Assistant Trader Interview Questions | Glassdoor

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Assistant trader interview questions shared by candidates

## Top Interview Questions

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Jan 1, 2011
 How much would you pay to play the following game: I roll a dice and pay you the amount showing (so if I roll a 4, I give you \$4). 5 Answers 1+2+3+4+5+6 / 6 = 3.5 If the EV is 3.5, then one would pay \$0.49 to play (assuming a 2% ROI is acceptable!) never mind, didn't read the ? properly! Show More Responses if you're expected over a prolonged period of play are 3.50; (determined in a similar fashion to the expected value on a roll of a die), then you would be willing to pay 3.49 to play. i would not think you'd play for 3.50 bcause you're expected returns are 0, and you would have to factor in the oppurtunity cost of the time spent winning nothing assuming risk neutral, any amount smaller than or equal to 3.5. for the second part, 4.25

Dec 3, 2011

Jan 27, 2010
 square root of .16 5 Answers .4 -- it's basically the square root of 16 and you move back a decimal point. +.4 or -.4 nign on the -.4 square root by convention must be positive. Show More Responses the answer above is incorrect. indeed the answer can be -0.4. the square itself may not be negative however If x^2 = y, Then x = + or - square root of y square root of any real number is positive

Mar 2, 2010
 What is the chance that at least two people were born on the same day of the week if there are 3 people in the room? 6 Answers ~89.8% incorrect. 1-364/365*363/365 = 1-.9918 = .0082...something like this for 2 people, for 2+ add in the chance of 3 of 3. So first guy then gets 365/365 choices, next two get 1/365 Also incorrect... They are asking if they were born the same day of the week not the exact same day as Chase alluded to. In this case the easiest way to do the problem is 1- probability that no one was born on the same day. 1-(1)(6/7)(5/7)+ 19/49 or 38.78% Show More Responses My bad =19/49 Or you can calculate it directly: The chance of SELECTED two people(but not three) that are born on the same day is 1* 1/7 - 1/7*1/7 =6/49 Now we have 3 chooses 2 = 3 possible pairs, so the probability that any PAIR(but not all three) that are born on the same day is 3* 6/49 = 18/49. However, since the question asks "at least 2", we will need to add the possibility that all three are born at the same day, which is 1/7 * 1/7 = 1/49. So the final result is 18/49 + 1/49 = 19/49 This obviously is a worse way of calculating it, but it's an alternative for some people who wants to think "straight". 1-(7/7*6/7*5/7) = 19/49 = 0.39 # 1 minus the opportunities that all 3 people were born in different day of the week.

Jan 12, 2012
 Hospital has three boys and an unknown number of girls. A mother has a baby. A nurse picks up a baby at random and it is a boy. What is the probability that the mother had a boy? 5 Answers 4/7 There are only 2 outcomes, the probability is always 50%. P(F) = .5, probability of giving birth to a female P(M) =.5, probability of giving birth to a male P(B) = ?, probability of a nurse choosing a boy x = number of babies in hospital after mother gave birth so the question wants you to find P(M|B) = conditional probability of mother giving birth to a male given that the nurse picked up a boy. so P(M|B) = P(M∩B)/P(B) If mother gave birth to a boy, then odds of picking a boy is 4/x If mother gave birth to a girl, odds 3/x Both these cases have a 50% chance of happening to total probability of picking a boy is: .5(4/x) + .5(3/x) = 3.5/x = P(B) P(M∩B) = probability that the woman gave birth to a boy AND that the nurse picked it which is .5 * 4/x = 2/x so P(M|B) = P(M∩B)/P(B) = (2/x)/(3.5/x) = 4/7 Show More Responses how do you get 3/x? Explanation has the best answer on the net Anon (above): one gets 3/x as this is the prob of selecting boy if new birth is a girl (as opposed to 4/x if mother gives birth to boy)

Feb 7, 2011
 We each flip three fair coins. I offer to pay you \$1 if we do not get the same amount of heads, if you agree to pay me \$2 if we do (get the same amount of heads). Will you agree to play this game? 5 Answers Play the game. EV = 1/16. Screwed up again, EV is -1/4, don't play Agreed. EV is - 0.25 Show More Responses would you mind sharing the math behind the answer Rob. Thank you 1/2^6 + 9/2^6 + 9/2^6 + 1/2^6 = 20/64 the probability to get same number of heads. Expected value : 20/64 X (-2) + 44*/64 * (1) = 1/16 play the game

Nov 29, 2010

May 4, 2011
 If we play a game in which Player 1 picks a number 1-11, and then player 2 can add 1-11 to that (i.e. player 1 picks 5, player 2 can add to make it 6-16), what is the strategy to win this game if Player 1 wants to make 60? 5 Answers 54 Use induction on each step. You win with certainty if your opponent has to add to 12j, j=0,1,2,3,4 Let them go first so they will say anything between 1-11, then you say 12, then they go, then you say 24, then they go, then you say 36, then they go, then you say 48, then they have to go for a number between 49-59 and then you say 60 Show More Responses You should start with number 10. 12 - 24 - 36 - 48 => 60 == you win