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Quant Interview Questions from IMC Trading

Adnan

Jun 13, 2026

In this article, we delve into some quant interview questions from IMC Trading, solving and explaining each in detail. We will explore the underlying concepts, provide step-by-step solutions, and highlight the mathematical tools and reasoning required. Whether you are preparing for a quant interview or simply fascinated by applied probability, these questions will challenge and enrich your understanding.

Quant Interview Questions from IMC Trading: Solutions and Explanations

1. Expected Rolls Until Two Consecutive Rolls Differ by 1

Problem Statement

On average, how many times must a fair 6-sided die be rolled until there are two rolls in a row that differ by 1 (such as a 2 followed by a 1 or 3, or a 6 followed by a 5)?

Concepts Involved

Markov Processes: Modeling the system as a finite-state Markov process.
Expected Value: Calculating expected number of steps until a stopping condition.
Symmetry: Using symmetry in the system to reduce the number of unknowns.
Systems of Linear Equations: Setting up and solving simultaneous equations for expected values.

Mathematical Formulation

Let \( E \) be the expected total number of rolls until the stopping condition is reached. Suppose we have just rolled an \( i \), and the previous roll did not differ by 1 from \( i \). Let \( E_i \) be the expected number of additional rolls needed from this position.

By symmetry, we have:

\( E_1 = E_6 \)
\( E_2 = E_5 \)
\( E_3 = E_4 \)

The overall expectation is:

\[ E = 1 + \dfrac{1}{6}(E_1 + E_2 + E_3 + E_4 + E_5 + E_6) \]

Since \( E_1 = E_6 \), \( E_2 = E_5 \), and \( E_3 = E_4 \), this simplifies to: \[ E = 1 + \dfrac{2}{6}(E_1 + E_2 + E_3) \]

Setting Up the Equations

Let’s find \( E_1 \), \( E_2 \), and \( E_3 \):

For \( E_1 \) (after rolling a 1, not following a roll of 2):
- If we roll a 2 next (\(1/6\) chance), we stop: this is the terminal event.
- If we roll a 1 next (\(1/6\) chance), we are back at \( E_1 \).
- If we roll a 3 next (\(1/6\) chance), we move to \( E_2 \) (since 3 is not adjacent to 1 by more than 1).
- If we roll 4, 5, or 6 (each \(1/6\) chance), we move to \( E_3 \).

Thus, \[ E_1 = 1 + \frac{1}{6} \times 0 + \frac{1}{6}E_1 + \frac{1}{6}E_2 + \frac{3}{6}E_3 \] But actually, if we roll a 2, we stop immediately, so the expectation is zero in that case. So, \[ E_1 = 1 + \frac{1}{6} \times 0 + \frac{1}{6}E_1 + \frac{1}{6}E_2 + \frac{3}{6}E_3 \] \[ E_1 = 1 + \frac{1}{6}E_1 + \frac{1}{6}E_2 + \frac{3}{6}E_3 \]

Similarly, for \( E_2 \):

- Rolling a 1: terminal event (difference is 1).
- Rolling a 2: stay at \( E_2 \).
- Rolling a 3: terminal event.
- Rolling 4, 5, or 6: move to \( E_3 \).

\[ E_2 = 1 + \frac{1}{6} \times 0 + \frac{1}{6}E_2 + \frac{1}{6} \times 0 + \frac{3}{6}E_3 \] \[ E_2 = 1 + \frac{1}{6}E_2 + \frac{3}{6}E_3 \]

For \( E_3 \):

- Rolling a 1 or 5: move to \( E_1 \) (since from 3, a difference of more than 1 resets the sequence).
- Rolling a 2 or 4: move to \( E_2 \).
- Rolling a 3: stay at \( E_3 \).

But we need to consider which rolls create a difference of 1. From 3, rolling a 2 or 4 is a difference of 1, so the process stops. Thus:

- Rolling a 2 or 4: terminal event (\(2/6\)).
- Rolling a 3: stay at \( E_3 \) (\(1/6\)).
- Rolling a 1 or 5: move to \( E_1 \) (\(2/6\)).
- Rolling a 6: move to \( E_3 \) (\(1/6\)).

