Classic brainteasers: interview questions and key results

Monty Hall, coin patterns, symmetry. The results to know for quant trading interviews, then a worked example of each of the 7 question types. Free, no sign-up.

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Key results

Look for symmetry before you calculate.
Coin patterns: E[wait] = Σ 2^k over self-overlaps k.
K special cards among N: E[position of first] = (N+1)/(K+1).
Say your reasoning out loud; interviewers score the process.

Worked examples (7)

Try each one before opening the solution. In the app, every type generates fresh numbers each time.

Monty Hall (n doors)

Monty Hall with 10 doors: one car, 9 goats. You pick a door; the host, who knows where the car is, opens one other door showing a goat. You switch to one of the 8 remaining closed doors at random. What is P(win)?

Answer and worked solution

Answer: 0.1125

P(first pick wrong) = 9/10. If wrong, the car is behind one of the 8 other closed doors, so switching wins with probability 1/8. Total: (9/10)·(1/8) = 0.1125, versus 1/10 = 0.1 for staying.

Coin-pattern waiting times

You flip a fair coin until the sequence HTH first appears. What is the expected number of flips?

Answer and worked solution

Answer: 10

Use Conway's overlap rule: add 2^k for every k where the first k letters equal the last k letters of the pattern. For HTH that gives 10. Self-overlapping patterns (like HH) take longer, because a near-miss throws away progress.

First special card

You turn over cards from a shuffled 52-card deck one at a time. What is the expected position of the first king or queen?

Answer and worked solution

Answer: 53/9 ≈ 5.8889

The 8 special cards split the other 44 into 9 gaps of equal expected size 44/9. The first special card comes after the first gap: 1 + 44/9 = 53/9 ≈ 5.8889.

Points in a semicircle

3 points are placed uniformly at random on a circle. What is the probability they all lie within some semicircle?

Answer and worked solution

Answer: 0.75

For each point i, P(all others lie in the semicircle clockwise from i) = (1/2)². These 3 events are disjoint, so P = 3/2² = 0.75.

Uniform sums

You draw independent Uniform(0,1) numbers until their running sum exceeds 0.5. What is the expected number of draws?

Answer and worked solution

Answer: 1.6487

P(sum of n uniforms ≤ t) = tⁿ/n! for t ≤ 1. Then E[N] = Σ_n≥0 P(N > n) = Σ tⁿ/n! = e^0.5 ≈ 1.6487.

Airplane seat

100 passengers board a plane with 100 assigned seats. The first has lost their boarding pass and sits in a random seat. Each later passenger takes their own seat if it's free, otherwise a random free seat. What is the probability the last passenger gets their own seat?

Answer and worked solution

Answer: 0.5

The last free seat is always either seat 1 or seat 100: any displaced passenger is equally likely to pick either one, and picking seat 1 resolves everything. By symmetry the answer is 1/2 for any n ≥ 2.

Broken-stick triangle

A stick is broken at two independent uniformly random points. What is the probability the three pieces can form a triangle?

Answer and worked solution

Answer: 0.25

A triangle needs every piece shorter than ½. With break points (x, y) uniform on the unit square, the valid region is two triangles each with area 1/8, so P = 1/4.

Classic brainteasers: interview questions and key results

Key results

Worked examples (7)

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