Trading intern Interview Questions
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I was asked about if a machine failed and a frustrated production worker was on that machine when it failed, how do you deal with the production worker while you working to fix that machine as they tend to get more frustrated the longer the machine is down? 1 AnswerI said that I would give the production worker the knowledge of knowing what was wrong with the machine, and an approximate time as to how long it may take. I would then try to conversate and diffuse the situation with humor. |
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 Answers10? 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. |
I have 10 cards face-down numbered 1 through 10. We play a game in which you choose a card and I give you the corresponding dollar amount. a) What is the fair price of this game? b) Now, after picking a card you can either take the dollar value on the card or $3.50. Also, cards worth less than 5 are now valued at $0. What is the maximum price you are now willing to pay for the game? 10 AnswersFair value is 5.5 2nd game: FV = (3.5 x 5 + 6 + 7 + 8 + 9 + 10)/10 = 4.75 You decide on taking the dollar value on the card or $3.50 *before* seeing the number on the card. And you get $5, if you open that card, not $0. So: 2nd game: (1/2)*3.5 + (1.2)*(1/10*[5+6+7+8+9+10]) = 4 3.5(.4) + (4.5)*.6 = $4.10 3.5 because this is what 40% will select, 4.5 as this is the avg of 5-10*1/10. There is not a 50% chance of selecting 3.5 @Your are WRong @ EASY. There is not a 10% chance of receiving $3.5 as numbers under 5 are worthless this will /should change your perspective. Show More Responses FJM is correct. "Cards worth less than 5 are now valued at $0" - that does not include card number 5. how does 4.10 make sense if 4.5 is the ev of taking the card..ii think you are incorrect. can you explain your reasoning. using 1/10 to take your average and then multiplying by .6 seems like you are double counting Total amount=3.5*4+(5+6+~+10) Expected value=total/10=5.9 @FJM no idea where your 4.5 coming from OK agreed with xinzhuo. Should be 7.5 instead of 4.5 in FJM's calculation, which gives 5.9. Agreed. I double counted. 3.5*.4 + 7.5*.6 3.5 was given 7.5 =(5+6+7+8+9+10)/6 "You decide on taking the dollar value on the card or $3.50 *before* seeing the number on the card." 2nd game: Obviously, the game price should be no less than 3.5, otherwise there is an opportunity for arbitrage. Now that the price is higher than 3.5, it makes no sense for a rational player to choose just take 3.5 dollar and exist the game. The player should always choose to bet. And the expectation for betting is (5+6+...+10)/10 = 4.5. I think question was phrased as such: " You can take 3.50 before seeing the value of the card, or pay an amount to see the value of the card". We have established that the EV of seeing the card is 4.5, but why would we pay 4.5 to get an EV of 4.5? That nets 0 profit per play on average as opposed to taking the risk free 3.5 Instead, we can see that we would at most bet 1 dollar per play. This way, the total EV of the game with the bet is 3.5 = the arbitrage free price. We can show it in terms of "plays" as follows, with first number as taking 3.5 and second as seeing the game Play 1: 3.5, 4.5-B Play 2: 7, 2(4.5-B) ... We can see that we would at most pay 1 dollar for this game |
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 Answers3/4 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? |
Trading Intern at Wolverine Trading was asked...
