B. Probability Distributions · Discrete Probability Distributions

Making Decisions with Expected Value

B1.2 B1.7 B1.4About 70 minutes

Learning goals

I can set up a payoff (profit or cost) random variable for a decision and find its expected value.
I can compare options with expected value and choose the one that is best in the long run.
I can combine expected value with binomial probabilities in applications such as overbooking and group testing.
I can explain the limits of expected value, including risk and one-time decisions.

Words to know

payoff
The amount gained (positive) or lost (negative) in a particular outcome of a decision.

expected payoff
E(payoff) = Σ payoff × probability: the average result per decision if the same decision were made many times.

risk
The chance of a bad outcome and how bad it is. Two options can have the same expected value but very different risk.

From games to real choices

In lesson 4 you used expected value to judge games. The same tool helps businesses, governments and you make real decisions. The method:

List the options.
For each option, list the possible outcomes, their probabilities and their payoffs (profit, cost, time, points).
Compute the expected payoff of each option.
Choose the option with the best expected payoff, then think about risk and anything the numbers leave out.

The probabilities often come from the distributions in this unit.

Worked example

Example 1: A food truck at a summer festival

A food truck owner can go to an outdoor festival or work an indoor market. At the festival, profit is $1200 if it is sunny and a loss of $200 if it rains. The forecast gives a 60% chance of sun. The indoor market gives a sure profit of $500. Which should she choose?

Festival: E = 1200(0.6) + (−200)(0.4) = 720 − 80 = $640.
Indoor market: E = $500.
Difference: the festival is better by $140 on average.

In the long run, the festival earns more ($640 vs $500 on average). But on a rainy day she loses $200. If she could not afford a loss, the sure $500 might be the wiser choice for her. Expected value is the starting point, not the whole decision.

Worked example

Example 2: Guessing on a test with a penalty

Some contests give +1 for a correct answer, −0.25 for a wrong answer and 0 for a blank. Each question has 5 choices. (a) Should you guess at random? (b) What if you can rule out one choice?

(a) Random guess: P(correct) = 1/5. E = 1(0.2) + (−0.25)(0.8) = 0.2 − 0.2 = 0. Guessing and leaving it blank are equal on average.
(b) With 4 choices left: E = 1(0.25) + (−0.25)(0.75) = 0.25 − 0.1875 = 0.0625 > 0.

(a) The penalty is set to make random guessing worth exactly 0 on average. (b) Once you can eliminate even one choice, guessing has a positive expected value, so you should guess.

Overbooking: expected value plus the binomial distribution

In the last lesson an airline selling 22 tickets for 20 seats had about a 34% chance of too many passengers. Should it overbook at all? To decide, it needs payoffs as well as probabilities.

Suppose each passenger who flies brings in $150, no-shows are refunded, and each passenger who is bumped (no seat) costs the airline $400 in compensation. Each ticket holder shows up with probability 0.9, independently, so X = number who show up ~ B(n, 0.9) when n tickets are sold.

Worked example

Example 3: How many tickets to sell?

Find the expected profit if the airline sells 20 tickets and if it sells 21 tickets.

20 tickets: nobody is ever bumped. E(X) = 20 × 0.9 = 18, so E(profit) = 150 × 18 = $2700.
21 tickets: X ~ B(21, 0.9). If X ≤ 20, profit = 150X. If X = 21 (probability 0.9²¹ ≈ 0.1094), 20 fly and 1 is bumped: profit = 150(20) − 400 = $2600.
E(profit) = 150 × E(X) − 150 × P(X = 21) − 400 × P(X = 21), because when X = 21 one of the 21 does not fly (−150) and is compensated (−400).
= 150(18.9) − 550(0.1094) ≈ 2835 − 60.18 ≈ $2774.82.

Selling 21 tickets raises expected profit from $2700 to about $2775. The graph shows that 22 tickets gives about $2729 and more tickets make things worse, so 21 tickets is best under these assumptions.

Expected profit rises with one extra ticket, then falls as the cost of bumped passengers grows. Each bar is an expected value computed from a binomial distribution B(n, 0.9).

Write the payoff as a random variable

The key step in every decision problem is writing the payoff for each value of X. A small table with columns x, P(X = x) and payoff(x) keeps everything organized. Then E(payoff) = Σ payoff(x) × P(X = x).

Worked example

Example 4: Group (pooled) testing

A lab tests water samples for a rare contaminant; each sample is contaminated with probability 0.02, independently. Instead of testing 10 samples separately, the lab mixes part of each into one pooled test. If the pool is negative, all 10 are clear (1 test). If it is positive, the lab tests all 10 separately (11 tests in total). Find the expected number of tests.

The pool is negative only if all 10 samples are clean: P = 0.98¹⁰ ≈ 0.8171 (binomial, X = 0).
P(pool positive) ≈ 1 − 0.8171 = 0.1829.
E(tests) = 1(0.8171) + 11(0.1829) = 1 + 10(0.1829) ≈ 2.83.

About 2.83 tests per 10 samples instead of 10: a saving of more than 70% on average. This idea, first suggested for blood testing in the 1940s, has been used to screen large numbers of samples quickly.

Pooling works because most pools come back negative. When the contaminant is rare, one test usually clears 10 samples at once.

