
Jane Street Quantitative Researcher Interview Question: Expected Value of a 10-Sided Die Game
The Jane Street Quantitative Researcher interview is renowned for its challenging probability and statistics problems. A classic example involves calculating the expected value in games of chance, such as those featuring dice. In this article, we will analyze a well-known interview question involving a fair 10-sided die, exploring the required concepts, step-by-step solutions, and practical implications. This comprehensive guide will not only help you approach such questions in interviews but also deepen your understanding of expected value and related quantitative reasoning skills.
Jane Street Quantitative Researcher Interview Question: Expected Value of a 10-Sided Die Game
Introduction to the Problem
Suppose you are presented with a fair 10-sided die, with faces numbered from 0 to 9. The die is rolled once, and your payoff depends on the outcome according to certain rules. The interviewer asks you:
- How do you compute the expected value of your payoff?
- What underlying mathematical concepts are involved?
- How would you approach and solve this, step by step?
Let’s break down the question, analyze the scenario, and provide a detailed solution. Along the way, we will explain key probability concepts, the notion of expected value, and several variations of payoff rules commonly encountered in quant interviews.
Understanding the 10-Sided Die
Definition and Fairness
A 10-sided die (sometimes called a d10 in gaming) is a polyhedral die with faces labeled 0 through 9. If the die is fair, each face is equally likely to appear on any given roll. Therefore, each outcome has a probability of \( \frac{1}{10} \).
| Die Face | Probability |
|---|---|
| 0 | \( \frac{1}{10} \) |
| 1 | \( \frac{1}{10} \) |
| 2 | \( \frac{1}{10} \) |
| 3 | \( \frac{1}{10} \) |
| 4 | \( \frac{1}{10} \) |
| 5 | \( \frac{1}{10} \) |
| 6 | \( \frac{1}{10} \) |
| 7 | \( \frac{1}{10} \) |
| 8 | \( \frac{1}{10} \) |
| 9 | \( \frac{1}{10} \) |
Notation and Sample Space
Let the random variable \( X \) denote the outcome of a single roll, with possible values \( X = \{0, 1, 2, ..., 9\} \).
What is Expected Value?
The expected value (often denoted as \( E[X] \)) is a fundamental concept in probability and statistics. It represents the long-term average or mean value of a random variable after many trials of the same experiment. In the context of a game, it tells you the amount you can expect to win (or lose) per play, on average.
Mathematical Definition
If \( X \) is a discrete random variable with possible values \( x_1, x_2, ..., x_n \) and corresponding probabilities \( p_1, p_2, ..., p_n \), then the expected value is given by:
\[ E[X] = \sum_{i=1}^{n} x_i \cdot p_i \]
Interpretation
- Positive Expected Value: The game favors the player in the long run.
- Negative Expected Value: The game favors the house or organizer.
- Zero Expected Value: The game is fair — no advantage to either side.
Step-by-Step Solution to the Die Game
Step 1: Define the Payoff Function
The expected value calculation depends on the payoff function defined for the game. The most common scenario is where the payoff equals the number shown on the die.
Let’s define:
- Payoff \( = X \) (the number rolled)
Step 2: List All Possible Outcomes and Probabilities
As previously established, the die outcomes \( X = 0, 1, 2, ..., 9 \) each occur with probability \( \frac{1}{10} \).
Step 3: Write the Expected Value Formula
Plugging into the expected value formula:
\[ E[X] = \sum_{k=0}^{9} k \cdot \frac{1}{10} \]
This simplifies to:
\[ E[X] = \frac{1}{10} \sum_{k=0}^{9} k \]
Step 4: Compute the Sum
The sum \( \sum_{k=0}^{9} k \) is the sum of the first 10 non-negative integers.
- Sum = \( 0 + 1 + 2 + ... + 9 \)
- There are 10 terms.
Recall the formula for the sum of the first \( n \) integers (from 0 to \( n-1 \)):
\[ \sum_{k=0}^{n-1} k = \frac{n(n-1)}{2} \]
Here, \( n = 10 \):
\[ \sum_{k=0}^{9} k = \frac{10 \times 9}{2} = 45 \]
Step 5: Calculate the Expected Value
Now, substitute back:
\[ E[X] = \frac{1}{10} \times 45 = 4.5 \]
Interpretation and Insights
The expected value of the payoff in this 10-sided die game is 4.5.
This means that, on average, each roll is worth 4.5 units (dollars, points, etc.), assuming your payoff is the number rolled.
Why Is This Important?
- Expected value helps you evaluate the fairness of a game or investment.
- It guides optimal decision-making under uncertainty.
- Understanding the distribution around the expected value (i.e., variance) is also crucial for risk assessment.
Variations of the Payoff Rule
Interviewers may tweak the payoff rule to test deeper understanding. Let's explore some common variations and their expected values.
Variation 1: Payoff is Double the Number on the Die
Payoff function: \( P = 2X \)
\[ E[P] = E[2X] = 2 \cdot E[X] = 2 \cdot 4.5 = 9 \]
Variation 2: Payoff is 1 if the Die Shows an Even Number, 0 Otherwise
Let’s denote \( P = 1 \) if \( X \) is even, \( 0 \) if odd.
