# Summer Interview Questions

Summer interview questions shared by candidates

## Top Interview Questions

What are the methods of valuing a company? 1 AnswerIt's DCF(Discounted casfhflow) & Relative method, where FCFE(Free cash flow to Equity) is vital for equity shareholders & FCFF(Free cash flow to Firm) is vital for the company. |

Tell me about a time when you were working in a team and your opinion was challenged. 1 AnswerI asked the team member to explain their opinion. No one is perfect, so keeping an open mind when listening to a team member’s response is key. There were times a team member could convince me they were right and other times when I could explain why my opinion may be the better option. |

Flip a coin until either HHT or HTT appears. Is one more likely to appear first? If so, which one and with what probability? 13 AnswersLet A be the event that HTT comes before HHT. P{A} = P{A|H}P{H} + P{A|T}P{T} = .5P{A|H} + .5P{A|T} P{A|T} = P{A} therefore, P{A|H} = P{A|T} P{A|H} = P{A|HH}P{H} + P{A|HT}P{T} = (0)(.5) + P{A|HT}(.5) Therefore, 2P{A|H} = P{A|HT} P{A|HT} = P{A|HTT}P{T} + P{A|HTH}P{H} = (1)(.5) + P{A|H}(.5) 2P{A|H} = .5 + P{A|H}(.5) P{A|H} = 1/3 and P{A|H} = P{A}, therefore, P{A} = 1/3 So, HHT is more likely to appear first and it appears first 2/3 of the time. P{A|H} = P{A|HH}P{H} + P{A|HT}P{T} = (0)(.5) + P{A|HT}(.5) Need help - - why is P{A|HH} = 0 ? P(A|HH) = 0 because after a sequence of consecutive heads, you can no longer achieve HTT. The moment you get a tail, you will have the sequence HHT. This the reason HHT is more likely to occur first than HTT. Show More Responses P(A|HH) = 0 because after a sequence of consecutive heads, you can no longer achieve HTT. The moment you get a tail, you will have the sequence HHT. This the reason HHT is more likely to occur first than HTT. HHT is more likely to appear first than HTT. The probability of HHT appearing first is 2/3 and thus the probability of HTT appearing first is 1/3. Indeed, both sequences need H first. Once H appeared, probability of HHT is 1/2 (b/c all you need is one H), and probability of HTT is 1/4 (b/c you need TT). Thus HHT is twice is likely to appear first. So, if the probability that HTT appears first is x, then the probability that HHT appears first is 2x. Since these are disjoint and together exhaust the whole probability space, x+2x=1. Therefore x=1/3. You guys seem to be mixing order being relevant and order being irrelevant. If order is relevant (meaning HHT is not the same as HTH) then this has a 1/8 of occuring in the first 3 tosses. Also HTT has a 1/8 chance of occurring in the first 3 tosses, making them equally likely. Now, if order is not relevant. (so HHT and THH are the same), then this has a (3 choose 2) * (1/8) probability of happening in the first 3 tosses. The same goes for HTT (which would be the same as THT etc and others) so this has a (3 choose 2) * 1/8 probability of happening in the first 3 tosses as well. Either way they come out to being equally likely, please comment on my mistake if I am doing something wrong. Ending up with HHT more likely (with probabilty 2/3). HHT is more likely (2/3) probability. People with wrong answers: Did you not Monte Carlo this? It takes 5 minutes to write a program, and you can then easily see that 2/3 is correct empirically. I don't get it. Shouldn't P{A|HH} = P{A} in the same sense that P{A|HTH} = P{A} from both HH and HTH we have get the first H from HTT and so it should be P{A|HH} = P{A|HTH} = P{A} Am i wrong? sorry, i meant: I don't get it. Shouldn't P{A|HH} = P{A|H} in the same sense that P{A|HTH} = P{A|H} from both HH and HTH we have get the first H from HTT and so it should be P{A|HH} = P{A|HTH} = P{A|H} Am i wrong? Above link is the best solution I have seen for this problem http://dicedcoins.wordpress.com/2012/07/19/flip-hhh-before-htt/ Apologies, Below* Here's my answer. Let x = probability of winning after no heads (or a tail). y=probability after just one heads. z=probability after two heads. w=probability after HT. Thus x=(1/2)x+(1/2)y, y=(1/2)z+(1/2)w, z=1/2 + (1/2)z, w=(1/2)y. Therefore, z=1, y=2/3, w=1/3, x=2/3. We wanted x at the beginning, so it is 2/3 that HHT comes up first. |

