2024 Citadel Quant Trading Interview with Analysis from Real Quants

[@AndrewHa-23]

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.

[@TheQuantGuide]

Thanks for the kind note, and glad to hear it. Congrats! - Alex, Quant Researcher

[@blubanister]

Hello, where can I find the Quant Program? Thank you.

[@TheQuantGuide]

@@blubanister Visit www.quantblueprint.com to get immediate access!

[@UCHECHI.]

Hi please can you tell me how to prep for an internship for such a position. Would love to start learning on my own

[@stevecase6168]

"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)

[@mykeegetsit]

😂 my thoughts exactly

[@MVN_29]

Broooo😂😂😂😂😂😂😂😂

[@BeyondAbstraction]

Ted Jiang : [to camera] Actually, my name's Jiang and I do speak English. Jared likes to say I don't because he thinks it makes me seem more authentic. And I got second in that national math competition.

[@lucngadi9482]

😂😂😂

[@846nick]

"thats pretty racist" - mark baum

[@jrbn4026]

yea ion think ima get the job

[@ramsesp5009]

"Ion" and "ima" is pretty much a giveaway

[@jrbn4026]

@@ramsesp5009 yeah the comment was a joke, you can't be that stupid to not detect that

[@lipca]

@@ramsesp5009 RACIST

[@BusinessSimon-t5j]

@@ramsesp5009 wdym?

[@aliceryan7053]

​@@ramsesp5009bro thinks he's the citadel hiring manager 😂😂😂

[@InfiniteQuest86]

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

[@nki111-pa]

Especially with the multi colour markers😂

[@pietdenolder]

I am flattered that the algorithm gave me this video but I'm really more of a memecoin person. Thank you though.

[@bigchonkers47]

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

[@gabestrenk5471]

you got this bro. good luck!

[@Amylin20112]

Thanks for making this video! This is super helpful.

[@TheQuantGuide]

Glad it was helpful!

[@mybeautifuldarktwistedfagotree]

​@@TheQuantGuide Are you on linkedin?

[@AMA_RILDO]

I’m still listening at the 1st question for the 21th times, like convincing myself that if I understand the question I can instantly find the answer

[@nu-tral]

there is a much simpler way of approaching it using the naive definition of probability. How many possible "good" pairings are there? well we have the following: { (4,4), (5,5), (6, 6), (5, 6), (5 , 4) }. Since we're only looking at unique sets, our denominator should be the total possible number of ways we can make distinct sets of 2 from 6 objects. This is by definition the binomial coefficient (the choose operator). We have 6 Choose 2 = 15. 5/15 = 1/3

[@collinspo]

​@@nu-tralgreat explanation. Can you do same with the 2nd question. Much appreciated

[@jonatansvensson8346]

For the last problem, introduce two states. One where we have balls in one of the buckets and one where we have balls in two of them. We know before flipping the coin that we will have at least one ball so put it in any bucket, we will then be in the first state described above. Now the game can be formulated as follows, whenever we toss a tails randomly add a ball to one of the three buckets and whenever we flip a heads we stop and see if all three buckets are full. The probability of winning this game (the game ending with all buckets full) is equivalent to the formulation in the video. We now see that the possible moves from state 1 are: to losing, back to state 1 or to state 2 and possible transitions from state 2 are: to losing, back to state 2 or winning. Now assign probabilities, p1 and p2, of us winning given that we are in state 1 and 2 respectively (this makes sense as the game has no memory, i.e the probability of winning from a given state is always the same). We know that p1 = 1/2*0 + 1/2*1/3*p1 + 1/2*2/3*p2. (Half of the time we flip a head and lose (only 1 bucket is full), the other half we add a ball randomly to our buckets. 1/3 of the time this will be to the bucket that is already full and 2/3 into one of the other 2 moving us to state 2). Similarly we get that p2 = 1/2*0 + 1/2*1/3 + 1/2*2/3*p2 (here one could talk about a winning state but it's not really necessary. The point is once we reach it we know we will win the game) implying p2=1/4. From the first equation p1 =2/5*p2 and we finally get p1=1/10. This might seem complicated when formulated in a comment like this but the methodology is really powerful and can be used for a wide range of problems. Almost always when you get an infinite sum in your solution the problem could've been formulated with states and implicit formulas would've given you the answer in a much neater fashion.

