Trainee sales trader Interview Questions | Glassdoor.co.uk

# Trainee sales trader Interview Questions

366

trainee sales trader interview questions shared by candidates

## Top Interview Questions

Sort: RelevancePopular Date

### Trader at Jane Street was asked...

27 Apr 2011
 was asked this question today...you want buy a car at an auction whos price is uniformly distributed between 0-1000 if you bid more than the car you win it at the price you bid.if you bid less that the cars proce you dont win it but dont lose anything. you can sell the car for 1.5 time the value of which you bought it. what should you bid on the car to maximumise your profit?12 Answersi told the interviewer that the expected value is negitve so you should bid 0 to maximum your profit.(0 is greater that any negitive number) but the answer which i was guided towards wasE(X)= X-(X/2)*1.5 letting X = your bid so if you bid 500 E(x) = 500 - 250(1.5) = 125 pretty weird question if you ask me. comming up with a formula for expected value when i was orgionaly asked what would be my bid to maxiumise my profit. any1 have an thoughts? maybe this was obvious?If you bid X then your expected gain is (X/2)*1.5 So (X/2)*1.5 - X should be your profit... I.E. negative expected value. Could it be the interview just wanted to see how you can mathematically back up your original statement?yeah good point! that is most likely what he was testing. i suppose i just assumed it was obviously neg EV.Show more responsesim sorry but can u guys explain why is it X/2 *1.5 and not x*1.5 Also, where have you guys taken into acct that the gain is only when your bid is successful. If its not successful u gain ntn. Hence in my opinion it should be E(X) = (X/1000) * (1.5*X - X)its x/2 because the expected value of all the other possible outcome when you win is x/2.the expected profit is negative. say you bid 500 and win then the expected value is 0(1/501) + 1(1/501).....500(1/501) that where the x/2 come fromShouldn't it be bid 1000? then you will win the car for sure, and you can make a 1000*50%=\$500 profit. (theoretically, assuming you can sell the car for 1.5 times the purchase price no matter what the purchase price is, you should bid as large as possible, then your profit is purchase price*50%)Looks like an options contract.I agree with "e". This doesn't seem like it should be more complicated than it is. if you can sell the car for 1.5 times the value of which you bid for, then no matter what the bid is, your profit will be 50%. You want to maximize profit, which implies the fact that you "want to win the bid" so that you can then sell it for more. Bidding less wins you nothing, and so you might as well bid the largest amount, i.e. \$1000, to ensure you win, so you can then spin it off for a 50% profit. Granted, this explanation is only valid if it's a "sure" thing that no matter what you pay for the car, you can 100% sell it for 1.5 times the price you paid. Thoughts?You dont sell it for 1.5 times you paid you sell it for a 1.5 times it's true value (a number uniformly distributed between 0 and your bid (which expected value 0.5 times your bid).assuming the question is wrongly worded, and we can sell it for 1.5 * its actual value. bid x dollars, between 0 and 1000: expected profit = probability you get the car * (1.5 * expected value of car - bid) p = (x/1000) * (1.5 * 500 - x) maximise w.r.t x to get p = 140.6 at x = 375a=actual price b=bidding price ------------------ If profit = 1.5b - a answer is easy, bid 1000 ------------------- If question was wrongly worded & profit = 1.5a - b to win, b>a to make profit, b<1.5a so to win and make profit: ab is close to the right answer, however, the expected value of the car isn't locked @ 500, it is dependent on x (the amount bid). If one were to bid 1, the expected value of the car isn't going to be 500. Consider a fixed x, [0,1000], the expected value of the car will be x / 2. In essence, we are shrinking the bounds of a uniform distribution. Thus, profit = (x/1000) * ( (x/2) * (3/2) - x) profit = -x^2 / 4000 Thus, the correct bid is 0. A quick sanity check would be to bid 1. Half the time, the car is worth 0, the other half it is worth 3/2. Thus, if we would win the car we would pay 1 dollar and expect to get 3/4 of value back. This is negative EV. It can be seen that as the bid increases, the EV becomes further negative.

