Cards & combinatorics: interview questions and key results

Hands, arrangements, birthday problems. The results to know for quant trading interviews, then a worked example of each of the 5 question types. Free, no sign-up.

Practise endless questions like these

Key results

Unordered selection: C(n,k). Arrangements with repeats: n!/(n₁!n₂!…).
Count sequentially (12/51 · 11/50…) or with combinations, and use one to check the other.
Birthday: P(no match) ≈ exp(−n²/2d).

Worked examples (5)

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

Same-suit hands

You draw 4 cards from a shuffled 52-card deck. What is the probability they are all the same suit?

Answer and worked solution

Answer: 0.010564

Pick a suit (4 ways), then 4 of its 13 cards: 4·C(13,4) = 2860. Total hands: C(52,4) = 270725. P = 44/4165 ≈ 0.010564. Sequential check: 1 × 12/51 × 11/50 × …

At least one ace

You are dealt 4 cards. What is the probability you get at least one ace?

Answer and worked solution

Answer: 0.28126

P(no aces) = C(48,4)/C(52,4) ≈ 0.71874. So P(at least one) ≈ 0.28126.

Arrangements with repeats

How many distinct arrangements are there of the letters in HEDGE?

Answer and worked solution

Answer: 60

5 letters, with repeats E×2. Arrangements = 5! / (2!) = 60.

Birthday problem

There are 44 people in a room. What is the probability at least two share a birthday? (365 equally likely days.)

Answer and worked solution

Answer: 0.93289

P(all different) = (365/365)(364/365)…(322/365) ≈ 0.067115. So P(shared) ≈ 0.93289. Quick estimate: P(all different) ≈ exp(−n(n−1)/730) = 0.074887.

Committees with a chair

From 9 people, how many ways can you form a committee of 2 with one member named chair?

Answer and worked solution

Answer: 72

Choose the committee, C(9,2) = 36, then the chair (2 ways): 72. Equivalently, pick the chair (9) then the rest, C(8,1): 9 × 8 = 72.

Cards & combinatorics: interview questions and key results

Key results

Worked examples (5)

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