[@peeper2070]

We were made to hunt boars and die at 40. How do I uninstall this dlc

[@yeetboi268]

start by counting your hunting success rate and maybe everything would start to make sense

[@full-timepog6844]

xD i think you can still do those things

[@geethanithi4171]

same ugh

[@thang1144]

Born to jest, forced to joust kinda energy

[@Albedo_ase]

that's my quant

[@airheadmrng]

MY QUANTITATIVE

[@jrbn4026]

@@airheadmrng MY MATH SPECIALIST

[@moneekdilawari4508]

LOOK AT HIS EYES 😂

[@tushrchaudhary]

YOU NOTICE ANYTHING DIFFERENT ABOUT HIM

[@ajinkyanaik4870]

I'll give you a hint, his name is Yang !

[@Hello-pl2qe]

Jesus no wonder Ive lost all my money

[@sebastiansantolalla9184]

lmaoooo

[@ericjondahl9418]

Lmao!!😂

[@Rainy_Day12234]

Citadel makes most of its money buying then front running retail order flow. No math needed.

[@nedhenry639]

@@Rainy_Day12234ur stupid if u think this is true

[@doniyorakramov706]

@@Rainy_Day12234 Can you explain this in detail, please?

[@lorcanoconnor6274]

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.

[@Ten_Eleven_Twelve_Thirteen]

I dont doubt your logic, but how do you arrive at the 3/4 and 3/5 probs?

[@lorcanoconnor6274]

@@Ten_Eleven_Twelve_Thirteen P(event X happens before event Y) = P(event X happens | event X or event Y happens) = P(X) / [P(X) + P(Y)], so e.g. the 3/4 comes from 3/6 / [3/6 + 1/6] = 3 / [1+3]. I think I meant to write P(end before filling B) in the above btw.

[@justlogical9846]

How does the answer change if we think that balls are identical? Or does it change at all?

[@thomasstanton1111]

You could say there are four possible arrangements with three producing those 3 acceptable pairs (ap). Therefore, 75% of the time you will get a pairing that satisfies the ap requirement. Another way to explain this is using the 1/3 fraction they used in the film, however I would write it 3/1. This is because I consider this like a step-up transformer formula. There are 3 conditions that work and 1 that doesn't. The 1st condition (1) is the primary coil and the 2nd condition (3) is the secondary coil. A step up transformer boosts power and because there are 3 positive outcomes to 1 negative outcome we know it is more likely to get the ap we want. Always divide the secondary coils by primary for the step-up calculation. So, 3/1 = 3. It all depends on how you look at things.

[@secretnobody6460]

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

[@artvandalay13]

i already do. You can join my team. Can you handle a spatula? Then you are hired. That's our only requirement.

[@Mimi1356-s4m]

*Wendy's

[@blindyogi4997]

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

[@ssbsnb1200]

Real quants don't blink, answer immediately, and then slap the table to summon more questions

[@YlliNici-e7n]

Q1 can be done in an easier way. the number of possible pairs is we pair the first shoes with 5 shoes, then the next one with 3 shoes and last 1 with the remaining shoe. (so there are5x3x1=15 total pairs). now the allowed ones are (using the provided notation) a1a2,b1b2,c1,c2 and then a1a2,b1c1,b2c2 or a1a2,b1c2,b2c1 (and for these last 2 we multiply them by 2 so that we get the cases where a's and b's mix). so in total 5/15 = 1/3 are allowed.

[@sparshailawadi7129]

Are we ignoring the fact that each left shoe must be paired with only a right shoe?

