Average each part separately

The expected total is the total of the expectations, one part at a time. Each card in a shuffled deck has a 1/52 chance of landing in its own spot, so the expected number of fixed cards is 52 x (1/52) = 1. The parts may interact however they like; averaging never asks about independence.

Words you need

Expected value (EV): The average result if you could repeat the bet many times: each outcome times its probability, added up.e.g. A die pays its face in dollars: EV = 3.5.

Why it works

Expectation distributes over addition: the average of a sum is the sum of the averages, for any joint behavior of the parts.

E[X + Y] = E[X] + E[Y], with no independence requirement

Spot it: A question asks for the expected sum, count, or total.

Technique reference and sources

One fair die averages 3.5: the faces 1 through 6 balance around the middle. Three dice average 3.5 + 3.5 + 3.5 = 10.5. You never listed all 216 outcomes or asked whether the dice influence each other. Each part contributes its own average, and the total is the sum.

When this method fits

A question asks for the expected sum, count, or total.
The total splits into many small contributions with easy averages.
The parts are dependent (cards without replacement, shared positions).
You see indicator structure: count how many of many events happen.

Split the total into parts. Write the total as X1 + X2 + ... + Xn, one part per card, die, or trial.
Find one part's expectation. One die: 3.5. One card landing home: 1/52. One coin: 0.5 heads.
Multiply by the count. Three dice: 3 x 3.5 = 10.5. Fifty-two cards: 52 x (1/52) = 1.
Skip the joint distribution. Dependence between parts never enters. Do not list joint outcomes.

Indicator variables: For a count of events, give each event a 0/1 variable and add their probabilities.

52 fixed-point indicators x 1/52 = 1
5 hand positions x 4/52 ace chance = 5/13

Per-trial average times trials: Identical trials contribute the same average; multiply once.

3 dice x 3.5 = 10.5
20 flips x 0.5 heads = 10

Common traps

Demanding independence before adding expectations. Linearity of expectation needs no independence. Add the averages regardless of how parts interact.

Adding variances the same way: Var(X + Y) = Var(X) + Var(Y). Variance of a sum adds the covariance term: Var(X + Y) = Var(X) + Var(Y) + 2 Cov(X, Y). It splits plainly when covariance is zero; independence is sufficient but not necessary.

Multiplying expectations for a product: E[X x Y] = E[X] x E[Y]. Independence guarantees E[X x Y] = E[X] x E[Y] when the required expectations exist. Dependent variables can also satisfy this equality, so check the numbers instead of assuming.

Another example: Expected number of aces in a dealt 5-card hand

One position holds an ace with chance 4/52.
Five positions: 5 x (4/52) = 20/52 = 5/13.
As a decimal: about 0.385.

Source notes

These explanations and questions are written for QuantArc. Links show the underlying mathematics and learning design.

Harvard Stat 110: Probability (Joe Blitzstein)
Free course site with videos and the open companion book Introduction to Probability; the topic spine for counting, Bayes, and linearity. Lesson examples and questions are original.

IES: Organizing Instruction and Study to Improve Student Learning
Supports worked examples, retrieval, explanations and spaced study. Does not establish QuantArc's exact lesson length, mastery threshold or review intervals.