r/learnquant

▲ 0 r/learnquant+1 crossposts

Citadel Securities for incoming high school freshman

Hi guys my lifelong goal is to become a quant researcher at a top firm like citadel securities. Currently, I am an incoming high school freshman and I have been grinding AMC 10/AIME and F=ma/USAPhO, math and physics olympiad, and my target school is MIT. Ik what skills are needed for the interview, but I’m also going to be majoring in physics just in case quant doesn’t work out for me. I also have very good network with people from other quant firms. What should I do now to get a job at my dream firm?

reddit.com
u/Sad_Tangerine_9069 — 4 days ago

Three random numbers from 0 to 1. What's the expected value of the biggest? [Probability · Easy]

Draw three numbers independently and uniformly at random from the interval [0, 1]. What is the expected value of the largest of the three?

Drop your answer with your reasoning in the comments. Solution tomorrow.

Hint: >!don't just average 1/2 with itself. the maximum is pulled upward toward 1.!<

reddit.com
u/anykash — 7 days ago

📄Quant Interviews Resume Megathread (Actual Quants Will Review It)

Hey all,

Drop your resume's below in the following format:

Location: USA/UK/EU/etc

Graduation Date: June 2029

Degree + School tier: e.g. BS Math, target/semi-target/non-target (people can keep the school anonymous if they want)

Preferred Role: QT/QR/QD/QA

Targeting: internship or full-time, and which cycle (Summer 2027 etc)

Relevant coursework/skills: e.g. probability, stochastic calc, C++, Python

Experience/projects: 1-2 lines

What you want feedback on: e.g. "is my project section strong enough," "am I cooked for QR without a PhD"

Notes: anything else

Resume PDF/JPEG

Important: Please redact all personal info!

reddit.com
u/Select-Angle-5032 — 8 days ago

Do quant interviews test whether you can do data analysis in python?

It seems like everyone focuses only on math, stats, and LeetCode-type questions, or at least that is the impression i get everytime I visit some quant questions site. Are more relevant skills, like whether you can actually code? Or at least do projects and stuff tested at all?

reddit.com
u/Fearless_Screen_4288 — 7 days ago

Is Anyone Else Struggling with AI-Generated Code?

Does anyone else run into this problem when using AI (especially Claude) for quantitative finance projects?

I use AI mainly to write code for my projects. The frustrating part is that even relatively simple ideas, sometimes involving only basic mathematics, quickly turn into huge codebases full of bugs, inconsistencies, and unexpected issues.

It often feels like the model can't keep all the different parts of the project connected. Fixing one problem creates two more somewhere else, and after a while you're stuck in an endless debugging loop.

Instead of converging toward a working implementation, the project seems to drift further away from a correct solution with every iteration.

It's like an infinite loop where each project never sees a proper end.

Is this a limitation of current AI coding models, or am I approaching these projects in the wrong way? How do you structure your workflow to avoid this?

reddit.com
u/Correct_Hedgehog_612 — 7 days ago

On average, how many die rolls to see every face at least once? [Probability · Easy]

You roll a fair six-sided die over and over. On average, how many rolls does it take until you've seen every face at least once?

Drop your answer with your reasoning in the comments. Solution tomorrow.

Hint: >!the answer is bigger than you'd guess. the final missing face alone takes 6 rolls on average.!<

EDIT

The solution:

>!The expected number of rolls is 14.7.!<

>!Break the process into stages by how many distinct faces you've already collected. Once you've seen k different faces, any roll lands a new face with probability (6−k)/6, so the number of rolls to get the next new face is geometric with expected value 6/(6−k). Sum these waits across all six stages: 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 1 + 1.2 + 1.5 + 2 + 3 + 6 = 14.7. Notice the last term: when you're down to one missing face, each roll only has a 1/6 chance of being it, so that final face alone takes 6 rolls on average, more than the first four faces combined. That heavy tail is why most people lowball this at 6 or 12. The general formula n·Hₙ is the classic Coupon Collector result, and it shows up everywhere from cache-miss analysis to estimating how long an A/B test needs to run to cover every variant.!<

More puzzles like this at myntbit.com

u/anykash — 9 days ago

Why is Quant so hard to break into?

There are a lot of you that keep asking how to break in, what step does it take, and even the body answers it the same way: it's a hard field so study harder and get smarter. People act like it's some impossible secret club, but the difficulty is honestly pretty simple to explain, it's a tiny number of seats, absurd money, and almost no barrier to applying, so the funnel is brutal. 

But for this reddit post I’ll dive deeper into why this is the system we work with. A top firm might take a few dozen new grads a year and get tens of thousands of applications from basically every strong math, CS, physics, and stats student on earth, plus PhDs, plus people leaving other finance and tech jobs. When the comp is $300-500k out of school, everyone smart enough to have a shot takes the shot, so the bar isn't just are you good. It's that you’re better than everyone that is good. 

However, I think it's understated how good you have to be. The skills actually are demanding, you need real probability and stats, fast quantitative reasoning under pressure, and for the dev side serious low-latency engineering, and the interviews are designed to be hard to fake your way through. Brainteasers, mental math, market-making games, and live coding all exist specifically to filter hard. 

