10 Brainteasers That Actually Appear in Quant Research Interviews
While gone are the days of pure trick questions, modern quant interview brainteasers are carefully selected to probe your logical reasoning, mathematical creativity, and market sense. If you’re interviewing for a quant research role, here are 10 brainteasers that not only frequently appear, but also reveal the underlying skills that top quant firms value.
10 Brainteasers That Actually Appear in Quant Research Interviews
1. The Dice Game: Optional Stopping and Dynamic Programming
Brainteaser: You roll a standard six-sided die. You can take the dollar value shown and stop, or you can choose to roll again (up to 3 rolls total). What is your optimal strategy, and what is the expected dollar value from playing optimally?
What This Tests
- Dynamic programming and backward induction
- Understanding of optional stopping
- Ability to optimize expected value under constraints
Solution Walkthrough
Let’s denote the expected value with n rolls remaining as \( V_n \).
With 1 roll left (\( V_1 \)), you must accept the value shown:
\[ V_1 = \mathbb{E}[\text{Die}] = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = 3.5 \]
With 2 rolls left (\( V_2 \)), after rolling, you can either keep the result or roll again with the expected value \( V_1 \):
\[ V_2 = \frac{1}{6} \sum_{i=1}^{6} \max(i, V_1) \]
Plug in \( V_1 = 3.5 \):
- If the roll is 1, 2, or 3, it’s better to roll again (since 3.5 > 1, 2, 3)
- If the roll is 4, 5, or 6, take it
\[ V_2 = \frac{1}{6}(3 \times 3.5 + 4 + 5 + 6) = \frac{10.5 + 15}{6} = \frac{25.5}{6} \approx 4.25 \]
With 3 rolls left (\( V_3 \)), the logic is identical, but now \( V_2 \) is your threshold:
\[ V_3 = \frac{1}{6}\sum_{i=1}^{6} \max(i, V_2) \]
- If the roll is 1, 2, 3, or 4, roll again (since 4.25 > 1, 2, 3, 4)
- If the roll is 5 or 6, take it
\[ V_3 = \frac{1}{6}(4 \times 4.25 + 5 + 6) = \frac{17 + 11}{6} = \frac{28}{6} \approx 4.67 \]
Summary: Your optimal stopping rule is: after your first roll, roll again unless you see a 5 or 6. After your second roll, roll again unless you see 4 or higher.
2. The Ant on a Rubber Rope
Brainteaser: An ant starts at one end of a 1-meter rubber rope. Every second, the rope is uniformly stretched by 1 meter, and the ant moves forward at 1 cm per second (relative to the rope). Will the ant ever reach the other end?
What This Tests
- Infinite series and convergence
- Understanding of limiting behavior and non-intuitive probability
Solution Walkthrough
Let’s model the problem. Let \( x(t) \) be the position of the ant at time \( t \) (in meters), and \( L(t) = 1 + t \) the rope length at time \( t \).
The ant covers 0.01 meters per second relative to the rope, but the rope stretches, so the ant’s position as a fraction of the rope is \( y(t) = \frac{x(t)}{L(t)} \).
The change in \( y \) per second is:
\[ \frac{dy}{dt} = \frac{0.01}{L(t)} \]
Integrate both sides from \( t = 0 \) to \( T \):
\[ y(T) = \int_0^T \frac{0.01}{1 + t} dt = 0.01 \ln(1 + T) \]
Set \( y(T) = 1 \) to find when the ant reaches the end:
\[ 0.01 \ln(1 + T) = 1 \implies T = e^{100} - 1 \]
So the ant does eventually reach the end (for enormous \( T \)), and the solution demonstrates surprising convergence.
3. Unfair Coin to Fair Bernoulli
Brainteaser: You have a coin with unknown bias \( p \) (probability of heads). How can you simulate a fair 50/50 coin flip using only this coin?
What This Tests
- Cleverness with probability
- Ability to extract symmetry from asymmetry
- Algorithmic thinking
Solution Walkthrough
Use the Von Neumann Trick:
def fair_flip(unfair_coin):
while True:
first = unfair_coin()
second = unfair_coin()
if first != second:
return first
Explanation: Toss the coin twice. If you see Heads-Tails, call it "Heads"; if you see Tails-Heads, call it "Tails". Ignore repeats (HH or TT). Since \( P(\text{HT}) = P(\text{TH}) = p(1-p) \), this is perfectly fair regardless of \( p \).
4. The Market Maker’s Dilemma
Brainteaser: You’re a market maker quoting a bid and ask on a binary event with a 50% true probability. You believe 10% of traders are informed; the rest are noise. How do you set your spread?
