Quantitative finance interviews at elite firms like Akuna Capital, Morgan Stanley, Jane Street, and WorldQuant are renowned for their challenging brainteasers, probability puzzles, and statistics riddles. These questions are designed to test not only your mathematical prowess but also your logical reasoning, problem-solving speed, and ability to explain complex ideas clearly. In this article, we’ll dive deep into some real quant interview questions reportedly asked at Akuna Capital and other top firms, exploring their detailed solutions, approaches, and optimal strategies. Whether you’re preparing for your own interview or are just curious about the type of thinking these roles demand, read on for a comprehensive breakdown of each problem.

Quant Interview Questions – Akuna Capital

1. Baby Gender Probability Problem

Question

There are 5 babies in a room: 2 boys and 3 girls. One baby with unknown sex is added. Then, you randomly choose one baby, and it turns out to be a boy. What is the probability that the added baby is a boy?

Solution & Explanation

Let’s define the events:

• B: The added baby is a boy
• C: The randomly chosen baby is a boy

We are asked for \( P(B | C) \): the probability the added baby is a boy, given that the randomly chosen baby is a boy.

Step 1: Possible Configurations

After adding the sixth baby, total babies: 6.
If the added baby is a boy: 3 boys and 3 girls.
If the added baby is a girl: 2 boys and 4 girls.

Step 2: Using Bayes’ Theorem

Bayes’ theorem gives:

\[ P(B | C) = \frac{P(C | B) P(B)}{P(C)} \]

Assume the added baby is equally likely to be a boy or a girl, so \(P(B) = 0.5\).

Step 3: Find \( P(C|B) \) and \( P(C|\lnot B) \)

• If added baby is a boy, then out of 6 babies, 3 are boys: \(P(C|B) = \frac{3}{6} = 0.5\).
• If added baby is a girl, then out of 6 babies, 2 are boys: \(P(C|\lnot B) = \frac{2}{6} = \frac{1}{3}\).

Step 4: Find \( P(C) \)

By the law of total probability:

\[ P(C) = P(C|B)P(B) + P(C|\lnot B)P(\lnot B) = (0.5)(0.5) + \left(\frac{1}{3}\right)(0.5) = 0.25 + 0.1667 = 0.4167 \]

Step 5: Plug into Bayes’ Formula

\[ P(B|C) = \frac{0.5 \times 0.5}{0.4167} = \frac{0.25}{0.4167} \approx 0.6 \]

Final Answer

The probability the added baby is a boy is 60% (or \( \frac{3}{5} \)).

2. Optimal Stopping: White and Black Balls Game

Question

There are 3 white balls and 4 black balls in a box. Drawing a white ball yields +1 payoff, black -1. You can stop anytime, and you can play the game infinitely many times. Should you play this game?

Solution & Analysis

This is an optimal stopping problem—specifically, a variation of the gambler’s ruin or expected value maximization.

Step 1: Compute the Expected Value If You Draw Once

The probability of picking white: \( p = \frac{3}{7} \)
Probability black: \( q = \frac{4}{7} \)

Expected value for a single draw:

\[ E[\text{single draw}] = 1 \cdot \frac{3}{7} + (-1) \cdot \frac{4}{7} = \frac{3}{7} - \frac{4}{7} = -\frac{1}{7} \lt 0 \]

Step 2: What If You Stop Early?

If you stop before drawing, your payoff is 0 (better than expected value of drawing).

Step 3: What If You Play Multiple Times?

Since you can repeat the game infinitely, but each time you start over with the same negative expectation, playing the game will always yield a negative expected value.

Step 4: Optimal Strategy

The optimal strategy is: Never play the game. The expected value is always negative, so the rational choice is to refuse to play.

3. Uniform Random Variables – Probability of Average Greater than 1

Question

Let \( X_1, X_2, ..., X_n \) be independent random variables drawn from a uniform distribution on [0, 1]. What is the probability that their average is greater than 1?

Solution & Explanation

Step 1: Range of the Average

Each \( X_i \) is in [0, 1]. Therefore, their average is also in [0, 1].

Step 2: Can the Average Be Greater than 1?

No, because the sum of the \( X_i \)'s is at most \( n \), so the average is at most 1. In fact, the only way the average equals 1 is if every \( X_i = 1 \), which has probability 0 (since continuous distributions have zero probability at points).

Step 3: Probability Calculation

\[ P\left( \frac{1}{n} \sum_{i=1}^{n} X_i > 1 \right) = 0 \]

Final Answer

The probability is 0. The average of uniform [0,1] variables can never exceed 1.