[@ashugoyal8919]

one of the approach for the first problem (maybe the easiest approach for a generalised case and this particular imo) p(n) respresent to number of ways to pair 1,2,3..,n among 1,2,3...n such that they all are acceptable, so p(n) = p(n-1)/(2*n-1)+2*p(n-2)/(2*n-1)(2*n-3) because there is a 1/n-1 probability to pair n->n(the remaining independent probability would be p(n-1)) and 2/(2*n-1)(2*n-3) probability to pair n->n-1 and n-> n-1 (remaining probabilty depends on independent probability of p(n-2)) the base case for this reccurence woould be p(0) = 1 p(1) = 1 this would result in p(3) = 1/3 which is the answer hope this helps!!!!

[@jacksdu13]

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

[@DmitryPesegov]

What always helps me when dealing with events is to always imagine a branching graph of possible transitions over all combos and resulting outcomes. Kinda limited markovian chain. When you see that graph it much easier to understand the simple idea of arithmetic for probability finding.

[@BisonsTrackClub]

does the quant think shorting GME is the play?

[@UnorthodoxKnox]

"Question 1: You find yourself in a position where you need to short $GME due to market demand from peasant retail investors, and there are no shares available in the market to open short positions on legally to offset this buy pressure, but this must be done otherwise our entire business model will be liquidated. Do you know how to hit 'F3' on a keyboard?"

[@kurtisbobke9000]

GME doesn’t trade based off fundamentals. That is a stupid question.

[@BisonsTrackClub]

@@kurtisbobke9000 tick tock

[@ps-dh8ef]

Lmfao

[@yeetboi268]

@@kurtisbobke9000 quant is not stupid

[@m3talHalide-rt2fz]

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).

[@bobthebuilder9416]

Yeah really can't stand people in the industry they're over paid & can't stand people not in the industry cause they have no desire to learn the basic arithmetic that would allow them to escape their indefinite indentured servitude.

[@zzz-oc4fe]

@@bobthebuilder9416 can you elaborate please?

[@ShubhamKumar-gd4sn]

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

[@MrS690]

That is kinda racy(st)

[@justjason888]

this video just called me dumb for a solid 23:22 mins

[@scamperooniespanker8736]

i worked as a quant at optiver for 3 years in the us office in chicago. These questions are entirely expected at an interview. You won't be doing this shit for work tho. These foundational skills help give you the ability to seek out patterns and signals within data but at the end of the day, your whole job wont just be maths -- you'll have actual projects to complete. Honestly just make sure you can program pretty decently -- especially with asynchronous programming, data science libs (like numpy, pandas, etc), and the like

[@average391]

Like...?

[@arachnid4910]

I remember reading Michael’s Lewis’ book about SBF and FTX. The section about SBF’s interview at Jane street was pretty fascinating. Quant’s have an interesting mind.

[@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

[@MichaelHamilton-d4k]

Yeah i did something similar and got 1/3rd too. The jane street questions seem way more absurd than this.

[@phageofficial404]

Does being a Quant Trader require a CS degree or a MBA? Quant trader develops complex trading algorithms, so is it CS that is mandatory or since it involves heavy quant and mathematics, it needs MBA?

[@syiqahsuratman7736]

Now Im not sure if it’s math or English that I’m terrible at

[@AYUSHSHUKLA2793]

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

[@MindMorphMedia]

Citadel will hire him and make him naked short GameStop Stock

[@markfaynboym5419]

I'm a bit rusty on these but dont we need to exclude k = 1 and k = 2? If we have less than 3 balls then the probability of having an empty bin is 100%. So in the geometric series we should exclude the first 2 values in our summation

[@WallStreetVikings]

Closed my eyes and came up with 1/3 as the answer to the first question. When he started talking, I thought to myself wow I’m way off I didn’t do any of that. Then he finalized the answer at 1/3. Either he over complicated it, or I got lucky.

[@michaelscott33]

Nope, it’s just more of that “show your work” bullsh*t.

[@SandraNelson-r6o]

Very good video! I have USDT in my OKX wallet and I have my recovery phrase. 「pride」-「pole」-「obtain」-「together」-「second」-「when」-「future」-「mask」-「review」-「nature」-「potato」-「bulb」 How do I transfer them to Binance?

[@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.

