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Quant Research Interview Questions - Graviton Research and Jane Street

Quantitative research interviews at elite trading firms like Graviton Research Capital and Jane Street are renowned for their rigor, depth, and the mathematical sophistication of their questions. These interviews probe not only your technical mastery but also your problem-solving agility and ability to communicate complex concepts clearly. In this article, we’ll dissect some recent and classic quant research interview questions from Graviton Research, Jane Street, and other top firms. We'll not only solve these problems, but thoroughly explain the underlying concepts—giving you the insight and preparation you need to excel in your own quant interviews.

Quant Research Interview Questions: Graviton Research and Jane Street

Game of Nim Winning Strategy (Graviton Research)
Birthday Problem: 5 in the Same Month (Jane Street)
Deriving the Black-Scholes Equation (Belvedere Trading)
Expected Longest Streak of Heads (Five Rings)

Game of Nim: Winning Strategy (Graviton Research Capital)

Question Statement

In a variation of the Game of Nim, you are presented with a single pile of rocks. On your turn, you can pick any number of rocks from 1 to 6. What is your winning strategy?

Understanding the Game

There is a single pile.
Players take turns removing 1–6 rocks.
The player to take the last rock wins (normal play).

Step 1: Identifying Winning and Losing Positions

Let \( n \) be the number of rocks remaining. We define:

Winning position: If there exists a move that leaves the opponent in a losing position.
Losing position: If every move leaves the opponent in a winning position.

Base Cases

If \( n = 0 \): you lose (no rocks left).
If \( n = 1 \) to 6: you win by taking all the rocks.

Step 2: Recursive Solution

A position \( n \) is losing if for every \( k \in \{1,2,3,4,5,6\} \), \( n-k \) is a winning position.

A position \( n \) is winning if there exists at least one \( k \in \{1,2,3,4,5,6\} \) such that \( n-k \) is a losing position.

Pattern Finding

Let’s tabulate the first few positions:

n	Winning/Losing	Reason
0	Losing	No rocks left
1–6	Winning	Can take all rocks
7	Losing	Any move leaves 1–6 (winning) for opponent
8–13	Winning	Can reduce to 7 for opponent
14	Losing	Any move leaves 8–13 (winning) for opponent

So, losing positions occur at \( n = 7, 14, 21, ... \).

Step 3: The Winning Strategy

Always leave a multiple of 7 for your opponent.

If the number of rocks at your turn is not a multiple of 7, remove enough rocks to make it so.
If it’s already a multiple of 7, you cannot win if your opponent plays optimally.

Explicit Strategy

On your turn, compute \( n \mod 7 \). If it is 0, you're in a losing position. Otherwise, remove \( n \mod 7 \) rocks.

Code Example


def optimal_move(n):
    if n % 7 == 0:
        return "No guaranteed win. Hope opponent errs!"
    else:
        return f"Remove {n % 7} rocks."

Concepts Involved

Combinatorial Game Theory: Analysis of impartial games like Nim.
Modular Arithmetic: Identifying losing positions via modulus.
Recursion: Defining positions in terms of smaller positions.

How Many People Before 5 Share a Birth Month? (Jane Street)

Question Statement

How many people must be in a room before you are guaranteed that at least 5 people have been born in the same month?

Understanding the Problem

This is a classic Pigeonhole Principle question.
There are 12 months, i.e., 12 "pigeonholes".
We want the minimal number \( N \) such that in every possible assignment of birthdays to months, at least one month has 5 or more people.

Step 1: Applying the Pigeonhole Principle

The general form: If you have \( n \) pigeonholes and \( k \) items, the maximum number per pigeonhole without exceeding \( m \) in any is \( n \cdot m \). To guarantee at least \( m+1 \) in some pigeonhole, you need \( n \cdot m + 1 \) items.

Here,

\( n = 12 \) months
\( m = 4 \) people per month (if we want to avoid 5 in any)

So, with \( 12 \times 4 = 48 \) people, you could in theory have 4 in each month. But with 49 people, the 49th must create a month with at least 5.

