Summer Trading Intern Interview Questions | Glassdoor

# Summer Trading Intern Interview Questions

7

Summer trading intern interview questions shared by candidates

## Top Interview Questions

Sort: Relevance Popular Date

Mar 11, 2011
 You are playing a game where the player gets to draw the number 1-100 out of hat, replace and redraw as many times as they want, with their final number being how many dollars they win from the game. Each "redraw" costs an extra $1. How much would you charge someone to play this game? 10 Answers 10? redraw 10 times and get the payoff around 77? the average draw will pay out$50.50 Show More Responses Every time when you are deciding whether to play once more, we consider the two options: stop now, then you get current number (the cost of $1 is sunk cost); continue, then the expectation of benefit would be$50.50-1=49.50. This means, as long as you have get more than $50 (inclusive), then you should stop the game. Suppose the game ends after N rounds (with probability (49%)^(N-1) x 51%, and in the last round, the expected number is (50+100)/2=75, and thus the expected net benefit would be 75-N . This shows N<=74. Then we take the sum:$Sigma_{N=1}^{74} (49%)^(N-1) x 51% x (75-N), which is 73. Every time when you are deciding whether to play once more, we consider the two options: stop now, then you get current number (the cost of $1 is sunk cost); continue, then the expectation of benefit would be$50.50-1=49.50. This means, as long as you have get more than $50 (inclusive), then you should stop the game. Suppose the game ends after N rounds (with probability (49%)^(N-1) x 51%, and in the last round, the expected number is (50+100)/2=75, and thus the expected net benefit would be 75-N . This shows N<=74. Then we take the sum:$Sigma_{N=1}^{74} (49%)^(N-1) x 51% x (75-N), which is 73. All of the above answers are way off. For a correct answer, see Yuval Filmus' answer at StackExchange: http://math.stackexchange.com/questions/27524/fair-value-of-a-hat-drawing-game Yuval Filmus proves that the value of the game is 1209/14=86.37 and the strategy is to stay on 87 and throw again on 86 and below.. Let x be the expected value and n be the smallest number you'll stop the game at. Set up equation with x and n, get x in terms of n, take derivative to find n that maximizes x, plug in the ceiling (because n must be integer) and find maximum of x. ceiling ends up being 87, x is 87.357, so charge $87.36 or more I guess the question asks for the expected value of the game given an optimal strategy. I suppose the strategy is to go on the next round if the draw is 50 or less. Hence, the expected value of each round is: (1) 1/2*1/50(51 + 52 + ... + 100) (2) 1/2*1/2*1/50(51 + 52 + ... + 100) - 1/2 (3) 1/2^3 * 1/50 (51 + 52 + ... + 100) - 1/4 .... Sum all these up to infinity, you'd get 74.50. This is all very interesting and I'm sure has some application...but to trading? I don't think so. I own a seat on the futures exchange and was one of the largest derivatives traders on the floor. Math skills and reasoning are important but not to this level. I would associate day trading/scalping more to race car driving i.e. getting a feel for what's going on during the day, the speed at which the market is moving and the tempo for the up and down moves. If I were the interviewer at one of these firms, I throw a baseball at your head and see if you were quick enough and smart enough to duck. Then if you picked it up and threw it at my head I'd know that you had the balls to trade. I know guys who can answer these questions, work at major banks, have a team of quants working for them and call me up to borrow money from me because they're not making money. At the end of the day, if you want to be a trader then...be a trader. If you want to be a mathematician then be a mathematician. It's cool to multiply a string of numbers in your head, I can do this also, but never in my trading career did I make money because in an instant I could multiply 87*34 or answer Mensa questions which...realistically the above answer is: it depends on the market as the market will dictate the price. You may want to charge$87 to play that game but you'd have to be an idiot to play it. In trading terms this means that when AAPL is trading at $700 everyone would love to buy it at$400. Now that it's trading at $400 everyone is afraid that it's going to$0. Hope this helps. No offense to the math guys on this page, just want to set the trading record straight.

