A Working List of Quant Finance Interview Questions
Introduction
This series is part of my passive preparation for interviews in the quant finance sector. I aim to publish a weekly post featuring problems, brain teasers, or theory questions. These will cover topics from probability and counting to algorithms and machine learning, compiled from diverse sources including Glassdoor quant research interview questions, Quant Questions, LeetCode, HackerRank, and books like “A Practical Guide to Quant Finance Interviews” and “Heard on the Street”.
Vision
The aim of this series is to engage like-minded individuals passionate about quant finance, offering a platform to test their skills and share problem-solving approaches. The vision is to evolve this series into an email newsletter, should there be sufficient demand.
Weekly Problems
So, without further ado, let’s dive deep into this week’s 5 problems, focusing on testing your skills in probability and counting.
Problem 1
Let 20 people, including exactly 3 women, seat themselves randomly at four tables (denoted 𝐴, 𝐵, 𝐶, 𝐷) of five persons each, with all arrangements equally likely. Let 𝑝 be the probability that no women sit at table 𝐴. Find 𝑝. Please round your answer 𝑝 to four decimal places
Answer 1
All possible arrangements in all four tables are 20! without any restriction. Now, given that table A should not have any women, the possible arrangements for table A are (17*16*15*14*13) — basically enumerating all possible arrangements excluding women, which are 3 in total-. Moving forward to tables B,C and D the possible arrangements are the remaining number from table A plus the available women (3), thus it will continue as (15 * 14 * 13 …… * 1) = 15! Therefore, the final equation for finding the probability of no women sitting at table A is:
Problem 2
There are 2 drawers. The first drawer contains only black balls. The second contains 50% black balls and 50% white balls. There are an equal number of balls in each drawer. I pick a ball at random and it is black. What is the probability that the ball came from the first drawer?
Answer 2
This is a classic problem testing our knowledge on Bayes Rule and the Law of Total Probability. Translating the problem into a mathematical formula we are asked to determine:
Applying the Bayes Rule that is defined as:
We end up with:
And utilizing the Law of Total Probability:
Now we know all the numbers, so it’s just a matter plugging in the correct values:
Problem 3
(Jane Street Capital)
What’s the probability that the NBA finals go to game 7? The East team has probability p of winning each game.
Answer 3
In order for a series to end up in game 7, each team needs to have 3 wins. Assuming that we are looking the problem from the East team’s perspective a possible combination of 6 games in order to end up in a game 7 is WWWLLL. However, that’s not the only one case where we will end up in a game 7. Another possible scenario would be LLLWWW. Generalizing this pattern we can see that all possible combinations for ending up in a game 7 are 6C3. Thus, the probability of NBA finals going to a game 7 would be:
Problem 4
(Morgan Stanley and Goldman Sachs)
We start with 1 amoeba. It and every subsequent amoeba can
equiprobably and independently split to either 0 amoebas (dies), to 1,
or to 2. What is the probability of eventual extinction of the
population?
Answer 4
This problem becomes easier once we explicitly name all possible events
- A: the event that the population goes extinct
- B0: the event that amoeba dies(0)
- B1: the event that amoeba remains the same (1)
- B2: the event that amoeba doubles (2)
So we can define P(A) utilizing the Law of Total Probability:
We know from the problem statement that all scenarios are equiprobable thus P(B0) = P(B1) = P(B2) = 1/3. Moreover, it’s important to note that there is no overlap between the events, thus P(B0) + P(B1) + P(B2) = 1 and thus they cover the whole sample space. Now, delving deeper into the equation above we can tackle each conditional probability seperately:
- P(A|B0) = This is the probability of eventual extinction given that the amoeba dies (0). This is trivially 1.
- P(A|B1) = This is the probability of eventual extinction given that the amoeba stays the same (1). Staying the same means that in the next round/era it will have the same choices again of B0, B1, B2, thus it becomes apparent that P(A|B1) = P(A)
- P(A|B2) = This is the probability of eventual extinction given that the amoeba doubles (2). This means that in the next round there will be two amoebas having choices B0,B1,B2, thus it becomes apparent that P(A|B2) = P(A) * P(A). Putting everything together we get:
So, P(A) = 1 which means that eventual extinction is a certainty eventually.
Problem 5
(DE Shaw)
A family has exactly two children. At least one is a boy. What’s the probability that both are boys?
Answer 5
So, the original sample space for a family that has two children is {BB, BG, GB, GG} where B = Boy and G = Girl. We are asked to find P(BB|One is a boy). With the information above we can deduct that the sample space now reduces to {BB, GB, BG}. By simple observation we can conclude that this probability is 1/3.
Conclusion
This is it for this week! We worked on various interview questions testing our fundamentals in probability and counting. Please feel free to engage with this series by commenting other possible solutions on the problems and letting me know questions or topics that you would like to be present on future posts!
See ya next week!
