It also allows an endless pissing contest of comparing new brainteasers among your colleagues, and telling stories of helpless candidates who couldn't even manage the first gasp of a hint of an idea towards the solution.
I tend not to like extremely complex brainteasers as interview questions, and especially not ones that require an "aha" moment to solve them correctly, because mostly what you're testing for is whether the candidate has seen that question before (or one like it).
On the other hand, I am a big fan of asking relatively easy math questions (i.e. first year undergraduate) in interviews, because if I'm going to be paying someone $100k+ to work on mathematical models, I'm damn well going to make sure that they understand basic probability, statistics, calculus and linear algebra. These are a bit like fizz-buzz for quants - if the candidate can answer them it doesn't prove anything, but if they can't answer them then you can stop the interview right there.
These are not very tricky brainteasers that depend on getting a particular insight. They're actually somewhat pedestrian. However, if you can't manage to competently work out a simple, well defined problem, you're going to struggle with the more complex issues we deal with.
I honestly have no idea.
chollida1's comment is spot on and I oscillate between 1 and 2.
"Give me an overview of a numerical method used to solve differential equations."
The more time you spend asking brainteasers, the less time you have to devote to your actual skills of interest. You may even find that quizzing someone on mathematics may help refresh and solidify your own understanding.
Many candidates will answer this question correctly and yet be totally unable to do anything when they're confronted with a non textbook case. To be clear the brain teasers I ask are mathematical problems, not the type of brain teasers used in consulting interviews. For instance:
We play a game where we each draw a secret random number uniformly between 0 and 1. We each may re-throw if dissatisfied with our first throw, or me may keep it. We do not know whether or not the other has chosen to re-throw. We then compare our results and he who holds largest number wins $1. What is the best strategy to follow?
That's the type of brainteaser I'd ask. It's accessible to a good high school student. I interviewed a PhD candidate in applied mathematics from a top Ivy league university who:
- wouldn't believe that maximizing the expected value of the number obtained wasn't optimal until shown an explicit counterexample
- was unable to write the equations properly or model the problem
- was unable to solve the equations after I handed them out to them
He was however able to talk about his thesis work. Your questions wouldn't have caught that at all. His thesis work was in game theory.
So, an explicit counterexample would be my opponent picking a strategy of only re-throwing above .25. His expectation is then .25 * (0 + .25)/2 + .75 * (0 + 1)/2 ~= .40, so I should not rethrow if I get above .40 and below .50, even though it would raise my own expectation.
Am I thinking about that correctly?
How do we maximize EV? A single throw's pdf is 1, for x in [0,1], so its EV is 0.5. The question is how to improve on a single throw by deciding to re-throw. A re-throw is independent and gives the same EV. We want a strategy that gives us higher cumulative EV. Say our strategy is that we have a threshold A, where we re-throw any result below A. Because x is uniform, the probability that we re-throw is also A. The cumulative EV of the strategy is A * EV(second_throw) + (1-A) * EV(keep_first_throw). Since we only keep the first throw for results in [A,1], the EV for that event (integrating x * pdf from A to 1) is (1+A)/2. So EV of the whole strategy is A/2 + (1-A) * (1+A)/2. It has max EV when A is 0.5, giving EV of 5/8.
So how do you do better?
Imagine we play a die game where whoever rolls the largest number wins. Which die would you rather play with?
1-1-1-1-1-10^250 or 2-2-2-2-2-2
The first die has an EV of about 1.67e249, the second die an EV of 2. Yet, the second die will win 5 out of 6 games against the other one.
To solve the problem, you must find the Nash equilibrium of the game. That is, you must find a strategy which the opponent cannot exploit, no matter what he does.
Nash's brilliance was in proving that under reasonable conditions, a mixed equilibrium always exists, which is far less obvious.
to get this I computed the function f(x,y) which is my expected value if I reroll if roll <= x and my opponent re-rerolls if roll <= y. let p be the optimal value. by symmetry f(p,p) = 0.5 so there must be a local minimum around that point. solving by taking a one sided derivative gives me (sqrt(5)-1)/2
(to be thorough, you also need to show that re-rolling based on a single threshold dominates other strategies, but it's fairly intuitive and not too hard to prove)
There's smart quants/programmers/traders/etc and people good at solving brainteasers. You need the former, but people who use brainteasers in the interview process typically only find the small sliver of overlap in the venn diagram.
If you're interviewing someone and you're lazy enough (or not confident enough) in your skills at assessing a candidate, that's a weakness.
They also liked taking Ivy grads with BAs and giving them receptionist jobs.
/s
Sales is tied to performance. Hedge funds' customers are even more conservative than they are.