Oct 4, 2011
 Say I take a rubber band and randomly cut it into three pieces. What's the probability that one of the pieces has length greater than 1/2 of the original circumference of the rubber band. 9 Answers 3/4 Suppose you have two cuts on the rubber band placed randomly. The probability of having one segment greater than half the circumference is the probability that the third cut will be inside the combined range of 90* to either side of the cuts. Since the average distance between the first two cuts is also 90*, the combined range is 270*, or 3/4 of the circle. You need 3 cuts to end up with 3 pieces. The first cut doesn't matter. The second cut can also be anywhere and the largest piece will still be at least half the circumference. What matters is the third cut, which should lie in the same half as the second cut. So the probability is actually 1/2. Show More Responses The correct answer is 3/4, as this problem is equivalent to the famous 3-points-on-semicircle problem. Why? If one of the pieces has length greater than 1/2 the circumference, then the three points of cutting must lie in the same semicircle. On other hand, if the three points of cutting lie on the same semicircle, then the longest piece must be at least 1/2 of the circumference. For reference to the 3-points-on-same-semicircle problem, see e.g., http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html 1/4 1 -3/4 suppose I have two points whose minor arc distance is t <= 1/2. Then the range of semicircles covering both points gives an arc length of (1/2+1/2)-t = 1-t. say we fix the first point, tracing the second point around gives minor arc lengths from 0 to 1/2 and then 1/2 to 0. Therefore the answer is 2*integral (1-t) from 0 to 1/2, which is 2(1/2-1/8) = 3/4 It's 3/4. Cut it into 1 piece make a line. Cut it close. Pretend the length is 100. If you cut the first at x=1, as long as it isn't between x=50-51, it will have a length greater than 50% so there's 99% chance. You can imagine that if the cut was infinitely close to the end it would be about 100%. Now cut at x=2 you can't do between x=50-52. For x=3 it's 50-53 etc. So when you get to right to infinitely close to 50 it is pretty much between x=50-100 so there is a 50% chance you hit your spot. (obviously 50-50 is 100%, but since this length is continuous there's little chance it lands on that point). Obviously since this is symmetrical you can see this pattern going from 50% to 100% at the other end. Since each point on the continuous line has the same probability of happening the answer is clearly 75%. This problem is also equivalent to the probability that, if you have a line segment from 0 to 1 and you make 2 random cuts on that line segment, what is the probability that the three resulting pieces do NOT make a triangle?

Aug 9, 2010
 How much distance is their between the hour and minute hand at 5:12? 3 Answers 42 degrees? How did you get that? The answer is 61.5 its 84

Jun 25, 2013
 One example question: if you dip a cube made of cubes with each edge having ten cubes into a paint vat, when you take it out how many cubes will have no paint on them? 1 Answer 8^3 = 512

Feb 8, 2014
 What is the sum of the square root of every number from 1 to 100? 3 Answers Take the square root of the middle number, around 49-50, and then multiply it by 100 for an approximate answer. Taking the square root of the middle number and multiplying by 100 isn't a good idea, since the numbers aren't uniformly distributed. A better idea would be to bound the answer via two integrals. We know that the answer, T, satisfies: \int_{1}^{101} \sqrt{x} dx < T < \int_{0}^{100} \sqrt{x} dx We get these bounds via the lower and upper sums. Evaluating the LHS and RHS integrals, we get: 1998/3 < T < 2000/3 So, I'd guess the answer to be 1999/3, with a confidence interval of +/- 1/3. x=sum(i^(0.5) for i = 1,..,100 log(x)=log(sum(i^(0.5) for i = 1,..,100) log(x)=[(0.5)^100]*log(sum(i) for i = 1,..,100) log(x)=[(0.5)^100]*log(2550) log(x)=log(2550^[(0.5)^100]) x=2550^[(0.5)^100]