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. |

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? |

3) Poker. 26 red, 26 black. Take one every time, you can choose to guess whether it’s red. You have only one chance. If you are right, you get 1 dollar. What’s the strategy? And what’s the expected earn? 12 Answersexpected earn is 25 cents. 1/2*1/2*1, prob of choosing to guess is 1/2, prob of guessing right is 1/2, and the pay is $1 I would start picking cards without making a decision to reduce the sample size. This is risky because I could just as easily reduce my chances of selecting red by taking more red cards to start, as I could increase my chances of selecting red by picking more black cards first. But I like my chances with 52 cards, that at some point, I will at least get back to 50% if I start off by picking red. Ultimately, I can keep picking cards until there is only 1 red left. But I obviously wouldn't want to find myself in that situation so I would do my best to avoid it, by making a decision earlier rather than later. Best case scenario, I pick more blacks out of the deck right off the bat. My strategy would be to first pick 3 cards without making a decision. If I start off by selecting more than 1 red, and thus the probability of guessing red correctly is below 50%, then I will look to make a decision once I get back to the 50% mark. (The risk here is that I never get back to 50%) However, if I pick more than 1 black card, then I will continue to pick cards without making a choice until I reach 51% - ultimately hoping that I get down to a much smaller sample size, and variance is reduced, while odds are in my favor that I choose correctly. The expected return, in my opinion, all depends on "when" you decide to guess. If you decide to guess when there is a 50% chance of selecting correctly, then your expected return is 50 cents (50% correct wins you $1 ; 50% incorrect wins $0 --- 0.5 + 0 = .5) If you decide to guess when there is a 51% chance of selecting red correctly, then the expected return adjusts to (0.51* $1) + (0.49 * $0) = 51 cents. So, in other words, your expected return would be a direct function of the percentage probability of selecting correctly. i.e. 50% = 50 cents, 51% = 51 cents, 75% equals 75 cents. Thoughts? There is symmetry between red and black. Each time you pull a card it is equally likely to be red or black (assuming you haven't looked at the previous cards you pulled). Thus no matter when you guess you odds are 50% and the expected return should be 50 cents. Show More Responses scheme: guess when the first one is black, p(guess) x p(right) x 1=1/2 x 26/51=13/51 0.5, just turn the first card to see if it's red. I think it's more about trading psychology. If you don't know where the price is going, just get out of the market asap. Don't expect anything. The problem should be random draw card and dont put it back. Every draw you have one chance to guess. So the strategy is after first draw you random guess it's red. If correct you get one dollar, next draw you know there is less red than black. So you guess black on next draw. Else if first guess you are wrong, you guess red on next round. It's all about conditioning on the information you know from the previous drawings This should be similar to the brainteaser about "picking an optimal place in the queue; if you are the first person whose birthday is the same as anyone in front of you, you win a free ticket." So in this case we want to find n such that the probability P(first n cards are black)*P(n+1th card is red | first n cards are black) is maximized, and call the n+1th card? The problem statement is not very clear. What I understand is: you take one card at a time, you can choose to guess, or you can look at it. If you guess, then if it's red, you gain $1. And whatever the result, after the guess, game over. The answer is then $0.5, and under whatever strategy you use. Suppose there is x red y black, if you guess, your chance of winning is x/(x+y). If you don't, and look at the card, and flip the next one, your chance of winning is x/(x+y)*(x-1)/(x+y-1) + y/(x+y)*x/(x+y-1) = x/(x+y), which is the same. A rigorous proof should obviously done by induction and start from x,y=0,1. The answer above is not 100% correct, for second scenario, if you don't guess, and only look, the total probability of getting red is indeed the same. However, the fact that you look at the card means you know if the probability of getting red is x/(x+y)*(x-1)/(x+y-1) or y/(x+y)*x/(x+y-1). Therefore, this argument only holds if you don't get to look at the card, or have any knowledge of what card you passed Doesn't matter what strategy you use. The probability is 1/2. It's a consequence of the Optional Stopping Theorem. The percent of cards that are left in the deck at each time is a martingale. Choosing when to stop and guess red is a stopping time. The expected value of a martingale at a stopping time is equal to the initial value, which is 1/2. My strategy was to always pick that colour, which has been taken less time during the previous picks. Naturally, that colour has a higher probability, because there are more still in the deck. In the model, n = the number of cards which has already been chosen, k = the number of black cards out of n, and m = min(k, n-k) i.e. the number of black cards out of n, if less black cards have been taken and the number of red cards out n if red cards have been taken less times. After n takes, we can face n+1 different situations, i.e. k = 0,1,2, ..., n. To calculate the expected value of the whole game we are interested in the probability that we face the given situation which can be computed with combination and the probability of winning the next pick. Every situation has the probability (n over m)/2^n, since every outcome can happen in (n over m) ways, and the number of all of the possible outcomes is 2^n. Then in that given situation the probability of winning is (26-m)/(52-n), because there are 26-m cards of the chosen colour in the deck which has 52-n cards in it. So combining them [(n over m)/2^n]*[(26-m)/(52-n)]. Then we have to sum over k from 0 to n, and then sum again over n from 0 to 51. (After the 52. pick we don't have to choose therefore we only sum to 51) I hope it's not that messy without proper math signs. After all, this is a bit too much of computation, so I wrote it quickly in Python and got 37.2856419726 which is a significant improvement compared to a basic strategy when you always choose the same colour. dynamic programming, let E(R,B) means the expected gain for R red and B blue remain, and the strategy will be guess whichever is more in the rest. E(0,B)=B for all Bs, E(R,0)=R for all Rs. E(R,B)=[max(R,B)+R*E(R-1,B)+B*E(R,B-1)]/(R+B). I don't know how to estimate E(26,26) quickly. |