Step 2: Final Answer and Justification

The minimum number of people is:

\[ N = 12 \times 4 + 1 = 49 \]

With 49 people, you are guaranteed that at least one month has 5 or more people. With 48 people, it is possible (if distributed evenly) that no month has more than 4.

Concepts Involved

Pigeonhole Principle: If more items than containers, at least one container holds at least two items.
Extremal Distributions: Constructing extreme cases to test tightness.

Derivation of the Black-Scholes Option Pricing Equation (Belvedere Trading)

Question Statement

Derive the Black-Scholes equation for option pricing.

Step 1: Model Assumptions

The underlying asset price \( S_t \) follows a geometric Brownian motion: \[ dS_t = \mu S_t dt + \sigma S_t dW_t \] where \( \mu \) is drift, \( \sigma \) is volatility, \( W_t \) is a standard Wiener process.
Risk-free rate: \( r \) (constant).
No dividends, frictionless markets, continuous trading.

Step 2: Option Price as a Function

Let \( V(S, t) \) be the price of the option as a function of stock price \( S \) and time \( t \).

Step 3: Constructing a Hedged Portfolio

Create a portfolio of:

Long 1 option (value \( V \)),
Short \( \Delta \) shares of stock (\( -\Delta S \)).

Portfolio value: \[ \Pi = V - \Delta S \]

Step 4: Applying Itô's Lemma

Apply Itô’s Lemma to \( V(S, t) \): \[ dV = \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial S} dS + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} (dS)^2 \] Recall \( (dW_t)^2 = dt \), so \( (dS)^2 = \sigma^2 S^2 dt \). \[ dV = V_t dt + V_S dS + \frac{1}{2} V_{SS} \sigma^2 S^2 dt \] Substitute \( dS \): \[ dV = V_t dt + V_S (\mu S dt + \sigma S dW_t) + \frac{1}{2} V_{SS} \sigma^2 S^2 dt \] \[ = [V_t + \mu S V_S + \frac{1}{2} \sigma^2 S^2 V_{SS}] dt + \sigma S V_S dW_t \]

Step 5: Portfolio Dynamics

The change in portfolio: \[ d\Pi = dV - \Delta dS \] \[ = [V_t + \mu S V_S + \frac{1}{2} \sigma^2 S^2 V_{SS}] dt + \sigma S V_S dW_t - \Delta (\mu S dt + \sigma S dW_t) \] \[ = [V_t + \mu S V_S + \frac{1}{2} \sigma^2 S^2 V_{SS} - \Delta \mu S] dt + [\sigma S V_S - \Delta \sigma S] dW_t \]

Choose \( \Delta = V_S \) to eliminate the stochastic component (\( dW_t \)): \[ d\Pi = [V_t + \frac{1}{2} \sigma^2 S^2 V_{SS}] dt \]

Step 6: No-Arbitrage Condition

Since the portfolio is riskless (no \( dW_t \)), it must earn the risk-free rate: \[ d\Pi = r \Pi dt = r (V - V_S S) dt \] Set equal: \[ V_t + \frac{1}{2} \sigma^2 S^2 V_{SS} = r (V - V_S S) \] \[ V_t + \frac{1}{2} \sigma^2 S^2 V_{SS} + r S V_S - r V = 0 \]

Step 7: The Black-Scholes PDE

\[ \boxed{ \frac{\partial V}{\partial t} + \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 V}{\partial S^2} + r S \frac{\partial V}{\partial S} - r V = 0 } \]

Step 8: Boundary Condition for European Call

At expiry \( t = T \), for a call: \[ V(S,T) = \max(S-K, 0) \] where \( K \) is the strike price.

Concepts Involved

Stochastic Calculus: Itô's Lemma for functions of stochastic processes.
Delta Hedging: Eliminating risk via portfolio construction.
No-Arbitrage Principle: Riskless portfolios earn the risk-free rate.

Expected Value of the Longest Streak of Heads (Five Rings)

Question Statement

30 people each flip a (different) unbiased coin 162 times. What is the expected value of the longest streak of Heads obtained by any one person?

