"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)
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.
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
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 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.
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).
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.
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
"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?"
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
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.
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.
@@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.
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.
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?
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.
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
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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.
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....
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.
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.
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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.
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?
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!?
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
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
@@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}]
Most people working with higher orders of mathematics know factorials up to 10 at the very least, as they pop up considerably in probability. Personally, my high school math teacher didn't let any of us leave class until we knew our "factorial tables" till 10 as well! It's similar to how most people know the multiplications tables or squared values up to a certain number, just a little less intuitive.
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
I must be really stupid because after she said the question I answered 32% with is 1/3 so I guess my math in my head is much better than this….why make thing complicated
I am 34 years old living in pakistan. I have retail branch banking experience of 9 years and two years experience working for a UK based asset management company that have a back office in pakistan as operations analyst overseeing the bank reconciliation and reporting area. I have a dream of working for a wall street firm. Could someone guide me to reach there?
The big banks have global operations. I'd suggest getting a job at one of them. Dont shoot for the moon. You wont join them in a front office role, but maybe an operations role like you do now if your CV is strong. If you are extraordinarily talented at programming with top tier math skills then apply to quant roles, if a big bank doesnt take you, a smaller firm might. Again you dont need to target WS firms directly. Dont listen to the person saying they dont hire non-Ivy/dont hire from Pakistan, that's not remotely true.
Is everyone usually like this when in interviews? He seems to be thinking a lot and still repeat phrases which doesn't make any sense. The question is simple yet he took a lot of time to solve it. Also completely ignoring the fact that shoes meant for left and right are different? It's just weird.
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.
Thanks for the kind note, and glad to hear it. Congrats!
- Alex, Quant Researcher
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Thank you.
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Hi please can you tell me how to prep for an internship for such a position. Would love to start learning on my own
"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)
😂 my thoughts exactly
Broooo😂😂😂😂😂😂😂😂
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.
😂😂😂
"thats pretty racist" - mark baum
yea ion think ima get the job
"Ion" and "ima" is pretty much a giveaway
@@ramsesp5009 yeah the comment was a joke, you can't be that stupid to not detect that
@@ramsesp5009 RACIST
@@ramsesp5009 wdym?
@@ramsesp5009bro thinks he's the citadel hiring manager 😂😂😂
It helps to have such a messy white board in the background to give the illusion you use it a ton.
Jesus no wonder Ive lost all my money
lmaoooo
Lmao!!😂
Citadel makes most of its money buying then front running retail order flow. No math needed.
@@Rainy_Day12234ur stupid if u think this is true
@@Rainy_Day12234 Can you explain this in detail, please?
He was supposed to close his eyes and answer the probability!!
these quant dudes are insanely intelligent, im about to take a part time masters in quant finance and the notes are looking tough af
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
that's my quant
MY QUANTITATIVE
@@airheadmrng MY MATH SPECIALIST
LOOK AT HIS EYES 😂
YOU NOTICE ANYTHING DIFFERENT ABOUT HIM
I'll give you a hint, his name is Yang !
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.
I dont doubt your logic, but how do you arrive at the 3/4 and 3/5 probs?
@@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.
Maam i just wanna work at McDonald's as part time
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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).
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.
@@bobthebuilder9416 can you elaborate please?
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
does the quant think shorting GME is the play?
"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?"
GME doesn’t trade based off fundamentals. That is a stupid question.
@@kurtisbobke9000 tick tock
Lmfao
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
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
Thanks for making this video! This is super helpful.
Glad it was helpful!
@@TheQuantGuide Are you on linkedin?
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.
Legit
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.
@@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.
@@yaboi608but in quant isn't accuracy stressed on, how come you're gonna say 1/3 instead of 32%
Look at him, that's my quant, my quantitative.
That is kinda racy(st)
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.
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?
There is always a HARD code test. Where is the HARD code test?
I got to admit, I'm confused on the last question with flipping a coin and bins being empty.
These problems feel like simple textbook exercises. I think real interview will require more clever tricks.
Citadel will hire him and make him naked short GameStop Stock
qtπ, what's your name?
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.
I think that you missed the word "random" in the task definition. You can't "chose", you toss the cube
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
@@mibli2935 he's not choosing. he just listed out every single possibility and of those possibilities, only 1/3 satisfy the condition.
@@mibli2935No she’s right this is math not philosophy
@@mibli2935shes basically writing out all the cases in the bayes approach. shes right, its essentially the same as the second solution.
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Wow. I'm a bit perplexed seeing her been mentioned here also Didn’t know she has been good to so many people too this is wonderful, I'm in my fifth trade with her and it has been super.
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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.
Hey,
The second part of your logic would be correct if we are replacing the shoes after picking them.
What is the point of all these questions? You hardly apply this.
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....
Idk man all i know is that i can spam the F5 button the fastest 😂
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.
lol That's what I thought too. When he said factorial I was like...this guy is an idiot.
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.
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.
Nope, it’s just more of that “show your work” bullsh*t.
can someone tell me why such calculations are important for trading ?
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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.
I think this is something I should do, but I've been stalling for a long time now. I don't really know which firm to work with; I feel they are all the same but it seems you’ve got it all worked out with the firm you work with so i surely wouldn’t mind a recommendation.
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.
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Bros first question should have been how many shoes are left foot and how many are right.