4. Expected Length of the Smallest Piece When Cutting a Rope

Question

A rope of unit length is cut into \( n \) pieces at random. What is the expected length of the smallest piece?

Solution & Analysis

Step 1: Model the Problem

Cutting a rope into \( n \) pieces means placing \( n-1 \) cuts randomly along [0, 1]. The lengths between cuts (plus the ends) are the pieces.

Step 2: Distribution of Piece Lengths

Let’s denote the cut points as \( 0 = X_0 < X_1 < X_2 < \dots < X_{n-1} < X_n = 1 \), where the \( X_i \) are the order statistics of \( n-1 \) independent uniform random variables on [0,1].

The lengths of the pieces are \( L_i = X_i - X_{i-1} \) for \( i = 1, ..., n \).

Step 3: Expected Value of the Smallest Piece

Let \( M = \min\{ L_i \} \). It’s a classic result that:

\[ E[M] = \frac{1}{n^2} \]

Proof Outline: The CDF of the minimum length \( M \) is:

\[ P(M \geq x) = P(\text{all } L_i \geq x) \]

But sum of all \( L_i = 1 \). The probability that all \( L_i \geq x \) is 0 for \( x > 1/n \), so we only consider \( x \leq 1/n \).

From combinatorics (see "broken stick problem"):

\[ P(M \geq x) = (1 - (n-1)x)^{n-1}, \quad 0 \leq x \leq 1/n \]

Then, \[ E[M] = \int_0^{1/n} P(M \geq x) dx = \int_0^{1/n} (1 - (n-1)x)^{n-1} dx \]

Let \( u = 1 - (n-1)x \), \( dx = -\frac{du}{n-1} \):

When \( x=0, u=1 \). When \( x=1/n, u=1-(n-1)/n=1/n \).

So, \[ E[M] = \int_{u=1}^{u=1/n} u^{n-1} \left(-\frac{du}{n-1}\right) = \frac{1}{n-1} \int_{u=1/n}^{u=1} u^{n-1} du \]

\[ = \frac{1}{n-1} \left[ \frac{u^n}{n} \right]_{u=1/n}^{u=1} = \frac{1}{n(n-1)} \left( 1 - (1/n)^n \right) \]

For large \( n \), \( (1/n)^n \to 0 \), so \[ E[M] \approx \frac{1}{n(n-1)} \]

For large \( n \), this is approximately \( \frac{1}{n^2} \).

Final Answer

The expected length of the smallest piece is approximately \( \frac{1}{n^2} \).

5. Probability Game: Best of 3 vs. "Ahead by 2"

Question

You’re playing a game where the probability of winning each round is 0.6. Would you rather play “best of 3” (first to 2 wins), or “play until someone is ahead by 2 points”?

Solution & Analysis

Let’s compare the probability of you winning under each rule.

Best of 3

You need 2 wins before your opponent does. Each round is independent, with probability 0.6 of you winning.

You can win in two ways:

• Win in 2 straight games: \( (0.6)^2 = 0.36 \)
• Win, lose, win: \( 0.6 \times 0.4 \times 0.6 = 0.144 \)
• Lose, win, win: \( 0.4 \times 0.6 \times 0.6 = 0.144 \)

Sum:

\[ P_{\text{best of 3}} = 0.36 + 0.144 + 0.144 = 0.648 \]

So, you have a 64.8% chance of winning.

"Ahead by 2" Rule

Play continues until one player is ahead by 2. This is a classic "gambler’s ruin" with absorbing barriers at +2 (you win) and -2 (you lose).

Let \( p = 0.6 \) be your probability of winning a round, \( q = 0.4 \) for opponent. Let’s model the difference in score as a random walk starting at 0, with absorbing barriers at +2 (you win) and -2 (you lose). Your probability of winning is:

\[ P_{\text{ahaed by 2}} = \frac{1 - (q/p)^2}{1 - (q/p)^4} \]

Plug in \( q/p = 0.4/0.6 = 2/3 \):

\[ P_{\text{ahead by 2}} = \frac{1 - (2/3)^2}{1 - (2/3)^4} = \frac{1 - 4/9}{1 - 16/81} = \frac{5/9}{65/81} = \frac{5}{9} \times \frac{81}{65} = \frac{405}{585} \approx 0.692 \]

So, you have about a 69.2% chance of winning.

Final Recommendation

It is better to play until someone is ahead by 2 points (69.2% win probability) than best of 3 (64.8% win probability).

Summary Table

Detailed Explanations and Further Insights

1. Baby Gender Probability – Deeper Dive

This conditional probability puzzle is a favorite in quant interviews because it tests your grasp of Bayes’ Theorem and your ability to set up the problem without making unwarranted assumptions.

