Are you preparing for a Jane Street Quant Trading Intern interview? One of the classic probability puzzles you might encounter involves expected value and dice games. Understanding how to calculate the expected value of simple games is crucial for quant trading interviews, as it tests your grasp of probability, combinatorics, and logical reasoning. In this comprehensive guide, we’ll break down a popular Jane Street interview question involving a dice game, walk you through every step of the solution, and explain all relevant concepts in detail.

Jane Street Quant Trading Intern Interview Question: Dice Game Expected Value

Understanding the Question

Here’s the problem statement:

Two players roll a fair six-sided die. You win $1 if your roll is strictly higher than your opponent’s, otherwise $0. What is the expected value?

Let’s dissect the problem and see how to approach it from a quantitative perspective.

Breaking Down the Problem

Step 1: Defining the Experiment

We have:

• Two players: Player A (you) and Player B (opponent).
• Both roll a fair six-sided die, independently.
• If Player A’s number is strictly higher than Player B’s, Player A wins $1.
• Otherwise (tie or Player B wins), Player A wins $0.

Our goal is to compute the expected value (EV) of this game for Player A.

Key Concepts Involved

• Probability: The chance that one event happens out of all possible events.
• Expected Value: The average payout (win or loss) over many repetitions of the game.
• Combinatorics: Counting the number of ways various outcomes can occur.

Let’s define these terms more formally.

Probability

Probability of an event is defined as:

\( P(\text{event}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \)

Expected Value

Expected value is the sum of possible outcomes, each weighted by its probability:

\( \text{EV} = \sum_{i=1}^{n} P_i \cdot V_i \)

Where:

• \( n \) = number of possible outcomes
• \( P_i \) = probability of outcome \( i \)
• \( V_i \) = value (payout) of outcome \( i \)

Enumerating All Possible Outcomes

Each player rolls a die, so both can roll any integer from 1 to 6. Since the dice are independent:

• Total number of possible outcomes = 6 (Player A) × 6 (Player B) = 36

But we don’t need to write out all 36 combinations. Instead, let’s count how many result in a win for Player A.

Calculating the Number of Winning Outcomes

Systematic Counting

For every possible value Player A rolls, let’s count how many possible values Player B can roll that are strictly less.

• If Player A rolls 1: Player B can roll 1 (tie), but not less. So, 0 wins.
• If Player A rolls 2: Player B can roll 1 (win). 1 win.
• If Player A rolls 3: Player B can roll 1 or 2 (wins). 2 wins.
• If Player A rolls 4: Player B can roll 1, 2, or 3. 3 wins.
• If Player A rolls 5: Player B can roll 1, 2, 3, or 4. 4 wins.
• If Player A rolls 6: Player B can roll 1, 2, 3, 4, or 5. 5 wins.

Let’s sum these up:

Total number of winning outcomes = 0 + 1 + 2 + 3 + 4 + 5 = 15

Calculating the Probability of Winning

Total number of possible outcomes = 36.

Probability that Player A wins:

\( P(\text{A wins}) = \frac{15}{36} = \frac{5}{12} \)

Calculating the Expected Value

Recall, Player A wins $1 with probability \( \frac{5}{12} \), and $0 otherwise.

\( \text{EV} = 1 \cdot P(\text{A wins}) + 0 \cdot P(\text{A does not win}) \)

\( \text{EV} = 1 \cdot \frac{5}{12} + 0 = \frac{5}{12} \)

So, the expected value of the game for Player A is \( \frac{5}{12} \).

Alternative Approach: Symmetry and Intuition

Let’s check our answer with another logical approach.

• Each pair (A, B) is equally likely.
• For any pair where A ≠ B, one of A or B wins.
• Number of possible (A, B) pairs where A ≠ B: 36 - 6 (ties) = 30.
• For each non-tie, A has a 50% chance of being the higher die (since dice are fair and independent).
• Therefore, A wins in half of the non-tie cases: \( \frac{30}{2} = 15 \).

This matches our previous calculation.

Full Probability Distribution Table

Generalizing the Problem

Suppose the dice had \( n \) sides instead of 6. For each possible roll by Player A (\( i \)), Player B can roll any value less than \( i \): \( i-1 \) possibilities.

