Expected Value

What you need to know

• Expected value is the long-run average. Multiply each possible value by its probability, then add. A game is fair when your expected net winnings (what you win minus what you paid) are $0$. Example: a spinner pays $1$, $4$, or $10$ points with probabilities $\frac12$, $\frac14$, $\frac14$. Its expected value is $\frac12\cdot1+\frac14\cdot4+\frac14\cdot10 = 4$, so $4$ points is a fair price to play.
• Linearity of expectation. Write $E(X)$ for the expected value of a random quantity $X$. Then $E(X+Y) = E(X)+E(Y)$, always, even when $X$ and $Y$ affect each other. Example: one die averages $3.5$, so the sum of two dice averages $7$.
• The indicator trick. The expected number of events (out of $n$) that happen is the sum of their probabilities. Why: score each event $1$ if it happens and $0$ if not. The count is the sum of the scores, and each score averages out to that event's probability. Example: $5$ letters go randomly into $5$ addressed envelopes. Each is right with probability $\frac15$, so on average $5\cdot\frac15 = 1$ is right.
• Values to know. A fair die averages $\frac{1+2+\dots+6}{6} = 3.5$. If each try succeeds with probability $p$, the expected number of tries until the first success is $\frac1p$ (a $6$ takes $6$ rolls on average).
• Restart equations. Let $T$ be the expected number of steps for a process that can go back to the start. One step always happens, and with probability $1-p$ it fails and you are back at the start, so $T = 1+(1-p)\,T$. Example: $p = \frac16$ gives $\frac16T = 1$, so $T = 6$.

How AMC 10 tests it

• Problems 8–15: the expected value of a die, a spinner, or a game with listed payoffs; add up value times probability, or find a ticket's fair price.
• Problems 14–22: the expected number of something (neighboring beads of the same color, matches between two shuffled decks, runs of heads), meant for indicators and linearity.
• Problems 18–24: the expected number of steps until a random process stops, using a restart equation or states.
• Sometimes mixed with counting: the expected size of a random subset.

Standard approaches

1. With few outcomes, list each value with its probability and add up value times probability.
2. If the quantity counts how often something happens, write it as a sum of $0$-or-$1$ scores and use linearity. You never need the full list of probabilities.
3. Use symmetry: positions that look alike contribute the same amount.
4. For "expected number of tries until," set up $E = 1 + (\text{probability of going on})\cdot E$, or a small system with one unknown per state.
5. Check that the answer is between the smallest and largest possible values.

Worked example

Ten cards numbered $1$ through $10$ are shuffled and laid out in a row. What is the expected number of positions $i$, with $1 \le i \le 9$, for which the cards in positions $i$ and $i + 1$ differ by exactly $1$?

(A) $\frac{9}{10}$ (B) $1$ (C) $\frac{9}{5}$ (D) $2$ (E) $\frac{5}{2}$

Solution.

Step 1: scores. For each $i$ from $1$ to $9$, let $X_i = 1$ if the cards in positions $i$ and $i+1$ differ by $1$, and $X_i = 0$ if not. The number we want is $X_1+X_2+\dots+X_9$.

Step 2: linearity and symmetry. The expected total is the sum of the nine expected scores, and each is the probability that its score is $1$. All nine neighboring pairs look alike in a random shuffle, so the answer is $9$ times the probability for positions $1$ and $2$.

Step 3: one probability. The cards in positions $1$ and $2$ are a random ordered pair of different cards: $10\cdot9 = 90$ equally likely pairs. The pairs that differ by $1$ are $\{1,2\}, \{2,3\}, \dots, \{9,10\}$, each in $2$ orders: $9\cdot2 = 18$. The probability is $\frac{18}{90} = \frac15$.

Step 4: multiply. $9\cdot\frac15 = \frac95$.

The answer is $\boxed{\textbf{(C)}\ \frac{9}{5}}$.

Pitfalls

• Thinking linearity only works when the events do not affect each other. It always works.
• Averaging the possible values without weighting them by probability. The spinner above averages $4$, not $\frac{1+4+10}{3} = 5$.
• Answering $3$ or $3.5$ for "the expected number of rolls until a $6$." It is $\frac1p = 6$.
• Confusing the expected value with the most likely value. It need not even be a possible outcome: a die averages $3.5$, which you can never roll.

Traps that recur

• Counting 'exactly one die matches' as 3/16 instead of 6/16, forgetting that either die could be the matching one. (2007 AMC 10B #22)
• Treating the three shapes 3-0-0, 2-1-0 and 1-1-1 as equally likely and averaging 3, 2, 1 to get 2, choice (E). (2025 AMC 10A #16)
• Counting only the case where neither swap touches the ball (9/25) and forgetting the swap-and-swap-back case, giving 1.8. (2021 AMC Fall 10B #16)
• Stopping after two tosses (probability 3/4 heads, giving 48) or adding 1/2 + 1/4 + 1/8 without recognizing it equals 7/8. (2015 AMC 10B #18)