### Summer Analyst at Goldman Sachs was asked...

how many times the hour and minute hands of the clock form right angle during one day? 10 Answerseach hour creates 2 right angles...2 x 24 = 48 times a right angle is formed in one day Wrong. Think what's happening around 3 and 9 o'clock 4 times? Show More Responses Let me show you a mathematical approach. Common sense dictates that the minute hand moves at a faster rate of 5.5 degrees a minute (because the hour hand moves 0.5 degrees a min and the minute hand moves 6 degrees a minute). We start at 12 midnight. The hands are together. For subsequent 90 degree angles to occur, the minute hand must "overtake" the hour hand by 90 degrees, then 270 degrees, then 360 + 90 degrees, then 360 + 270 degrees, then 360 + 360 +90 degrees.. and so on. This can be re-expressed as: (1)90, 3(90), 5(90), 7(90), 9(90), 11(90)... n(90). The number of minutes this takes to happen can be expressed as (1)90/5.5, 3(90)/5.5, 5(90)/5.5, 7(90)/5.5, 9(90)/5.5, 11(90)/5.5... n(90)/5.5. In one day, there are 24 hr * 60 mins = 1440mins To find the maximum value of n, n(90)/5.5 = 1440 n = 88 but as seen from above, n must be an odd number (by pattern recognition and logic) hence n must be the next smallest odd number (87) counting 1,3,5,7,9,11......87, we see that the number of terms = (87-1)/2 +1 = 44. In other words, the minute hand "overtakes" the hour hand on 44 occasions in 24 hours in order to give a 90 degree angle. Therefore the answer to your question is 44. i agree with right_ans. although i got lost in his explanation, though i'm sure it is correct. i found another answer here: http://brainteaserbible.com/ Each hour has 2 occurrences of 90 degrees. In 12 hrs, it overtakes 24 times. BUT hrs 2 to 3 has only 1 NOT 2. Also hr 8 to 9 has only 1. So subtract 2 from 24. You get 22. In half a day (12 hrs) you get 22 times. Therefore in 1 day ie 24 hrs, it cross 22 * 2 = 44 times. 42 Relative speed is 5.5 degree/min. Time is 24*60 mins. Total distance is 5.5*24*60 degrees. How many full circles it is? 5.5*24*60/360 = 5.5*4 = 22. Each full circle contains 2 right angles (90 and 270). So answer is 22*2 = 44. 44 Calm down, we first must convert time to angle, two different units. The hour hand completes one full revolution each 12 hours (considering a 12 clock). So, theta_h = 360 x h/12, where theta_h is the angle that the hour hand makes with 12 hs mark and h is the number of hours since 0hs. So at 0 hs, the angle is 0, and at 12, the angle is 360 = 0. Since h = m/60, where m is the number of minutes since 0h, we have: theta_h = 360 x m/60 / 12 = 360 m / 12 x 60 = 0.5m Now, given the number of minutes since 0h, we can tell the angle of hour hand using theta_h. The minute hand angle, theta_m is: theta_m = 360 x m / 60 = 6m So the difference between theta_h and theta_m is |theta_h - theta_m| = 5.5m. Now given the minutes since 0h, we can tell the angle between hour and minute hand. Now, how many minutes we need to make |theta_h - theta_m| = 5.5m = 90? About 16.36. Since we have 12 x 60 minutes in a day, we have 12 x 60 / 16.36 gaps that satisfy 90 degrees, which is 44. |

