Jane Street Capital Interview Questions and Solutions
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SCLG 2632
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Mathematics
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Jul 17, 2023
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Jane Street Capital Interview Questions Nikolaos Rapanos March 8, 2010 First Round. The first round phone interview took place on Tuesday, 16 February 2010. I was asked the following questions: The Questions 1.Mental Arithmetic.Questions like 1 million - 101, 342, 54 percent of 110 etc. 2. (a) What is the expected value of a fair die? (b) What will be the expected value of the same die if for each dot you are given 100 GBP? (c) Now you are given the chance to roll the die twice, but if you choose to do so then you are obliged to keep your second outcome. What is your strategy in order to maximize the expected value and what is the new expected value? (d) Same as (b) but now you are allowed to re-roll twice and obliged to keep your third roll. 3. You roll two dice and you are told that one of them is six. What is the probability that both are six? 4. You hold a ball 10m above the ground. You leave the ball and you are told that each time the ball hits the ground, it will bounce up to the half of the height it was left from. So the first time the ball hits the ground will bounce up to 5m, the second time to 2.5m and so on. What is the total distance the ball will travel? How confident are you about your answer? The Answers 1. In general, we can speed up calculations usingVedic Mathematicsor other simple tricks. For instance, 342= 302+ 2·30·4 + 42 = 900 + 240 + 16 = 900 + 256 = 1156. 2. (a) The expected value of a fair die is6 Xi=1 1 6 ·i= 3.5 (simply the values weighted by the corresponding probabilities). (b) We only need to multiply each of the values by 100, and so the expected value will be 100 times greater than the previously computed expected value. Thus the new expected value turns out to be 350 GBP. (c) Obviously if the first outcome is less than or equal to 3, we choose to roll the die again. Therefore, the new expected value will be the average of the expected values of the first and the second round. The expected value of the first round will be the expected value among the values that we choose to keep, namely 4, 5 and 6. Hence the expected value of the first round is 5 and the expected value of the second round is 3.5 as derived above. Therefore, the new expected value will be 0.5(5 + 3.5) = 4.25. Note that as this is an extension of question (b) we also have to multiply by the factor 100, and so the expected value will become 425. What this says, is that if we play this game then we should expect to gain in average 425 GBP. (d) In this case we stop on the first roll if we get 5 or 6 (higher than 4.25). Also we stop on the second roll if we get 4, 5 or 6 (higher than 3.5). Hence the expected value is 100·135.5 +234.25=140 3 = 466.7 GBP. 1
Comments on Problem 2 Comment 1.Now we are told that we have to pay a (one-off) ticket in order to play the game in (b), and by definition thefair valueof this ticket is the value that would force the expected profit to be zero. As follows, from the analysis above the fair value of this game would be 425 GBP. This can be interpreted as the maximum amount of money we are willing to spend in order to play the game (i.e. take the risk with the hope to make profit).The logic can be conversed.If for example we are the "owner" of this game, then we can set the ticket price for the game to be equal to the fair value of the game plus possibly a small marginfor profit. Comment 2.In order to demonstrate the concept of thefair valueconsider the following example appeared in a past Jane Street interview: You pay a ticket in order to roll a die. If the result is 6, then you get 10 GBP. What is the fair value of this game? Letxbe the fair value of this game.The expected profit for this game isEP= (10-x)·16-x·5 6 . DemandingEP≥0 we getx≤10 6 = 1.67 GBP. Comment 3.Now suppose that we are told that we have to pay a fixed value ticketeachtime we decide to roll a die, then this value would be the half of the aforementioned fair value. In fact, if we follow the logic demonstrated above we can explicitly deduce this intuitive result. Indeed, ifxis the fair value of the ticket then denoting byEP1andEP2 the expected profits for the first and second roll respectively, we get the following expressions: EP1 = (100·5-x)12-x 1 2 EP2 = (100·3.5-x)12-x 1 2 HenceEP=EP1+EP2 , and by demanding this total expected profit to be non-negative we get x≤425 2 = 212.5 3.Solution 1.We know that all the possible combinations (outcomes) of two dice are 6·6 = 36.From these combinations, we are interested in those which contain at least one 6. This is equivalent to counting the outcomes which contain no 6 and then subtracting them from the total number of possible outcomes which is 36. The outcomes which contain no 6 are 5·5 = 25, and hence the outcomes which contain at least one 6 are 36-25 = 11. Therefore the probability of having exactly two 6 isP=1 11 . Solution 2.Alternatively, we can work this out with conditional probabilities.ByAdenote the event that both will face up a 6, and byBdenote the event that at least one will face up a 6. Now, callB1 the event that the first die is a 6, andB2 the event that the second die is a 6. It is more than obvious that B=B1∪B2and thereforeP(B) =P(B1) +P(B2)-P(B1∩B2 ). Using the definition of conditional probability we have: P(A|B) =P(A∩B)P(B)=136P(B1) +P(B2)-P(B1∩B2 )=1 36 16+16- 136 = 1 11 Solution 3.A slight deviation from the solution discussed above, makes use ofBayes Theoremand worths to be presented here. Following the above notation we have: P(A|B) =P(A)·P(B|A)P(B)by Bayes Theorem, and plugging in numbers we getP(A|B) =1 36 ·11136 = 1 11 Solution 4.Finally, we can also tackle this problem usingBinomial Distribution(see Appendix A). We roll two dice, son= 2 and the probability of each outcome is a sixth, thereforep=1 6 . Let us denote by Pn (x) the probability ofxsuccessesinntrials and in our case asuccesscan be interpreted as a rolling a die and getting a six. Following the aforementioned notation, we conclude that the probability of having exactly two 6 can be calculated as the probability of two successes in two trials divided by the probability of having at least one success in two trials (which in turn is the sum of the probabilities of one success in two trials and two successes in two trials). Mathematically, we can write 2
P(A|B) = P2 (2) P2(1) +P2 (2) =22· 1 62 21·16·56+22· 1 6 2= 1 11 Comment.Especially the last solution to the problem makes evident that the problem can be naturally generalized and also a universal approach can be developed and adopted in similar situations. Obviously a prerequisite for this, is a deep understanding of Binomial Distribution. Consider for example the following problem which appeared in another Jane Street interview: You flip four coins and at least two are heads. What is the probability that exactly three are heads? This type of questions should ring the bell. The answer isBinomial Distribution forn= 4andp=1 2 . Using the notation introduced in Solution 4, we compute the requested probabilityPas follows: P=P4 (3)P4(2) +P4(3) +P4 (4)= 4 11 4. In this problem we have to be careful to take into account that the ball will travel once downwards and once upwards in each bounce. Therefore the total distance traveled by the ball can be calculated as d= 10 + 2·10·12+ 10·122+...!= 2·10 + 10·12+ 10·1 2 2 +...! -10 = =20·limn→∞ (1 2 ) n -1 1 2 -1 ! -10 = 20·2-10 = 30. Regarding the confidence question, I was told that there is no correct answer but the whole point is that after taking the decision to implement a trading idea, we first have to allocate some time debating on how confident we are and consequently how much risk we are willing to take. I find it helpful to think that I bet my own money on this, so even if I am sure about my answer I have to be quite reserved since I risk my own money. 3
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