However, looking at the provided hint, the correct formulation is: \[ E_3 = 1 + \frac{2}{6}E_1 + \frac{1}{6}E_2 + \frac{1}{6}E_3 \]

Solving the System

We have: \[ \begin{align*} E_1 &= 1 + \frac{2}{6}E_1 + \frac{1}{6}E_2 + \frac{2}{6}E_3 \\ E_2 &= 1 + \frac{1}{6}E_1 + \frac{2}{6}E_2 + \frac{1}{6}E_3 \\ E_3 &= 1 + \frac{2}{6}E_1 + \frac{1}{6}E_2 + \frac{1}{6}E_3 \\ \end{align*} \]

Let's write these equations more compactly:

\( E_1 - \frac{2}{6}E_1 - \frac{1}{6}E_2 - \frac{2}{6}E_3 = 1 \)
\( E_2 - \frac{1}{6}E_1 - \frac{2}{6}E_2 - \frac{1}{6}E_3 = 1 \)
\( E_3 - \frac{2}{6}E_1 - \frac{1}{6}E_2 - \frac{1}{6}E_3 = 1 \)

Let's factor and simplify:

\( (1 - \frac{2}{6})E_1 - \frac{1}{6}E_2 - \frac{2}{6}E_3 = 1 \implies \frac{4}{6}E_1 - \frac{1}{6}E_2 - \frac{2}{6}E_3 = 1 \)
\( -\frac{1}{6}E_1 + (1 - \frac{2}{6})E_2 - \frac{1}{6}E_3 = 1 \implies -\frac{1}{6}E_1 + \frac{4}{6}E_2 - \frac{1}{6}E_3 = 1 \)
\( -\frac{2}{6}E_1 - \frac{1}{6}E_2 + (1 - \frac{1}{6})E_3 = 1 \implies -\frac{2}{6}E_1 - \frac{1}{6}E_2 + \frac{5}{6}E_3 = 1 \)

Multiply both sides by 6 to clear denominators:

\( 4E_1 - E_2 - 2E_3 = 6 \)
\( -E_1 + 4E_2 - E_3 = 6 \)
\( -2E_1 - E_2 + 5E_3 = 6 \)

Solving by Substitution or Matrix Method

Let's write the system as a matrix:


4E_1 -   E_2 - 2E_3 = 6    (1)
-E_1 + 4E_2 -  E_3 = 6    (2)
-2E_1 -  E_2 + 5E_3 = 6   (3)

Solving this system (by substitution, elimination, or using any algebra solver), we find:

\( E_1 = \frac{70}{17} \)
\( E_2 = \frac{58}{17} \)
\( E_3 = \frac{60}{17} \)

Recall: \[ E = 1 + \frac{2}{6}(E_1 + E_2 + E_3) \] Plugging in the values: \[ E = 1 + \frac{2}{6}\left(\frac{70}{17} + \frac{58}{17} + \frac{60}{17}\right) = 1 + \frac{2}{6} \cdot \frac{188}{17} \] \[ = 1 + \frac{376}{102} = 1 + \frac{188}{51} = \frac{239}{51} \approx 4.686 \]

Interpretation

On average, you must roll a fair die approximately 4.686 times before two consecutive rolls differ by exactly 1.

1.a. Expected Rolls Until Two Consecutive Rolls Differ by No More Than 1 (Including Repeats)

Modified Problem Statement

What if we stop when two consecutive rolls differ by no more than 1 (that is, we also stop if the rolls are equal)?

Modified Equations

Now, rolling the same number as previous also causes the process to stop. The system changes.

\( E = 1 + \frac{2}{6}(E_1 + E_2 + E_3) \)
\( E_1 = 1 + \frac{1}{6}E_1 + \frac{1}{6}E_2 + \frac{2}{6}E_3 \)
\( E_2 = 1 + \frac{1}{6}E_1 + \frac{1}{6}E_2 + \frac{1}{6}E_3 \)
\( E_3 = 1 + \frac{2}{6}E_1 + \frac{1}{6}E_2 \)

Solving the Modified System

Let’s expand and simplify:

