Expected Value

## What you need to know
- Definition: $E[X] = \sum_x x \cdot P(X = x)$; a game is fair when the expected net winnings are $0$.
- Linearity of expectation: $E[X + Y] = E[X] + E[Y]$ always, even when $X$ and $Y$ are dependent.
- Indicator trick: if $X$ counts how many of $n$ events occur, then $E[X] = \sum_{i=1}^{n} P(\text{event } i)$.
- Standard values: a fair die averages $3.5$; the expected number of trials until a success of probability $p$ is $1/p$.
- A restarting process gives an equation such as $T = 1 + (1 - p)\,T$ for the expected number of steps.

## How AMC 10 tests it
- Problems 8–15: expected value of a die, spinner, or game with listed payoffs; compute $\sum x\,p(x)$ or find the fair price of a ticket.
- Problems 14–22: expected number of something (adjacent same-colored beads, matches between two shuffled decks, runs of heads), intended for indicators and linearity.
- Problems 18–24: expected number of steps until a random process stops, via a restart equation or states.
- Occasionally combined with counting: expected size of a random subset.

## Standard approaches
1. With few outcomes, tabulate values and probabilities and sum $x\,p(x)$.
2. If the quantity counts occurrences, write it as a sum of indicators and use linearity; do not find the full distribution.
3. Use symmetry: identically distributed positions contribute equally.
4. For "expected trials until," set up $E = 1 + (\text{probability of continuing}) \cdot E$ or a small state system.
5. Check that the answer lies between the minimum and maximum 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. Let $X_i = 1$ if the cards in positions $i$ and $i+1$ differ by $1$, and $0$ otherwise. By linearity and symmetry, $E = \sum_{i=1}^{9} E[X_i] = 9\,P(X_1 = 1)$. The cards in positions $1$ and $2$ form a uniformly random ordered pair of distinct cards, $10 \cdot 9 = 90$ possibilities. Pairs differing by $1$ are $(k, k+1)$ and $(k+1, k)$ for $k = 1, \dots, 9$, so $18$ of them. Thus $P(X_1 = 1) = \frac{18}{90} = \frac{1}{5}$ and $E = \frac{9}{5}$. Answer $\boxed{\textbf{(C)}\ \frac{9}{5}}$.

## Pitfalls
- Believing linearity requires independence; it does not.
- Averaging the possible values without weighting by their probabilities.
- Answering $3$ or $3.5$ for "expected rolls until a $6$" instead of $1/p = 6$.
- Confusing expected value with the most likely value; the expectation need not be an attainable outcome.

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)