Dice Probability Problems for Quant Interviews
Dice problems test your ability to think systematically about sample spaces, expected values, and geometric distributions — core skills assessed at every major quant firm. Below are four worked problems that regularly appear in first-round screens.
Key Strategy: Look for linearity of expectation opportunities and geometric distribution patterns before enumerating all outcomes.
Problem 1: Expected Sum of Two Dice
Problem
You roll two fair six-sided dice. What is the expected value of their sum?
Key insight: Linearity of expectation. You do not need to enumerate all 36 outcomes. The expected value of each individual die is (1 + 2 + 3 + 4 + 5 + 6) / 6 = 3.5, and E[X + Y] = E[X] + E[Y] regardless of whether X and Y are independent.
Solution: E[sum] = E[die 1] + E[die 2] = 3.5 + 3.5 = 7.
This warm-up appears constantly as the first step of a harder problem — interviewers want to see you invoke linearity of expectation without hesitation rather than listing cases.
Problem 2: Probability the Sum Equals 7
Problem
You roll two fair six-sided dice. What is the probability the sum equals 7?
Key insight: Enumerate the favorable outcomes explicitly. The total sample space has 36 equally likely outcomes.
Solution: The pairs (d₁, d₂) that sum to 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — exactly 6 outcomes. So P(sum = 7) = 6/36 = 1/6.
Note that 7 is the most likely sum from two dice — more so than any other value — because it has the most representations. Interviewers sometimes follow up by asking you to compare P(sum = 6) = 5/36 vs. P(sum = 7) = 6/36 to confirm you understand why.
Problem 3: Rolling Until You Get a 6
Problem
You roll a single fair six-sided die repeatedly until you get a 6. What is the expected number of rolls?
Key insight: This is a geometric distribution. Each roll is an independent Bernoulli trial with success probability p = 1/6. The expected number of trials until the first success is 1/p.
Solution: E[rolls] = 1/p = 1/(1/6) = 6.
Let E be the expected rolls. On roll 1 you either get a 6 (prob 1/6, done in 1 roll) or not (prob 5/6, now you are back to the start having used 1 roll). So E = 1/6 · 1 + 5/6 · (1 + E). Solving: E = 1/6 + 5/6 + 5E/6, so E/6 = 1, giving E = 6.
The two approaches agree, and being able to derive it both ways impresses interviewers.
Problem 4: Yahtzee — Five of a Kind Probability
Problem
In Yahtzee you roll five fair six-sided dice simultaneously. What is the probability of getting five of a kind (all five dice showing the same face)?
Key insight: Count the favorable outcomes and divide by the total. The total number of outcomes for five dice is 6⁵ = 7776. There are exactly 6 five-of-a-kind outcomes (all 1s, all 2s, …, all 6s).
Solution: P(five of a kind) = 6 / 6⁵ = 6 / 7776 = 1 / 6⁴ = 1/1296 ≈ 0.077%.
Fix the value of the first die (any face, probability 1). Each remaining die must match it, and each has probability 1/6. So P = (1/6)⁴ = 1/1296. This conditional-then-multiply approach is often faster than counting in combinatorics problems and generalises cleanly.
Why These Problems Matter
Quick Facts
Dice problems are approachable enough to solve under interview pressure but deep enough to reveal whether you think in terms of sample spaces and conditional probability — or just guess. Every major quant firm uses them as filters.
Key Patterns to Remember
Use E[X + Y] = E[X] + E[Y] rather than brute-force enumeration whenever possible. This saves time and shows mathematical sophistication.
Recognize "trials until success" patterns instantly. Expected trials = 1/p where p is success probability per trial.
Count carefully by fixing one variable and conditioning on the rest. Avoid missing cases by being methodical.
Ready to practice more probability problems? Try our full probability question bank with detailed solutions and firm-specific examples.