Trade Assistant Interview Questions | Glassdoor

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Trade assistant interview questions shared by candidates

## Top Interview Questions

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Sep 28, 2010
 What is the expected value to you of the following game: You and another person have 3 coins each. You both flip all of your coins and if your three coins show the same number of tails as the other person, you pay them \$2, otherwise they pay you \$1.4 Answers1/16Here's the explanation to the above (Since none was given): There are four different number of Tails that can be shown when you flip 3 coins: 0,1,2,3. The Probability of 0 Tails, P(0T) = 1/16, P(3T) = 1/16, P(1T) = 3/16, P(2T)=3/16. Since your flips and your opponent's flips are independent, we know that the Probability that I roll 0 tails and you roll 0 tails = P(0T)*P(0T). Thus, square each of the above probabilities and sum them: you get 1/64+1/64+9/64+9/64=20/64. Thus P(same number of Tails) = 20/64 = 5/16. Now it is asking you for expected value, so plug the above probability into an expected value equation: (5/16)(-2) + (11/16)(1) = 1/16 ie, 5/16 of the time you roll the same number of tails and lose two dollars, 11/16 of the time you don't and you win a dollar, thus your expected value of each play of the game is 1/16 of a dollar. (In the long run if you play many times you should expect to win that much per play)The above has the correct answer, but the probabilities are 1/8, 1/8, 3/8, 3/8. Then square them....and so on.Show More ResponsesProbabilities are: p(0)=1/8, p(1)=3/8, p(2)=3/8, p(3)=1/8.

Sep 28, 2010
 What is 29^2?4 Answers841Mental math. You could do it mentally. 29 is close to 30. So the upper bound is 30*30 = 900. You would have to subtract 30 and 29. 900- 59 = 841.or, take 29 x 30 to get 870. Take away one 29, get 841. Same idea, but more elegant. You've multipled 29 thirty times, you only need 29 of them, so take one away.Show More ResponsesEasiest way I can think of is to mentally calculate (30-1)(30-1) and then mentally foil the answers. You end up getting 900 - 30 - 30 + 1 = 841 which is pretty easy to do in your head.

Jul 3, 2009
 If you have a deck of cards split into 4 piles and was offered 1:1 odds to draw a face card (J Q K A) from at least one of the piles, would you take the game? Why or why not?4 AnswersNo, basic statisticsI'm no expert but I don't see it that way. You lose the game if you draw non-face cards from all 4 piles. The probability of that is 39/52*38/51*37/50*36/49 = 30%. So you have a 70% chance of winning the game.Yes, play the game at 1:1 odds. There are 16 face cards and 4 equal piles of 13 cards. By adding the probabilities of drawing a face card from at least 1 pile, you get 16/13. (1/13 + 1/13 + 1/13 + 13/13, or 4/13 + 4/13 + 4/13 + 4/13, and so on)...Doesn't matter how the face cards are arranged in the 4 piles. Thus, you would be willing to play the game at 1:1 odds.Show More ResponsesIf there are 4 piles and you can look at all of them then it's obvious that you're going to take it - all the cards are there. If the question is to take the top card of each of the piles, then it's equivalent to finding the probability of a face card in 4 cards dealt. That is equal to 78.25%. So you would take the bet. (Total = 52 choose 4, Not Wanted = 36 choose 4, P(A face card) = 1-Total/Not Wanted) This is different from the answer given by the previous person because the numerator is wrong: he started at 39 when there are in fact 36 non-face cards (16 face cards - 52-16=36) If, however, the question is to pick one of the piles and then see if it has a face card, it is equivalent to randomly shuffling the deck and seeing if there is at least one face card in the first 13 cards. The probability of not having a face card is 0.363% and thus, the probability of having a face card is 99.637% - this means you should take the bet. So while the question can be interpreted in different ways, you should find it that taking the bet is a good proposition. Another way to look at it is as follows: if we assume sampling with replacement, to make calculations easy, we basically have 4 (or 13, depending on how you view it), chances to pick at least one card with probability 4/13 of the deck. This means that the probability of what is asked is almost equal to 1-(9/13)^4=77.04% (really close to the first case) or almost equal to 1-(9/13)^13=99.16% (also really close to the first number).

Jun 12, 2012

Feb 20, 2012
 Roll a fair dice, if it's a prime number, you pay me that number times \$100, if it's not a prime number, I pay you that number times \$100. Consider 1 as a prime number. How much willing to pay to play this game.7 Answersanything below \$387.50Don't be surprised if you have not been employed. The question is "what is the expected return of this game?": E[X]=Sum(P(Xi)*Xi)=1/6*(-1*100+2*100+...)=50\$ Esco910, we make a deal when you want, I have always dreamed about to buy a ferrari2 is a prime number, and the expected value is less than 0. So I wouldn't play this game.Show More Responses\$50 is correct, the others are wrong. You would play this infinitely many times because on average, you are printing money.If you consider 1 to be a prime number then \$50 is the expected value.. but mathematicians do not consider 1 to be prime. Since 1 is NOT a prime number then the EV = (1/6)(100 + 200 - 300 + 400 - 500 + 600) = 83.3333333. But to pay less than that amount for 1 try is not very practical. In the interview it is wise to ask "how many times do i get to play this game since the EV is high?" If they say once, you'd be a fool to pay \$83 to play this game only once.. What would you really pay to play once? Twice? 10 times? 100 times? 100,000 times? 1,000,000 times? Infinite? It's about perception, but clearly as the number of times you play goes up then you would pay more to play each time. Mathematics is not always the solution in trading. This is a very good question to ask... anyone can calculate EV but it's definitely not about that. EV only tells you what's going to happen over time.2 is definitely a prime number... the expectation is less than 0Since there seems to be some confusion here is a list of all outcomes and there values 1: -100 2: -200 3: -300 4: +400 5: -500 6: +600 Since they are all equally weighted (1/6) we can average. -100/6= -16.66666 You wouldn't pay to play this game.

Oct 14, 2010
 How many degrees separate the hands on an analog clock at 3:15?3 AnswersI would say it's 5 degrees, though momentarily I would be tempted to say zero. Basically my maths is as follows: 1) Since 15mins also point at 3. We need to find out the small shift in the Hour Hand that was made to represent this 15mins. 2) 360d/12 = 20d (this represent the degress between each hour) 3) 20d/4 = 5 degrees (to represent the 15mins shift). Hope i'm correct. Open to corrections if any :)the answer is 7.5. the minute hand will be on the 3, the hour hand will be 25% of the way from 3 to 4. there are 30 degrees in each hour (360/12), so 30/4 = 7.5Adam is right. 7.5 degrees.

Mar 6, 2012
 What are odds of getting EXACTLY two (2) heads if you flip a fair coin 10 times?2 Answerscombinatorial problem: let T = total number of flips, n= number of heads you want; then the number of possible ways= T!/[(T-n)!*n!]. thus 10!/(2!*8!)=45 the total number of flips= 2^10 (assuming fair coin, even odds). Thus the odds are 45/2^10~ 4.3%The question asked for you to calculate the odds, not the probability.

Feb 6, 2014
 What angle does that clock make at 4:15? If you have 100 red and white marbles and had to place them in two separate jars - how would you distribute the marbles to give you the highest chance (EV) of getting a red marble when you are blindfolded? You have to use all the marbles. 1 Answer360 degrees in a clock, divide amongst the hours - you can guess the rest. 1 red marble in one jar and 49 red marbles and 50 white marbles in the second jar. 1/2*1 + 49/99*1/2 = answer.