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# Quant Interview Questions

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Nov 22, 2010

Mar 17, 2013

### Quant Internship at Jane Street was asked...

May 6, 2011
 You have 100 marbles, 50 are blue, 50 are red. You want to distribute them between two drawers, in such a way that none is left outside and no drawer is left empty. After distributing them you are gonna select a drawer randomly and from that drawer you are gonna remove one marble randomly. How do you distribute the marbles in such a way that the probability of getting a red marble is maximized?6 AnswersPut one red marble in one drawer and all the others in the other drawer.It doesn't matter how you distribute the marbles. The fact that they are split into two drawers doesn't affect the sample space. The probability of choosing either draw is 1/2 (i.e. the same).The first answer is correct. Say you put one red marble in drawer 1, and drawer 2 has 24 red and 25 blue. The chances of choosing drawer 1 and a red marble are .5 or 49/98 since all the marbles are red in drawer 1. The chances of choosing drawer 2 and a red marble are .5*(24/49) so 12/49 or 24/98. The total chances of choosing a red marble in this case are 73/98, much higher than 1/2.Show More ResponsesIn drawer #1: 1 RED 0 BLUE In drawer #2: 49 RED 50 BLUE 50% chance of selecting drawer #1 - where there is a 100% chance of selecting RED = .5 * 1.00 = .5 50% chance of selecting drawer #2 - where there is a 49/99 or .4949% chance of selecting RED = .5 * .4949 = .2474 .5 + .2474 = .7475Stephen is correct. The answer is indeed 1/2*(1/1) + 1/2*(49/99) = 1/2*(148/99) = 74/99 = 74.75%. "P" has the right approach but has assumed the wrong number of marbles, as there should be 100 marbles in total, not 50 :)Here's a really intuitive way to see derive the solution without any actual math. You can obviously achieve 50% odds by putting all the red marbles in one drawer and all the blue marbles in the other. If you were to take a blue marble from the blue drawer and put it into the red drawer, then you slightly decrease the odds overall. Conversely, if you were to take a red marble from the red drawer and put it into the blue drawer, then you slightly *increase* the odds overall, since if you pick the red drawer you still have a 100% chance (at that point) of getting a red marble, and if you pick the blue drawer you now have a nonzero chance of getting a red marble. Repeating this, we arrive at the optimal solution posted by others: one drawer should contain one marble while the other drawer should contain all other marbles.

### Quant at Jane Street was asked...

Dec 21, 2010
 find a subset of {1,2,3,...,30} so that numbers in that subset is coprime to each other. 6 Answers2,3,5,7,11,13,17,19,23,29MMXMV, you forgot about 1: 1 is coprime to all intergers Should be 1,2,3,5,7,11,13,17,19,23,29I believe this question was supposed to be "find a subset such that the sum is the greatest" so the answer would be 1,2,7,11,13,17,19,23,25,27,29Show More Responsessorry, delete 2 add 16woops, take out 16 and 7 and add 28! since 16+7 < 281 11 13 17 19 23 25 27 28 29 Here is the steps: Start from list of primes and 1 1 2 3 5 7 11 13 17 19 23 29 Now from the left switch out to increase sum. 28 in, 2 and 7 out 1 3 5 11 13 17 19 23 29 27 in, 3 out 1 5 11 13 17 19 23 27 29 25 in, 5 out 1 11 13 17 19 23 25 27 29 23 already in. 21 < 28. 15 < 25. Can stop here.

### Quant at Jane Street was asked...

Nov 22, 2010
 suppose you live in middle town, your mom lives in downtown while your girl friend lives in uptown. You take subway to have dinner with either your mom or your girl friend every night. You always the first train that arrives at the station. why do you end up with have dinner with your girl friend 90% of the times? assume the uptown and downtown train run at the same intervals.6 AnswersThis would happen if suppose the downtown train always arrives slightly after the uptown. Suppose they each arrive in 10 minute intervals. Suppose the uptown arrives at 6:00, 6:10, 6:20, etc, and the downtown arrives at 6:01, 6:11, 6:21, etc. Then supposing you arrive at a random time (uniformly distributed), it is highly you arrive before an uptown train rather than before a downtown.And there is one more reason because boys will always prefer to stay with their girl friends than their mum.The downtown train should arrive six minutes after the uptwon train to visit mom 10% of the time.Show More Responsesif the trains arrive once an houri would see the girlfriend 100% of the timesAssuming the interval is X minutes. If uptown train arrive at t, t+X, t+2X...., the downtown train should arrive at t+0.1X, t+X+0.1X,....

### Quant Research Intern at Susquehanna International Group was asked...

Mar 17, 2013
 Suppose you have 100 GB of data that you want to sort, but you only have 1 GB of memory. How would you sort this data?8 AnswersHint: This isn't really a difficult question (just was an unexpected one for me). You don't really need to know the answer to figure this out. As it turns out, the obvious thing actually works here (and it is a known sorting algorithm).Can you expand on this? What is sorting algorithm?Sorting algorithm = a computer algorithm to sort a list of objects. Well pretend you just have 2 GB of data (for simplicity, assume they are integers) and 1 GB of memory, since the technique is the same. And pretend you want to sort these integers in increasing order. What would you do? Like, what's the first idea that comes to your mind?Show More ResponsesYou do an on disk merge sort, bring chunks in to memory and sort using quick sort, then had the sorted data in to buckets (files). When your done merge them using a merge sort.Yep, exactly.External sortbucket sort. Sort each bucket, then merge.Mark

### Quant at Jane Street was asked...

Apr 1, 2012
 Unfair coin with P(H) = 1/3 and P(T) = 2/3. a) How to make an event with 50% probability? b) Expected number of flips until a realization occurs? c) Can you create a strategy to reduce the number of flips necessary? d) Can you create a strategy to reduce the number of flips necessary for an unfair coin with any bias?7 Answersa) Event 1 = {T,H}, Event 2 = {H,T}, if any other outcome, then re-roll b) The probability of Event 1 or Event 2 occurring is 1/9+1/9=2/9. The expected number of 2-roll "tries" is 9/2. And each "try" consists of two rolls so 9 expected rolls for a realization. c) Event 1 = {T,T}, Event 2 = any other combination. Probability of either event is 4/9. d) Many solutions. Trick is to not discard any rolls. Use strategy from part (a), but if you roll {T,T}, then continue and put {T,T,H,H} in Event 1 and {H,H,T,T} in Event 2.the c) is wrong. probability of event 2 is 5/9 not 4/9mokhlos is right. Discard HH.Show More ResponsesInterview Candidate, P(H) = 1/3 and P(T) = 2/3 so a H-T combo will have a 2/9 prob. For a), the probability of Event 1 or Event 2 occurring is 2/9 + 2/9 = 4/9. Therefore, the expected number of 2-roll "tries" is 9/4. Each try consists of two rolls - so an expected 9/2 rolls until a realization.best answer: TT is counted as case 1 -> p = 4/9 TH or HT is case 2 -> p = 2/9 + 2/9 = 4/9 only discard HH Expected # of flips E = 8/9*2 + 1/9*(2 + E) -> E = 9/4 or 2.25 flipsWhat is the best (minimum number of expected tosses) possible for any strategy ?Using Information Theory, Expected number of coin-tosses required is at least (1/ H(1/3))=1.08

Nov 22, 2010