- Even numbers from 0 to 9: 0, 2, 4, 6, 8 (total 5)
- Probability of rolling an even number: \( \frac{5}{10} = 0.5 \)
\[ E[P] = 1 \cdot 0.5 + 0 \cdot 0.5 = 0.5 \]
Variation 3: Payoff is 10 Minus the Number Rolled
Payoff function: \( P = 10 - X \)
\[ E[P] = E[10 - X] = 10 - E[X] = 10 - 4.5 = 5.5 \]
Variation 4: Payoff is the Square of the Number Rolled
Payoff function: \( P = X^2 \)
\[ E[P] = \sum_{k=0}^{9} k^2 \cdot \frac{1}{10} \]
First, compute \( \sum_{k=0}^{9} k^2 \):
\[ \sum_{k=0}^{n-1} k^2 = \frac{(n-1) n (2n-1)}{6} \]
For \( n = 10 \):
\[ \sum_{k=0}^{9} k^2 = \frac{9 \cdot 10 \cdot 19}{6} = \frac{1710}{6} = 285 \]
\[ E[P] = \frac{285}{10} = 28.5 \]
Generalizing to Any n-Sided Die
The expected value formula can be generalized for an \( n \)-sided die with faces numbered from 0 to \( n-1 \):
\[ E[X] = \frac{1}{n} \sum_{k=0}^{n-1} k = \frac{1}{n} \cdot \frac{n(n-1)}{2} = \frac{n-1}{2} \]
For a 10-sided die:
\[ E[X] = \frac{10-1}{2} = 4.5 \]
This formula is powerful: it allows you to instantly compute the expected value for any fair die labeled from 0 to \( n-1 \).
Python Code for Expected Value Calculation
Let’s write a Python snippet to compute the expected value for any custom payoff function for a 10-sided die.
# Define the die faces
faces = list(range(10))
# Define the payoff function (identity: payoff is the number rolled)
def payoff(x):
return x
# Compute expected value
expected_value = sum(payoff(x) for x in faces) / len(faces)
print("Expected Value:", expected_value)
To compute for squared payoff, change the payoff function to return x ** 2.
Probability Distribution Table
| Outcome (X) | Payoff | Probability | Contribution to EV |
|---|---|---|---|
| 0 | 0 | \( \frac{1}{10} \) | 0 |
| 1 | 1 | \( \frac{1}{10} \) | 0.1 |
| 2 | 2 | \( \frac{1}{10} \) | 0.2 |
| 3 | 3 | \( \frac{1}{10} \) | 0.3 |
| 4 | 4 | \( \frac{1}{10} \) | 0.4 |
| 5 | 5 | \( \frac{1}{10} \) | 0.5 |
| 6 | 6 | \( \frac{1}{10} \) | 0.6 |
| 7 | 7 | \( \frac{1}{10} \) | 0.7 |
| 8 | 8 | \( \frac{1}{10} \) | 0.8 |
| 9 | 9 | \( \frac{1}{10} \) | 0.9 |
| Total | 4.5 | ||
Variance and Standard Deviation
Understanding the spread of outcomes is as important as knowing the expected value, especially in risk-sensitive domains like quantitative trading.
Variance Formula
The variance of \( X \) is:
\[ Var(X) = E[X^2] - (E[X])^2 \]
From earlier, \( E[X] = 4.5 \) and \( E[X^2] = 28.5 \). Therefore:
\[ Var(X) = 28.5 - (4.5)^2 = 28.5 - 20.25 = 8.25 \]
Standard deviation:
\[ SD(X) = \sqrt{8.25} \approx 2.872 \]
Common Interview Follow-Ups and Extensions
Jane Street interviews may build on this question with more challenging twists:
- Conditional Payoff: What is the expected value if you are paid the square of the number only if the number is even?
- Compute \( E[P] = \sum_{k=0,2,4,6,8} k^2\cdot \frac{1}{10} \)
For even numbers (0, 2, 4, 6, 8): their squares are 0, 4, 16, 36, 64.
The expected value is:\[ E[P] = \frac{1}{10}(0 + 4 + 16 + 36 + 64) = \frac{120}{10} = 12 \]
- Repeated Game: What is the expected total payoff if you roll the die 100 times?
- Since each roll is independent, the expected value for 100 rolls = \( 100 \times 4.5 = 450 \).
- Threshold Payoff: You receive $1 if the number rolled is greater than 5, $0 otherwise.
- Numbers greater than 5: 6, 7, 8, 9 (4 outcomes)
- Probability = \( \frac{4}{10} = 0.4 \)
- Expected value = \( 1 \times 0.4 + 0 \times 0.6 = 0.4 \)
- Rolling Until a Certain Condition: What is the expected number of rolls to get a 0?
- The probability of rolling a 0 is \( p = \frac{1}{10} \).
- The expected number of rolls to get the first 0 is \( \frac{1}{p} = 10 \) (a classic geometric distribution result).
