Math Core

Expected value is the long-run average of a random quantity. It shows up on almost every AMC 10 and 12, and the problems that look hardest often collapse in two lines thanks to one fact: linearity of expectation. This module teaches the definition, then shows how to break a messy count into simple pieces.

The definition

If a random quantity $X$ takes values $x_1, x_2, \dots$ with probabilities $p_1, p_2, \dots$ , its expected value is

$E[X] = x_1 p_1 + x_2 p_2 + \cdots.$

For a fair die, $E[X] = \dfrac{1 + 2 + 3 + 4 + 5 + 6}{6} = \dfrac{7}{2}$ . The expected value doesn't have to be a possible outcome; it's an average.

Linearity of expectation

For any random quantities $X$ and $Y$ and constants $a$ , $b$ ,

$E[X + Y] = E[X] + E[Y], \qquad E[aX + b] = aE[X] + b.$

This holds even when $X$ and $Y$ are dependent.

The expected sum of three dice is $3 \cdot \dfrac72 = \dfrac{21}{2}$ , with no need to list $216$ outcomes.

The real power comes from indicator variables. To find the expected number of times something happens, write the count as a sum of $0$ -or- $1$ variables, one for each place it could happen:

$X = I_1 + I_2 + \cdots + I_n, \qquad I_k = \begin{cases} 1 & \text{if event } k \text{ happens} \\ 0 & \text{otherwise.} \end{cases}$

Since $E[I_k] = P(\text{event } k)$ ,

$E[X] = P(\text{event } 1) + P(\text{event } 2) + \cdots + P(\text{event } n).$

You never need the distribution of $X$ itself, only the probability of each small event. The events can overlap and depend on each other in complicated ways; it doesn't matter.

Waiting times

If each trial succeeds with probability $p$ , independently, then the expected number of trials up to and including the first success is $\dfrac{1}{p}$ . For example, you expect to roll a die $6$ times to get a $6$ .

Why: let $E$ be the expected number of trials. The first trial always counts. With probability $p$ you're done; otherwise you're back where you started. So $E = 1 + (1 - p)E$ , which gives $E = \dfrac1p$ .

Worked example: Direct definition

You roll a fair die and win $n^2$ dollars, where $n$ is the number rolled. What is your expected winnings?

$E = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6} \approx 15.17 \text{ dollars}.$

Note that $E[X^2] = \tfrac{91}{6}$ is not $(E[X])^2 = \tfrac{49}{4}$ . Linearity works for sums, not for squares or products of dependent quantities.

Worked example: Indicators: adjacent heads

A fair coin is flipped $10$ times. What is the expected number of places where two consecutive flips are both heads? (In HHHT, there are $2$ such places.)

There are $9$ pairs of consecutive flips. Let $I_k = 1$ if flips $k$ and $k + 1$ are both heads. Each has probability $\tfrac14$ . So the expected count is $9 \cdot \tfrac14 = \tfrac94$ .

The indicators overlap (pairs $1$ – $2$ and $2$ – $3$ share a flip), but linearity doesn't care.

Worked example: Indicators: distinct values

Three fair dice are rolled. What is the expected number of different values that appear?

For each face value $v$ from $1$ to $6$ , let $I_v = 1$ if $v$ appears at least once. By the complement, $P(I_v = 1) = 1 - \left(\tfrac56\right)^3 = \tfrac{91}{216}$ . So the expected number of distinct values is

$6 \cdot \frac{91}{216} = \frac{91}{36} \approx 2.53.$

Common mistake

Don't try to find the full distribution when linearity will do. Students often spend ten minutes computing $P(X = 0), P(X = 1), \dots$ for a count $X$ . If the question only asks for $E[X]$ , look for indicators first: "the expected number of ___" almost always means "sum the probabilities of each ___."

Worked example: Collecting every face

A fair die is rolled until every face has appeared at least once. What is the expected number of rolls?

Split the process into stages. Once you've seen $k$ different faces, each roll shows a new face with probability $\tfrac{6 - k}{6}$ , so that stage takes $\tfrac{6}{6 - k}$ rolls on average. By linearity, the total is

$\frac66 + \frac65 + \frac64 + \frac63 + \frac62 + \frac61 = 6\left(1 + \tfrac12 + \tfrac13 + \tfrac14 + \tfrac15 + \tfrac16\right) = \frac{147}{10} = 14.7.$

Practice

Practice 1

Three fair dice are rolled. What is the expected value of their sum?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A bag holds $4$ red marbles and $6$ blue marbles. Three marbles are drawn without replacement. What is the expected number of red marbles drawn?

Practice 3

Two fair dice are rolled. What is the expected value of the larger of the two numbers? (If they are equal, that number is the larger.)

Practice 4

Seven people each put their name in a hat, and each person draws one name at random (all $7!$ assignments are equally likely). What is the expected number of people who draw their own name?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Every subset of $\{1, 2, 3, \dots, 10\}$ , including the empty set, is written on a card, and one card is chosen at random. What is the expected value of the sum of the numbers in the chosen subset? (The empty set has sum $0$ .)

Practice 6

The numbers $1$ through $8$ are arranged in a random order $a_1, a_2, \dots, a_8$ . A descent is a position $i$ with $a_i \gt a_{i+1}$ . What is the expected number of descents?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

A fair coin is flipped repeatedly until two heads in a row appear. What is the expected number of flips?

Practice 8

Four fair dice are rolled. What is the expected number of different values that appear?

Practice 9

Five men and five women sit at random around a round table with $10$ seats. What is the expected number of pairs of neighbors consisting of one man and one woman?

Real contest practice

2021 Fall AMC 10B, Problem 16: the expected number of balls back in their starting spots after two random swaps.
2021 Fall AMC 12B, Problem 23: an expected count of consecutive pairs, a natural fit for indicators.