30 Probability Interview Questions with Solutions

These 30 problems cover the probability concepts that appear most frequently in quant trading and prop firm interviews. Each has a complete worked solution and notes on the underlying concept. Work through them in order — the concepts build.

Expected Value

1. A fair die is rolled. You win $X where X is the number shown. How much would you pay to play?

EV = (1+2+3+4+5+6)/6 = 21/6 = $3.50

You'd pay up to $3.50 for fair value. Anything less is +EV.

2. A coin flip pays $2 for heads and loses $1 for tails. What's the EV? What's the Kelly fraction?

EV = 0.5 × $2 + 0.5 × (−$1) = $1 − $0.50 = +$0.50 per flip.

Kelly: f* = (bp − q)/b where b = 2, p = 0.5, q = 0.5.
f* = (2 × 0.5 − 0.5)/2 = 0.5/2 = 25% of bankroll per flip.

3. You can roll a die once and receive the face value in dollars, or roll again after seeing the first result (and must keep the second roll). What's your optimal strategy?

If you roll and see X, you re-roll if E[second roll] > X. E[second roll] = 3.5.
Re-roll if X ≤ 3. Keep if X ≥ 4.

E[game] = P(X ≤ 3) × 3.5 + P(X ≥ 4) × E[X | X ≥ 4]
= 0.5 × 3.5 + 0.5 × (4+5+6)/3 = 1.75 + 0.5 × 5 = $4.25

4. A game pays $n if a fair coin first lands heads on flip n. How much would you pay to play? (St. Petersburg Paradox)

EV = Σ 2⁻ⁿ × n → this actually diverges. But in practice, with any finite bankroll, you'd pay roughly $20–30. The paradox illustrates that EV alone isn't always the right decision criterion — expected utility matters.

Concept: EV divergence, risk aversion, expected utility.

5. You draw a card from a standard deck. You win $13 if it's an ace, lose $1 otherwise. Should you play?

EV = (4/52) × $13 + (48/52) × (−$1) = $1 − $48/52 = $1 − $0.923 = +$0.077 per play.

Positive EV — yes, play.

Conditional Probability and Bayes' Theorem

6. A test is 95% accurate. 2% of the population has a disease. You test positive. What's the probability you have the disease?

P(disease | positive) = P(positive | disease) × P(disease) / P(positive)

Numerator: 0.95 × 0.02 = 0.019
Denominator: 0.95 × 0.02 + 0.05 × 0.98 = 0.019 + 0.049 = 0.068

P(disease | positive) = 0.019/0.068 ≈ 28%

The base rate (2%) dominates. Even with a 95% accurate test, a positive result is only 28% likely to be true.

7. You roll two dice. Given that the sum is at least 8, what's the probability both dice show the same number?

Doubles with sum ≥ 8: (4,4), (5,5), (6,6) → 3 outcomes
Total outcomes with sum ≥ 8: (2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6) → 15 outcomes

P(doubles | sum ≥ 8) = 3/15 = 1/5 = 20%

8. A bag contains 3 red and 5 blue balls. You draw 2 without replacement. What's the probability both are red?

P(both red) = (3/8) × (2/7) = 6/56 = 3/28 ≈ 10.7%

Concept: drawing without replacement — events are NOT independent.

9. A fair coin is flipped 3 times. Given that at least one is heads, what's the probability all 3 are heads?

P(all heads | at least one heads) = P(all heads and at least one heads) / P(at least one heads)
= P(all heads) / P(at least one heads)
= (1/8) / (7/8) = 1/7 ≈ 14.3%

10. You have 2 children. Given that at least one is a boy, what's the probability both are boys?

Sample space: BB, BG, GB, GG (equal probability)
At least one boy: BB, BG, GB → 3 outcomes
Both boys: BB → 1 outcome

P(both boys | at least one boy) = 1/3

Classic: many candidates say 1/2. The conditioning changes the space.

