Are you preparing for a quant research interview at Jane Street? These interviews are renowned for their rigorous probability, combinatorics, and problem-solving questions. In this article, we dive deep into a second set of real quant research interview questions previously asked at Jane Street. For each question, we provide a detailed solution, intuition, and step-by-step reasoning to help you master the concepts and ace your upcoming interview.

Jane Street Quant Research Interview Questions - Part 2: Solutions & Explanations

1. Two Decks of Cards - Probability of Drawing Same Color

Question: You have two decks of cards:

• A 52 card deck (26 black, 26 red)
• A 26 card deck (13 black, 13 red)

You randomly draw two cards and win if both are the same color. Which deck would you prefer? What if the 26 card deck was randomly drawn from the 52 card deck? Which deck would you prefer then?

Solution: Calculating the Probability for Each Deck

a) 52 card deck (26 red, 26 black)

We are to compute the probability that when two cards are drawn at random, both are of the same color.

• Number of ways to draw 2 cards from 52: \( \binom{52}{2} \)
• Number of ways to draw 2 red cards: \( \binom{26}{2} \)
• Number of ways to draw 2 black cards: \( \binom{26}{2} \)

So, the total ways to draw 2 cards of the same color:

$$ \text{Total ways (same color)} = \binom{26}{2} + \binom{26}{2} $$

Therefore, the probability is:

$$ P_{52} = \frac{2 \times \binom{26}{2}}{\binom{52}{2}} $$

Now, plug in the values:

• \( \binom{26}{2} = \frac{26 \times 25}{2} = 325 \)
• \( \binom{52}{2} = \frac{52 \times 51}{2} = 1326 \)

So,

$$ P_{52} = \frac{2 \times 325}{1326} = \frac{650}{1326} = \frac{325}{663} \approx 0.490 $$

b) 26 card deck (13 red, 13 black)

Similarly, we calculate for the 26 card deck:

• Number of ways to draw 2 cards from 26: \( \binom{26}{2} = 325 \)
• Number of ways to draw 2 red: \( \binom{13}{2} = 78 \)
• Number of ways to draw 2 black: \( \binom{13}{2} = 78 \)

So,

$$ P_{26} = \frac{78 + 78}{325} = \frac{156}{325} \approx 0.480 $$

c) Which deck is preferable?

Comparing:

• 52 card deck: \( \approx 0.490 \)
• 26 card deck: \( \approx 0.480 \)

So, you should prefer the 52 card deck.

d) 26 card deck randomly drawn from the 52 card deck

Now, the 26 card deck is constructed by randomly selecting 26 cards from the 52 card deck. Now, when drawing 2 cards from this new deck, what is the probability both are the same color?

Let X be the number of red cards in the new 26 card deck. X follows a hypergeometric distribution:

$$ P(X = k) = \frac{\binom{26}{k}\binom{26}{26-k}}{\binom{52}{26}} $$

Given X, the probability that two cards drawn from this deck are the same color:

$$ P(\text{same} | X) = \frac{\binom{k}{2} + \binom{26-k}{2}}{\binom{26}{2}} $$

The overall probability:

$$ P = \sum_{k=0}^{26} P(X = k) \cdot P(\text{same} | X) $$

But, by linearity of expectation, the expected value of the probability that two randomly chosen cards have the same color is equal to the probability that two randomly chosen cards from the initial 52 card deck have the same color. This is a crucial and deep property.

So, the probability is still \( \frac{325}{663} \approx 0.490 \), same as the 52 card deck.

Final Preference

• Original decks: Prefer 52 card deck (slightly higher chance).
• Random 26-card deck from 52: No difference, same as 52 card deck.

2. Coin Toss Game: B Wins if He Has At Least as Many Heads as A

Question: A tosses \( n+1 \) coins. B tosses \( n \) coins. B wins if he has at least as many heads as A. What is the probability that B wins?

Solution: Understanding the Problem

• A: \( n+1 \) coins
• B: \( n \) coins
• B wins if \( H_B \geq H_A \)

Let’s denote \( H_A \) = number of heads by A, \( H_B \) = number of heads by B.

Symmetry Argument

Let’s consider a clever approach. If both toss the same number of coins, the probability is 0.5 by symmetry. But here, A has one more coin.

Let’s denote \( X \) = number of heads in \( n \) tosses (B), \( Y \) = number of heads in \( n+1 \) tosses (A).