The limits of expected value

Expected value is the right tool when a decision is made many times (an airline's daily flights, a lab's thousands of samples, an insurer's millions of policies). For a one-time decision with a large possible loss, people and businesses also consider:

Risk: how bad is the worst case, and how likely is it? (The spread of the distribution, measured by standard deviation, helps here.)
Values that are not money: a bumped passenger is an unhappy customer who may not fly with you again.
Whether the probabilities are trustworthy: a model built on a wrong p gives a wrong expected value.

Just for fun

Expected value says the food truck should go to the festival. Expected value has never had to scrape wet ketchup off a picnic table in the rain.

Hands-on activity

Plan a fundraiser with expected value

You need: paper, pencil and a calculator (a spreadsheet is helpful).

Your club is choosing between two fundraisers for a winter carnival.

Option A: outdoor hot chocolate stand. Expected sales depend on weather: cold and clear (probability 0.5) → profit $400; mild (probability 0.3) → $250; storm cancels the carnival (probability 0.2) → loss of $100 for supplies you cannot return.
Option B: indoor bake sale. Profit $260 for sure.

Compute the expected profit of option A and compare it with B.
Your club will run this fundraiser once, and your account has only $50 in it. Which would you choose, and why?
Suppose you could run the fundraiser at 10 different winter events. Which option would you pick then?
Change one probability or payoff so that the two options have the same expected profit.

Model answer: E(A) = 400(0.5) + 250(0.3) − 100(0.2) = 200 + 75 − 20 = $255, slightly less than B's $260. B is better on average and has no risk, so it is the clear choice once; over 10 events it still wins on average. To make them equal, A needs $5 more expected profit, for example a mild-day profit of about $266.67, or a storm loss of $75 instead of $100.

Your turn

A farmer can plant crop A (profit $8000 in a good year, $2000 in a poor year) or crop B (profit $5000 either way). P(good year) = 0.6. Which has the higher expected profit?

A test gives +4 for a right answer and −1 for a wrong answer on 5-choice questions. Is a random guess worth it?

A contestant can take $500 now or answer a question worth $1200 (else $0). She thinks she has a 45% chance of answering correctly. What does expected value suggest?

In the pooled-test example, what is the probability that the pool comes back positive?

A company can buy a $300 service plan for a machine, or pay for repairs: $0 (prob. 0.7), $500 (prob. 0.2), $2000 (prob. 0.1). Compare expected costs.

Using the overbooking model, what is the expected number of passengers who show up when 22 tickets are sold?

Why is expected value a better guide for an airline's daily overbooking policy than for a one-time decision by a family?

More practice: warm-up

Option A pays $100 with probability 0.5, else $0. Option B pays $40 for sure. Which has the higher expected value?

What is the expected value of a payoff of −$10 with probability 1?

A 5-choice question: +1 right, −0.25 wrong. E of a random guess?

If p(contaminated) = 0.02, what is P(a sample is clean)?

Name one thing expected value does not measure.

More practice: on your own

A store sells a $60 accessory. If it orders 10 and sells X ~ B(10, 0.7) of them, with profit $25 per sale and a loss of $10 on each unsold item, find the expected profit.

In group testing, find the expected number of tests per sample for pools of 5 with p = 0.02.

Compare pools of 5 and pools of 10 (2.83 tests per 10 samples) when p = 0.02. Which uses fewer tests per sample?

A student who knows the answer to 60% of 5-choice questions guesses randomly on the rest (+1 right, −0.25 wrong). On a 40-question test, what is the expected score?

For the overbooking model, compute the expected number of bumped passengers when 22 tickets are sold. (P(21) ≈ 0.2407, P(22) ≈ 0.0985.)

Challenge

Verify the overbooking expected profit for 22 tickets: E(profit) = 150[E(X) − E(bumped)] − 400 E(bumped), with E(X) = 19.8 and E(bumped) ≈ 0.4377.

In group testing with pools of size g and contamination probability p, show that the expected number of tests per sample is 1/g + 1 − (1 − p)ᵍ. Evaluate it for g = 20, p = 0.02.

If the bumping cost rose to $2000 per passenger, would selling 21 tickets still beat selling 20? Use P(X = 21) ≈ 0.1094 for B(21, 0.9).

Check your understanding

Quiz

Option A: $1000 with probability 0.3, else $0. Option B: $250 for sure. Which statement is true?

On a 4-choice question scored +1 right and −1/3 wrong, what is the expected value of a random guess?

Ten samples are pooled; each is contaminated with probability 0.02. What is P(pool negative)?

For which kind of decision is expected value most reliable as the only guide?

A store expects to sell X ~ B(20, 0.6) items, earning $10 per sale. What is the expected revenue?

An airline sells 21 tickets for 20 seats; each holder shows up with probability 0.9. What is P(someone is bumped)?

For parents and teachers

Big idea (B1.2, B1.7): expected value turns probabilities and payoffs into one number for comparing options. Businesses use it with the binomial distribution for overbooking, inventory and testing decisions. It is most reliable for decisions made many times; for one-time, high-stakes decisions, risk matters too.

How to help: talk through a real family decision (an extended warranty, a fundraiser, a weekend plan that depends on weather) and estimate the probabilities and payoffs together. Ask your teen what expected value leaves out.

Common difficulties:

Forgetting to include every outcome (for example, the loss on rainy days).
Mixing up the probability of an outcome with its payoff.
Treating the option with the highest expected value as automatically "right" without thinking about risk.