What This Tests
- Market microstructure fundamentals
- Understanding adverse selection
- Risk management
Solution Walkthrough
Set bid/ask to balance profit from uninformed flow against losses to informed traders. If your bid/ask is too tight, you lose money to informed traders. Too wide, you lose to competitors.
Let spread be \( S \). Assume fair value is 0.5. If you quote bid = \( 0.5 - S/2 \), ask = \( 0.5 + S/2 \):
- Noisy traders buy/sell with equal probability; your expected profit per trade is \( S/2 \).
- Informed traders buy only when event will occur (or sell if not); you lose \( S/2 \) per trade.
Your expected profit per trade is:
\[ \text{Profit} = (1-\alpha)\frac{S}{2} - \alpha\frac{S}{2} \]
where \( \alpha = 0.1 \).
Set this to zero for a break-even spread:
\[ (1-0.1)\frac{S}{2} - 0.1\frac{S}{2} = 0 \implies S = 0 \]
But in practice, you must quote a nonzero spread to compensate for the informed traders and operational risk, so you’d set \( S \) such that expected profit is slightly positive. This is the core intuition behind the Glosten-Milgrom model.
5. Expected Number of Cards Until an Ace
Brainteaser: From a shuffled deck, turn over cards one by one until you see an ace. What is the expected number of cards you’ll turn over?
What This Tests
- Linearity of expectation
- Breaking complex problems into smaller parts
Solution Walkthrough
Let’s denote \( X \) as the position of the first ace in the deck. The expected position of the first ace is:
\[ \mathbb{E}[X] = \sum_{k=1}^{52} k \cdot P(X = k) \]
But a shortcut is to use symmetry: In a randomly shuffled deck, each card is equally likely to be any card. The four aces split the deck into five segments. The expected number of cards before the first ace is:
\[ \frac{52 + 1}{4 + 1} = \frac{53}{5} = 10.6 \]
But for the expected number of cards turned over until (and including) the first ace: Each card has a 4/52 chance of being the first ace. So:
\[ \mathbb{E}[X] = \frac{53}{5} = 10.6 \]
Or, more generally, for \( n \) aces in \( N \) cards:
\[ \mathbb{E}[X] = \frac{N+1}{n+1} \]
6. The Broken Stick Triangle Problem
Brainteaser: Break a stick at two random points. What is the probability the resulting three pieces can form a triangle?
What This Tests
- Geometric probability
- Visualization and setup of inequalities
Solution Walkthrough
Let’s pick two points \( x, y \) uniformly at random along a stick of length 1, with \( 0 < x < y < 1 \).
Let the pieces be of lengths \( x, y-x, 1-y \). For these to form a triangle, the sum of any two must exceed the third:
- \( x + y - x > 1 - y \implies y > 0.5 \)
- \( x + 1 - y > y - x \implies x > y - 0.5 \)
- \( y - x + 1 - y > x \implies x < 0.5 \)
Graphing these inequalities in the unit square and finding the area that satisfies all, you find the answer is \( \frac{1}{4} \).
Probability: \( \boxed{1/4} \)
7. The Clock Hands Problem
Brainteaser: How many times do the hour and minute hands of an analog clock overlap in a 24-hour period?
What This Tests
- Systematic reasoning
- Pattern recognition
Solution Walkthrough
In 12 hours, the hands overlap 11 times (not 12, because after 11:00, the next overlap is at 12:00). In 24 hours, they overlap \( 11 \times 2 = 22 \) times.
Answer: 22 times per day.
8. The Pirate Game
Brainteaser: Five pirates (A, B, C, D, E) have 100 gold coins to divide. The highest-ranking pirate proposes a split; if 50% or more agree (including proposer), it’s accepted. Otherwise, the proposer is thrown overboard, and the next in rank proposes. Pirates are rational and want to maximize coins, but also want to avoid death. What does the first pirate propose?
What This Tests
- Backward induction
- Strategic reasoning and incentives
Solution Walkthrough
Let’s work backwards:
- 2 pirates: Pirate D proposes, keeps all 100 coins (he has the tie-breaker)
- 3 pirates: Pirate C must get one vote; D knows that if C is thrown overboard, D gets nothing (since with 2 pirates, the other keeps all coins). So C offers 1 coin to E, keeps 99 for himself
- 4 pirates: Pirate B needs 2 votes; offers 1 coin to D (who would get nothing if B is killed), keeps 99
- 5 pirates: Pirate A needs 3 votes; offers 1 coin to C and 1 to E (both would get nothing if A is thrown overboard). So, A keeps 98, C and E get 1 each, B and D get 0.
