# 414

Quantitative Trader interview questions shared by candidates

## Top Interview Questions

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### You bid for a coin. You're confident that the price of the coin is between 0 and 100, if your bid is greater than the price, you win and sell it to your friend at the price of 1.5 times price. what's your bid to max your profit?

I loved this question and want to renew this debate. What do you guys think about my two approaches to solve it: 1) If we can only play this game once AND our goal is to maximize profit (as the question states). I agree with above that expected value of a coin is 50. Given that we bid 51 to win auction and pocket 24. Problem is we only win if coin is (0:50) which gives us new expected value of 25, and so we lose. We can deduct this way all the way to zero bid. 2) Nothing beats little Monte Carlo experiment. I created a matrix of 100X1000000. Where 100 is the number of possible bids given certain price. 1M is the number of random uniformly distributed prices 0-100. Calculated expected gain at each bid level 0 to 100. I wish I could post a MATLAB graph here. It looks as downward facing 1/2 of parabola with max value of 0 and min of -25. Results: best gain of 0 achieved at 0 bid, worst average gain of -25 is at 100 bid. Comments appreciated! Less

I am 95% sure it is 0. The people saying 100 are not correctly counting other numbers. If you bid 100, then it could be 0, meaning you sell it for 0. Or 1, and you sell it for 1.5(losing 98.5), et cetera until 100. You make a profit if &gt;66 gets drawn, but it doesn't outweigh the loss from what I can see from doing trials on 1,2,3,10 Less

### There are 3 coins. One coin has heads on both sides, one coin has tails on both sides, the third one has head on one side and tail on the other side. Now I pick up one coin and toss. I get head. What is the chance that the coin I picked has heads on both sides?

2/3

Because you have 1/3 chance to get double head coin and you will surely get head, 1/3 chance to get single head coin and then 1/2 chance to get head. So the probability of choosing double head coin and get head is 1/3, while choosing single head coin and get head is 1/6. Then, given you get head after tossing, then chance that you chose double head coin is (1/3)/(1/3+1/6) = 2/3 Less

### N points lie on a circle. You draw lines connecting all the points to each other. These lines divide up the circle into a number of regions. How many regions is this? Assume that the points are scattered in such a way as to give the maximum number of regions for that N.

it is "Moser's circle problem" 1+nC2+nC4

This is a famous example in mathematics; it's often used as a warning against naive generalization. Here are the answers for the first six natural numbers: (# points) : (# regions) 1 : 1 2 : 2 3 : 4 4 : 8 5 : 16 6 : 31 Yes, 31. You can see, e.g., Conway and Guy's "The Book of Numbers" for an account of this. Less

Err, 2*n-1+summation(n-1C-&gt;n-1)

### There are 25 horses. Each time you can race 5 horses together. Now you need to pick the top three horses among them. How many races do you need to conduct?

I think it is 7. Kannp, the top five are not established after 5 because the second in one race might be better than the first in another race. Have five races of five each and keep the top 3 in each races. Then, take each of the winners and race them against eachother. The two bottom and the 4 who lost to them are discarded. The two who lost to third place is discarded. And, the one who got 3rd in the race with the horse who gets second in the 6th race is discarded. Now, there are six horses left. Race all but the horse who won twice and keep the top two, combined with the horse who sat out. Now you are done in 7 races. Less

He gave a very good explanation. I'll decline to explain why prm is incredibly wrong. Less

7 times.

### If we flip a coin 100 times, what is the probability of getting even number of heads?

1/2. Whether even or odd heads is ultimately determined by the nth flip, for any n. Probability even to odd or vice versa is n. Less

( (2^99)+1) / (2^100) Even numbers include 0 heads to 100 heads. There exist 50% even complements and 50% odd complements from 1 heads to 100 heads.. So half our sample space. Now we do our zero case which is no heads and there is only 1 outcome. Less

1. 0-99 flips: 50% even; 50% odd 2. P(even after 100 flips) = P(odd after 99)*0.5 + P(even after 99)*0.5 = 0.5*0.5+0.5*0.5 = 0.5 Less

### The only difficult question I was asked went as follows: &#034;You and your opponent flip a coin. If the first player gets heads, the second player pays him $30. If he flips tails, the coin goes to the second player. If the second player flips heads, he wins$30 from the first player. How much should you pay to go first?&#034;

Let x be the expected value of the game. Then, we can see that the following recursion holds: x=0.5(30)+.25(-30)+.25(x) It follows that x=10, so I'd pay no more than \$10.00 to go first. Less

no more than 7.5?

The question is not very clear : does the game continue until someone gets heads ? Less

### Imagine I flip 100 coins in a sequence, and you need to guess the sequence. You are allowed to ask one yes/no question. What do you ask to maximize the probability of guessing the sequence?

The highest information gain is when you split all possible sequences into two equal sized groups, so the simple question "is the first flip heads?" yields the best chance. Asking if there are more heads than tails is not correct since it does not split the possibilities into two equal groups (one group contains the 50/50 split sequences) Less

I was also asked this question..... Totally lost..

I think it's asking if there are more heads than tails. Once you know there are more heads than tails, you know you need at least 50 of them, and you have 100 choose 50 places to put them. Then you have 2^50 choices for the remaining coins, so your total number of possible combinations is (100 choose 50) + 2^50. If you only ask if the first toss is heads, then you have 2^99 possible combinations. Using wolfram alpha, 2^99 is bigger than (100 choose 50) + 2^50 so asking if there are more heads than tails is better. If I had to compute 2^99 and (100 choose 50) + 2^50 without using wolfram alpha I would be screwed. I also don't think a rough approximation would work since they're of the same order of magnitude. Less

### You flip four coins. At least two are tails. What is the probability that exactly three are tails? Do this in your head, you are not allowed to write anything down for this question.

At least two are tails, no HHHH neither THHH, HTHH, HHTH, HHHT. 2^4-5=11 possible scenarios. Three tails, HTTT, THTT, TTHT, TTTH = 4 cases P=4/11 Less

its 4/11... it'd only be one half if they told you the first two coins were tails. Use bayes rule or actually write out all the possibilities Less

can you show me the math behind it? i got P(B|A) x P(A) / P(B) = (1x 1/4) / (5/8)=2/5... Less

### What is the probability of getting a sum of 10 if you roll three dice?

To turn this into a probability, note the conditioning: Rolled 1, 9 left: 4 ways (out of 36) 8: 5 ways 7: 6 ways 6: 5 ways 5: 4 ways 4: 3 ways Now each of the first rolls is 1/6 probability, hence you have total probability: P(10) = P(9 from 2 dice|1 from first) * P(1 from first) + ... Which is 1/6 * 27/36 = 1/8 Less

The combinations is not 6*3 but 6*3+3*3=27, which will give the probability 1/8

You will have the following possibilities with 3! permutations: (1, 3, 6) ; (1, 4, 5) ; (2, 3, 5). Besides, there are other 3 vectors resulting into 10 but with only 3 permutations each: (2, 2, 6) ; (2, 4, 4) ; (3, 3, 4). That way, we have 6⋅3+3⋅3 = 27 possible combinations. This divided by 6^3, finally gives us the Probability{Sum(a,b,c) = 10}= 1/8. # Less