You have three barrels. One barrel is filled with apples, one with oranges, and one with both apples and oranges. Each barrel is mislabeled. You can take out as many fruit as you like form each barrel and look at it. What is the minimum number of fruits you need to remove to correctly label the barrels? 8 AnswersJust one. Take a fruit from the barrel labeled apples and oranges. You know this barrel is not the apples and oranges barrel, so whatever fruit you take must be the only fruit in that barrel. The other two barrels are mislabeled, so you can figure out the other two as well. Two. Take a fruit from A+O, if its an O then this barrel is O ...Now take out a fruit from A, if its an O then this barrel is A+O...and the 3rd is by default barrel A One. Start of from A+O, the fruit u pick say O should be the barrel name. now that means that the barrel labelled A can be A+O or A... But since we know that barrels are mislabelled it cant be A. hence, we concluded it to be A+O and the last remaining barrel is A. Show More Responses 1 Trick question, just look inside the barrel... One. Since it's mislabeled, A can be labeled as O or A+O. Similarily, O = {A, A+O} and A+O = {O, A}. Case 1: Take 1 from A+O label, it turns out to be A. Relabel it. We have A, O mislabels left, and A+O, O real labels left. Mislabeled ones can ot be the same as real labels. So the mislabeled O should be A+O and A should be O. Case 2: also choose from A+O, but it turns out to be O. Solving the same as case 1. None, you don't need to move any of the fruit to correctly label the barrels. You need to change the labels. All three are mislabeled and you cannot look inside the barrels. If you take one out of the apples+oranges barrel, whatever fruit you pull out is the fruit of the barrel. And Since you know the other two are mislabeled, you would switch the labels. So you need to take out only one fruit |
What the probability of getting 2 consecutive heads in a total of N tosses (I found this one pretty hard and I didn't figure out the right answer at the time.) 7 AnswersDo you mean Prob(at least 1 observation of 2 consecutive heads appears) ?? Then since the subsets can be: 1. all tails 2. 1 head 1 tail 1 head 1 tail .... 3. at least 2 heads consecutively appear once So Pr(3) = 1- Pr(1) - Pr(2) = 1- (1/2)^N - (1/2)^N = 1 - power(1/2,N-1) I think the answer should be this: Since the probability of getting two heads in a row and getting two tails in a row is the same, we only need to figure out the probability of the total. The remaining event is: 1 head 1 tail......due to different order, events should be 2 Therefore the probability=0.5(1-2(1/2)^N)=0.5(1-(1/2)^(N-1)) Only solution I could think of, which took quite some time, was to actually start by drawing a tree of possible outcomes. From there, I could see only three different states at each node. Either I am in a state similar to my initial state (when I toss a tail), I can also be in tilt state (I have just flipped heads after flipping a tail) or I am in the done state (I have just flipped heads after having flipped heads before). I consider that someone would only continue flipping if he/she has not flipped two heads in a row. That is, is not in the done state. That is, for any toss level N, it wrong to assume that number of nodes is equal to 2^N. It will be strictly smaller as of level N = 3. Anyway, after 3-4 levels drawn, one should see a pattern, that is, the number of tails, or initial state nodes (is), at a level N is equal to the number of is's I had at level N-1 + the number of is's I had at level N-2. We can thus write: is(N) = is(N-1) + is(N-2). we then realize that at every level N, the number of tilt nodes (t) is equal to the number of initial states from the previous level. That is: t(N) = is(N-1) That makes sense, because on the level N-1, all initial states output one tail (becoming a new initial state at level N) and a heads (becoming a tilt state at level N). the other states of level N-1 are either done (so no continuation) or tilt. The possible outcomes of a tilt state are either done or initial state (if I flip heads again, I am done, so I can't go from tilt to tilt.) One then realize that following the same logic, the number of done (d) nodes for a level N is given by: d(N) = t(N-1) = is(N-2) that is, at every level (as of the second level, or two tosses), I will have a number of done outcomes equal to the number of initial states to levels (or tosses) ago. So the general recursive formula to get the number of done nodes for a certain level really is: d(N) = is(N-2) = is(N-3) + is(N-4) We note, by definition: is(N) = 0, is N<0 is(N) = 1, if N = 0, or 1 is(N) = is(N-1) + is(N-2) otherwise. Furthermore, the probability of being at any node of a level N is simply (1/2)^N. So if I want to find the probability that I observed at least one occurrence of 2 heads in a row. I simply add the number of done nodes from level 1 to N multiplying them by the probability of them occurring at each level. That is: P(hh after N tosses) = sum(from i = 1 to N) (d(i).