[@Landlord_3]

Legit

[@yaboi608]

This is incorrect. First, the shoes are not replaceable, meaning after every acceptable pair is drawn the probability of drawing another acceptable pair decreases. So it cannot be closer to one than zero. The odds of 3 acceptable pairs being drawn in a row in your example with 15 pairs is ~ 32%. Math: (11/15) x (9/13) x (7/11). 2 removed from numerator and denominator every time an acceptable pair is drawn. So basically 1/3. Also, there are only 9 acceptable pairs. You are viewing this from the perspective that each individual shoe is unique. Ex: A 45 pair can have 2 variations with the first 4 shoe being with the first 5 or the second 5. This is redundant as they are the same. Imagine I laid out for you 4 pens. 2 of them are red, 2 of them are blue. Using your way of thinking, there would be 6 pairs. But really, there are only 4 distinctive pairs you could create. You are focused more on the individual pieces, rather than the overall pairs.

[@zxyjulzeeeks]

@@yaboi608 This is also incorrect. The answer is exactly 1/3. Firstly, your way of removing 2 from numerator and denominator is incorrect. Suppose the first pair picked was (4,5), then remaining shoes are (4, 5, 6, 6). These now have 4 acceptable pairs ((4,5), (5,6), (5,6), (6,6)) and 5 total pairs (and now if we had picked (5,6) then there are 0 acceptable pairs for the last choice, but there would be 1 acceptable pair had we chosen (4,5) which also shows why this subtraction would not work). I have provided my calculation below which can be verified by a simple python script: Total Number of ways to sample 3 pairs from the given 6 shoes: 6C2 * 4C2 * 2C2 = 90 Now, let us find the number of ways to sample 3 pairs such that at least 1 pair is unacceptable. The only unacceptable pair is (4,6). Number of ways to get (4,6) is 2C2 * 2C2 = 4. For the remaining 2 pairs we have 4C2 * 2C2 = 6 options. This (4,6) pair can be permuted within the 3 pairs in 3C1 = 3 ways. Thus, total is 4 * 6 * 3 = 72 such ways. But this overcounts samples like ((4,6), (4,6), (5,5)). So we must subtract number of samples where at least 2 pairs are unacceptable. In this sample we will have 2 pairs of (4,6). Ways to generate the first (4,6) is 4 as above. The second (4,6) has only 1 way to be generated. And for the last pair we have 2C2 = 1 option again. These two (4,6) pairs can be permuted in 3C2 = 3 ways. Thus, total is 4 * 3 = 12. (Technically using PIE we should add back number of ways to draw 3 samples such that at least 3 of them are (4,6) but that number is 0) Thus total number of ways to draw 3 pairs where at least one is unacceptable is 72 - 12 = 60. Thus, number of ways to draw 3 pairs where all are acceptable is 90 - 60 = 30. And the probability is 30/90 = 1/3.

[@Rakshay223]

​@@yaboi608but in quant isn't accuracy stressed on, how come you're gonna say 1/3 instead of 32%

[@ThePipojp]

Quant Trading isn't even a thing in my country, but I keep watching this 😂

[@YufanLi-b8f]

These problems feel like simple textbook exercises. I think real interview will require more clever tricks.

[@vassiliguillemin6560]

For the first question, I have an issue with the answer proposed by the student : he decides to put an order on how shoes are picked. For example, in his reasoning, he distinguishes (a1,a2) with (a2, a1). But, to me, such pairs are the same. Am I wrong ? Should such cases be distinguished ?

[@williamwarren5234]

You can count either way assuming each shoe of a size is different or not, as long as you stay consistent. Total combinations is 6!/2^3, as opposed to permutations of 6!

[@KeerikkadanJose007]

For the Q1 for shoe pairs, the provided answer doesn't look to be correct. There is a total of 90 combinations - 6C2 * 4C2 * 2C2 and 8 pairs are not accepted which are (a being left foot and b right foot) - 6a with 5a, 4a, 4b And 6b with 5b, 4a, 4b And 5a, 4a And 5b, 4b. So all the combination with these 8 pairs will have to discarded as unacceptable, which gives 8 * 4C2 * 2C2= 48 total combinations with at least 1 pair going wrong. So total acceptable combinations = (90-48)/90 = 46%

[@kjkang11]

We did this math in high school. Classic AMC / KMC questions

[@vokoaxecer]

qtπ, what's your name?

