Expected Value Problems for Quant Interviews

What Expected Value Is

The expected value of a random variable X is the probability-weighted average of all its possible outcomes: E[X] = Σ x · P(X = x). It answers the question “what does this random quantity average to over many repetitions?” Quant interviewers care about EV because trading and market-making are fundamentally about finding situations where E[payoff] > E[cost].

Linearity of Expectation: The Most Powerful Tool

Linearity of expectation states that E[X + Y] = E[X] + E[Y] for any two random variables X and Y — regardless of whether they are independent. This is the most powerful and most frequently applicable tool in quant probability. It lets you decompose complicated random variables into simple parts and compute each part separately.

Example: the expected sum of two dice is E[die 1] + E[die 2] = 3.5 + 3.5 = 7. No need to enumerate all 36 outcomes. The same idea extends to any number of terms: E[X₁ + X₂ + … + Xₙ] = E[X₁] + … + E[Xₙ].

5 Classic Expected Value Problems

1. Expected Value of a Fair Die

E[die] = (1 + 2 + 3 + 4 + 5 + 6) / 6 = 21 / 6 = 3.5. This is the canonical warm-up. An interviewer who asks this wants to see you write the sum immediately rather than fumble.

2. Expected Number of Coin Flips for the First Head

This is a geometric distribution with p = 1/2. E[flips] = 1/p = 2. Derivation via the recursive trick: let E be the answer. E = 1/2 · 1 + 1/2 · (1 + E), so E/2 = 1, giving E = 2. Recognizing geometric distributions instantly is essential — they appear in disguise constantly.

3. The Coupon Collector Problem

Break the problem into phases. When you have k distinct coupons, the probability the next one is new is (n − k) / n, so the expected draws to get a new one is n / (n − k). Summing over all phases:

E[total] = n/n + n/(n−1) + n/(n−2) + … + n/1 = n · (1 + 1/2 + 1/3 + … + 1/n) = n · Hₙ

where Hₙ is the n-th harmonic number. For n = 6 (like rolling all faces of a die): E = 6 · H₆ = 6 · (1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6) ≈ 6 · 2.45 ≈ 14.7 rolls. Jane Street loves this problem in various disguises.

4. The Hat-Check Problem (Expected Fixed Points)

Define indicator variable Iᵢ = 1 if person i gets their own hat, 0 otherwise. P(Iᵢ = 1) = 1/n for each i. By linearity of expectation:

E[fixed points] = E[I₁ + I₂ + … + Iₙ] = n · (1/n) = 1

Remarkably, the expected number of fixed points is exactly 1 regardless of n. The indicator variable technique is what makes this tractable — computing this by direct counting would be a combinatorial nightmare. This is the canonical example of why linearity of expectation is so powerful.

5. The St. Petersburg Paradox

E[payoff] = Σₖ₌₁^∞ P(heads on flip k) · 2^k = Σₖ₌₁^∞ (1/2)^k · 2^k = Σₖ₌₁^∞ 1 = ∞.

The expected value is infinite — yet no rational person would pay more than a few dollars to play. This is the St. Petersburg paradox. It illustrates why raw EV is insufficient: you also need to account for diminishing marginal utility, risk preferences, and the practical impossibility of a counterparty paying 2^k for large k. Interviewers ask this to see if you can argue beyond the formula.

How to Apply EV in Interviews

Always define your sample space explicitly. State what the outcomes are and their probabilities before computing anything. Interviewers penalise candidates who skip this step and land on the right answer by luck.

Default to indicator variables. Whenever you are computing expected counts — expected number of X that occur — define Iᵢ = 1 if event i happens and apply linearity. This avoids combinatorial sums in almost every case.

Check limiting cases. After computing an EV, verify it makes intuitive sense. If E[X] = n · ln(n) and n = 1 gives E = 0, that is a good sign. If n = 1 gives a nonsensical answer, recheck your derivation.