# Financial Analytics & Research interview questions

## Interview Questions

How would you get in contact with someone who doesn't answer the phone and has no voicemail? 1 AnswerCall employer. If unemployed, family members. |

I got a few C++ questions, then a question on sorting algorithms then a brainteaser. The brainteaser went as follows: Three people are given hats. Each hat is either red or blue, chosen at random. Each person can see the other 2 hats, but not their own. They each must simultaneously either guess their own hat's color, or pass. No communication is allowed, although they can agree on a strategy ahead of time. What strategy will give them the best chances of at least one person guessing right, and nobody guessing wrong? 12 AnswersIf the 2 hats you see are the same, guess the opposite color, otherwise pass. If all 3 players use this rule, it works 75% of the time. It fails only when all the hats are the same color. You might want to start by listing all the possibilities. One person can assume there is an odd number of blue hats, another person can assume there is an even. By counting the number of blue hats amount the other two people each person arrives at a unique prediction for the color of there own hat. Sine the number of blue hats must be either even or odd, one of the guesses must be right. Yes, but the other one will be wrong. The question says that all the people who talk must give the correct color. The others can pass. However, if one answers wrong, they loose, so your strategy wouldn't work S. Show More Responses Here's a strategy: each member passes to the left if they see the same color (assume they are standing in a triangle) and to the right if they see different colors. If the first person sees two blue hats, and the second sees a red and a blue, the first will know their hat is red. If all hats are the same color, after one pass to the left, the second person will know their hat is the same as the third person. At most this will require two passes. I'm confused by the answers. We know the following: State: 1. There are three people 2. Each person is given one - and only one - hat 3. The colour will be either red or blue - completely random 4. No communication is allowed. 5. Each person can see the other two hats - but not their own. Behaviour: Each must simultaneously either guess their color or pass. Proposed Strategy: I see no rule against each individual passing his/her hat in a uniform direction - either left or right. This way, no communication has taken place and you now know the colour of your hat. It can obviously be considered as a form of communication after the start of the game. Just assume that every person can do one and only one action and that is pass or say a color. Also, that single action has to be done at the same time as the action or the two others, that is, there is only one pass. The strategy can be simple: Each person can see the color of the other two hats, so the strategy before the game could be, in the second vote the person should mention the color of the third persons hat. Assuming he tells the truth. So the first person can just tell any color, now only two people are left, the 2nd person will the tell the color of third persons hat, and the third person would repeat it. This will ensure every time that at least one answer is correct. Your answers are all horrible. Yeah, they really all are. The answer I posted is the correct one. It's actually on the internet if you google that problem, and it's really well explained. I mean the question is stated in an understandable manner. I really don't get all the ridiculous answers that are posted. two colors and allow the number of players to be of the form 2k − 1, where k 2, then we can devise a strategy for winning that—perhaps surprisingly—results in increasing probability of a win as the parameter k get large. Even better, this probability of winning approaches unity as k ! 1 ! In fact, we’ll show how to devise a strategy whose probability of a win is Prob(win) = 1 − 2 ^ (2k−k−1)/ 2 ^ (2k−1) . Note that if k = 2 (three players), this sets the probability of winning at 1 − 2 8 = 3 4 . The reason for this somewhat forced constraint on the number of people is that it allows a strategy based on the so-called Hamming codes, which we now briefly describe. We let F be the binary field, i.e., the field with two elements (0 and 1). All vector spaces shall be over the field F. Let k be a fixed positive integer ( 2) and let W be a fixed vector space of dimension k over F. Therefore we see that W contains exactly 2k − 1 nonzero vectors; we label these vectors w1,w2, . . . ,w2k−1. Now set n = 2k − 1 and let V = Fn: 1 2 V = {(a1, a2, . . . , an) | all ai 2 F}. The tautological map is the surjective linear transformation : V ! W defined by (a1, a2, . . . , an) = Xn i=1 aiwi 2 W. Since this is surjective, the kernel H (null space) has dimension dimH = dim V − dimW = n−k = 2k−1−k. We call H the (2k − 1, 2k − k − 1)-Hamming code; note that if k = 3, then H has dimension 4; when k = 4, then H has dimension 11. The only result that we need concerning the Hamming codes is the following. We define the “unit vectors” e1, e2, . . . , en 2 V by setting ei = (0, 0, . . . , 0, 1, 0, . . . , 0), where there is a “1” in the i-th position and 0s elsewhere. For convenience, set e0 = 0 2 V . Lemma. Let H be the (2k − 1, 2k − k − 1)- Hamming code. Then V = [n i=0 (ei + H) (disjoint union), where we set e0 = 0. Proof. Note first that if i 6= j, 1 i, j n, then (ei + H) \ (ej + H) = ;. For otherwise, it would happen that ei+ej 2 H; since (ei+ej) = wi + wj 6= 0 (set w0 = 0 2 W), this assertion follows. Therefore, S (ei + H) has cardinality (n + 1) · |H| = 2k(22k−k−1) = 22k−1 = |V |, and the result follows. Now assume that k 2 is fixed and that we have n = 2k − 1 members on the team. Label these people 1, 2, . . . , n. In preparation for the “hat trial,” we shall agree on the following strategy. First of all, let us represent the colors “black” and “white” by the elements 0 and 1 of F, respectively. Therefore, the random placing of hats on the team members’ heads is tantamount to the specification of a random vector v = (a1, a2, . . . , an) 2 V where person i has been given a hat with color ai. Note that it may or may not be the case that v 2 H (in fact it usually won’t be—this is an easy calculation). Therefore, if the distribution vector describing the placement of hats is given by the vector v = (a1, a2, . . . , an) 2 V, then person i can only infer that the distribution vector is of the form v = (a1, a2, . . . , ai−1, 0, ai+1, . . . , an) + biei = vi + biei, for some bi 2 F, and where the vector vi is defined above by what person i sees. In terms of the above notation, here’s the strategy: (a) If vi + biei 62 H for either choice of bi 2 F, then player i shall pass; (b) If vi + biei 2 H, then player i shall guess that his hat color is 1 + bi. We proceed to prove that the above is a welldefined strategy and is, in fact, a winning strategy for all hat distributions v such that v 62 H. First of all, the well-definedness issue stems from (b), above; namely we must show that we cannot have vi +biei 2 H for both choices of bi 2 F. Indeed, were this to happen then we would have ei = vi + vi + ei 2 H. Since (ei) = wi 6= 0 we have a contradiction. Therefore the strategy is well-defined. Next we show that if v 62 H, then the above strategy result in a win, i.e., at least one person will guess his hat color correctly and no one will guess incorrectly. Thus, assuming that v 62 H, we have, by the above lemma that v+ej 2 H for some unique index j, 1 j n. Therefore, we see immediately that person j will necessarily correctly guess his hat color. We contend that everyone else must pass. To see this, fix i 6= j 3 and write v = vi + aiei. For person i to do anything but pass, it must happen that vi+biei 2 H for some bi 2 F. Were this the case, then since v = vi+aiei, and since vi+aiei+ej = v+ej 2 H, we infer that (ai+bi)ei+ej = (vi+biei)+(vi+aiei+ej) 2 H. Since ((ai + bi)ei + ej) = (ai + bi)wi + wj 6= 0 this contradiction proves the result. We could be content to stop here as we’ve already shown that the above strategy results in a win with probability at least 1 − |H| |V | = 1 − 2−k. However, this is the exact probability as it’s clear that when v 2 H, then everyone guesses wrong. What about just telling the colour of hat of next person in cycle. Like A will tell of b , b of c and c of a. It guarantees atleast one will be correct Come on, the most important rule is they must response together! You miss that |

