2024 Citadel Quant Trading Interview with Analysis from Real Quants

Breakdown of the interview:
0:00 You work at a shoe factory, and you’re working on creating boxes with pairs of shoes. Currently in front of you, imagine there are 3 pairs of shoes (for a total of 6 individual shoes) with the following sizes: 2 size 4s, 2 size 5s, 2 size 6s. The factory defines an “acceptable” pair as 2 shoes that differ in size by a maximum of 1 size — so a shoe with size 5 and a shoe with size 6 would count as an “acceptable” pair. If you close your eyes, and randomly pick 3 pairs of shoes, without replacement, what is the probability that you end up drawing 3 acceptable pairs?
0:43 The candidate asks clarifying questions
1:17 The candidate breaks down the question and starts brainstorming solutions
2:06 Our instructor analyzes the candidate's initial response to the question and points out what he did well
2:54 The candidate walks through the methodology for his solution, and solves the question correctly.
10:10 The interviewer asks the second question. Say you're flipping a fair coin until you obtain the first H. If the first H occurs on the k'th flip, you're given k balls. We're going to randomly put these k balls into 3 bins, labeled 1 2 and 3. Find the probability that none of these 3 bins end up empty.
10:34 The candidate dissects the question and asks clarifying questions.
11:03 The candidate works through some examples and logically breaks the question down to answer the question effectively.
17:55 The candidate has answered the question correctly, and now summarizes his approach.
20:37 Our instructor breaks down the approach the candidate used and whiteboards the fundamental probability theory behind this question.

[AndrewHa-]

I took y'alls program last year and landed a quant internship this summer at a tier 1. Def the best interview prep content out there.

[stevecase]

"Look at him! That's my quant. My quantitative! My math specialist. Look at him. Do you notice anything different about him? Look at his face...look at his eyes! His name is Yang. He won a national math competition in CHINA and he doesn't even speak English! Yeah, I'm sure of the math..."

- Jared Vennett (the Big Short 2015)

[InfiniteQuest]

It helps to have such a messy white board in the background to give the illusion you use it a ton.

[bigchonkers]

these quant dudes are insanely intelligent, im about to take a part time masters in quant finance and the notes are looking tough af

[HotPepperLala]

I found another way that I used. Basically for the shoe problem if there is a difference of 1, you can take 2/6C2 and for the same size you have only 3 outcomes so it is 3/6C2 and they are both equal to 2/15 and 3/15 respectively. Their sum isj ust 5/15 = 1/3

[lorcanoconnor]

More elegant way to do the last question is to view each toss as going to bin A, B, C or ending the game with probss 1/6, 1/6, 1/6, 3/6. wlog A gets the first ball. P(B empty) = P(fill B before ending) = 3/4. P(B, C both empty) = 3/5. Inclusion-exclusion -> P(B or C empty) = 2*3/4 - 3/5 = 9/10.

[blindyogi]

He was supposed to close his eyes and answer the probability!!

[AYUSHSHUKLA]

Total cases = 6!. We can select (a1, c1) as one of the pairs. This would give us 3*2*4! ways to select atleast 1 invalid pair. Similarly, (a1, c2) would give us 3*2*4!. This would give a total of 2/5. Now, we are only left with 2 conditions that have not been taken into account and where there is possibility of invalid pairs being selected: (a1, b1) and (a1, b2). In each of these conditions we would get 4*4! ways to select at least 1 invalid pairs. With (a1, b1) we must have either (a2, c1) or (a2, c2). Similar cases would be there for (a1, b2). Thus, we get (2*4*4!)/6! = 4/15. Only valid pairs probability = 1 - (2/5 + 4/15) = 1/3

[Amylin]

Thanks for making this video! This is super helpful.

[secretnobody]

Maam i just wanna work at McDonald's as part time

[jacksdu]

First question: 1/5 * 2/3 + 1/5 * 1/3 = 1/5 (left and right shoes).
3 pairs of 6 shoes, no replacement. Fix the first shoe, you have 5 options left. Only 2 of those make a valid pair, and once chosen fist valid pair, remaining has 2/3 and 1/3 prob for each choice

[TheQuantGuide]

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[BisonsTrackClub]

does the quant think shorting GME is the play?

[ShubhamKumar-gdsn]

Look at him, that's my quant, my quantitative.

[mtalHalide-rtfz]

This seemed like a really weird way to say how many invalid sets can be made and whats the probability randomly selecting a valid set from all possible sets. The selection bit trips people up a lot. It doesnt matter because everything gets paired. Its really just asking of all possible sets of {445566}, imagine all the valid {(45), (54), 66)} and invalid sets {(64), (64), (55)} (*2 ways this could happen*) - were written on a scraps of paper and thrown in a bag, whats the chance of picking out a scrap that had a valid set, or what is the probability of not picking a scrap that has (46) or (64) in any of the 3 pairings. If the bag had every variation of A1A2.. how many scraps would have AC pairings in any of the 3 pairs (A1C1, A1C2, A2C1, A2C2).

[TopCygamer]

For the first problem the answer can not be 1/3 because the total acceptable combinations is 11 and the total combinations is 15 so the correct answer should be closer to 1 than to 0. The correct answer is 11/15.

[rozaepareza]

Another way to do the first problem: there are really only 15 unique ways of choosing pairs. Let's label the shoes 4a, 4b, 5a, 5b, 6a, 6b. First we choose the partner for 4a; there are 5 choices. Now there are 4 shoes left. Now take the first one of those and choose it's partner; there are 3 choices. Now there are 2 shoes left, so there are no more choices to make. 5*3=15. So there are few enough possibilities to just think through them all. If 4a is paired with 4b, we are guaranteed to succeed, so that is 3 ways to succeed, depending on how the other shoes are paired. If 4a is paired with 5a, then 4b must be paired with 5b. And if 4a is paired with 5b, then 4b must be paired with 5a. That is 2 more ways to succeed. If 4a is paired with 6a or 6b, we have already failed. So there are 5 ways to succeed out of 15, or a 1/3 chance.

[zondberg]

I solved q1 by realizing that left shoes 4l, 5l, 6l match with the right shoes only when 6r follows 4r - 4r5r6r, 5r4r6r, 4r6r5r. So the chance of getting three acceptable pairs is 1/2. Where did I go wrong?

[andrewjones]

the solution to the first problem i can understand, but no way would i be able to do it under time pressure;

you just count all possible 6 shoe configurations, 6! in total (720) and since they are all equally likely you just count the valid ones. But once you start counting i feel like with the stress of an interview itd be very easy to mess it up and get nervous and lose track of what youre doing.

[mitchkiss]

I got to admit, I'm confused on the last question with flipping a coin and bins being empty.