Dice & coins: interview questions and key results

Counting outcomes, complements, binomials. The results to know for quant trading interviews, then a worked example of each of the 5 question types. Free, no sign-up.

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Key results

Complement for "at least one": 1 − P(none).
Sum of two dice: P(k) = (6 − |k − 7|)/36.
One die: mean 3.5, variance 35/12.
Binomial: P(k) = C(n,k)p^k(1−p)^n−k.

Worked examples (5)

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

Two-dice sums

Two fair six-sided dice are rolled. What is the probability their sum is 9?

Answer and worked solution

Answer: 1/9 ≈ 0.11111

36 equally likely outcomes. Sum 9 occurs in 6 − |9 − 7| = 4 of them, so P = 4/36 = 1/9 ≈ 0.11111.

Complement rule

You roll a fair die 7 times. What is the probability of at least one six?

Answer and worked solution

Answer: 0.72092

Complement: P(no six) = (5/6)⁷ ≈ 0.27908, so P(at least one) ≈ 0.72092. "At least one" almost always means: use the complement.

Binomial counts

A fair coin is flipped 6 times. What is the probability of exactly 4 heads?

Answer and worked solution

Answer: 0.23438

Each sequence has probability 1/64. There are C(6,4) = 15 sequences with 4 heads, so P = 15/64 ≈ 0.23438.

Three-dice tails

Three fair dice are rolled. What is the probability the sum is at least 11?

Answer and worked solution

Answer: 0.5

The sum of 3 dice is symmetric about 10.5, so P(S ≥ 11) = P(S ≤ 10). Counting outcomes gives 108 of 216, so P = 1/2 ≈ 0.5.

Variance of sums

What is the variance of the sum of 2 fair six-sided dice?

Answer and worked solution

Answer: 35/6 ≈ 5.8333

One die: E[X] = 3.5 and E[X²] = 91/6, so Var(X) = 91/6 − 3.5² = 35/12. Variances of independent variables add: 2 × 35/12 = 35/6 ≈ 5.8333.

Dice & coins: interview questions and key results

Key results

Worked examples (5)

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