! Since the carnival owner would rather not lose money by paying players over the long run we need to make sure to charge players enough to offset the average payout.

If the carnival owner charges exactly $2.08 to play, the game is considered a fair game since the expected winnings would be$0. In a fair game, the player isn’t expected to win anything, nor is the owner expected to earn anything over the long run. However, if the carnival owner charges the player more than $2.08 to play, they will earn money over the long run. Suppose you suggest charging$5 to play. We can determine the net winnings by subtracting the $5 the player has to pay from their expected winnings. This gives us: \begin{gather*} \text{Net player winnings}=\$2.08-\$5=-\$2.92 \end{gather*}

This means that over the long run, players can expect to lose an average of $2.92 each game they play, and the carnival owner can expect to earn an average of$2.92 per game over the long run. Here’s another example.

###### Example4.3.2.

Pick4 is a game by the Oregon Lottery that costs $1 to play. In this game you pick 4 numbers in a specific pattern. If you get the exact sequence, you can in theory earn a lot of money. Suppose that the payouts are as follows. Determine the player’s expected net winnings. Prize ($) Probability
$250 $$\sfrac{1}{417}$$$500 $$\sfrac{1}{1833}$$
$1000 $$\sfrac{1}{1667}$$$1500 $$\sfrac{1}{2500}$$
Solution.

This table is not quite a complete probability distribution since it is missing one important outcome: when the player loses. In that case the prize is $0. We need to add a line for this. The prize for this missing outcome is$0, and since losing is the complement to winning something, the probability will be:

\begin{align*} P(\text{Win }\0)\amp=1-(\frac{1}{417}+\frac{1}{1833}+\frac{1}{1667}+\frac{1}{2500})\\ \amp=1-0.0039\\ \amp=0.9961 \end{align*} Adding this information to the table gives a complete probability distribution. Now we can see that players are going to lose more than 99% of the time, so the expected value will be heavily weighted toward winning0.

Prize ($) Probability$250 $$\sfrac{1}{417}$$
$500 $$\sfrac{1}{1833}$$$1000 $$\sfrac{1}{1667}$$
$1500 $$\sfrac{1}{2500}$$$0 0.9961
\begin{align*} \text{Expected winnings}\amp=\$250(\frac{1}{417})+\$500(\frac{1}{1833})+\$1000(\frac{1}{16773})+\$1500(\frac{1}{2500})+\$0(0.9961)\\ \amp=\$2.07 \end{align*}

Therefore, the player’s expected winnings are $2.07, on average, over the long run. To find the expected net winnings, we subtract the cost to play. Since it costs$1 to play,

\begin{gather*} \text{Net Expected Winnings}=\$2.07-\$1.00=\$1.07. \end{gather*} Assuming the given payouts are correct, this would be one game you would want to play for investment purposes since you can expect to earn$1.07 per game, on average, over the long run. Play a million times, and you just might become a millionaire!

In general, if the expected value of a game is negative, it is not a good idea to play, since in the long run you will lose money. It would be better to play a game with a positive expected value (good luck trying to find one!), although keep in mind that even if the average winnings are positive it could be the case that most people lose money and one very fortunate individual wins a great deal of money.

Not surprisingly, the expected value for casino games is always negative for the player, and therefore positive for the casino. It must be positive for the casino, or they would go out of business! Players just need to keep in mind that when they play a game repeatedly, they should expect to lose money. That is fine so long as you enjoy playing the game and think it is worth the cost, but it would be wrong to expect to come out ahead.

Expected value is not only used to determine the average amount won and lost at casinos and carnivals, it also has applications in business and insurance, just to name a few. Let’s look at a couple of those applications.

###### Example4.3.3.

For 3 months, a coffee shop tracked their morning sales of coffee, between 6am and 10am. The following results were recorded:

 Number of cups sold Probability 145 150 155 160 170 0.15 0.22 0.37 0.19 0.07

How many cups of coffee should they expect to sell each morning?

Solution.

In this case the table tells us that 15% of the time they sell 145 cups of coffee between 6am and 10am, 22% of the time they sell 150 cups, 37% of the time they sell 155 cups, 19% of the time they sell 160 cups, and 7% of the time they sell 170 cups. Since the highest probability is associated with 155 cups, the expected value should lie somewhat close to this.

To find the expected number of coffees sold, we multiply each number of cups of coffee by its respective probability and then add the products.

\begin{align*} \text{Expected number of coffees sold}\amp=145(0.15)+150(0.22)+155(0.37)+160(0.19)+170(0.07)\\ \amp=154.4\text{ cups of coffee} \end{align*}

This means that over the long run, the coffee shop can expect, on average, to sell around 154 cups of coffee each morning. This is an important tool for businesses since it helps inform them how much stock they should keep on hand.

###### Example4.3.4.

On average, a 40-year-old man in the US has a 0.242% chance of dying in the next year 1 . An insurance company charges $275 annually for a life insurance policy that pays a$100,000 death benefit. What is the expected value for the insurance company on this policy?

Solution.

The first thing we want to do is organize the probabilities and outcomes in a probability distribution table. There are two outcomes – either the policy holder dies, and the insurance company pays the benefit, or the policy holder does not die, and they do not pay anything.

The probability of paying the death benefit is equal to the chance of the person dying in the next year, and the probability of paying nothing is equal to the complement of the chance of dying in the next year.

Insurance payout Probability
$100000 0.00242$0 1-0.00242=0.99758

Then we can calculate:

\begin{gather*} \text{Expected Payout}=\$100000(0.00242)+\$0(0.99758)=\$242 \end{gather*} So, the expected payout for the insurance company is$242, but they are charging $275 for the policy. Their net revenue would be \begin{gather*} \text{Net Value to Insurance Company}=\$275-\$242=\$33 \end{gather*}

The insurance company is making, on average, $33 per policy per year. It shouldn’t be too surprising because there are overhead costs and the insurance company can only afford to offer policies if they, on average, make money on them. But how much money should they make? As a consumer it is important to know about expected value. ### Exercises4.3.2Exercises ###### 1. A bag contains 3 gold marbles, 6 silver marbles, and 28 black marbles. Someone offers to play this game: You randomly select on marble from the bag. If it is gold, you win$3. If it is silver, you win $2. If it is black, you lose$1.

1. Make a probability model for this game.

2. What is your expected value if you play this game?

###### 13.

Create a problem using the concept of expected value. Possible topics include insurance policies, financial decisions, gambling and lotteries. Determine the expected value of the situation you created.