EXPECTED VALUE

Overview

The expected value (or expectation) of a random variable $X$ is the long-run average outcome if the experiment were repeated many times. It is defined as

$E[X] = \sum_{\text{all outcomes}} (\text{value}) \cdot P(\text{outcome}).$

You can think of it as a weighted average where each possible value is weighted by its probability. If you roll a fair die many thousands of times, the average result will get closer and closer to $E[X] = 3.5$ — that is what "expected value" means in practice.

AMC10 loves expected value because it appears in games, random walks, and counting problems that would be nearly impossible to solve by brute force. The most powerful tool is linearity of expectation, which lets you compute complex expectations by breaking them into tiny, easy pieces — each piece being an indicator variable that equals 1 when some event happens and 0 otherwise.

Key facts

• Basic formula: $E[X] = \sum_{k} k \cdot P(X = k)$ Use when you can list all values $X$ can take with their probabilities.

• Linearity of expectation (the star tool): $E[X + Y] = E[X] + E[Y]$ This holds for any random variables $X$ and $Y$ — they do not need to be independent.

• Scaling: $E[aX + b] = a\,E[X] + b$ for constants $a$ and $b$.

• Indicator variable trick: For any event $A$, define the indicator $\mathbf{1}_A$ which equals 1 if $A$ occurs and 0 otherwise. Then $E[\mathbf{1}_A] = P(A).$ To find the expected number of objects satisfying a property, write the count as a sum of indicators and apply linearity: $E\!\left[\sum_i \mathbf{1}_{A_i}\right] = \sum_i P(A_i).$

• Independent variables: When $X$ and $Y$ are independent, $E[XY] = E[X]\cdot E[Y].$ Note: this requires independence; linearity holds without it.

• Expected value in a game / fair price: A game is fair if its expected payout equals the cost to play. The fair price to pay for a game with payout $V$ is $E[V]$.

• Geometric expectation: If each trial succeeds independently with probability $p$, the expected number of trials until the first success is $1/p$.

• Expected counts via indicators: The expected number of "good" objects among $n$ is $n \cdot p$ where $p$ is the probability that any single object is "good" — even if the objects are not independent.

Worked example 1

A bag contains 4 red and 6 blue marbles. A marble is drawn at random. Red pays \5$, blue pays$$2$. What is the expected payout?

List the outcomes:

| Outcome | Value | Probability | |---------|-------|-------------| | Red | \5$|$\dfrac410 = \dfrac25$| | Blue |$$2$|$\dfrac610 = \dfrac35$ |

$E = 5 \cdot \frac{2}{5} + 2 \cdot \frac{3}{5} = 2 + \frac{6}{5} = \frac{16}{5} = 3.2.$

The expected payout is \\dfrac165$.

Worked example 2

Five students line up at random. What is the expected number of students standing in their original position (i.e., student $k$ is in spot $k$)?

It would be painful to count all derangement-related cases. Instead, use linearity of expectation with indicators.

For each student $k$ (where $1 \le k \le 5$), define: $I_k = \begin{cases} 1 & \text{if student } k \text{ stands in spot } k \\ 0 & \text{otherwise.}\end{cases}$

The total number of "fixed" students is $X = I_1 + I_2 + I_3 + I_4 + I_5$.

By symmetry, $P(I_k = 1) = \dfrac{1}{5}$ for each $k$ (student $k$ is equally likely to occupy any of the 5 spots).

By linearity: $E[X] = E[I_1] + \cdots + E[I_5] = 5 \cdot \frac{1}{5} = 1.$

The expected number of students in their original position is $\boxed{1}$ — regardless of how many students there are! The same calculation works for any $n$ students: $E = n \cdot \frac{1}{n} = 1$.

Key insight: We never needed to worry about whether $I_j$ and $I_k$ are independent — linearity works anyway.

Common traps

• Confusing $E[X^2]$ with $(E[X])^2$. These are equal only when $X$ is constant. In general $E[X^2] \ge (E[X])^2$ (Jensen's inequality). For a fair die: $E[X] = 7/2$ so $(E[X])^2 = 49/4 = 12.25$, but $E[X^2] = (1+4+9+16+25+36)/6 = 91/6 \approx 15.2$.

• Requiring independence for linearity. Linearity of expectation $E[X+Y]=E[X]+E[Y]$ holds always. Only the product rule $E[XY]=E[X]E[Y]$ requires independence.

• Forgetting to include all outcomes. If a game can end with "no payout," that \0$outcome still contributes to the average — just with a coefficient of$0$. You must account for all outcomes summing to probability 1.

• Mixing up "expected value" and "most likely value." The expected value of a die roll is $3.5$, which is not even a possible outcome. "Expected" means the long-run average, not the mode.

• Double-counting with indicators. When setting up indicator variables, make sure each object gets exactly one indicator — do not define $I_{ij}$ for both an ordered pair $(i,j)$ and the reversed pair $(j,i)$ if they represent the same event.