I am graduating soon and I started interviewing with the Big I banks for a Quant Strategist position. I had close to 8 phone interviews over the past 3 weeks all with Russian Theoretical physicists. Being a mere master’s student, you can imagine how intimidating it was. Anyways, I thought it would be fun to share some of the insanity I’ve seen:
One liners:
Question 1: How many digits does 2^71 has? How confident you are in your answer?
Question 2: How many primes are there less than 1000? Provide a confidence interval.
More involved games:
Question 3: A frog sits on an infinite line. It can jump in steps of 2 to the right and in steps of 1 to the left ( both with probability of 1/2). What is the probability that it returns its initial position.
Question 4: You have a standard deck of cards with 26 reds and 26 blacks placed on a table face down. You and a player take turns flipping the cards , one by one, if it’s red you win 1 dollar if black you loose 1 dollar. What is the optimal strategy to play this game? i.e When should you stop.
I have plently more if folks are interested.
Surprisingly , I managed to do well . I’ll find out in 2 weeks if I got the job 
January 8th, 2010 at 6:54 pm
Solving the y= 2^x = e^(x ln 2).
1og10(y)+1 will give you number of digits in y.
log10(y) = x ln2/ln10 = x log10(2) ~ 0.3 x.
=> # of digits = 0.3*71 + 1 = 22.
January 8th, 2010 at 7:25 pm
Before leaving, i’ll write the answer to the frog questions:
pr = prob of moving right
pl = prob of moving left
for i integer. When k = 3i
p_(3i) = (3i)choose(i) pr^i pl^2i
otherwise p_k = 0
January 9th, 2010 at 5:45 am
you are right about the fist one, it is simply 72/log2(10) rounded up. Your approach for the second is the long and tedious way and doesn’t lead you to a closed form answer. You can get a number if you solve it by induction. Basically if you are at n, there are only 2 ways you could have gotten here, by taking +2 step from n-2 or by taking -1 from n+1 so you can say p(n)= pr*p(n-2)+pl*p(n+1). This is a difference equation that you can solve easily. it has a solution of the form c^n. You replace in equation and solve for c and then use boundary conditions to get the final answer.