Intuitive Insights for Interview Success
In Jane Street interviews, clarity of thought and the ability to generalize are highly valued. Here’s how to impress:
- State your assumptions: Clearly define the payoff rule as soon as the question is posed.
- Lay out the mathematical steps: Use the expected value formula explicitly and show all substitutions.
- Generalize your findings: Discuss how the formula would change for an n-sided die, or for other payoff functions.
- Discuss risk: Briefly mention variance and standard deviation, showing awareness of risk, not just average outcome.
- Anticipate extensions: Prepare for follow-ups involving conditional probabilities, repeated trials, or more complex payoffs.
Applications in Quantitative Finance and Beyond
Expected value calculations aren’t just for dice and games—they are at the heart of quantitative finance and risk management. Examples include:
- Option Pricing: The expected value of a derivative’s payoff under a risk-neutral measure determines its fair price.
- Portfolio Management: Portfolio expected return is a weighted sum of asset returns, similar to the die example.
- Trading Strategies: Assessing the long-term profitability of a trading strategy is fundamentally an expected value calculation.
- Insurance: Premiums are set based on the expected payouts to policyholders.
Common Pitfalls and How to Avoid Them
Even experienced candidates sometimes make mistakes in these seemingly simple problems. Watch out for:
- Omitting zero as a possible outcome: For a die labeled 0–9, including 0 is crucial.
- Incorrect sum formulas: Remember the sum from 0 to n-1 is \( \frac{n(n-1)}{2} \), not \( \frac{n(n+1)}{2} \).
- Misapplying the probability: Each outcome’s probability must be considered, especially if the die is not fair or faces are not equally likely.
- Neglecting variance: In risk-sensitive environments, always mention or compute the variance.
Practice Problems
To strengthen your understanding and impress in interviews, try solving these related problems:
- If a fair 10-sided die is rolled, what is the expected value if your payoff is $2 for rolling an odd number and $0 otherwise?
- If you roll two 10-sided dice, what is the expected value of their sum?
- What is the variance of the sum of two independent rolls?
- Suppose the die is biased so that rolling a 0 has probability 0.2 and all other numbers have equal probability. What is the expected value of the outcome?
- If the payoff equals the absolute difference between the number rolled and 5, what is the expected value?
Solutions to Practice Problems
- Payoff: $2 for odd numbers (1,3,5,7,9):
5 odd numbers: probability = 0.5. Expected value = \( 2 \times 0.5 = 1 \). - Sum of two dice:
Each die has expected value 4.5, so sum = \( 4.5 + 4.5 = 9 \). - Variance of sum:
Variance of one die = 8.25 (from above). For independent dice, variance adds. So total variance = \( 8.25 + 8.25 = 16.5 \). - Biased die (P(0)=0.2, P(1–9)=0.088888... each):
Let \( p_0 = 0.2 \), \( p_{1-9} = \frac{0.8}{9} \approx 0.088888 \).
\[ E[X] = 0 \cdot 0.2 + \sum_{k=1}^{9} k \cdot \frac{0.8}{9} \] \[ \sum_{k=1}^9 k = \frac{9 \cdot 10}{2} = 45 \] \[ E[X] = 0 + 45 \cdot \frac{0.8}{9} = 4 \text{ (since } 45 / 9 = 5, 5 \cdot 0.8 = 4 \text{)} \] - Payoff = |X – 5|:
Compute for each outcome and average:Sum = 5 + 4 + 3 + 2 + 1 + 0 + 1 + 2 + 3 + 4 = 25.X |X–5| 0 5 1 4 2 3 3 2 4 1 5 0 6 1 7 2 8 3 9 4
Expected value = \( \frac{25}{10} = 2.5 \).
Summary Table: Common Payoff Rules for a 10-Sided Die
| Payoff Rule | Expected Value |
|---|---|
| Number rolled (\(X\)) | 4.5 |
| Double the number (\(2X\)) | 9 |
| 10 minus number (\(10-X\)) | 5.5 |
| Square of number (\(X^2\)) | 28.5 |
| 1 if even, 0 otherwise | 0.5 |
| 1 if \(X>5\), 0 otherwise | 0.4 |
| Absolute difference from 5 (\(|X-5|\)) | 2.5 |
Conclusion
The Jane Street Quantitative Researcher Interview Question: Expected Value of a 10-Sided Die Game is a fundamental yet versatile problem that tests your grasp of probability, expected value, and mathematical reasoning. By thoroughly understanding how to enumerate outcomes, apply the expected value formula, and adapt to different payoff structures, you can confidently tackle this and similar interview challenges.
Remember, the key steps are:
- Define the sample space and payoff clearly.
- Apply the expected value formula with proper probabilities.
- Generalize your approach for any n-sided die or arbitrary payoff function.
- Consider risk (variance) in addition to average outcomes.
- Practice with variations to build intuition and speed.
With these principles, you’ll not only ace the Jane Street interview but also strengthen your quantitative toolkit for a successful career in finance, data science, or any analytic field.