### Summer Analyst at Goldman Sachs was asked...

Why 6 AnswersYour subjects are more from hardware side so how could you help us? but isn't it a bit early to say that you didn't get the second round? It's been a week, i wasnt sure how much time it takes... Show More Responses I'll leave a comment here again once I get a reply. I hope you get one too :-) Thanks Bro yo, i got it. |

### Software Engineer at LinkedIn was asked...

Java: how do you make n threads run at the same time? 6 AnswersI think it defeats the purpose of having threads. Why would you want that? Anyway, the correct answer is using join. Actually, there are needs for multiple threads to run at the same time, so the system can waiting for all parties are ready before start. For example, if we want to measure how long the tests take when all threads run "Concurrently' (see the example in the book "Java Concurrent in Practice" by Brian Goetz etc. The join only works for waiting 1 thread. If you have 5-10 threads, join is not the good option. You can use CountDownLatch as gate or CyclicBarrier for this purpose. Here is the example from the above mentioned book. public long timeTasks(int nThreads, final Runnable task) throws InterruptedException { final CountDownLatch startGate = new CountDownLatch(1); final CountDownLatch endtGate = new CountDownLatch(nThreads); for (int i = 0; i < nThreads; i++) { Thread t = new Thread() { public void run() { try { startGate.await(); try { task.run(); } finally { endtGate.countDown(); } } catch (InterruptedException ignored) {} } }; t.start(); } long start = System.nanoTime(); startGate.countDown(); endtGate.await(); long end = System.nanoTime(); return end-start; } weird question - threads do run at the same time, unless you do something special to prevent it. this code makes threads run at the same time: for (int i = 0; i < n; ++i) new Thread().start(); the answer posted by Anonymous is to a different question - how to synchronize certain parts of threads (that do run at the same time :). This is very legit question and CountDownLatch / CyclicBarrier are standard mechanisms to achieve that Show More Responses Ugh, how about using ExecutorService and creating a fixed sized pool. That ensures that only a maximum of pool size threads run at a given time. 1. In the main thread we can create an instance of java.util.concurrent.CyclicBarrier class using the constructor Barrier b = CyclicBarrier(n); 2. Then each thread in their first line do b.await(), All threads until n of them have called it will block on this call 3. When the last thread does b.await() all of them will move past this call; Simple answer would be fixed size threadpool.This type of pool always has a specified number of threads running; if a thread is somehow terminated while it is still in use, it is automatically replaced with a new thread. |

