The Indicator-Variable Trick

The problem. In a shuffled deck of 52 cards, how many cards, on average, are in their original "sorted" position (card 1 in slot 1, card 2 in slot 2, and so on)? The distribution of that count is genuinely ugly — computing it directly means wading through inclusion-exclusion over all 52 positions. The expected value, though, is one line, if you know the trick.

The trick: split a hard sum into simple 0/1 pieces

Instead of thinking about "the number of matches" as one complicated random variable, define 52 tiny ones: let $X_i = 1$ if card $i$ lands in position $i$, and $X_i = 0$ otherwise. The total number of matches is just their sum, $X = X_1 + X_2 + \cdots + X_{52}$. This looks like it hasn't helped — $X$ is exactly as hard to describe as before — but linearity of expectation says $E[X] = E[X_1] + E[X_2] + \cdots + E[X_{52}]$ regardless of whether the $X_i$ are independent. That "regardless of independence" clause is the entire trick: you never have to worry about how one card's position affects another's, because expectation of a sum is always the sum of expectations.

Each $X_i$ is an indicator variable, and for an indicator, $E[X_i] = P(X_i = 1)$ — the expectation of a 0/1 variable is just the probability it equals 1. For a uniformly random shuffle, $P(\text{card } i \text{ in position } i) = \frac{1}{52}$ for every $i$, by symmetry. So:

$$E[X] = \sum_{i=1}^{52} \frac{1}{52} = 52 \times \frac{1}{52} = 1.$$

In plain English: on average, exactly one card ends up in its original spot, no matter how the deck is shuffled — a strikingly clean answer to a question whose full distribution is genuinely messy.

Worked example: expected number of fixed points in general

The argument never used the number 52 in an essential way. For $n$ cards shuffled uniformly at random, define $X_i = 1$ if card $i$ is in position $i$. Then $P(X_i = 1) = \frac{1}{n}$ by the same symmetry argument, so $E[X] = \sum_{i=1}^n \frac{1}{n} = n \times \frac{1}{n} = 1$ — the expected number of matches is exactly 1 for any deck size $n$, from a 2-card deck to a million-card one. That the answer doesn't depend on $n$ is itself the tell that indicator variables were the right tool: a direct combinatorial approach would need $n$-dependent inclusion-exclusion terms that all happen to cancel down to this constant.

The general technique: when to reach for indicators

Whenever a question asks for the expected count of something — matches, collisions, empty bins, fixed points, inversions — and the something can be described as "does event $i$ happen," define one indicator per potential occurrence, sum them, and apply linearity. You never need independence, you never need the joint distribution, and you rarely need more than a one-line probability calculation per indicator. This is the single most reusable trick in probability interviews precisely because it converts "count something complicated" into "add up simple probabilities."