[@johnoldfield2390]

I drew a blank on the probably question I’ve forgotten the formula for probability!

[@TheQuantGuide]

Get immediate access to our #1 quant interview preparation course at www.QuantBlueprint.com and break into firms like Jane Street, Citadel, HRT, and more.

[@andrewjones9018]

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.

[@jeddyxie224]

The complement strategy the guy uses is pretty terrible. The most efficient way to do the complement strategy is to use inclusion-exclusion e.g. P(A or B) = P(A) + P(B) - P(A and B). Denote A as a1 paired with c1 or c2, denote B as a2 paired with c1 or c2. Thus P(A) = P(B) = 1/5+1/5 = 2/5. Then for P(A and B), utilize Bayes theorem in the way that is noted at the end of problem 1, e.g. P(a1 paired with c1 or c2) * P(a2 paired with remaining c1 or c2 | a1 paired with c1 or c2) which evaluates to 2/5*1/3 = 2/15. The solution then is 2/5+2/5 - 2/15 which = 10/15. Then to get P(acceptable pair), you just do 1-P(unacceptable pair) which = 1 - 10/15 = 5/15 = 1/3. Elegant solution.

[@andrewjones9018]

​@@jeddyxie224 my comment was more about myself than anything else. i dont do too well under time pressure (unfortunately, since i think its quite a good thing if you can manage it), so i was just stating that id probably fail that test. im sure theres all sorts of elegant ways of solving these problems, but in this context i think its most important to solve them within the time limit.

[@luisneuner8506]

can someone tell me why such calculations are important for trading ?

[@DamonNeal]

Because successful trading is about finding statistical edges in the market. Every professional trading system or model has probability as a key parameter

[@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.

[@mibli2935]

I think that you missed the word "random" in the task definition. You can't "chose", you toss the cube

[@usagibutt]

They didn’t miss anything. Their solution is the cleanest solution out of all the ones discussed here, including the one used by the person in the video

[@templarknight7]

@@mibli2935 he's not choosing. he just listed out every single possibility and of those possibilities, only 1/3 satisfy the condition.

[@ibrahimalssoussi7106]

@@mibli2935No she’s right this is math not philosophy

[@Weltbummler23]

@@mibli2935shes basically writing out all the cases in the bayes approach. shes right, its essentially the same as the second solution.

[@gh-sb1dy]

but what about 4 an d6 possibility of . what of the 6 isd nto the 4 and the coreALtion of sizes ar e nto the corelation of pairs

[@BlueLegacies]

What does this have to do with stocks. Ask why social media platforms were shunned and blocked people’s from talking about certain subjects.

[@bloodyblase3074]

Idk man all i know is that i can spam the F5 button the fastest 😂

[@hcasavantes]

I think they missed a pair of two lefts (or two rights) as invalid. So, if the number of permutations of 6 shoes taking 2 at a time are 15: Pick 1. "L4" "L4" "L4" "L4" "L4" "R4" "R4" "R4" "R4" "L5" "L5" "L5" "R5" "R5" "L6" Pick 2. "R4" "L5" "R5" "L6" "R6" "L5" "R5" "L6" "R6" "R5" "L6" "R6" "L6" "R6" "R6" The valid ones are those who meet the minimum size difference, but also are from a different foot. There are 7 possibilities: Pick 1. "L4" "L4" "R4" "L5" "L5" "R5" "L6" Pick 2. "R4" "R5" "L5" "R5" "R6" "L6" "R6" Which makes the probability of getting a valid pair to 46% I believe you must also have the context onto consideration.

[@mehg8407]

lol That's what I thought too. When he said factorial I was like...this guy is an idiot.

[@seunghyeoklee9043]

Exactly, that's what I was thinking. They should've used something other than shoes if they intended the question to be answered this way.

[@handavid8465]

this might be a dumb question but doesn't the probability change if the first pick if 4/6 vs 5. just because the two 5s matched doesn't mean that the next two pairs are acceptable.