We are interested in \( P(X \geq Y) \).

Direct Computation

Let’s fix the outcome of the extra coin A tosses (call it coin #0). For each value of \( X \), the probability that \( X \geq Y \) is the probability that, in the \( n \) shared coins, B gets \( X \) heads, and in the same \( n \) coins, A gets \( k \) heads, plus the extra coin.

But if we look at all possible outcomes, for any configuration of the \( n \) coins, the probability that B gets at least as many heads as A is the same as the probability that, tossing \( 2n+1 \) coins, the number of heads in the first \( n \) is greater than or equal to the number in the next \( n+1 \).

But here's a standard result: The probability that in tossing \( n \) coins and \( n+1 \) coins, the first has at least as many heads as the second is \( \frac{1}{2} \).

Alternative Approach: Explicit Calculation

Let’s write:

$$ P = \sum_{k=0}^{n} P(B \text{ gets } k) \cdot P(A \text{ gets } \leq k \text{ heads}) $$

\( P(B \text{ gets } k) = \binom{n}{k}2^{-n} \)

\( P(A \text{ gets } \leq k \text{ heads}) = \sum_{j=0}^{k} \binom{n+1}{j}2^{-(n+1)} \)

So,

$$ P = \sum_{k=0}^{n} \binom{n}{k}2^{-n} \cdot \left( \sum_{j=0}^{k} \binom{n+1}{j}2^{-(n+1)} \right) $$

But if you expand this, you get \( \frac{1}{2} \).

Final Answer

• The probability that B wins is \( \boxed{0.5} \) (i.e., exactly half).

3. Dart Game: Estimating Number of 3-Point Darts

Question: You throw 1000 darts. Each one has a 50% chance to score. For the first 500 darts each is worth 1 point, for the second 500 darts each is worth 3 points. If you score 1500 points, most likely how many 3-point darts have you scored?

Solution: Setting Up the Problem

• Let \( x \) = number of successful 1-point darts (out of 500)
• Let \( y \) = number of successful 3-point darts (out of 500)
• Total points: \( x + 3y = 1500 \)
• Each of \( x \), \( y \) is binomial(500, 0.5)

Expectation

• Expected \( x \) = 250
• Expected \( y \) = 250
• Expected total points = \( 250 + 3 \times 250 = 1000 \)

But the actual score is 1500, which is 500 above the mean.

Most Likely Values: Maximum Likelihood Estimation

We want to find the most likely value of \( y \) given \( x + 3y = 1500 \), with both \( x, y \leq 500 \).

For fixed total \( x + 3y = 1500 \), the probability of observing \( x, y \) is:

$$ P(x, y) = \binom{500}{x} 0.5^{500} \cdot \binom{500}{y} 0.5^{500} $$

We seek \( y \) that maximizes \( P(x, y) \) subject to \( x + 3y = 1500 \).

Let’s solve for \( x \): \( x = 1500 - 3y \).

So, the function to maximize is:

$$ f(y) = \binom{500}{1500 - 3y} \cdot \binom{500}{y} $$

Take logarithms and use Stirling's approximation for large \( n \).

Log-likelihood Function

$$ \log f(y) \approx 500 H\left(\frac{1500 - 3y}{500}\right) + 500 H\left(\frac{y}{500}\right) $$

Where \( H(p) = -p\log p - (1-p)\log(1-p) \) is the binary entropy function (natural logarithm).

But more simply, for large \( n \), the joint likelihood is maximized when both \( x \) and \( y \) are as close as possible to their expected value, given the constraint \( x + 3y = 1500 \).

Set up the constraint:

• Expected \( x \) = 250, expected \( y \) = 250, total = 1000
• We need \( x + 3y = 1500 \)

Let’s solve for \( y \):

• \( x = 1500 - 3y \)
• But also, \( x + y = 500 \Rightarrow x = 500 - y \)

Setting equal:

$$ 500 - y = 1500 - 3y \implies 2y = 1000 \implies y = 500 $$

But if \( y = 500 \), then \( x = 0 \). This would mean you scored all the 3-point darts and none of the 1-point darts.

Let’s check if that's possible:

• All second 500 darts scored (unlikely), first 500 all missed.

Alternatively, let's use the method of Lagrange multipliers for maximizing under constraints.

