ProbabilityProbabilityHigh school

Expected Value

Long-run average of a random variable.

The expected value of a random variable is its long-run average outcome: each possible value weighted by its probability, $E(X) = \sum x_i p_i$ . It need not be a value the variable can actually take. A game with expected value $-\$0.50$ loses fifty cents per play on average over many plays, even though no single play loses exactly fifty cents.

Expected value is how insurers set premiums, how casinos guarantee profit and how decision-makers compare risky options. Variance $\sum (x_i - \mu)^2 p_i$ then measures how much outcomes spread around that average.

Key terms

Expected value: Probability-weighted sum of outcomes.
$E[X] = \sum x\,P(X = x)$

Key ideas

$E(X) = \sum x_i P(x_i)$ ; check the probabilities sum to 1.
Linearity: $E(aX + b) = aE(X) + b$ and $E(X + Y) = E(X) + E(Y)$ .
A fair game has expected value 0.
Binomial: $E(X) = np$ .

Worked example

A raffle sells tickets for $\$5$ . One ticket in 200 wins $\$500$ . What is the expected net gain per ticket?

Win: net $+\$495$ with probability $\frac{1}{200}$ .
Lose: net $-\$5$ with probability $\frac{199}{200}$ .
$E = 495 \cdot \frac{1}{200} - 5 \cdot \frac{199}{200} = \frac{495 - 995}{200}$ .
$E = -\$2.50$ .

Answer: $-\$2.50$ per ticket

Common mistakes

Forgetting to subtract the ticket price from the prize.
Using probabilities that do not sum to 1.
Expecting the expected value to be the most likely outcome.

Practice questions

Try each one, then open the answer. Numerical answers on this page are verified by an automated calculation check.

1.A fair die: find the expected value of one roll.
Show answer
$3.5$
2. $X$ is binomial with $n = 20$ , $p = 0.3$ . Find $E(X)$ .
Show answer
$6$
3.If $E(X) = 4$ , find $E(3X - 2)$ .
Show answer
$10$

Key terms

Key ideas

Worked example

Common mistakes

Practice questions

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