Another was a puzzle: A king orders 100 bottles of wine for a celebration. A courtier who's angry with the king over something puts poison in one of those bottles. The king has a way of identifying the poisoned bottle by giving a few drops of wine to a monkey. Since the poison is fast acting, the monkey will die immediately. Whats the minimum number of monkeys needed to find the poisoned bottle? 5 AnswersAssuming that there is only one poisoned bottle, then you only need one monkey- because as soon as it dies, you found the bottle. However, if there is more than one bottle, or suspicion of more than one bottle, you will need at least two monkeys. you need only one monkey since you will keep giving him wine till he dies and as soon as he's dead you know that was the bottle. I think it means we can only test once. So the minimum # of monkeys is 99, is it? Show More Responses The answer is 7. Let monkeys be numbered 1-n. Each number less than or equal to 100 can be written as a 7 bit number. Hence, bottle one(0000001) is given to monkey 1. bottle three(0000101) is given to monkeys 1 and 3 and so on. Now say monkeys 1 3 5 died, it means that the number is 0010101 which means the bottle 41 is the poison! The way it's written, it seems like one would be the answer, or perhaps considering a monkey's tolerance for wine, N = 100 / (number of drops of wine a monkey can drink before passing out)... When asked with the condition that the monkey will die some time later, i.e. not immediately, the binary number technique described by Sri Krishna is best. |