### Assistant Trader at Jane Street was asked...

29 Dec 2012
 We consider numbers from 1 to 1 million. How many digits 2 are there??10 Answers600 000940951654321: The total number of digits 1-999,999 is 5888889 Total 1-9 = 9 10-99 = 2*9*10 and so on so number of 2s is 5888889/9Show more responses654321: The total number of digits 1-999,999 is 5888889 Total 1-9 = 9 10-99 = 2*9*10 and so on so number of 2s is 5888889/9Each time you fix a digit 2 (in units, tens...), then you change the other 5 digits. So you'll have 100 000 number. Having that for the six placements, the result would be 600 000.600,000 6C6 with 6 digits, 9(6C5) with 5, (9^2)(6C4) with 4...etc. Add them up 600,000.11111110^6-9^6observe the recurrence relation. 1-9 there is 1 from 1-99 there is 10 1-9 intervals plus 10 digits from 2x. and then it continuous in exactly the same fashion add one 0 you get 10 times as many plus 10^(n-1) from the nth digit you add. (((((10+10)*10+100)*10+1,000)*10+10,000)*10+100,000)=600,000Consider from 0000000 to 9999999, the number of 0,1,2,3,4,5,6,7,8,9 present must be the same since they are all symmetric. prove: the position can be filled with 0 can be filled with 1, so the number of 0=the number of 1; the number of 1=the number of 2 for the same reason. ... so the number of 0-9 are all the same. There are 6*1million positions in total, thus 6 million positions, divided by 10, is 600,000.

### Trader at Jane Street was asked...

6 Oct 2013
 If I write down all of the numbers from 1 to 1,000,000 on a page, how many times do I write down the digit 2?10 Answers(base 10 )log(x)/10*xI got 468599600000 (verified through brute force)Show more responsesimagine the numbers written as 000001 000002 000003.....999998 999999, then we 1 million numbers containing 6 digits, each digit of 10 digits we have, is used as much as any other number. Therefore, you write down the digit 2 10% * 6 million - 600,000the minus must be an equal sign on the end, so 600,000 is the right number[1 x 10 x 10 x 10 x 10 x 10] + [10 x 1 x 10 x 10 x 10 x 10] - 1 + (bcos 222222 covered above) [10 x 10 x 1 x 10 x 10 x 10] - 2 + (bcos 222222 & 22222 covered) [10 x 10 x 10 x 1 x 10 x 10] - 3 [10 x 10 x 10 x 10 x 1 x 10] - 4 [10 x 10 x 10 x 10 x 10 x 1] - 5 = 600000 - 15 = 599985Continuation from above from 1 upto 1 with n 'zeros' (e.g. 10 has 1 zero, 100 has 2 zero, etc), answer = n(10^[n-1]) - 0.5n(n-1) So upto 1000000 would give 6 x 10 ^ 5 - 0.5 x 6 x 5 = 599985The idea is to exploit the symmetry among 0,1,2...9. Instead of 1->10^6, one should think of 000000 -> 999999, where 0-9 occurs with the same probablity . Altogether 000000 -> 999999 has 10^6 numbers, 6*10^6 digits; divided by 10 this is 600,000.The idea is recursion, let x_n be number of times that digit 2 appear in 0 to the largest n digit number. For example, x_1 = number of times that 2 appears in 0-9. x_2 is up until 99, etc. you work out to see x_1 = 1. x_2 = 10*x_1 + 10 etc. finally, I think you will get 600000From 0 to 999,999 there are 10^6 numbers. Each is formed by 6 digits. Hence we have a total of 6*10^6 digits. Now, numbers are equally spread, therefore the number of times each number appears is: 6*10^6/10 = 600,000

### Assistant Trader at Jane Street was asked...

19 Dec 2012
 If you extend the faces of a tetrahedron as planes infinitely in all directions, how many regions does this divide 3D space into?6 Answers15There is 4 planes. You are either left or right of each so you could be in one of 2^4 = 16 distinct regions.Did they specified if the subspaces had to be disjoint?Show more responsesThink of the tetrahedron as the edge of a cube. Think of extending the planes as extending the cube into a 2x2x2 cube. Think of the bottom of the tetrahedron as a plane slicing 7 of the 8 cubes (except for the one in the opposite corner). So 7x2 + 1 = 15.You might be able to visualise this, I think it helps. There will be one region for each vertex, edge and face and one region inside the tetrahedron. So there will be 4 + 4 + 6 + 1 = 15 regions.Use Barycentric coordinates, observe that if the coordinates are w, x, y, z then w + x + y + z = 1. Lying on a side of a plane is given by the variable being positive, and observe that other than all variables being negative, all other combinations are possible.

