16.2: Expected Value

Last updated
Save as PDF

Page ID: 95160

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Most things in life are uncertain, so we don’t have any choice but to base our decisions on our views about probabilities. But the costs and benefits, and the value and disvalue of outcomes, also play a role in our decisions. For example, imagine you are thinking about going to see a movie, but the weather report said that there was a 40% chance of rain tonight, and you don’t like driving on slick roads. Should you go? If you don’t want to see the show very badly you may stay home, but if this is your only chance to see something you’ve really been wanting to see, the trip may be worth the risk.

Odds of 2 to 1 may be enough for someone to bet a few dollars, but not to bet your life (as you would in a case of risky surgery). Both the probabilities and the values (and disvalues) of outcomes play quite a role in our decisions. The following examples should help us see how this should work if we are reasoning well.

Example 1: Three Point Shots

Wilma, one of the guards on the UCLA basketball team, hits 40% of her shots from less than three-point range and 30% of her shots from three-point range. It may be best for Wilma to take certain shots in certain cases (e.g., if two points will win the game, then she should go for two).

But in general, is it better for her to take two-point shots or three-point shots? The probability of hitting a three pointer is lower, but the payoff is higher. How do we weigh these two considerations?

The following table gives us the answer:

Over the long-haul Wilma will, on average, get 0.8 points for each twopoint shot she takes and 0.9 points for each three-point shot. We say that .8 is the expected value of Wilma’s two point shot and 9 is the expected value of her three-point shots. Over the course of a season this difference can matter, and other things being equal it is better for Wilma to attempt three pointers.

Example 2: Rolling Dice

Your friend asks you to play the following game. You roll a die. If you get a six, they pay you six dollars. If you don’t get a six, you pay them one dollar. Would this be a profitable game for you to play? To answer this question, we need to determine the expected value of this game.

The formula for this when two outcomes are possible is this:

In the case of two pointers and three pointers, we could leave out probability of failure since the payoff in such cases is zero points. When we multiply this by the probability of failure, the result is still zero, so it drops out of the picture. But in the present case there is a “negative payoff” for failure.

Plugging the numbers in for the game proposed by your friend, your expected value is determined by the following rule:

The expected value of this game for you is 1/6 of a dollar. Over the long run, your average winnings per roll will be 1/6 of a dollar, or about sixteen and a half cents. Over the short run this isn’t much, but it could add up over time. So, it’s a good game for you (though not for your friend—unless they enjoy losing).

Exercise: What payoffs should your friend propose if they want the game to be fair for both of you?

The treatment of expected values can be extended in a natural way to cover more than two alternatives at a time. Just list all the possible outcomes, and record the probability and the payoff for each (listing losses as negative payoffs). Multiply the probability for each outcome by the payoff for that outcome. Then add up all these numbers.

You should think a bit about expected value before you play the slot machines, buy tickets for a lottery, or the like. In all these cases, there is a positive expected value for those running the game, a “house advantage,” and a negative expected value for those playing it. A similar point holds for insurance premiums. The insurance company calculates the probabilities of various outcomes and then determines prices of policies and amounts of payoffs so that the company will have a sufficiently high expected value for each policy.

There is a subjective side to payoffs. Even in games of chance, dollars aren’t the only things that matter. Some people like gambling, and so even if they lose a little money over the long run, their enjoyment compensates for this loss. Other people dislike risk, so even if they win a bit over the long run, the overall value of the game is negative for them.

There are many other cases where payoffs involve a person’s own feelings about matters. Wilbur has a heart condition that severely limits the things he can do. The probability that a new form of surgery will improve his condition dramatically is about 50%, the chances he’ll die in surgery are 7%, and the chances the surgery will leave him about the same are 43%. Should he get the surgery?

The answer depends on how much various things matter to Wilbur. If being alive, even in a very unpleasant physical condition, is important to him, then his assessment of the payoffs probably means that he shouldn’t elect surgery. But if he can’t stand being bed-ridden, he may assess the payoffs differently.

Pascal’s Wager

Blaise Pascal (1623–1662) was one of the founders of probability theory. He was also a devout Catholic in seventeenth century France. He argued that we should believe in God for the following reasons. While we are on this earth, we can never really settle the matter of whether God exists or not. But either He does, or he doesn’t.

Case one: God exists

If God exists and I believe that He exists, then I get a very high payoff (eternal bliss).
If God exists and I do not believe in Him I get a very negative payoff (fire and brimstone for all eternity).

Case two: God does not exist

If God does not exist and I believe that He does, I made a mistake, but its consequences aren’t very serious.
If He doesn’t exist and I don’t believe in Him, I am right, but being right about this doesn’t gain me a lot.

Pascal uses these claims to argue that we should believe in God. What are the relevant probabilities, payoffs, and expected values in each case? Fill in the details of his argument. What are the strengths and the weaknesses of the argument?

Exercises

Edna hits 45% of her three-point shots and 55% of her two-point shots. Which shot should she be trying for?
Suppose your friend Wilma offers to play the following game with you. You are going to roll a pair of dice. If you get a 7 or 11 (a natural) she pays you $3. If you roll anything else, you pay her $15. What is the expected value of the game for you? What is it for her?
Wilbur and Wilma are on their first date and have gone to the carnival. Wilma is trying to impress Wilbur by winning a stuffed toy for him. Wilma is trying to decide between two games: the duck shoot and the ring toss. She can shoot 55% of the ducks, which are worth two tickets each, and she can make about 35% of the ring tosses, which are worth four tickets each. Assuming Wilma needs to accumulate 15 tickets to win the toy, which game should she play?
In an earlier chapter, we learned about roulette. Calculate the expected value for betting on the number 13 (recall that the true odds against this are 37 to 1, but the house odds are 35 to 1).