Given 1000 bottles of wine, 1 has poison, how many rats would it take to find the poison if you only get one timestep. I.E. You can only run your experiment once and check the rats once. 6 AnswersIt would take 10 rats It would take 5 rats 10 rats, I think. 2^10>1000 Show More Responses I get 10 rats with a carefully designed experiment...how to combine the bottles of wine so that each rat gets one tasting? I think the first rats wine gets a mixure of all the odd bottles, the second 2...6...10, third 4....12...20..., and so on. Correction...just the bottles numbers corresponding to a 1 for the decimal-binary conversion: 1 gets 1,3,5,7,..., 2 gets 3,6,7,10,11, so on 10 rats Think about using the rats to create bit representations of each number 0 to 1000. Administer wine to the group as per the corresponding set bits e.g., 1010 would correspond to bottle number 10. It would imply rats 2 and 4 drink the wine (starting from 0 bit). So if in such a scenario, only rat 2 and 4 die, we would know the answer. Express all 1000 bottles in bit equivalents and proceed (1024 - 2^10 ). We would therefore need 10 bottles. |

Tell me about your self? Some questions about my thesis. What is the expectation value by rolling a dice? What if you have the second chance if you feel the first result is not large enough? 4 AnswersExpected Value if die is to be rolled once = 3.5 Expected Value in second case: if output of first roll 1/6*(1+2+3)+1/6*(1+2+3+4+5+6) = 4.5 first part is for first roll and second part is for the second roll. Case doesn't rolls again: => 1/6(4+5+6) = 2.5 =>E[X] = 1/2*(4.5) + 1/2*(2.5) = 3.5 It is wrong. Should be 4.25. Indeed 4.25 Show More Responses El is wrong. Em and John are right. Here's the details- E[1roll] = 3.5 Roll second time only if you get 1,2,3 ie with probability 1/2. So get 4or5or6 with prob 1/2 (stop after first roll) and get expected value of 3.5 (2nd roll) with another prob 1/2 = 1/2 * 5 + 1/2 * 3.5 = 4.25 |

### Business Analyst at Deloitte was asked...

Behavioral: what is the similarity between a milk carton and a plane seat 10 AnswersI believe the answer to that question demonstrate a reasoning ability, and thought process. Looking at the two cases starting with a plane seat; plane seats are meant for comfort and support. Comfort in the sense as to soothe uneasiness in the travel. Support: to hold you in place, aligning to the motion of the plane. Second, a milk carton has more to do with support than comfort because milk are inanimate. I wouldn't think milk carton is made of the same material as plane seat because of the calculated risk involved, and value of the product: human life versus milk content. So, it has more to do with support; so that both don't fall out of place. Duh... Both are a little squishy! Both can float! Show More Responses Its a silly question and it deserve a silly answer as well. Answer: both have flat bottom ..... Both have expiration date. If they are not sold/used by the expiration date/flight time, they are useless Both are designed for efficient transportation both are some form of inventory They are both designed to be lightweight, which translates into saving on transportation costs and energy. They both are designed to hold things safely. They both are designed to fit the maximum 'product' in a given area.(milk cartons are square so you can fit more in a fridge than round bottles) They both have tons of federal regulation regarding their design and how they hold their respective contents. The contents of both have extremely low margins and often serve as loss leaders. ie milk is sold cheap to get customers in to buy other goods and seats are sold cheap to get customers to buy other goods like ryan air. The cost for both are a small proportion of the total price. They both are designed in a very utilitarian manner, but also have luxury versions. paper cartons vs glass milk bottles and coach vs first class seats Indirect complimentary objects that came with the purchase of a product! Those objects are both necessary for the sale of milk, and a flight. The customer is not considering those two objects when they are making their purchase. *Admittedly a seat on a flight has more importance than the carton |

### Analyst at Deloitte was asked...

If you get a 20% raise after your first year and now make $80K, how much did you make your first year? 6 Answers$80K divided by 1.2 (didn't ahve to actually do the math) easy to approximate in your head. Divide 80 not by 5, but by 6. That is 13 1/3. that's the amount of your raise. So 80 minus 13 1/3 is 66 2/3, your original pay. Check it, it works because 13 1/3 is 1/5th of 66 2/3, which is 20%. Elegant. 80/1.2 = 80/(6/5) = 80*5/6 = 400/6 = 66 + 4/6 = 66 + 2/3 = 66.6666667 Show More Responses It depends if "now" is the time after first year. Now might be 3rd of 4th year which will make it impossible to calculate the amount earned the first year as we don't know how much raise the person got the following years. who the devil knows Another simple way to do this is that 80k is know 120% of original salary. So divide 80k with 120 and then multiply by 100 which is 66666.6667 |