Final split: A: 98, B: 0, C: 1, D: 0, E: 1
9. Covariance of Maximum and Minimum of Correlated Normals
Brainteaser: Let \( X \) and \( Y \) be standard normals with correlation \( \rho \). What is the covariance of \( \max(X, Y) \) and \( \min(X, Y) \)?
What This Tests
- Advanced probability/statistics
- Handling of bivariate normals
Solution Walkthrough
Let’s denote \( M = \max(X, Y) \), \( m = \min(X,Y) \). We seek \( \text{Cov}(M, m) \). Start by noting a key identity: \[ M + m = X + Y \] \[ M \cdot m = X \cdot Y \] Therefore, \[ \text{Cov}(M, m) = \text{Cov}(M, X + Y - M) = \text{Cov}(M, X + Y) - \text{Var}(M) \] But since \( X \) and \( Y \) are standard normals with correlation \( \rho \), we have \( \text{Cov}(X, Y) = \rho \). Let’s use another useful identity: \[ \mathbb{E}[M \cdot m] = \mathbb{E}[X \cdot Y] = \rho \] Now, for any two random variables, \[ \text{Cov}(M, m) = \mathbb{E}[M m] - \mathbb{E}[M]\mathbb{E}[m] \] But also, \[ \mathbb{E}[M] + \mathbb{E}[m] = \mathbb{E}[X] + \mathbb{E}[Y] = 0 \] So, \[ \mathbb{E}[M] = -\mathbb{E}[m] \] Therefore, \[ \mathbb{E}[M m] = \rho \] And, \[ \mathbb{E}[M]\mathbb{E}[m] = -\mathbb{E}[M]^2 \] So, \[ \text{Cov}(M, m) = \rho + \mathbb{E}[M]^2 \] For \( \rho = 0 \), it's known that \( \mathbb{E}[M] = \mathbb{E}[\max(X, Y)] = \frac{1}{\sqrt{\pi}} \). So, \[ \text{Cov}(M, m) = 0 + \left(\frac{1}{\sqrt{\pi}}\right)^2 = \frac{1}{\pi} \approx 0.318 \] For general \( \rho \), the full answer is: \[ \boxed{\text{Cov}(\max(X, Y), \min(X, Y)) = \rho + \mathbb{E}[\max(X, Y)]^2} \] Calculating \( \mathbb{E}[\max(X, Y)] \) for arbitrary \( \rho \) is more involved, but for most interviews, explaining this approach is sufficient and demonstrates high-level quant reasoning.
10. The “No Math” Fermi Problem: S&P 500 Futures Daily Volume
Brainteaser: Estimate the total daily volume (number of contracts traded) in S&P 500 futures.
What This Tests
- Market intuition
- Ability to decompose and estimate complex quantities
- Comfort with making reasonable assumptions
Solution Walkthrough
Approach with Fermi estimation: break the problem into parts.
- S&P 500 futures are the most liquid equity futures in the world.
- Let’s estimate by considering the notional traded per day.
- The S&P 500 index is around 4,500 (as of 2024), and the E-mini contract multiplier is $50 per point.
- One contract = $50 × 4,500 = $225,000 notional.
- Suppose $200 billion notional is traded daily (reasonable for a major equity index future).
- Contracts traded = $200,000,000,000 / $225,000 ≈ 900,000 contracts per day.
In reality, the number is typically around 1 to 2 million contracts daily, so this estimate is in the right ballpark. The key is systematic decomposition and market familiarity, not exact accuracy.
How to Approach Any Quant Interview Brainteaser: A Framework
Brainteasers in quant interviews are not just about getting the answer—they’re about demonstrating your approach. Here’s a repeatable framework:
- Clarify assumptions: Ask clarifying questions. Are all outcomes equally likely? Are there constraints or edge cases?
- Start simple: Use small numbers, simple cases, or extreme values to build intuition.
- Talk through your logic: Explain your reasoning at each step, even if you’re unsure. Interviewers want to see your thought process.
- Generalize: Once you have a solution for a special case, try to extend it to the general case.
This approach shows you can structure problems, communicate clearly, and adapt to new information—all essential quant research skills.
Conclusion: The Real Goal of Quant Interview Brainteasers
If you’re preparing for a quant research interview, remember: the goal of quant interview brainteasers is not to catch you out with trivia, but to reveal how you reason under pressure. Top firms want to see how you clarify problems, break them down, and communicate your logic—especially when you don’t know the answer immediately.
Practice these classic brainteasers, but more importantly, practice explaining your approach out loud. Show you can think probabilistically, generalize from simple cases, and keep your cool when facing the unexpected. That’s what sets apart the best quant candidates—and why mastering these problems is about much more than just getting to the right answer.