Total winning outcomes for Player A:

\( \sum_{i=1}^{n} (i-1) = 0 + 1 + 2 + \ldots + (n-1) = \frac{n(n-1)}{2} \)

Total possible outcomes: \( n^2 \)

So, the expected value is:

\( \text{EV} = \frac{\text{Number of winning outcomes}}{\text{Total outcomes}} = \frac{\frac{n(n-1)}{2}}{n^2} = \frac{n-1}{2n} \)

For \( n = 6 \):

\( \frac{6-1}{2 \times 6} = \frac{5}{12} \)

Python Code to Simulate the Game

Let's verify our result with a simple simulation.

import random

def simulate_dice_game(trials=100000):
    wins = 0
    for _ in range(trials):
        a = random.randint(1, 6)
        b = random.randint(1, 6)
        if a > b:
            wins += 1
    return wins / trials

print(simulate_dice_game())

The output should be close to 0.4167, which is \( \frac{5}{12} \).

Common Mistakes and Interview Discussion Points

• Counting Ties as Wins: Remember, “strictly higher” does not include ties.
• Ignoring Independence: Both dice are independent; don’t assume otherwise.
• Not Normalizing Over All Outcomes: Always divide by the total number of possible outcomes.
• Generalization: Interviewers may ask you to generalize to \( n \)-sided dice, as above.
• Algebraic Manipulation: Be comfortable with sums like \( \sum_{i=1}^{n} (i-1) \).

How to Explain This in an Interview

When asked this question in a Jane Street Quant Trading Intern interview, structure your response clearly:

1. Restate the problem to show understanding.
2. Define all variables and assumptions (fair dice, independent rolls).
3. Count all possible outcomes (6 × 6 = 36).
4. Count winning outcomes for you, sum them up (0 + 1 + 2 + 3 + 4 + 5 = 15).
5. Compute the probability of winning (\( \frac{15}{36} \)).
6. Calculate expected value (\( 1 \times \frac{5}{12} \)).
7. Generalize to \( n \)-sided dice if prompted.
8. Check your answer with intuition or symmetry.

Why This Matters for Quant Trading

Quantitative trading is all about understanding probabilities, expected values, and risk/reward tradeoffs. This dice game question is a microcosm of those skills:

• Expected Value: Fundamental to evaluating trades and strategies.
• Probabilistic Thinking: Helps avoid cognitive biases and make optimal decisions.
• Combinatorics: Counting outcomes is essential for pricing derivatives and options.
• Generalization: Real trading involves more complex distributions and edge cases.

Additional Variations and Extensions

• Different Payouts: What if you win $2 for a win, lose $1 for a loss, and $0 for a tie? Then the expected value would be:
  \( \text{EV} = 2 \times \frac{5}{12} + (-1) \times \frac{5}{12} + 0 \times \frac{1}{6} = \frac{5}{12} \)
• Multiple Dice: What if each player rolls two dice and sums them? This introduces the distribution of sums (a triangular distribution) and more combinatorics.
• Weighted Dice: What if the dice are not fair? Each outcome must be weighted by its probability.
• Conditional Probability: What’s the probability you win given that you rolled a 4? (Player B must roll 1, 2, or 3: 3/6 = 0.5.)

Final Thoughts and Key Takeaways

• Expected value is a core concept in quant trading and interviews.
• Systematic counting and probability calculation are vital skills.
• Always check your answer with intuition or alternate methods.
• Be ready to generalize and expand the problem if prompted in an interview.
• When explaining, structure your reasoning clearly and justify each step.

Sample Interview Dialogue

Let’s look at how you might present your solution in a real Jane Street quant trading interview, incorporating the key steps and communicating your thought process:

Interviewer: “Two players each roll a fair six-sided die. You win $1 if your roll is strictly higher than your opponent’s, otherwise $0. What is the expected value?”

You: “Let’s clarify: both dice are fair and independent. For each roll, there are 6 possible outcomes for each player, so 36 outcomes in total.

I’ll count the number of ways I can win. If I roll a 1, I can’t win. If I roll a 2, I win if they roll 1 — that’s 1 win. If I roll a 3, I win if they roll 1 or 2 — 2 wins. That pattern continues up to rolling a 6, where I win if they roll 1 through 5, so 5 wins.

Summing these: 0+1+2+3+4+5 = 15 winning outcomes. My probability of winning is 15/36, which simplifies to 5/12.

The expected value is $1 × 5/12 + $0 × 7/12 = 5/12, or about 0.4167.

