Expected value.
Expected value is the average outcome per trial if the experiment were repeated indefinitely. The math of fair games, insurance premiums, and casino edges.
Should you play this game? A spinner has six equal slices. If it lands on 1 you win $1; on 2 you win $3; on 3 you win $5; on 4 you win $7. But if it lands on 5 or 6, you lose $8. The good outcomes look juicy — four out of six slices pay you money. The bad outcomes hit hard — lose $8 on either of the two remaining slices. So is the game in your favor?
The casino's question is sharper. They don't care whether you win on any single spin — they care what happens after millions of spins. The number that answers that question is expected value: the average payout per trial if the game were played indefinitely. Compute it by multiplying each possible payoff by its probability and summing. If the answer is positive, the player gains in the long run; negative, the player loses; zero, the game is fair.
For Debra's spinner, the four winning payoffs average to a gain, but the two $8 losses just barely balance them out. The expected value is exactly $0 — a fair game. Play it 10,000 times and you'll end up roughly where you started. That's the power of expected value: a single number that summarizes a complicated payoff structure in one shot.
Multiply each payoff by its probability; sum.
E(X) = x₁ · P(x₁) + x₂ · P(x₂) + ··· + xₙ · P(xₙ) = Σ x · P(x)
What it means. Imagine running the experiment a million times and averaging the outcomes. E(X) is the number that average will converge to. For a single trial, you almost never get the expected value — the spinner pays $1, or $7, or $-8, never $0. But averaged across many trials, the dollars settle in around E(X).
Three readings. Positive E(X) means the player gains money in the long run (would expect to win $0.50 per spin, say). Negative E(X) means the player loses (slot machines, lottery tickets, most casino games). Zero E(X) means the game is fair — no advantage to either side.
Left: the six-slice spinner. Right: the per-row calculation. Multiply each payoff by 1/6, sum all six. The four winning slices contribute +$2.67; the two $8-loss slices contribute −$2.67. They cancel exactly → E(X) = $0. A fair game.
The expected value of a discrete random variable X is E(X) = Σ x · P(x), summed over every possible value of X. It represents the long-run average outcome per trial.
Words you'll see on ALEKS
- Random variable (X) A quantity whose value depends on the outcome of a random experiment. The payoff on a spinner spin, the number of heads in 10 coin flips, the score on a die roll. Usually written X (or another capital letter).
- Expected value (E(X)) The mean (average) of the random variable, weighted by probability: Σ x · P(x). Also written μ in some textbooks. Has the same units as the random variable itself (dollars per spin, points per roll, etc.).
- Fair game A game with E(X) = 0. Neither player nor house has a long-run advantage. Pure fair games are rare in practice — most casino games have a slightly negative E for the player (the "house edge") and insurance premiums are set to give the insurance company a slightly positive E.
- Long-run interpretation E(X) is not what you should expect to win on any single trial — it's the average across many trials. If E(X) = $0.50, you should expect to be ahead by about $50 after 100 plays, $500 after 1,000 plays. Variance still matters in the short run.
- House edge When the operator of a game (a casino, a lottery, an insurance company) has a negative expected value from the player's side, the absolute value of that E(X) is the house edge. Roulette's edge is about $0.053 lost per dollar bet, every spin, on average. That's the long-run profit per trial that keeps the lights on.
Debra's spinner — full E(X) calculation.
"Debra is playing a game in which she spins a spinner with 6 equal-sized slices numbered 1 through 6. She wins $1 if the spinner stops on 1, $3 if it stops on 2, $5 if it stops on 3, $7 if it stops on 4. She loses $8 if the spinner stops on 5 or 6. Find the expected value of playing the game. What can Debra expect in the long run?"
List every outcome with its payoff and probability.
Six equal slices → each probability is 1/6. The payoffs are:
P(x) = 1/6 for each
Multiply each payoff by its probability.
Compute the six terms:
(+3)(1/6) = +3/6
(+5)(1/6) = +5/6
(+7)(1/6) = +7/6
(−8)(1/6) = −8/6
(−8)(1/6) = −8/6
Sum the six terms.
All six have denominator 6, so just sum the numerators:
= 0 / 6 = $0
Interpret.
E(X) = $0 means the game is fair — Debra can expect to break even in the long run. On any single spin she'll win or lose money, but averaged over many spins her account balance stays approximately where she started.
→ E = 0 → break evenThree E(X) computations.
You roll a fair six-sided die. You win $3 if it lands on 1, 2, or 3, and you lose $2 if it lands on 4, 5, or 6. What is the expected value per roll? (Decimal, rounded to 2 decimal places.)
A lottery ticket costs $2. There's a 1-in-500 chance of winning $200; otherwise you win nothing. What is the expected value of buying one ticket (net of the ticket cost)?
An insurance company sells a 1-year policy for $300. The probability of having to pay out the $10,000 benefit is 0.02 (2 in 100). What is the expected value to the company per policy?
Three fast questions before you move on.
Q1. If the expected value of a game is −$0.50 per play, what should the player expect after 1,000 plays?
Q2. A coin is flipped. You win $5 on heads, lose $3 on tails. What is E(X)?
Q3. Which interpretation of expected value is correct?
The single number behind every game and every premium.
Expected value is the math behind fair value. Insurance premiums are set so that E(X) is slightly positive for the company (otherwise they'd go bankrupt). Lottery tickets have E(X) sharply negative for the buyer (otherwise the state wouldn't run them). Casino games have small but consistent negative E for the player — the "house edge" is literally the absolute value of E.
In finance, the expected return on a stock drives every portfolio decision. In healthcare, expected utility (a probability-weighted average of outcomes) drives treatment choices. The formula stays the same: list outcomes, multiply by probabilities, sum.
Next: Lesson 6 closes the topic by comparing theoretical probability (what the math predicts) with experimental probability (what the data shows). As trials accumulate, the two converge — the Law of Large Numbers, stated in plain English.
Continue to Lesson 06Different angle? Need another rep? These are optional — tap any that look helpful.
Expected profit from lottery ticket
Sal applies E(X) = Σ value × probability to a lottery ticket, computes the expected profit, and shows why the EV is negative for the player (positive for the lottery). Hits the formula and the house-edge insight in one shot.
Expected payoff example — lottery ticket
2021 Khan reshoot of the same concept with cleaner audio. Functionally interchangeable with the above; pick whichever production quality fits.