Expected Value and Decisions

Is the game worth playing?

A booster club sells $500$ raffle tickets at $2$ dollars each. One ticket wins $300$ dollars and two more win $50$ dollars each. What is a ticket worth?

Each outcome gets a net gain — prize minus the $2$ paid — and a probability:

Outcome	Net gain	Probability
grand prize	$298$	$\tfrac{1}{500}$
second prize	$48$	$\tfrac{2}{500}$
no prize	$-2$	$\tfrac{497}{500}$

The expected value of the net gain is

$$298\cdot\tfrac{1}{500} + 48\cdot\tfrac{2}{500} - 2\cdot\tfrac{497}{500} = \frac{298 + 96 - 994}{500} = -1.20$$

On average a ticket loses $1.20$ dollars. That is how the club raises money: the $1{,}000$ dollars in sales pays out $400$ in prizes.

Fair games and fair prices

A game is fair when its expected net gain is $0$. The fair price of a game is its expected winnings, before paying.

Why expected value decides in the long run

Expected value is the average of many repetitions. Some choices are made over and over: running a raffle every season, pricing a carnival game, insuring thousands of cars. Those are judged well by their expected value, because the long-run average is what will actually happen to the total.

A choice made once needs one more question: what is the worst case, and can it be afforded? The average only emerges over many repetitions, and a single bad outcome can matter more than a good average.

Comparing strategies

Two car insurance plans cover the same repairs:

	Premium per year	Deductible
Plan A	$1{,}200$	$500$
Plan B	$800$	$2{,}000$

Suppose a driver has a $10\%$ chance of a claim in a year, and any claim is large enough to use the full deductible. The expected yearly cost of each plan is the premium plus the deductible times its probability:

$$\text{A: } 1200 + 0.1(500) = 1250 \qquad \text{B: } 800 + 0.1(2000) = 1000$$

Plan B costs $250$ dollars less per year on average. But in a bad year, Plan A costs $1{,}700$ and Plan B costs $2{,}800$. A driver who could not cover the larger bill might still reasonably choose A.

The plans cost the same on average when $1200 + 500p = 800 + 2000p$, which gives $p \approx 0.27$. A driver with a claim chance above about $27\%$ does better with Plan A on average too.

Making fair decisions

Probability also decides things fairly: when several people have equal claims to one prize, a random process gives each the same chance.

Choosing among	A fair method
two people	flip a coin
three people	roll a die: faces $1$–$2$, $3$–$4$, $5$–$6$
any number	draw names from a hat, or use a random number generator

A method is fair only if the probabilities really are equal. Flipping a coin twice has four equally likely results, and four does not split evenly into three. Assigning HH, HT and TH to three people leaves TT, and giving TT to one of them doubles that person’s chance. Flipping again on TT restores fairness.

Tests and false positives

A medical test catches $95\%$ of people who have a condition and correctly clears $95\%$ of people who do not. The condition affects $2\%$ of people tested. If someone tests positive, how likely is it that they have the condition?

Picture $10{,}000$ people, as in a two-way table:

	Test positive	Test negative	Total
Have the condition	$190$	$10$	$200$
Do not	$490$	$9{,}310$	$9{,}800$
Total	$680$	$9{,}320$	$10{,}000$

Of the $680$ positives, only $190$ have the condition:

$$P(\text{condition} \mid \text{positive}) = \frac{190}{680} \approx 0.279$$

About $28\%$. The test is accurate, but the condition is rare, so the $5\%$ false-positive rate among the $9{,}800$ healthy people creates more false alarms than there are true cases. That is why a positive screening result is usually followed by a second, different test. The idea is the conditional probability of the positive group, not of the whole population.

Worked examples

Common mistakes

Practice problems