Indicator Variables

Write a count as a sum of 0/1 indicators to turn combinatorial counting problems into simple probability calculations.

1. The idea

An indicator variable $I_{A}$ takes the value 1 if event $A$ occurs and 0 otherwise, so $E [I_{A}] = P (A)$ . Many counting quantities — fixed points, matches, collisions, occupied boxes — are naturally sums of such indicators.

The power of this trick is that $E [I_{A}]$ requires only understanding a single event $A$ in isolation, even when the individual indicators are highly dependent on each other. Linearity then adds them up for free.

This is the standard way to compute expected counts in permutations (fixed points), balls-in-bins problems (occupied boxes), and matching or collision problems — situations where the full joint distribution of the count is complicated or unnecessary.

2. Why it matters

Indicator variables are the most frequently tested combinatorial expectation trick in quant interviews, appearing in fixed-point, hashing-collision, and occupancy problems.

3. Core mathematics

Indicator expectation

E [I_{A}] = P (A)

Expected count

N = i = 1 \sum n I_{i} ⟹ E [N] = i = 1 \sum n P (A_{i})

Random permutation fixed points

E [fixed points] = n \cdot \frac{1}{n} = 1

4. How to recognize it

Think of this when…

The question asks for the expected number of matches, fixed points, or coincidences
You're counting occupied bins/boxes after randomly distributing items
Individual events overlap or are dependent, ruling out a simple product formula
The quantity of interest is a sum of yes/no outcomes across many positions or items

5. Worked example

Worked example

A random permutation of ${1, \dots, n}$ is chosen. Find the expected number of fixed points (positions $i$ where the permutation maps $i \to i$ ).

1
Let $I_{i} = 1$ if position $i$ is a fixed point, else 0. The total fixed points is $N = \sum_{i = 1}^{n} I_{i}$ .
$N = i = 1 \sum n I_{i}$
2
By symmetry, each position is equally likely to be any value, so $P (I_{i} = 1) = 1/ n$ .
$P (I_{i} = 1) = \frac{1}{n}$
3
Apply linearity — the $I_{i}$ are dependent, but that does not matter.
$E [N] = i = 1 \sum n \frac{1}{n} = n \cdot \frac{1}{n} = 1$
Answer: $E [N] = 1$ , independent of $n$

6. Common mistakes

Trying to find the full distribution of the count instead of just its expectation
Assuming indicators must be independent to sum their expectations
Miscounting the number of indicator terms (e.g., off-by-one in the number of adjacent pairs)
Forgetting that $E [I_{A}] = P (A)$ requires $I_{A}$ to be exactly 0/1, not a general count

7. Quick check

$n$ balls are thrown independently into $n$ boxes. What is the expected number of empty boxes?

In a random shuffle of a deck, what is the expected number of cards in their original position?

8. Practice

10 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Indicator Variables Linearity of Expectation

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