25 racehorses, no stopwatch. 5 tracks. Figure out the top three fastest horses in the fewest number of races. 11 AnswersWe can do it in seven races, we'll call them races A-F. For notation, we'll say that the Nth horse in race X is called X.N. Races A-E: Divide the horses into 5 groups of 5 each such that each horse races only once. We can eliminate the slowest 2 horses in each of the five races because there are definitely 3 horses faster in each case. As a result, we eliminate 5x2 = 10 horses: {A.4,A.5,B.4,B.5,C.4,C.5,D.4,D.5,E.4,E.5} Race F: Race the fastest horses in each race A-E: {A.1, B.1, C.1, D.1, E.1}. To simplify notation, we'll label F.1 as horse A.1, F.2 as horse B.1, and so forth. That means the winner of this race is A.1, and it is the fastest horse of all. We don't have to race A.1 anymore. We can eliminate D.1 and E.1 = 2 horses. Because they are not in the top 3. As a result, we know that all remaining horses from D and E are eliminated. This is D.2,D.3,E.2,E.3 = 4 horses. We know that A.1,B.1, and B.2 are all faster than B.3 (and similar for C.3) so they are not in the top 3.We can eliminate B.3 and C.3 = 2 horses. Finally, we know that A.1 is faster than B.1, which is faster than C.1, and thus C.2, so we can remove C.2 = 1 horse. Race G: We have removed 19 horses from competition and are sure that A.1 is the fastest horse of them all. This leaves just 5 horses: {A.2,A.3,B.1,B.2,C.1}. We race them and select the top 2 to join A.1 as the top 3 fastest horses. just run them all on the one track :) one race, and you get your 3 fastest horses in one go........or am I missing something! 6 races. Divide 25 horses into 5 groups. Each group races and the fastest is selected. The winner of each of the 5 races all race together. Pick Top 1,2 and 3. My only concern: Could the answer be this simple? Show More Responses B, you're mistaken. Imagine the top three fastest horses are Santa's Little Helper, Yojimbo, and I'm Number One. By random luck, in your first race, the five random horses you choose includes all three of those. I'm Number One wins and goes on to the final race; the other two do not. 8 5 top horses from each race of 5 races (25 / 5) 5 top contenders race; 1 wins--that's one top horse (5-1) 4 remaining top horses race, one wins; that's 2 top horses (4-1) 3 remaining top horses contend; winner is #3 That's 3 top horses from 8 races Race#1 Race#2 Race#3 Race#4 Race#5 A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 A4 B4 C4 D4 E4 A5 B5 C5 D5 E5 Race#6 A1 B1 C1 D1 E1 Let's Say ranking 1st 2nd 3rd 4th 5th Eliminate D1 E1 D2 E2 D3 E3 A4 B4 C4 D4 E4 A5 B5 C5 D5 E5 Left with B1 C1 A2 B2 C2 A3 B3 C3 Eliminate C3 as there are more than three faster horses C2, C1, B1, A1 Eliminate C2 as there are three faster horses C1, B1, A1 Eliminate B3 as there are three faster horses B2, B1, A1 Left with 5 horses for Race#7 B1 C1 A2 B2 A3 So 7 races 7 races. put 25 horses in 5 group. and we will have 5 sorted list of horses in each group. put 1st place horse in each group, and we will have a sorted list X. X_1 is the 1st place horse, and X_2 is 2nd place horse's candidate, X_3 is 3rd place's candidate. 2nd place horse in X_1's group is candidate for 2nd place, 3rd place one is candidate for 3rd place. and 2nd place horse in X_2's group is a candidate for 3rd place. that's 5 horses in total, 2 from X_1's group, 2 from X_2's group, X_3. race them, and 1st place is 2nd place, 2nd place is 3rd place horse. 8 the answer is one race as the question doesn't specify all the horses have to run in separate races. 0 Races. Ask the jockeys to rate the 10 fastest horses and take the average of the top 3. Nobody knows the horses that are fast better than the jockeys. Race results can vary from race to race but the jockeys truly know the fastest and besides because race results vary anyway you will not find the fastest horses with the least races you will find the fastest on that day. Either way your results will not be accurate without a larger dataset. The jockeys however already have a deeper dataset. Better yet find the local bookie for the track and ask for the odds on the horses. Why solve a problem that has already be solved. |

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