(1/2)^i). E.g. for N = 2, I get: d(1).1/2 + d(2).(1/4) = s(-1).1/2 + s(0).(1/4) = 0 + 1/4. We know that this is true. you can test it for a larger number of tosses and see it is true. as N goes to infinity, the sum will converge to 1, as I have theoretically no chance of never tossing 2 heads in a row if I throw a coin an infinity number of times. I haven't found a solution that didn't use recursions. There probably is one, I'd be interested in seeing one. Show More Responses sorry for all the typos, the solution should be understandable in spite of that. This can be solved by considering the number of sequences of heads/tails that do NOT have consecutive heads. If N coins are tossed, you can have 0 heads in 1 way. You can have 1 head in N ways--choose 1 from any of the N positions in the sequence. You can have 2 heads as long as they are not consecutive. Imagine you have N-2 tails and you need to decide where to place the 2 heads. You can insert them before any of the tails, or after the last tail--so you need to choose 2 out of a possible N-1 positions. Similarly, for any H less than or equal to N/2, you can have H heads by selecting H positions from a possible (N-H)+1 positions. So the number of sequences which do NOT have two consecutive heads can be found by the sum: 1 + nCr(N,1) + nCr(N-1,2) + nCr(N-2,3) + ... Evaluating this starting with N = 2 gives the values 3, 5, 8, 13, 21, 34, ... These are Fibonacci numbers. The number of sequences of length N without 2 consecutive heads is given by F_{N+2}, where F_1 = 1, F_2 = 1, and F_N = F_{N-1} + F_{N-2}. It follows that the probability for obtaining two consecutive heads in N flips of a fair coin is given by 1 - ( F_{N+2}/ 2^N). Note: j-dw has a correct solution. The solution given by Charles ignores the fact that many sequences will have BOTH two consecutive tails AND consecutive heads. He treats these as non-overlapping sets. Delusion ignores all cases such as HTTHTTHTT in which there is more than one T separating the heads. (N-1)/(4x2^(N-2)) One or more comments have been removed. |
Trading Analyst at Virtu Financial was asked...
We play a game. You flip two coins, if you get both heads I give you $1. If you get both tails you flip again. If you get one of each, I pay you nothing. What's the expected value of this game? 6 AnswersDismiss TT since it is not part of the sample space (ie there is no outcome from it) You are twice as likely to get one of each (H-T or T-H) versus getting both heads answer: (2/3)($1) + (1/3)(0) = 66.7 cents E=0.25x1+0.5x0+0.25xE 0.75xE=0.25 E=0.33 interviewer told me .667 is correct Show More Responses HT and TH add up to 1/2 of the probability with zero payout. HH is 1/4 probability with $1 payout. TT results in a "reset" so we can disregard. After we normalize the probabilities, we get: E[payout]= (1/3)*$1 + (2/3)*$0 = $0.33 You can also see it a geometric series. By writing E=0.25x1+0.5x0+0.25xE, Kerzane could as well have written E?0.25x1 + 0.5x0 + 0.25x(0.25x1 + 0.5x0 + 0.25x(0.25x1 + 0.5x0 + 0.25x(....))) This is quite simply the geometric sum: (n-> infinity) E=0.25 + 0.25^2 + 0.25^3....0.25^n which is equal to 0.25/(1-0.25) = 0.25/0.75 = 1/3 or 0.333 Sample Space = H/H, H/T, T/H, T/T, except T/T isn't part of the sample space due to the fact that T/T loops back into the game. An outcome, part of the sample space, is only counted if you "win" or "lose". So the sample space is: H/H, H/T, T/H. H/H = 1/3 probability of getting; T/H, H/T = 2/3 probability of getting. Expected value, or E, is the theoretical value attained from the law of large numbers which is arrived at as the number of trails approaches infinity. Therefore, in probability theory we may say that E(X) = 1/3 for winning or 2/3 for losing. 1/3 = .33 chance of winning, and 2/3 =.667 chance of losing. |
Trading Analyst at Virtu Financial was asked...
