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Sig·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Apr 2026

Summary

SIG quant engineer interview, probability-heavy round that felt like they pulled questions straight from the Green Book. Two dice problems back to back, both requiring clean combinatorial arguments.

Questions Asked (2)

Q1

Using a fair 9-sided die rolled a fixed number of times, what is the probability of rolling a 'straight', meaning a run of consecutive distinct values across all rolls? Walk through your assumptions and counting argument.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

The setup sounds clean but I fumbled the assumptions part at first.

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AI HintsAI Generated

Suggested Approach

Clarify the problem by defining the number of rolls (n), the die size (9), and what constitutes a 'straight' (a set of n consecutive distinct values). Then compute the probability by counting favorable outcomes: for each possible starting value, count the permutations of the n distinct values, and divide by the total outcomes (9^n).

Pro tip: Explicitly state your assumptions (e.g., n ≤ 9, order matters, and the straight must be exactly the rolled values) and consider edge cases like n=1 or n=9 to validate your formula. This shows rigor and prevents misinterpretation.

1. Clarify assumptions and definitions

Confirm the number of rolls (n), that the die is fair and 9-sided, and that a 'straight' means the n rolls are all distinct and form a consecutive run (e.g., 3,4,5). Note that n must be between 1 and 9.

2. Determine total number of outcomes

Since each roll has 9 equally likely outcomes, the total number of sequences of n rolls is 9^n.

3. Count favorable outcomes

For a straight of length n, the set of values must be {k, k+1, ..., k+n-1} for some starting value k from 1 to 10-n. For each such set, there are n! permutations (orders) that yield a straight. So total favorable outcomes = (10 - n) * n!.

4. Compute probability and simplify

The probability is (10 - n) * n! / 9^n. Simplify if possible and verify with small cases (e.g., n=1 gives probability 1, n=9 gives 9! / 9^9).

5. Discuss edge cases and extensions

Mention what happens if n > 9 (probability 0) or if the straight can be longer than n (not possible). Optionally, discuss variations like allowing repeats or considering circular straights.

Key Points to Mention

  • Definition of a straight: n consecutive distinct values, order matters.
  • Total outcomes: 9^n for n rolls.
  • Number of possible starting values: 10 - n (since values range from 1 to 9).
  • Permutations of the n distinct values: n! ways to order them.
  • Probability formula: P = (10 - n) * n! / 9^n.
  • Edge cases: n=1 (probability 1), n=9 (probability 9!/9^9), n>9 (probability 0).

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Q2

Using the same 9-sided die rolled 5 times, compute the probability of a 'full house', meaning exactly three of one face value and two of another. State your assumptions and show the counting.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Felt more comfortable here than on the straight question.

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AI HintsAI Generated

Suggested Approach

First, clarify that the die is fair and rolls are independent, then compute the probability by counting favorable outcomes over total outcomes. Use combinatorics: choose the face for three-of-a-kind and the face for the pair, then count the arrangements of these multiset rolls.

Pro tip: State your assumptions upfront and consider edge cases like whether the die faces are labeled 1-9; this shows attention to detail and prevents ambiguity. Also, mention that the same logic applies to any n-sided die, demonstrating generalization.

1. Clarify assumptions

Confirm that the die is fair, 9-sided, and rolls are independent. Specify that faces are distinct and equally likely.

2. Define total outcomes

Calculate the total number of possible outcomes for 5 rolls: 9^5.

3. Count favorable outcomes

Choose the face for three-of-a-kind (9 ways) and the face for the pair (8 ways). Then count the number of sequences with exactly three of one face and two of another: 5!/(3!2!) = 10. Multiply: 9 * 8 * 10 = 720.

4. Compute probability

Divide favorable outcomes by total outcomes: 720 / 9^5. Simplify if possible.

5. Verify and present

Double-check the counting logic and present the final probability as a fraction or decimal, stating assumptions clearly.

Key Points to Mention

  • Assumptions: fair die, independent rolls, faces labeled 1-9.
  • Total outcomes: 9^5 = 59049.
  • Favorable outcomes: choose triple face (9 ways), pair face (8 ways), arrangements (10 ways) = 720.
  • Probability = 720/59049 = 80/6561 ≈ 0.0122 (1.22%).
  • Generalization: For an n-sided die, probability = n*(n-1)*10 / n^5 = 10(n-1)/n^4.
  • Alternative approach: multinomial coefficient and probability mass function.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.