Counting and Combinatorics

11. How many ways can 5 people be seated in a row?

5! = 5 × 4 × 3 × 2 × 1 = 120 ways

12. In how many ways can you choose 3 people from a group of 8?

C(8,3) = 8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56 ways

13. A password requires exactly 3 letters followed by 3 digits. How many passwords are possible?

26³ × 10³ = 17,576 × 1,000 = 17,576,000 passwords

14. What's the probability of getting a full house in a 5-card poker hand?

Full house: choose the triple rank (13 ways), choose 3 of 4 suits for the triple (C(4,3)=4), choose the pair rank (12 ways), choose 2 of 4 suits for the pair (C(4,2)=6).

Favorable outcomes: 13 × 4 × 12 × 6 = 3,744
Total 5-card hands: C(52,5) = 2,598,960

P(full house) = 3,744/2,598,960 ≈ 0.144% (about 1 in 694)

15. You shuffle a deck of 52 cards. What's the probability the top 4 are all aces?

P = (4/52) × (3/51) × (2/50) × (1/49) = 24 / (52 × 51 × 50 × 49) = 24/6,497,400 ≈ 1 in 270,725

Random Walks and Sequences

16. What's the expected number of coin flips to get the first heads?

Geometric distribution with p = 0.5. E[flips] = 1/p = 2 flips.

17. What's the expected number of coin flips to get two consecutive heads?

Let E = expected flips from a fresh start.
Let A = expected flips after seeing one head.

From fresh start:

Flip tails (prob 1/2): back to start, cost 1 flip. E = 1 + E
Flip heads (prob 1/2): move to state A. E = A

E = 1/2 × (1 + E) + 1/2 × (1 + A)

From state A:

Flip heads (prob 1/2): done! Cost 1 flip. A contributes 1/2 × 1
Flip tails (prob 1/2): back to start. A = 1/2 × 1 + 1/2 × (1 + E)

A = 1/2 + 1/2 × (1 + E) = 1 + E/2

Substituting back: E = 1 + E/2 + 1/2 × (1 + 1 + E/2) = 1 + E/2 + 1 + E/4

E − E/2 − E/4 = 2 → E/4 = 2 → E = 6 flips

Classic SIG/Jane Street question. Most candidates guess 4.

18. A drunk man stands at position 0 on a number line. Each step he moves +1 or −1 with equal probability. What's the expected number of steps to reach +1 for the first time?

This is an absorbing random walk. By the optional stopping theorem: the expected number of steps to reach +1 from 0 is infinite (the walk is recurrent but steps to reach any specific point have infinite expectation).

Concept: symmetric random walks have infinite stopping times in expectation.

19. You roll dice until the sum exceeds 6. What's the expected number of rolls?

This requires computing P(sum > 6 in exactly 1 roll) = P(first roll > 6) = 1/6 (only a 7+ works, not possible — 0/6 = 0 actually for a standard die). Wait: a standard die has faces 1-6, so one roll can never exceed 6 if the max is 6. We need the sum to exceed 6, which requires at least 2 rolls.

This is a stopping time problem. The expected number of rolls until cumulative sum first exceeds 6 is approximately e ≈ 2.718 (it converges to e for "exceed 1" problems on the unit interval, extended here).

Concept: expected stopping time, geometric distribution.

20. A fair coin is flipped repeatedly. What's the probability of getting 3 heads before 3 tails?

By symmetry and the negative hypergeometric distribution: need to get to 3H before 3T.

Total sequences of length at most 5 (at most 4 flips to resolve if 2-2 after 4): This requires careful enumeration. By symmetry, the answer is 50% exactly (since the coin is fair, by symmetry P(3H first) = P(3T first) = 1/2).

Concept: symmetry argument. Always look for symmetry before computing.

Geometric and Continuous Probability

21. Two people agree to meet between 12:00 and 1:00 pm. Each waits 15 minutes for the other. What's the probability they meet?

Geometric approach: In a 60×60 square, meeting occurs when |X − Y| ≤ 15.

P(meet) = 1 − P(|X − Y| > 15) = 1 − 2 × (45²)/(60²)/2 = 1 − (45/60)² = 1 − (3/4)² = 1 − 9/16 = 7/16 = 43.75%

22. A point is chosen uniformly at random in a unit square. What's the probability it's within distance 0.5 from the center?

Circle of radius 0.5 in a 1×1 square: area = π × 0.5² = π/4 ≈ 0.785.

But the full circle fits inside the unit square (center at (0.5, 0.5), radius 0.5 just touches edges).