Gaussian Approximation (for Most Likely Value)

Let’s treat \( x \) and \( y \) as continuous and maximize the joint density:

$$ \text{Maximize } \exp\left( -\frac{(x-250)^2}{2 \cdot 125} - \frac{(y-250)^2}{2 \cdot 125} \right) $$

Subject to \( x + 3y = 1500 \).

Substitute \( x = 1500 - 3y \):

$$ F(y) = -\frac{(1500-3y-250)^2}{250} - \frac{(y-250)^2}{250} $$

Set derivative with respect to \( y \) to zero:

$$ \frac{d}{dy} F(y) = 0 $$

Compute \( F(y) \):

$$ 1500 - 3y - 250 = 1250 - 3y $$ $$ F(y) = -\frac{(1250-3y)^2}{250} - \frac{(y-250)^2}{250} $$

Take derivative:

$$ \frac{dF}{dy} = -\frac{2(1250 - 3y)(-3)}{250} - \frac{2(y - 250)}{250} = 0 $$ $$ \frac{6(1250 - 3y)}{250} - \frac{2(y - 250)}{250} = 0 $$ $$ 6(1250 - 3y) - 2(y - 250) = 0 $$ $$ 7500 - 18y - 2y + 500 = 0 $$ $$

Continuing from the previous calculation:

$$ 7500 - 18y - 2y + 500 = 0 \\ 8000 - 20y = 0 \\ 20y = 8000 \\ y = 400 $$

So, the most likely number of 3-point darts scored is 400.

Recall, there are 500 possible 3-point darts (second half), so scoring 400 means you got 80% of those to land.

Sanity Check

Let’s check how many 1-point darts you made:

• \( x = 1500 - 3y = 1500 - 3 \times 400 = 1500 - 1200 = 300 \)

You scored 300 out of the first 500 (60%). That’s well above the expected 250, but less than the 3-point darts.

Conclusion

• Most likely number of 3-point darts scored: 400.
• Most likely number of 1-point darts scored: 300.

This allocation maximizes the joint probability under the constraint \( x + 3y = 1500 \).

4. Vegas Gambling Game: Expected Value for Prime/Composite Numbers

Question: Suppose you want to gamble in Vegas. In a game, you win $x if the number is prime and lose $x/2 if composite. The number is uniformly randomly generated by a machine between 1 and 10 inclusive. Will you play this game? Follow up: What if you can play n number of times and then stop. Will you play it?

Step 1: List Out Possible Outcomes

Numbers from 1 to 10. Identify which are prime:

• Primes: 2, 3, 5, 7 (4 out of 10 numbers)
• Composites: 1, 4, 6, 8, 9, 10 (6 out of 10 numbers)

Note: 1 is neither prime nor composite by strict definition, but for this game, let's follow the standard convention and treat 1 as composite (since you do not win on it).

Step 2: Compute Expected Value per Game

Let’s denote the bet size as $x.

• Probability of prime: \( P_{prime} = \frac{4}{10} = 0.4 \)
• Probability of composite: \( P_{comp} = \frac{6}{10} = 0.6 \)

Expected value:

$$ E = P_{prime} \times x + P_{comp} \times \left(-\frac{x}{2}\right) $$ $$ E = 0.4x - 0.3x = 0.1x $$

So, your expected value per game is positive. For each $1 bet, you expect to win $0.10.

Step 3: Should You Play?

• Yes! The expected value is positive.

Step 4: What if You Can Play n Times and Then Stop?

• Since each game is +0.1x in expectation, the more you play, the more you expect to gain. There is no adverse effect for stopping at any point; you can play as many times as you like and expect to profit in the long run.
• There is no reason to stop early unless you’re maximizing utility or facing a risk constraint, but mathematically, playing more increases your expected profit.

Summary Table: Jane Street Quant Interview Questions (Part 2)

Conclusion

Quant interviews at elite firms like Jane Street probe your ability to think deeply about probability, combinatorics, and expected value. The problems above exemplify the kind of rigorous, subtle reasoning required. By working through detailed solutions and understanding the underlying thought process, you’ll be well-prepared for similar quant research interviews.

• Always break down problems into manageable pieces.
• Use symmetry and combinatorics to your advantage.
• Don’t forget the power of expectation and linearity!

If you found this guide helpful, check out Part 1 of our Jane Street Quant Research Interview Questions series and subscribe for more quant interview insights!

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