Assume that on a chessboard, the King piece is on tile E1 (bottom row, 5 columns across). It can move straight forward, forward diagonally to the left, or forward diagonally to the right on each move. How many different paths can it take to get to tile E8 (top row, five columns across) in exactly 7 moves? 5 Answers393 126 141 126 30 45 51 45 30 4 10 16 19 16 10 4 1 3 6 7 6 3 1 1 2 3 2 1 1 1 1 K K is the starting spot for the king. There is 1 way to get to each of the tiles in the row right above K. For the row above the row of 1-1-1, the number of paths to each tile is the sum of the possible paths to each tile that leads to it. (ie, for the first 2 in that row, you can get to it from either of the first two 1s in the row below). This is basically a 'trinomial' expansion as each number is the sum of the 3 numbers directly below it. This continues until the 8th row (where the diamond formation tapers inward since the number of left moves must equal the number of right moves in order to end up in the same column (E) where the King started) The final answer is 393 as demonstrated in the diagram above. It took me well over 30 minutes to get it, and the interviewer had lost all interest by that point. He was still nice in making sure to explain the methodology and ensuring that I arrived at the right answer, but was quick to hang up once the solution had been determined. I don't think IC is correct. I haven't gone through your methodology, but I think the answer is larger. The piece moves forward each turn. It moves either left, right or straight ahead. Take left=-1, right=+1, straight=0. The sum of 7 such numbers must be 0. Therefore, at least one move must be straight ahead. There are four types or routes. a 0x+1, 0x-1, 7x0 b 1x+1, 1x-1, 5x0 c 2x+1, 2x-1, 3x0 d 3x+1, 3x-1, 1x0 there is only 1 route of type a there are 7x6 of type b, etc. a 1 b 7x6=42 c 7x6x5x4=840 d 7x6x5x4x3x2=5040 a+b+c+d=5923 Am I wrong? Yes I was wrong. IC was right. there is only 1 route of type a there are 7x6 of type b, etc. there are ((7x6)/2)x((5x4)/2)=210 of type c there are ((7x6x5)/6)x((4x3x2)/6)=140 of type d giving a total of 393 Show More Responses Using permutations. You can see the problem as such: any move you make to the left has to be compensated by move to right. I have to go up 7 tiles total. With R representing a right move, L a left move and A a move straight ahead, my possible combinations are: 1) AAAAAAA 2) RLAAAAA 3) RRLLAAA 4) RRRLLLA One now has to compute all the permutations which is fairly straightforward: we have 1 + 7!/5! + 7!/(3!x2!x2!) + 7!/(3!x3!) which is: 1 + 42 + 210 + 140 = 393 This 4) RRRLLLA you cannot start with RRR as it will go beyond the chessboard |
Roll a fair dice, if it's a prime number, you pay me that number times $100, if it's not a prime number, I pay you that number times $100. Consider 1 as a prime number. How much willing to pay to play this game. 7 Answersanything below $387.50 Don't be surprised if you have not been employed. The question is "what is the expected return of this game?": E[X]=Sum(P(Xi)*Xi)=1/6*(-1*100+2*100+...)=50$ Esco910, we make a deal when you want, I have always dreamed about to buy a ferrari 2 is a prime number, and the expected value is less than 0. So I wouldn't play this game. Show More Responses $50 is correct, the others are wrong. You would play this infinitely many times because on average, you are printing money. 2 is definitely a prime number... the expectation is less than 0 One or more comments have been removed. |
How many ways can you arange five people at a round table, such that there they are in increasing age order (clockwise or anti-clockwise)? 5 AnswersTrick question, you can't. Oldest will always meet youngest Was asked this question (phrased as "what is the probability that 5 people with different birthdays will sit in increasing age order around a round table (clockwise or counter-clockwise), and the recruiter said that you should ignore the jump from the oldest to youngest. In other words, it's not a trick question -- you actually have to solve the problem. The answer is 1/12. The question is answer how many ways, nothing about probability. There are 10 ways to seat the people, 5 clockwise and 5 counterclockwise Show More Responses Assuming there are five seats at the table then there are ten different ways to seat the five people in increasing age order. There are (5-1)! ways of seating 5 people around the table. Of these, there are 2 permutations wherein there will be ordered sitting (one clockwise & the other anti-clockwise). Hence, the probability is 2/((5-1)!) i.e., 1/12 |
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