P = π/4 ≈ 78.5%

23. You break a stick at a uniformly random point. What's the expected length of the shorter piece?

If break point U ~ Uniform(0,1), shorter piece = min(U, 1−U).

E[min(U, 1−U)] = E[U | U < 0.5] × P(U < 0.5) + E[1−U | U > 0.5] × P(U > 0.5)
= 0.25 × 0.5 + 0.25 × 0.5 = 0.25

The expected length of the shorter piece is 1/4 of the stick.

Classic Puzzles

24. You have 12 balls, identical except one is either heavier or lighter. You have a balance scale. What's the minimum number of weighings needed to identify the odd ball AND determine if it's heavier or lighter?

3 weighings. This is a classic information theory problem. Each weighing has 3 outcomes (left down, right down, balanced). With 3 weighings: 3³ = 27 states, which covers 24 possibilities (12 balls × 2 possibilities: heavier or lighter).

Concept: information theory, binary search generalization.

25. You have 3 boxes: one contains 2 gold coins, one contains 2 silver coins, one contains 1 gold + 1 silver. You pick a box at random and draw a gold coin. What's the probability you drew from the 2-gold box?

P(2-gold box | drew gold) = P(drew gold | 2-gold) × P(2-gold) / P(drew gold)
= 1 × (1/3) / (1 × 1/3 + 0 × 1/3 + 1/2 × 1/3)
= (1/3) / (1/3 + 0 + 1/6)
= (1/3) / (1/2) = 2/3

Classic Bayesian puzzle. The answer is 2/3, not 1/2.

26. The Monty Hall Problem: Behind one of 3 doors is a car, behind the other two are goats. You pick door 1. The host opens door 3 (revealing a goat). Should you switch?

Yes. Conditional on the host opening door 3:

P(car behind door 2 | host opened door 3) = 2/3.

Switch. You win 2/3 of the time by switching, 1/3 by staying.

Concept: conditional probability + the host's action conveys information.

27. You have a 3-sided die (faces 1, 2, 3) and a 4-sided die (faces 1, 2, 3, 4). You each roll once. What's the probability you beat the 4-sided die?

P(3-sided wins) = Σ P(3-sided = k) × P(4-sided < k)

k=1: 1/3 × 0/4 = 0
k=2: 1/3 × 1/4 = 1/12
k=3: 1/3 × 2/4 = 2/12

P(win) = 3/12 = 1/4 = 25%

28. In a class of 30 students, what's the probability that at least 2 share a birthday?

P(no shared birthday) = (365/365)(364/365)...(336/365) ≈ 0.294

P(at least one shared birthday) = 1 − 0.294 ≈ 70.6%

The birthday problem: most people drastically underestimate this.

Markets and Finance

29. A stock is at $100. It rises or falls $10 with equal probability each day. After 4 days, what's the expected value?

E[stock after 4 days] = $100 (random walk, expected value = starting value for symmetric walk).

But: E[stock²] ≠ $100² — variance grows. The expected value is unchanged, but the distribution spreads.

30. You're market-making on a coin flip. Bid/ask on the number of heads in 10 fair coin flips. Where do you price it?

E[heads in 10 flips] = 10 × 0.5 = 5. Variance = 10 × 0.5 × 0.5 = 2.5. SD ≈ 1.58.

Fair value: 5. Your bid/ask spread should reflect your risk — perhaps 4.75/5.25 for a tight market.

This is the core of options market making: pricing expected value and quoting a spread that compensates for variance risk.

Patterns Across These Problems

Bayes' Theorem appears in problems 6, 9, 10, 25, 26. The pattern: always update on the information you have. The base rate dominates more than intuition suggests.

Symmetry arguments (problems 20, 29) save enormous computation time. Before counting, ask: is there a symmetry that tells you the answer directly?

Without replacement (problems 8, 14, 15) means events are dependent. Adjust probabilities as you draw.

Expected stopping times (problems 16, 17, 18) often have elegant recursive solutions. Set up the equation E = 1 + E×(probability of returning to start).

Practice these daily at Fermiq's probability category and in the daily drill.

30 Probability Interview Questions with Solutions (Quant & Prop Firm)