But here's the part that gets lost in the doom posting: hard to break into is not the same as gated or rigged. There's no required pedigree, no secret handshake, the application is open to anyone and the interviews are pretty meritocratic, they mostly test things you can actually study and get better at. The reason most people don't make it isn't that they were locked out, it's that the prep is long and specific and most people start too late, apply too few places, or treat it casually. The ones who get in usually started early, drilled the green book and other prep for months, did a competition or two, and applied to posting weekly rather than waiting. 

So all in all, it's hard, but it's actually-hard, not gatekeeping-hard, which means the difficulty is something you can attack with time and deliberate prep rather than something you're either born into or not.

As I said before, I don’t want this to be a doom post, but just speaking my mind of what is required to get into this field. You gotta do project and network like crazy. But other than that you just gotta optimize resume screens to get the shortlisted.

Finally, leaving you with an exercise, put yourself in the shoes of the employer, would you hire the person sitting across the table. If that answer is not yes with your eyes closed then… you get the point.

reddit.com
u/Wonderful-Bunch-3343 — 12 days ago

5 people check their hats, get them back at random. Odds nobody gets their own? [Probability · Easy]

Five guests check their hats at the opera. The clerk loses every tag and hands the hats back in a uniformly random order, with every possible arrangement equally likely. What's the probability that not a single guest gets their own hat back?

And the follow-up: as the number of guests grows very large, what does that probability approach?

Drop your answer with your reasoning in the comments. Solution tomorrow.

Hint: >!the probability barely changes whether it's 5 guests or 5,000. that's the real surprise.!<

EDIT:

The solution

>!The probability is 44/120 = 11/30 ≈ 0.367, and as the number of guests grows it approaches 1/e ≈ 0.368.!<

>!An arrangement where nobody gets their own hat is called a derangement. To count them, use inclusion-exclusion: start with all 5! = 120 arrangements, subtract the ones where at least one person gets their own hat, add back the overlaps you double-subtracted, and so on. That alternating process gives the derangement formula !n = n!·(1 − 1/1! + 1/2! − 1/3! + … + (−1)ⁿ/n!). For n = 5 that's 120·(1 − 1 + 1/2 − 1/6 + 1/24 − 1/120) = 44, so the probability is 44/120 = 11/30. The series in the parentheses is exactly the expansion of e^(−1), so as n grows the probability converges to 1/e. The surprise is how fast it gets there: it's already about 0.375 at n = 4 and 0.367 at n = 5, so adding more guests barely moves it, and it never decays toward 0. The usual traps are flipping the inclusion-exclusion signs, or assuming more guests makes a full mismatch less likely when it actually stabilizes almost immediately.!<

More puzzles like this at myntbit.com

u/anykash — 14 days ago

An ant random-walks the edges of a cube. Expected steps to reach the opposite corner? [Random Walks · Medium]

An ant sits at one corner of a cube. At every step it picks one of the three edges meeting its current corner, uniformly at random and independently each time (so it's allowed to backtrack), and walks along it to the next corner.

What is the expected number of steps before it first reaches the corner diagonally opposite its start?

Drop your answer with your reasoning in the comments. Solution tomorrow.

Hint: >!don't write eight equations. the cube's symmetry collapses it to three.!<

EDIT

The solution:

>!The expected number of steps is 10.!<

>!The trap is to set up an equation for all eight corners. Symmetry cuts that to three. Group the corners by their distance from the target corner D: the start A is distance 3, there are three corners at distance 2 (call the class B), three at distance 1 (class C), and D itself. By symmetry, every corner in a class has the same expected steps to reach D, so let a, b, c be those expected values.!<

>!Now read off the moves. From the start, all three edges lead to distance-2 corners, so a = 1 + b. From a distance-2 corner, one of its three edges goes back toward the start (to A) and two go forward to distance-1 corners, so b = 1 + (1/3)a + (2/3)c. From a distance-1 corner, two edges go back to distance-2 and one reaches D, so c = 1 + (2/3)b.!<

>!Solve the system: substituting gives b = 3 + (3/5)a, and since a = 1 + b, we get a = 4 + (3/5)a, so (2/5)a = 4 and a = 10. The whole problem cracks open the moment you collapse the eight vertices into four symmetry classes instead of grinding eight equations.!<

More puzzles like this at myntbit.com

u/anykash — 13 days ago
▲ 6 r/learnquant+1 crossposts

PLEASE HELP IM A BEGINNER

So im a 3rd year undergrad student from BITS pilani, in india.
my past 2 years havent gone well at all and i have bad cgpa.
i can pull my cgpa up a bit, but i have come to accept that breaking into quant right after college is as good as not possible, but i still want to eventually move into quant.

if someone has experienced a similar situation please please help me.
im willing to learn whatever and put in as much effort as i possibly can to get into quant in the next 4-6 years.

i need a mentor to guide, any help is appreciated, even if you have a little bit of idea, please give that advice.
i need anything to start with

EDIT:- Iforgot to mention but im pursuing B.E in Mathematics and Computing, so i think that i do have a shot at quant.

reddit.com
u/Electronic_Sound7641 — 14 days ago