Understanding the Problem

Each person performs 162 independent, unbiased coin tosses.
A streak is a sequence of consecutive Heads.
We want the expected longest streak observed by any one person (i.e., the maximum across all 30 people).

Step 1: Expected Longest Run for One Person

Let’s first consider one person. For \( n \) tosses, the expected length \( L \) of the longest run of Heads is approximately: \[ L \approx \log_2 n \]

For \( n = 162 \): \[ \log_2 162 \approx 7.35 \] So, the expected longest run is about 7 or 8.

Step 2: Adjusting for 30 People

Now, with 30 people, we want the expected maximum of 30 independent such random variables.

Let’s denote \( X \) as the longest run in 162 tosses for one person, and \( M = \max(X_1, ..., X_{30}) \).

Step 3: Probability That a Streak of \( k \) Occurs

The probability that a run of at least \( k \) occurs in \( n \) tosses is approximately: \[ P(\text{run} \geq k) \approx 1 - \left(1 - 2^{-k}\right)^{n - k + 2} \]

For large \( n \), the expected maximum streak \( m \) among \( N \) people is found by solving: \[ N \cdot P(\text{run} \geq m) \approx 1 \]

Plugging in Numbers

\( n = 162 \), \( N = 30 \)

Let’s estimate \( m \):

Set \( P(\text{run} \geq m) \approx \frac{1}{N} \), so \[ 1 - (1 - 2^{-m})^{n-m+2} \approx \frac{1}{30} \] \[ (1 - 2^{-m})^{n-m+2} \approx 1 - \frac{1}{30} \approx 0.967 \] \[ (n-m+2) \cdot 2^{-m} \approx 0.033 \quad \text{(using } (1 - x)^k \approx 1 - kx \text{ for small } x) \] \[ (162 - m + 2) \cdot 2^{-m} \approx 0.033 \] \[ (164 - m) \cdot 2^{-m} \approx 0.033 \]

Numerical Solution

Try \( m = 8 \): \[ (164 - 8) \cdot 2^{-8} = 156 \cdot \frac{1}{256} \approx 0.609 \] Try \( m = 10 \): \[ (164 - 10) \cdot 2^{-10} = 154 \cdot \frac{1}{1024} \approx 0.150 \] Try \( m = 11 \): \[ (164 - 11) \cdot 2^{-11} = 153 \cdot \frac{1}{2048} \approx 0.075 \] Try \( m = 12 \): \[ (164 - 12) \cdot 2^{-12} = 152 \cdot \frac{1}{4096} \approx 0.037 \] Try \( m = 13 \): \[ (164 - 13) \cdot 2^{-13} = 151 \cdot \frac{1}{8192} \approx 0.018 \] So \( m \) between 12 and 13.

Step 4: Final Answer

The expected value of the longest streak among the 30 people is approximately 12 (rounded to the nearest integer).

\[ \boxed{12} \]

Concepts Involved

Probabilistic Analysis of Runs: Using geometric distributions and approximations.
Extremes of Distributions: Estimating the maximum of several independent trials.
Approximation Techniques: Applying Taylor expansions and logarithms for estimates.

Conclusion: Mastering Quant Interview Questions

The quant research interviews at Graviton Research, Jane Street, and other top proprietary trading firms test far more than raw mathematics—they probe for structured thinking, deep understanding of probability and combinatorics, and the ability to communicate complex solutions. As seen above, questions range from combinatorial puzzles to stochastic calculus and probability extremes.

To excel:

Master the core concepts (combinatorics, probability, stochastic calculus, logic).
Practice explaining your reasoning clearly and concisely.
Understand the underlying structures in each problem—often, general principles like the Pigeonhole Principle or modular arithmetic unlock the solution.
Prepare for both classic puzzles and advanced finance questions (like Black-Scholes derivation).

By dissecting questions in the way we've done above, not only do you prepare for the specific types of problems that top quant firms ask—you also build the intuition and communication skills to stand out in interviews.