• Common Pitfall: Many candidates mistakenly think the answer should be 0.5, ignoring the information gained from the observed boy.
• Key Takeaway: Always condition on all available information. Here, seeing a boy makes it more likely the added baby was a boy.

This problem is a great illustration of the power of Bayesian reasoning for updating beliefs in light of new evidence—an essential skill in quantitative finance, especially in trading and risk estimation.

2. White and Black Balls – Optimal Stopping Theory

This is a disguised negative expectation game, and the optimal stopping point is crucial. In quant trading, the ability to recognize and avoid negative expectation scenarios is vital.

• Generalization: If the expected value per play is negative, even with the ability to stop, you should not start.
• Real-World Application: This mirrors the discipline needed to walk away from unfavorable trades, even if the structure of the game seems to offer flexibility.

3. Uniform Random Variables – Probability Insights

This problem tests your understanding of the support of a distribution and the properties of averages. Since no single draw can exceed 1, neither can the average.

• Common Trick: Sometimes, the question could ask for “greater than or equal to 1,” but even then, for continuous random variables, the probability that the average is exactly 1 is zero.
• Generalization: For any set of independent uniform [0, 1] variables, their average’s maximum is 1.

4. Rope Cutting – Order Statistics & Broken Stick Problem

The rope cutting problem is a classic of order statistics, frequently called the “broken stick problem.” It’s a staple in probability theory, and knowing the expected value of the smallest segment is a neat trick.

• Extension: If you’re asked about the expected length of the largest piece, for large n, it approaches \( \frac{\log n}{n} \).
• Code Example (Python Simulation):

  
  import numpy as np
  
  def expected_smallest_piece(n, simulations=100000):
      mins = []
      for _ in range(simulations):
          cuts = np.sort(np.random.uniform(0, 1, n - 1))
          pieces = np.diff(np.concatenate(([0], cuts, [1])))
          mins.append(np.min(pieces))
      return np.mean(mins)
  
  print(expected_smallest_piece(5))  # Example for n = 5
  

5. Best of 3 vs. Ahead by 2 – Markov Chains & Gambler’s Ruin

This question evaluates your knowledge of Markov processes and optimal game strategies. “Ahead by 2” is more favorable for the player with a higher probability of winning each round, because it reduces the variance and increases the impact of your skill advantage.

• Key Insight: Whenever you have an edge, longer matches or those requiring a greater margin increase your chances of winning.
• Mathematical Generalization: For probability of winning a single game \( p > 0.5 \), the probability of winning in a “first to k” (difference) series increases with k.
• Code Example (Recursive Calculation):

  
  def gambler_ruin(p, lead):
      # Probability of reaching +2 before -2, starting from "lead"
      if lead == 2:
          return 1.0
      if lead == -2:
          return 0.0
      return p * gambler_ruin(p, lead + 1) + (1 - p) * gambler_ruin(p, lead - 1)
  
  print(gambler_ruin(0.6, 0))  # Should be about 0.692
  

FAQs about Akuna Capital Quant Interview Questions

Are these questions typical for all quant interviews?

Yes, many top quant firms, including Akuna Capital, Morgan Stanley, Jane Street, and WorldQuant, use similar brainteasers to test mathematical and logical reasoning. They assess both technical ability and creative problem-solving.

What topics should I review for quant interviews?

• Probability theory (Bayesian methods, order statistics, Markov chains)
• Statistics (distributions, expectation, variance, optimal stopping)
• Brainteasers and logical puzzles
• Programming (Python, MATLAB, R for simulations and analysis)
• Market intuition and risk management

How should I approach these questions during an interview?

• Clarify all assumptions with the interviewer.
• State your reasoning clearly and out loud.
• Use examples or quick calculations to verify your intuition.
• If stuck, express your partial ideas—often, interviewers value your approach as much as your answer.

Conclusion

Quant interviews at Akuna Capital and other top firms are designed to challenge your understanding of probability, statistics, and logical reasoning. The five questions and solutions discussed here represent the types of problems that test both your technical proficiency and your ability to think creatively under pressure.

By practicing such problems, reviewing the underlying concepts, and simulating your solutions, you’ll be well-prepared for even the toughest quant interviews.

Good luck in your quant interview preparation! If you have more questions or want to see more detailed explanations, write to us on [email protected].

Related Articles

• Quant Research Interview Questions - Jane Street
• Quant Interview Questions - WorldQuant
• Quant Interview Question - Jane Street
• Quant Research Interview Test - Jane Street
• Quant Interview Question - JP Morgan