[@kushyglowy8409]

Fantastic

[@aecet9652]

Legendary

[@abhinaavsaikia7833]

I wonder why he didn't used permutations and combinations formula🤷‍♂️

[@TheCheukhin]

There is always a HARD code test. Where is the HARD code test?

[@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?

[@ishangupta2380]

Yeah their answer is wrong it would either be 1/2 or 1/5 ( if we count left left combo as an illegitimate pair for our calculations)

[@MichaelHamilton-d4k]

It was obvious that it was 1 third b.c of the complement but why did he reason it using the factorials instead of just fractionalizing it? Only possibility of picking the bad pair is 2/6 then just adjust. This took 45 seconds in my head or did i just get lucky on 33% and my reasoning was faulty?

[@MorebhNYC]

should not the last question be 2/2^k?

[@mitchkiss2456]

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

[@user-qk6th5jj2p]

Maybe someone can help me with this, but the answer to the first one doesn't make much sense to me. The total number of pairs one can make from 6 objects is 6 choose 2 (6C2), which is 15. This should be the size of our sample space. Finding the number of ways that we get an invalid set is trivial, so if we have the set configured like the instructor had it at 8:16 , then we have the set {a1, a2, b1, b2, c1, c2}. What we need to know is how many ways can we pair the a's to the c's to get our number of invalid set. We can easily see that this number is 4, or computationally it could be 4 choose 2, and then subtracting the instances where we get a1,a2 and c1,c2 in order to get 4 examples of an invalid pair. It then follows that our probability for an invalid pair is 4/15, or 0.26666 for a single pair-making trial. So then our complement of this event is 15/15 - 4/15 = 11/15, aka the probability of selecting a valid pair of shoes. The probability that when performing the pair selection process 3 times, with all three being valid pairs, then ought to be (11/15)^3, as the events are independent, pairs are being selected at the same time as one another. That figure comes out to be 0.39437037037, which in my mind is the correct answer to the question.

[@harshparikh7060]

Hey, The second part of your logic would be correct if we are replacing the shoes after picking them.

[@sennsita01]

For the first question, I don't think there's a need to subtract the "double counting" part. 3*8*4! = 576 is the number of ways AT LEAST 1 pair consists of a 4 and a 6. That's already the complement of what we want. By subtracting the number of ways you can have 2 pairs of a 4 and a 6, you end up with the number of ways EXACTLY 1 pair consists of a 4 and a 6. You end up getting that 1 pair of a 4 and a 6 is not legal but 2 pairs of them are legal....

[@Jeter23]

Am i the only one that did the first question differently ? Probably incorrect but i got the correct answer. If youre asked to select 50% of the population and you have a 33% chance of each shoe than you just take 33% of 50 in this case its 66% cause they said you can go up or down a size and it would still match .66•50%= 33% 😅 quicker easier and less panicked

[@lukep3022]

Can somebody who has gone through the interviews comment on if they are really this easy?

[@oscarromero9702]

I did Jump Trading internship interview and the questiona where this kind of difficult

[@joelwillis2043]

@@oscarromero9702 these are easy

[@saltymonke3682]

He definitely won that math olympiad in china

[@jonatansvensson8346]

Seems weird not to consider that a valid pair of shoes needs to consist of one left and one right....

[@pfever]

Asian vs Asian

[@moneekdilawari4508]

What’s the point of all this when you all just gonna front run with your algos and trade routing 😂

[@RobTheQuant]

truth spoken 😁

[@rawcircoking]

Algos need math and to figure out math you need to be a problem solver and have critical thinking skills.

[@ryanyoung8886]

how do u think they write the algos and trade routing logic my guy it doesnt fall out of the sky

[@yashgaurkar8686]

Here the shoes are not explicitly mentioned to be distinct ( 2 size 6 are assumed to be different, which I don't think is mentioned or clarified ) but while solving he is just assuming that to be the case. Let me know if I am thinking it wrong or missed a part where its clarified. Also the pairs are ordered too which is again not clarified. I think the answer will change if we say that 2 size 6 are identical and the pairs are not ordered.

[@Dan_1348]

It will not

[@aaatt268]

They're supposed to be different. They're left and right shoes.