Fantastic
And HFT is only as good as their being fast/first. There's zero sum and then there's quants.
Can somebody who has gone through the interviews comment on if they are really this easy?
I did Jump Trading internship interview and the questiona where this kind of difficult
@@oscarromero9702 these are easy
Legendary
Asian vs Asian
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ChatGPT could have answered all the questions in less than 10 seconds without as many “uh, ums”.
He definitely won that math olympiad in china
The only word I understood was probabilities, am I quant now hahahaha
A failing U.S. economy and elevated global tensions reduce the likelihood of prolonged inflation or higher long-term Treasury yields. I've seen folks amass up to $1m amid crisis, and even pull it off easily in a favorable economy. Unequivocally, the bubble/collapse is getting somebody somewhere rich
I do not disagree, there are strategies that could be put in place for solid gains regardless of economy or market condition, but such executions are usually carried out by investment experts or advisors I speak from experience.
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this is great! think your advisor would get on the phone with an unknown? i'm in dire need of proper portfolio allocation
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.
Thanks for this. could easily spot her website just after inputting her full name on my browser. she replied my inquiry and we scheduled a consulting session sometime tomorrow.
What’s the point of all this when you all just gonna front run with your algos and trade routing 😂
truth spoken 😁
Algos need math and to figure out math you need to be a problem solver and have critical thinking skills.
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.
It will not
They're supposed to be different. They're left and right shoes.
The efforts are great, but still i feel maximum audience didn’t get anything my man explained in the video
Why do they have to explain their reasoning throughout working on their answer. So many uhms and aahs. Cant they speak after finalizing the answer?
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?
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.
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!?
Wrong
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?
Interviewer: are you willing to steal from retail investors?
Quant: yes
Interviewer: you're hired
smart money does care about retail, retail makes up a tiny portion of the total trading volume
two out of three duhh!
I got 66
‘qualifications’ doesn’t buy genius
Harris William Smith Sarah Robinson Michelle
Whats going on here 😟
Students in elementary school in asia can answer this question in 3 seconds..
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
The answer to your problem is: 2
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
@@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}]
So many “um” and “uh”
Did he count 6 quickly! or did he just know? or is the interview rehearsed?
Most people working with higher orders of mathematics know factorials up to 10 at the very least, as they pop up considerably in probability. Personally, my high school math teacher didn't let any of us leave class until we knew our "factorial tables" till 10 as well!
It's similar to how most people know the multiplications tables or squared values up to a certain number, just a little less intuitive.
@@tejaskala516 Cool!! Thanks for the explanation!
@@deniswolf1846 6! = 30*24 = 720
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
Like...?
You lost me after 17 seconds 😂
Really? What the heck does this have anything to do with the market? Lol.
I must be really stupid because after she said the question I answered 32% with is 1/3 so I guess my math in my head is much better than this….why make thing complicated
yeah wtf
This guy spent way too much time on the first question. lol
is that a wig?
I love "quant math" (probability and complexity theory for eg) and quant programming but GME ma nibba
Wow this is boring
Does your mom still swallow?
Wendy’s. The answer is Wendy’s.
Seems like math nerd genius. Math and engineering top degree. I graduated 2.6 GPA from none target in history and communications. Possible get in?
not at all
@@liminfer8380 bro how you know? You have to be a genius math grad nerd?
bruv this paragraph is wild
@@tgp120 bro gotta be some spots for frat party dudes and not just 4.0 math and finance nerds from Ivy league. I need get in
@@davidc4408 dawg sue your midwife i think they tapped your head a bit too hard
How the fuck does this even matter in trading
0
Edwin lin son?
🤔🤔🤔🤔🤔 i really wonder why are they not bilioners yet they know the math.
' just because you know same thing doesn't mean you understand it'.
There is a much easier way to explain this lol
If you have 6 shoes, and you pick 3 pairs, you have all the shoes, Duh! Common sense is more important than education.
She lost me at probability.
So quants are just thr next level of math olympiads...
Not really these are basic probability concepts
I am 34 years old living in pakistan. I have retail branch banking experience of 9 years and two years experience working for a UK based asset management company that have a back office in pakistan as operations analyst overseeing the bank reconciliation and reporting area. I have a dream of working for a wall street firm. Could someone guide me to reach there?
No ..they only hire from Ive leagues not from some shitty third world country.
@@aliasone9827 why?
@@aliasone9827Are you trolling my troll?
@@aliasone9827lolllllllll
The big banks have global operations. I'd suggest getting a job at one of them. Dont shoot for the moon. You wont join them in a front office role, but maybe an operations role like you do now if your CV is strong. If you are extraordinarily talented at programming with top tier math skills then apply to quant roles, if a big bank doesnt take you, a smaller firm might. Again you dont need to target WS firms directly. Dont listen to the person saying they dont hire non-Ivy/dont hire from Pakistan, that's not remotely true.
Saying he works at a shoe factory… are they playing jokes or is that a hint on the racy(ist) side…. Really?….
I could out trade him just looking at a chart. This is painful to watch.
Is everyone usually like this when in interviews? He seems to be thinking a lot and still repeat phrases which doesn't make any sense. The question is simple yet he took a lot of time to solve it. Also completely ignoring the fact that shoes meant for left and right are different?
It's just weird.
Only chinese can understand her english that's why they both understand each other 😂