### Trader at Jane Street was asked...

16 May 2012
 2 fair dice. What is the probability of both showing six if I have observed at least one six.6 Answers1/6. Independent probabilities, so seeing one makes no difference to the second.Not quite... The subtlety is "at least one". From two dice rolls there are 6*6 = 36 distinct outcomes, of these, 11 contain at least one six, i.e. (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (5,6), (4,6), (3,6), (2,6), (1,6), therefore P( (6,6) | at least one 6 ) = 1 / 11.More formally: P(observe one 6 ) = P( 1st die is 6 OR 2nd die is 6 ) = P(1st is 6) + P(2nd is 6) - P(1st 6)*P(2nd 6) = 1/6 + 1/6 - 1/36 = 11/36, by reunion. Now, we want to calculate P(both are 6's | observed one 6 ) = P ( observe one 6 | both are 6's) * P ( both are 6's) / P ( observe one 6 ) = 100% * 1/36 * 36/11 , where last factor was calculated above. So, all together P (both 6's| obs. one 6) = 1/11Show more responses11 possibilities of getting at least one six (6 x 1 + 1 x 6 - 1) so P = 1/11I disagree with the above: P(rolling at least one 6) = 1 - p(rolling no 6). rolling no 6 twice in a row = (5/6)^2. = 25/36 therefore 1 - 25/36 = 11/36. I thought njm had answered it halfway through his answer but then somehow arrived at 1/11? I believe he looked at 11 outcomes, instead of the initial 36.Ah I do apologise, mis read the question! yes you are all right! ha

### Trader at Jane Street was asked...

16 May 2012
 n unbiased coins. What is the probability that half of them exactly are heads. Answer the question for n= 2, 3, 200006 AnswersImagine the sequence: HHTTTHTHHHTT .... , length n. It has n! different permutations, but some are indistinguishable because of subsets like HHH or TTT.. . So, we overcounted by a! times b!, where a = total #H and b = #T in the sequence. as a+b = n , a = n/2 , all together this means n!/((n/2)!*(n/2)!) distinguishable combinations. final answer is: Prob(half are heads ) = #success outcomes/total outcomes = n!/ ( n/2 ! * n/2 ! * 2^n ) Prob ( H | 2 flips ) = 2! / ( 2^2) = 2 / 4 = 50% Prob ( HH| 4 flips ) = 4! / ( 2! 2! * 2^4) = 24 / ( 4 * 16 ) = 6/16 = 37.5% Note that there is a simpler way to look the problem: the number of successes ( heads ) follows a binomial distribution. P( X = x number of H | 0.5 chance, n trials ) = nCx * 0.5^(x) * 0.5^(n - x ) = nCx / 2^n which is in agreement with my justification above.for high numbers n=20000 use the Central limit theorem to approximate the outcome by a Gaussian distribution, no way to calculate nC(n/2) for high numbersIndeed, for large numbers, one should use B(n,p) ~ N(np, np(1-p) ). Then, compute the integral of N() over 1 unit, centered around the mean. The exponential bit is 1, so only the normalization factor in front matters, and that is 1/ ( sqrt ( n p ( 1-p) 2Pi)). For n = 400, binomial gives 4.0%, whereas the gaussian integral gives 3.98%.Show more responsesP = 0 if n is odd P = [n! / (0.5*n)!^2] x (0.5)^n for n =2, P = 0.5 for n = 3, P = 0 for n = 20000, P = [20000! / (10000!^2)] x (1/2) ^ 20000The probability is exactly [ [ n / 2 ]! ]^2 / n! as explained above For large n, this can be approximated using Stirling's approximation: ln(n!) ~= n [ ln(n) - 1 ] where ln(n) is the natural log of n Therefore, the probability is approximately exp( [ n * ln(1/2) ] - 1 ) For n=2000, this gives a probability of 3.20*10^-603 which is close to the 'exact answer' of 4.88*10^-601 according to WolframAlphause stirling's approximation n! ~ sqrt(2 * pi * n) * (n/e)^n