[@ChipsChallenge95]

I literally had the answer 6 seconds after the question was asked

[@phamstone7150]

is 8/15 the answer for the first question, im just calculate it in my head, sorry if im wrong

[@jeddyxie224]

The most efficient way to do the complement strategy is to use inclusion-exclusion e.g. P(A or B) = P(A) + P(B) - P(A and B). Denote A as a1 paired with c1 or c2, denote B as a2 paired with c1 or c2. Thus P(A) = P(B) = 1/5+1/5 = 2/5. Then for P(A and B), utilize Bayes theorem in the way that is noted at the end of problem 1, e.g. P(a1 paired with c1 or c2) * P(a2 paired with remaining c1 or c2 | a1 paired with c1 or c2) which evaluates to 2/5*1/3 = 2/15. The solution then is 2/5+2/5 - 2/15 which = 10/15. Then to get P(acceptable pair), you just do 1-P(unacceptable pair) which = 1 - 10/15 = 5/15 = 1/3. Elegant solution.

[@bodinliberalart2462]

Are you sure trading is really about probability? I’ve heard of high-frequency trading companies using questionable tactics like naked shorting, buying retail trader data from platforms, or even stealing orders from big traders and selling them back immediately. It’s surprising because even highly intelligent people, like those with math or physics Olympiad backgrounds working in big financial institutions, sometimes struggle to make money trading on their own. Instead, they seem to rely on their positions to front-run their company’s large orders.

[@mrsudiwala3957]

I will just touch the shoes and calculate the size or different kind of pair category to answer the question 😂

[@ojuschowdhari1]

What is the point of all these questions? You hardly apply this.

[@ruydfubiudifudiuapoupou4121]

just IQ selection, but you can learn these types of problems by doing them over and over again so it doesn't really work anymore. so now they weed out indians by letting you do stuff like 80in8

[@maxaffe3195]

all of that to underperform SPY

[@zyvex128]

Students in elementary school in asia can answer this question in 3 seconds..

[@anirudhSiwach-q1o]

We cant wear 2 left sided shoes of suitable size 😅

[@gouravkohli4573]

The efforts are great, but still i feel maximum audience didn’t get anything my man explained in the video

[@crispytastychicken]

Doesnt treating this problem as a permutations of a string type problem only works if we are choosing one shoe at a time? If we choose two shoes at a time, wouldnt this change the answer?

[@George70220]

How can you choose 2 shoes at a time in a different way than choosing 2 separately? Either way there is no replacement and you end up with a pair.

[@nathan4678]

Without replacement random draws are considered to happen simultaneously.

[@MuayThaiPai]

A students are managed by B students who are trained and directed by C students who respond to the behind the scenes "get it done" D students who work closely with the F/dropouts that own the company.

[@tonyg6334]

This is just statistics?

[@colstoun4762]

Bros first question should have been how many shoes are left foot and how many are right.

[@VR_investments]

The only word I understood was probabilities, am I quant now hahahaha

[@purplerings1969]

Interviewer: are you willing to steal from retail investors? Quant: yes Interviewer: you're hired

[@CriminalTrades]

smart money does care about retail, retail makes up a tiny portion of the total trading volume

[@AKGuerilla]

I guessed the right answer to both of these questions in like 5 seconds just using common sense and I don't know all these mathematical terms. Too bad I don't have a college degree or else I'd be tempted to apply for a Quant interview.

[@m.h.f3350]

A:"Ok, so we are going to start off with a comaprison question: What is the difference to a duck?'"

[@jerrytomas3136]

For the first one, Why couldn’t it have been as simple as taking the average of the equally likely 2 conditions? 2/6 of picking a size 5. That goes with any of the remaining 4/4 shoes so 2/6 x 4/4 = 1/3 We don’t even need to compute after this. Since we know the situations are equally likely and we want an average, the probability of one valid scenario is the probability of all (both) valid scenarios For the sake of completion: 4/6 of picking a size 4 or 6. Either size can only go with 2/4 of the remaining shoes. So 4/6 x 2/4 = 1/3 (1/3+1/3) /2 = 1/3 Would it hurt me to use this simpler method in an interview!?