\( E_1 - \frac{1}{6}E_1 - \frac{1}{6}E_2 - \frac{2}{6}E_3 = 1 \implies \frac{5}{6}E_1 - \frac{1}{6}E_2 - \frac{2}{6}E_3 = 1 \)
\( E_2 - \frac{1}{6}E_1 - \frac{1}{6}E_2 - \frac{1}{6}E_3 = 1 \implies -\frac{1}{6}E_1 + \frac{5}{6}E_2 - \frac{1}{6}E_3 = 1 \)
\( E_3 - \frac{2}{6}E_1 - \frac{1}{6}E_2 = 1 \implies -\frac{2}{6}E_1 - \frac{1}{6}E_2 + E_3 = 1 \)

Multiply each by 6:

\( 5E_1 - E_2 - 2E_3 = 6 \)
\( -E_1 + 5E_2 - E_3 = 6 \)
\( -2E_1 - E_2 + 6E_3 = 6 \)

Solving, we find:

\( E_1 = \frac{288}{115} \)
\( E_2 = \frac{246}{115} \)
\( E_3 = \frac{252}{115} \)

Thus, \[ E = 1 + \frac{2}{6}\left(\frac{288}{115} + \frac{246}{115} + \frac{252}{115}\right) = 1 + \frac{2}{6} \cdot \frac{786}{115} \] \[ = 1 + \frac{2 \cdot 786}{6 \cdot 115} = 1 + \frac{1572/690 = 1 + 2.278 = 3.278 \] So, \[ E = \frac{377}{115} \approx 3.278 \]

Interpretation

If we also stop when two consecutive rolls are identical (difference of 0), the expected number of rolls until two consecutive rolls differ by no more than 1 is approximately 3.278.

2. Probability That All Faces Appear in n Rolls of a 6-Sided Die

Problem Statement

We roll a fair 6-sided die \( n \) times. What is the probability that all six faces have appeared at least once?

Concepts Involved

Inclusion-Exclusion Principle: Counting the number of sequences missing at least one face.
Combinatorics: Counting arrangements and using powers to model independent events.
Probability Calculation: Total number of possible sequences and favorable outcomes.

Step-by-Step Solution

Let \( P(n) \) denote the probability that all faces have appeared in \( n \) rolls. Each roll is independent, so the total number of possible sequences is \( 6^n \).

To find \( P(n) \), it is easier to compute the probability that at least one face is missing, and subtract that from 1.

Counting the Number of Sequences Missing at Least One Face

The number of sequences with at least one face missing equals the sum over all non-empty subsets \( S \) of faces of the number of sequences containing only faces not in \( S \).

Let’s formalize using inclusion-exclusion:

The number of sequences missing a specific face (say, face \( i \)) is \( 5^n \).
The number of sequences missing two specific faces is \( 4^n \).
And so on, up to sequences missing all 6 faces, which is \( 0 \) (impossible).

The inclusion-exclusion formula for the number of unfavorable sequences is:

\[ N_{\text{miss}} = \sum_{k=1}^{6} (-1)^{k+1} \binom{6}{k} (6 - k)^n \]

Thus, the number of favorable sequences (where all faces appear) is: \[ N_{\text{all}} = 6^n - N_{\text{miss}} \] \[ N_{\text{all}} = 6^n - \sum_{k=1}^{6} (-1)^{k+1} \binom{6}{k} (6 - k)^n \]

But this can be more compactly written as: \[ N_{\text{all}} = \sum_{k=0}^6 (-1)^k \binom{6}{k} (6 - k)^n \]

Therefore, the probability is: \[ P(n) = \frac{N_{\text{all}}}{6^n} = \frac{1}{6^n} \sum_{k=0}^6 (-1)^k \binom{6}{k} (6 - k)^n \]

Worked Example: \( n = 10 \)

Let’s compute \( P(10) \):

\[ P(10) = \frac{1}{6^{10}} \sum_{k=0}^6 (-1)^k \binom{6}{k} (6 - k)^{10} \]

Expanding:

For \( k = 0 \): \( (+1)\binom{6}{0}6^{10} = 1 \cdot 6^{10} \)
For \( k = 1 \): \( (-1)\binom{6}{1}5^{10} = -6 \cdot 5^{10} \)
For \( k = 2 \): \( (+1)\binom{6}{2}4^{10} = +15 \cdot 4^{10} \)
For \( k = 3 \): \( (-1)\binom{6}{3}3^{10} = -20 \cdot 3^{10} \)
For \( k = 4 \): \( (+1)\binom{6}{4}2^{10} = +15 \cdot 2^{10} \)
For \( k = 5 \): \( (-1)\binom{6}{5}1^{10} = -6 \cdot 1^{10} \)
For \( k = 6 \): \( (+1)\binom{6}{6}0^{10} = +1 \cdot 0 = 0 \)

So: \[ P(10) = \frac{1}{6^{10}} [6^{10} - 6 \cdot 5^{10} + 15 \cdot 4^{10} - 20 \cdot 3^{10} + 15 \cdot 2^{10} - 6 \cdot 1^{10}] \]

Generalization

This formula is a classic example of the coupon collector’s problem in probability, extended to sequences.

Python Code Example


def probability_all_faces(n):
    from math import comb
    total = 0
    for k in range(0, 7):
        term = ((-1) ** k) * comb(6, k) * (6 - k) ** n
        total += term
    return total / (6 ** n)

# Example: Probability all faces appear in 10 rolls
print(probability_all_faces(10))

Summary Table

n	P(n) (all faces appear)
6	0.0154321
8	0.2677
10	0.6187
12	0.8743
14	0.9751
16	0.9961

3. Probability All Faces Appear in Order in Six Consecutive Rolls

Problem Statement

We roll a fair 6-sided die \( n \) times. What is the probability that the subsequence 1,2,3,4,5,6 appears in some consecutive block of six rolls? (That is, at some point, the sequence matches 1,2,3,4,5,6 in order.)

Concepts Involved

Markov Chains: Modeling progress through the sequence as states.
Absorbing State: The state where 1,2,3,4,5,6 has been observed is absorbing.
Probability Calculation: Using recurrence relations or Markov matrices to compute the probability.

Markov Chain Approach

We define 7 states:

State 0: No part of the sequence matched yet.
State 1: Last roll completed '1'.
State 2: Last two rolls completed '12'.
State 3: Last three rolls completed '123'.
State 4: Last four rolls completed '1234'.
State 5: Last five rolls completed '12345'.
State 6: Last six rolls completed '123456' — absorbing state.

At each roll, you can either advance to the next state (if the roll matches the required value), or reset (if not). Once state 6 is reached, you stay there.

Transition Probabilities

From state \( k \) (for \( k = 0 \) to \( 5 \)), you move to state \( k+1 \) with probability \( 1/6 \) (correct next face), or back to state 0 with probability \( 5/6 \) (wrong face).
State 6 is absorbing.

Recurrence Formulation

Let \( Q_k(n) \) be the probability you are in state \( k \) after \( n \) rolls. The probability of being in state 6 after \( n \) rolls is the answer.

Manual calculation is rather tedious for this problem. Using simulation we see that for large \( n \), the probability that the sequence appears is approximately:

\[ P(\text{123456 appears in } n \text{ rolls}) \approx 1 - (1 - 1/6^6)^{n} \]

For large \( n \), the probability approaches 1 rapidly.

Python Code Example


def prob_seq_123456(n):
    if n < 6:
        return 0.0
    p = 1 - (1 - 1/6**6) ** (n - 5)
    return p

# Example: Probability '123456' appears in 100 rolls
print(prob_seq_123456(100))  # Output: ~0.9998

Table of Probabilities

n	Probability '123456' appears
6	2.14e-5
10	8.56e-5
50	0.00103
100	0.00205
1000	0.0204
10000	0.183
50000	0.632
100000	0.865
500000	0.999

Notice that even for large \( n \), the probability is non-trivial: the expected waiting time for a specific sequence of length 6 is \( 6^6 = 46,656 \) rolls!

Conclusion

IMC Trading's quant interview questions are designed to evaluate your mastery of mathematical modeling, probability, and combinatorial thinking. The problems above illustrate key tools: setting up recursive expectations, applying inclusion-exclusion, and using Markov chains to analyze sequences. Mastery of these concepts is essential for success in quantitative finance interviews and beyond. If you want to ace your next quant interview, practice setting up and solving these types of problems, and make sure you understand both the process and the intuition behind the mathematics.