### Options Trader at Liquid Capital was asked...

21 Jul 2010
 If there was a drawer in front of me containing 4 socks, either black or white, and I know that I have a 50% probability of pulling out two white socks in a row, then what is the probability of pulling out two black socks?6 AnswersThere is zero probability of pulling two black socks.Assuming there are 2 whites and 2 blacks then getting 2 socks of the same colour is the sameAbove meant to say: Has the same probabilityShow more responsesYou can't assume there are two white socks and two black socks. This problem is an easy enumeration problem actually. From the problem, you have at least 2 white socks, otherwise it would be impossible to take out 2 in a row. So you have either 2 white socks, 3 or 4. Just compute the three prob of getting two white socks in a row for all case and see which is 50%. P(two w in a row| 2 white socks) = (2/4) * (1/3) = 1/6 => not two socks. P(two w in a row| 3 white socks) = (3/4) * (2/3) = 1/2 => this is our answer. Obviously, if there were 4 socks, we would have a 100% chance of getting two whites in a row. So as calculated, there must be 3 white socks.The probability is 0 as there is only 1 black sock. Solution: Assume there are k white socks and 4-k black ones in the drawer. => 0.5 = P(2 white socks in a row) = k/4 * (k-1)/3 => k^2 - k - 6 = 0 => k = 3 is the number of white socks. qed."50% probability of pulling out two white socks in a row" 0.5 = 3/4 * 1/3 (3/4 of a circle has 3 slices, so take 1 slice out and you have half a circle :p) "containing 4 socks , either black or white" So we can deduce that we have 3 white socks and 1 black sock. There is no chance we can pull out more than 1 black sock.

### Assistant Trader at Jane Street was asked...

3 Sep 2012
 Pretty standard question that has been asked before... keep flipping a coin until a winning combination appears (either HHT or HTT). Which strategy would you pick given the choice and why. Find the probabilities of winning associated with each strategy.5 AnswersEssentially HHT is the better strategy. The probabilities are 2/3 HHT and 1/3 HTT. To see this if you draw a tree diagram (best to draw 4-5 iterations if you can't see it) and look at all the possible ways of winning. It turns out HHT is twice as likely to win hence the 2/3, 1/3 split.By drawing 5 iterations, both HHT and HTT seem to make an appearance 7 times. How is HHT twice as likely to win?The mathematical expectation of hte number of flipping for both compbinations is 25/2. In any case the probabilities of appearance of these compbinations are the same, so one can use both strategies with the same result. 50% 50%Show more responsesBecause you can throw h h h and all you need to throw is a tails to get HHt, but there is no fail, please play again throw for htt, so yes, on a throw of 3, they have th same probability, but in a continued series you can play again on h h t if you fail on th 3rd throw.Note that both HHT and HTT combination starts with H, so you can almost neglect H, so when you first observe an H, you wait for HT or TT both of which are equally as likely given the order matters i.e. (HT and TH are different), hence both should be good strategies.

### Assistant Trader at Jane Street was asked...

15 Jul 2015
 If I role a dice and can keep the value on the dice face up, what should my strategy be if I can have more than one role and can keep the value of all the roles but if I go 7 or above I go bust.5 AnswersI said EV is 3.5 so roll a 1,2 or 3 and roll again. 4,5 and 6 stop.This is the wrong answer. The correct answer is not to roll again!Both of the answers above are wrong. If you roll 1. Then you would certainly reroll, since you have only 1/6 probability of getting busted, and you lost \$1. But you have 1/6 chance of getting 1,2,3,4,5 dollars. E[earning] - E[losing] = 15/6 - 1/6 if you roll 2. You have 1/6 chances of wining 1,2,3,4, 1/3 of chance getting busted. E[earning] - E[losing] = 10/6 - 1/3*2 > 0, you should always reroll If you roll 3. You have ........ 1,2,3 1/2 chance of losing 3 dollars. Then E[earning] - E[losing] = 1 - 1/2*3 = i. ps. n is the sides on the fair dice, in this case, n=6Show more responsesFor my last post, I tried to give a mathematical equation of the stopping criterion. sum_1^(n-i) - i^2/n < 0. When this equation is satisfied, we should record that i. and that i is the number when you stop reroll. if you roll some number ksum_1^(n-i) i/n - i^2/n < 0

### Assistant Trader at Jane Street was asked...

15 Jul 2015
 If I choose randomly 6 numbers from a list of the first 12 prime numbers, then what is the probability that the sum of those six numbers will odd?6 AnswersI said that one number has to be the number two so that there is an odd number of odd numbers plus two which would make the sum odd. I can't remember the mathsCorrect, only if 2 is included in the set of 6 will the sum be odd. Half of the subsets include 2, so the answer is a 1/2Show more responses11c6 = 11!/6!*5! 12c6= 12!/6!*6! so a/b= 11!/12! * 6!/5! = 1/2 If people are wondering why Aug18th is wrong. My logic would just be the prime 2 has a 50% chace of being chosen (6/12)Precisely when there is a 2 included. Number of subsets including a 2: 11 choose 5. Number without a 2: 11 choose 6=11 choose 5. Thus 1/2.P(2 has been chosen) = (6 choose 1) / 12 = 0.5 One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.
110 of 366 Interview Questions

More