[@brisso_cs]

Wrong

[@mr.vladislav5746]

How does this work - there are not 4 but 5 shoes left when one is picked. And the 4s and 6s go with 3/5 of these?

[@ttn8477]

I'm sorry but I don't think that this guy took in accountability that there's a left and a right shoe, if a left is paired with a left the pair isn't correct right ?

[@aarsethdiane853]

Given the persisting global economic crisis, it's essential for individuals to focus on diversifying their income streams independent of governmental reliance. This involves exploring options such as stocks, gold, silver, and digital currencies. Despite the adversity in the economy, now is an opportune moment to contemplate these investment avenues.

[@masangogabriel3356]

I agree, that's the more reason I prefer my day to day investment decisions being guided by an advisor, seeing that their entire skillset is built around going long and short at the same time both employing risk for its asymmetrical upside and laying off risk as a hedge against the inevitable downward turns, coupled with the exclusive information/analysis they have, it's near impossible to not out-perform, been using my advisor for over 2years+ and I've netted over 2.8million.

[@masangogabriel3356]

I definitely share your sentiment about these firms. Finding financial advisors like Monica Lisa Payne who can assist you shape your portfolio would be a very creative option. There will be difficult times ahead, and prudent personal money management will be essential to navigating them.

[@emmapriscilla6132]

_thanks FOR clearing that up, I curiously searched for Monica Lisa Payne on the internet and thankfully, I came across her my goal is to retire in 5years time.

[@ixlness]

I don’t know why I was recommended this lol, but cool I guess

[@terryschwartz4257]

Two left feet was not taken into consideration

[@ColeAnderton-pw7qp]

Can someone help me with the second probability question? the video is a little convoluted and I can't follow it. My solution is summing from k=3 to infinity : 1/2^k * (1-3k/(k+2 choose 2)) where 1/2^k is the probability that k happens and 3k is the number of ways for at least one bin to be empty and k+2 choose 2 is the number of ways to organize k balls into 3 bins. According to chat gpt it converges to 1/10, but I don't know how to evaluate it for myself

[@johnvak7390]

The answer to your problem is: 2

[@rozaepareza]

The way he sets it up is 1-∑((1/2^k)*(3*2^k-3)/(3^k)). 3*2^k-3 is the number of ways where at least one bin is empty, where 3 is choosing which bin should be empty, and 2^k is arranging the balls in the remaining bins, and -3 is compensating for the fact that we double counted the cases where all balls are in one bin (e.g. we counted the case where all balls are in bin A both by choosing bin B to be empty and then putting all balls in bin A, and also by choosing bin C to be empty and then putting all balls in bin A). 3^k is the number of ways to put balls in 3 bins. The sum is simplified by first distributing the fraction to get ∑(3/3^k-3/6^k), then splitting the sum to get ∑(3/3^k)-∑(3/6^k), then applying the rule for the sum of a geometric series to both sums to get 3/2-3/5, which is 9/10, and 1-that is 1/10

[@lorenzot.7045]

@@rozaepareza Thank you man, i had the same problem but i solved it in a similar but different way: 1-3/4-3∑((1/2^k)*(2/3)^k(1-2/2^(k)) + (1/2^k)*(2/3)^k (k from 3 to +inf) = 1/4-3(Sum[1/3^k,{k,3,inf}]-Sum[1/6^k,{k,3,inf}]

[@Dbenji29]

Bro's a bit anxious. :D

[@grv-tw]

I hate interview where they read the question to you, just give him the question and wait for the answer.

[@cpte.3800]

I got 66

[@Jump-2-the-moon]

ChatGPT could have answered all the questions in less than 10 seconds without as many “uh, ums”.

[@harrypotterobamasonic10inu]

Whats the whole point of this. We are going to mars and build a base on the moon.

[@salembeats1875]

two out of three duhh!

[@justjason888]

erm.... good to learn, but i dont think i can qualify to ever be a quant hahahah

[@MergenBattur]

that seemed too elementary for a quant interview no?

[@ronaldquan8007]

that wig is insane