If you’d like, I can generalize this to n-sided dice as well.”

This answer shows clarity, logical progression, and readiness to generalize, which are highly valued in quant interviews.

Further Practice: Dice Game Variations

Let’s deepen your understanding with some related problems. Practicing these will strengthen your probability and expected value skills for quant trading interviews.

• Variation 1: What if you win $1 for strictly higher, lose $1 for strictly lower, and $0 for a tie? The expected value becomes:
  \( \text{EV} = 1 \times \frac{5}{12} + (-1) \times \frac{5}{12} + 0 \times \frac{1}{6} = 0 \)
  This makes sense since the game is fair under these rules.
• Variation 2: What is the probability of a tie?
  \( P(\text{tie}) = \frac{6}{36} = \frac{1}{6} \)
• Variation 3: What’s the expected value if you win $2 for a win, lose $3 for a loss, and get $0 for a tie?
  \( \text{EV} = 2 \times \frac{5}{12} + (-3) \times \frac{5}{12} + 0 \times \frac{1}{6} = \frac{10}{12} - \frac{15}{12} = -\frac{5}{12} \)
• Variation 4: Suppose both players roll two dice and sum the results. What is the expected value?
  This requires calculating the probabilities for sums from 2 to 12, then counting the number of winning pairs. This is a good exercise in distribution of sums and combinatorics.

Advanced Generalization: n-sided Dice

Let’s formalize the general case for n-sided dice, which is a common interview extension.

For dice with \( n \) sides numbered 1 through \( n \):

• For each roll \( i \) by Player A, Player B can roll \( 1 \) through \( i-1 \), so \( i-1 \) winning outcomes for Player A.
• Summing over all possible rolls:
  \( \sum_{i=1}^{n} (i-1) = 0 + 1 + 2 + \ldots + (n-1) = \frac{n(n-1)}{2} \)
• Total outcomes: \( n^2 \)
• Probability Player A wins: \( \frac{n(n-1)/2}{n^2} = \frac{n-1}{2n} \)
• Expected value: \( \frac{n-1}{2n} \)

This formula is useful in interviews, as it demonstrates your ability to generalize and work algebraically.

Visual Representation: Probability Matrix

Let’s visualize the possible outcomes using a matrix, where rows are your rolls and columns are your opponent’s rolls:

Count the number of “Win” cells to verify the previous calculation.

Common Follow-up Questions in Interviews

Interviewers often build on this question with further challenges:

• What if the dice are not fair?
  Assign probability weights to each face and compute the expected value accordingly.
• What is the variance of your winnings?
  Calculate \( \text{Var}(X) \) where \( X \) is your payout:
  \( \text{Var}(X) = E[X^2] - (E[X])^2 \)
  Here, since the payout is 1 with probability \( p \) and 0 otherwise:
  \( E[X^2] = E[X] = \frac{5}{12} \)
  So,
  \( \text{Var}(X) = \frac{5}{12} - \left(\frac{5}{12}\right)^2 = \frac{5}{12} - \frac{25}{144} = \frac{60-25}{144} = \frac{35}{144} \)
• Suppose you can choose your die after seeing the opponent’s roll. What is your expected value?
  If you can react, your expected value jumps to 1, since you always beat their roll (assuming you have a die with numbers higher than theirs).

Practical Applications in Quantitative Trading

This dice game is more than just a puzzle; it models real-world trading decisions:

• Edge calculation: Determining the probability-weighted average return of a trade.
• Risk assessment: Understanding the distribution and variance of outcomes.
• Scenario analysis: Considering variations in rules or payoffs models real-life trading strategies.
• Generalization: Ability to extend simple models to more complex, realistic ones is vital in trading.

Summary Table: Key Results

Conclusion

The Jane Street Quant Trading Intern Dice Game Expected Value question elegantly tests a wide range of skills: probability, combinatorics, expected value, and the ability to generalize and communicate clearly. The answer, \( \frac{5}{12} \), is a simple yet powerful result, and the process of reaching it demonstrates essential reasoning for any aspiring quantitative trader. Practicing questions like this will prepare you for both interviews and the real-world challenges of trading, where probability and expected value underpin every decision.

Key takeaway: Mastering problems like the dice game expected value will sharpen your quantitative intuition and set you apart in competitive interviews at top firms like Jane Street.