Trader Interview Questions in New York City, NY | Glassdoor

# Trader Interview Questions in New York City, NY

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Trader interview questions shared by candidates

## Top Interview Questions

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Apr 7, 2011

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Jan 12, 2011
 Russian Roulette - 4 blanks 2 bullets, all in a row. If someone shoots a blank next to you, would you take another shot or spin 12 Answerstake another shot, 3/4 chance of surviving vs 2/3 if you respinHere is my answer: the prob. of survival after re-spin is: 3/5. the prob. of survival with on re-spin is: 1/15 * 6/4 = 1/10.correction: the prob. of survival after re-spin is: 4/6.Show More Responsesmy final correction: the prob. of survival after re-spin is: 4/6 = 2/3 the spin with no spin is: 2/5 = c(4, 2) / c(6, 2).Denoting the blanks by 0's and live bullets by 1's, and adjoining the left and right edges (denoted by ~) , so as to make a cylinder, the following diagram illustrates how the bullets are arranged in the cylinder of the revolver: ~000011~ . Denote the bullet that is to be fired if the trigger is pulled by enclosing it in parenthesis. Denote an empty chamber by *. Assume that when you pull the trigger, the cylinder rotates clockwise (to the right in the diagram above). If when you pulled the trigger for the first time, the bullet was a blank, then before and after you pulled the trigger, the cylinder was and is in one of the following states: 1) Before: ~(0)00011~ After: ~*0001(1)~ 2) Before: ~0(0)0011~ After: ~(0)*0011~ 3) Before: ~00(0)011~ After: ~0(0)*011~ 4) Before: ~000(0)11~. After: ~00(0)*11~ If you pull the trigger again without spinning the cylinder, then only in case one will the bullet be a live bullet, yielding a 3/4 probability of the bullet being a blank . If you spin the cylinder and then pull the trigger, then you have a 4/6=2/3 probability of the bullet being a blank. Clearly, you should not spin the cylinder.don't spin, 3/4 prob survival if not spinnedrespin - 4/6 = 2/3 (obvious) don't spin - you only die if the blank was the "last" one, which is 1/4 chance. hence 3/4 chance of live. since 3/4>2/3 don't spin.Wouldn't play.Spin Spin: 4/6=0.6667 to survive Not Spin: C(4,2)/C(5,2)=3/5=0.6condition on 2 bullet is in consecutive position and someone survives the 1st shot, we have: spin it-> survival rate is 4/6=2/3=67% not spin it-> survival probability=3/4=75% so...not spin it will have a bigger chance of survival.the question did not say that the bullets are next to each other so in this case you should spinI. Using Bayes' Theorem: P = probability of a blank : 4/6 P = probability of a bullet : 2/6 P[1|0] = probability of a bullet given a blank : find this P[0|1] = probability of a blank given a bullet If a bullet occurs then the next pull can be a bullet or a blank and so P[0|1] = 1/2 P[1|0] = P[0|1]*P / P = 1/4 So there is a 25% chance that the next pull is a bullet, or a 75% chance that it is a blank. The first pull had a probability of 8/12 of being a blank. Given a blank on the first pull, the second pull has a 9/12 probability of being a blank. II. From a frequentist perspective: So when the first pull is made and it is a blank the event space decreases from 6 possible outcomes to 4 possible outcomes. The first pull was on the last bullet or the the first pull was on the second to last blank. Out of the four possible events there is one where after the pull the chamber is sitting on the blank right before the bullet. Out of the four events 3 are blanks (3/4) and 1 is a bullet (1/4).