### Business Technology Analyst at Deloitte was asked...

A company has acquired many other companies resulting in multiple transaction and report generating systems and technologies within the main company resulting in a $10 million maintenance bill. As a consultant how would you integrate all the technologies and reduce the maintenance costs. 5 AnswersFeel free to ask the interviewer clarifying questions. Stick to the frameworks provided in the website and DO NOT GET NERVOUS! What framework did you use in this case? I'm pretty new to case interview, and feel the frameworks are mostly about profit or market. How to use them in tech case interview? Thank you!! Tech interview frameworks are pretty simple as they are few in number. For this question I applied the implementation strategy framework which turned out to be right. But please do not jump to a conclusion right away. I had to ask many questions and I eventually managed to whittle down the problem to that of an effective and common data management solution where the interviewer offered me multiple choices and I decided to go with the ERP system. And then she asked me how I would go about installing it in multiple offices across the world and in a given time frame etc....It was pretty straightforward from that. The important takeaway is frameworks are important but do not be bound by them. And they do not look for a right or wrong answer from you, but instead study your thought process and your approach to problems you have no idea about...Hope this helped. Best of luck!! Show More Responses What other frameworks are there in tech interviews? Chronos-- you mentioned tech interview frameworks are simple and few in number. I'm having a hard time finding resources for these frameworks, however. Could you mention/explain the different tech frameworks you've come across, or point me to somewhere I can learn more information? I really appreciate it! |

### Trading at Tower Research Capital LLC was asked...

Expected number of coin tosses to get heads/heads. 3 AnswersI assume you mean 2 heads in a row? Then: E(H,H) = 2 + .5*E(H,H) ==> p(H,H) = 4 flips I'm getting 6 flips. Let x = E(H,H) x = 1/4 * 2 + 1/2 * (x + 1) + 1/4 * (x + 2) x = 6 The first part --- 1/4 * 2 --- of the right hand side of the above equation refers to the outcome in which we get two heads in the first two coin tosses. The probability of this happening is 1/4 and the resulting number of tosses is 2. The second part refers to the outcome in which we get a T on the first toss. We are then back to our original problem of trying to get two heads in a row but one coin toss is done. The third part refers to the outcome in which we get a HT. This happens with probability 1/4 and we are then back to our original problem with 2 coin tosses done. A difference equation is often useful here. Let a = expected number of throws to first head. We must make 1 throw at least and we have probability 1/2 of a head and probability 1/2 of returning to a, so a = (1/2)1 + (1/2)(1 + a) (1/2)a = 1 a = 2. Let E = expected number of throws to 2 consecutive heads. Consider that we have just thrown a head and what happens on the next throw. We are dealing with the (a + 1)th throw, with probability 1/2 this is not a head and we return to E. So E = (1/2)(a + 1) + (1/2)(a + 1 + E) (1/2)E = a + 1 E = 2(a + 1) and now putting in the value a = 2 we get E = 2(3) = 6 Expected throws to 2 consecutive heads is 6. |

There are 10 red, 11 blue, 12 green chameleons. Sometimes, two chameleons meet. If they are the same color, nothing happens. If they are different colors, they will both change to the third color. Can all chameleons ever be the same color? 5 AnswersNo. Set of amounts mod 3 is always {0, 1, 2} I'm stating a particular case where this is possible let us assume if 9 green combines with 9 red to form 9 more blues.. then now 3 green and 1 red are left. if 1 green combines with 1 more blue then 1 red will be generated and now the number of red and green are equal which is 2 So just combine these 2 and get all blues I think you misunderstood Tanuj. 9 Green will combine with 9 Red to form 18 Blues. And 1 Green will combine with 1 Blue to make 2 Reds (not 1). Show More Responses The interview candidate's answer is horribly unclear. I don't understand how "No. Set of amounts mod 3 is always {0, 1, 2}" answers this question at all. Here's a cleaner solution: Suppose that we start off with (a, b, c). A transition could be (a', b', c') = (a - 1, b - 1, c + 2). Now, note that each time we take a transition, we see that b' - a' = b - a mod 3 c' - b' = c - b mod 3 a' - c' = a - c mod 3 This is an invariant in the problem! Specifically, this means that in our case, b' - a' = c' - b' = 1 mod 3 and a' - c' = 2 mod 3, always. Now, we want to get to the state (0, 0, 33). This isn't possible, since b' - a' = 0 mod 3 (this violates our invariant). So, this isn't possible. I just think the problem this way and I am not sure if it makes sense. Since each time we can only convert any two chameleons into the third color, so definitely the final case cannot be all red or all green, since 11 blue ones cannot be paired with others, that means it finally reaches the state (0, 1, 32). So the only possible case is all blue finally. But as the # of red and green ones are imbalanced, so it only to arrive at the state (0, 2, 30), which cannot reach (0,0,33) neither. |