Apr 7, 2011

Oct 4, 2010
 You have a box filled with cash. Cash value is uniformly randomly distributed from 1 to 1000. You are trying to win the box in an auction: you win the box if you bid at least the value of the cash in the box; you win nothing if you bid less (but you lose nothing). If you win the box, you can resell it for 150% of its value. How much should you bid to maximize the expected value of your profit (resale of box minus bid)?13 AnswersMy bad, I meant "0 to 1000" not "1 to 1000". Not that it affects the answer (hint hint).the expectation is always negative?bid 0. The expectation is negative.Show More Responsescan someone explain why it is the case mathematically?? I thought your expected payoff is (0+1000)/2 x 1.5=750.. so if you bid 500. you will get expected 250 profit...Anon, If you bid 500 and it is worth 1, you only get paid $2 so you lose 498.$0 is correct. Generalize this for 1,2,5,10 maybe and find the EV of all those, or just 1,2,10, and then you will be able to "guess" that it isn't getting more and more negative and you should bid 0you should bet 1, that is the minimum in the box, so you will win 0 or .5if you bet one you'll never win. if you bet 500, you're missing out if value(box) >500 and <750 since those are +EV too. believe the answer is 749.The answer is bid 0 dollars. Soln: If you bid X dollar, you get the box if if it contains Y dollars, where Y < X. Or you nothing, which doesn't hurt anyway. Now think of possible values of Y in PAIRS, (0, X), (1, X-1), (2, X-2)... in each pair (a, b), expected payoff is P(Y being a or b) * E(payoff given Y = a or b). but the second term E(payoff given Y = a or b) = 1/2 (3/2 a - X) + 1/2 (3/2 b - X) = 3/4 (a + b) - X = (-1/4) X, which is always negative. If we sum over all possible pairs, we should get a negative number. Ofocz you need to handle the edge case, where X is small and the pairing does not really work.I think that since we are dealing with expected profit, we should ignore the case when n (our bid) is less than X (cash in the box) as nothing happens in this case. So we create a new distribution where X is uniformly distributed on [0, n]. Thus, E(Profit) = E(1.5X - n) = 1.5E(X) - n = 1.5*n*(n+1)/(2*(n+1)) - n = 0.75n - n = -0.25n <= 0. So the expected profit is maximised (=0) if n=0.Sorry, guys. I seem to have made a mistake in my solution above. Let X = cash value, n = our bid price. Then Profit = 1.5X - n. We want to find E(1.5X - n) = 1.5*E(X) - E(n). Now in order to find E(X), we notice that that X is uniformly distributed. So every outcome has a probability 1/1001. Also we only need to sum up over the values from 0 to n. Hence, E(X) = n(n+1)/(2*1001). For E(n), we see that we will pay the amount n with probability (n+1)/1001 and 0 otherwise. So E(n) = n(n+1)/1001. Therefore, E(Profit) = 1.5*n(n+1)/(2*1001) - n(n+1)/1001 = -0.25*n(n+1)/1001<=0. In order to maximize profit, we should bid zero.The answer is 0 if you ask me and here is the reason. let V be the value of the box and let x be your bid then you profit P is P=(1.5V-x)1_{V<=x}. we take expected value of this and use conditional expectation to get E(P)=E(1.5v-x|v<=x)p(vThe answer is 2 with EV 1/3996. Let x, t, w be your pick, the cost, and your winnings. If x 1000 we should just pick 1000. Otherwise 1 x we have E(w) = 1/999 int (1.5t - x)dt from 1 to x, which equals 1/999(3/4 x^2 - x^2 - 3/4 + x)=1/3996(-x^2+4x-3) = 1/3996(1-(x - 2)^2) <= 1/3996 with equality iff x = 2.Answer is 2. The key is that the amount of cash X is uniform on (1,100) rather than (0,100). If we bid y money, then are expected value of profit P is E[P | bid y]=(\int_1^{y}(1.5-y)dx)/99=-(y^2-4y+3)/(4*99). Here the limits of integration are because P will be 0 if X is smaller than y, and we will get profit 1.5x-y if X=x and x<=y. Solving the maximum of this function gives y=1, with an expected profit of 1/(4*99)=0.0025. If the value was in fact from (0,100), then our expected value for any y would just be -y^2/(4*99) so we wouldn't want to play that.