WorldQuant Quantitative Researcher Intern Interview Experience — Round Two Expected-Value Brainteasers

The whole process has three rounds of interviews. This is round two.

Interview content

Behavioral

• Self-introduction
• How much I know about WorldQuant
• Walk through my past internship experience
• What I know about finance and investing
• Can you talk about your own investing experience?
• Why did you start investing?
• What did you learn from investing?

Math questions

Problem 1: Four boxes

There are four boxes, and only one of them has a 100 yuan bill inside.

Rules of the game:

• Each time you start the game, you pay X yuan
• Each turn you can only open one box that hasn't been opened yet
• You can stop the game at any point
• The game ends once you find the 100 yuan

Question 1: What's the most you'd be willing to pay to start this game?

Answer: 40 yuan.

One way to think about it: let the price of the game be X.

On the first opening:

• There's a 1/4 chance of getting the 100 yuan directly
• There's a 3/4 chance you move on to the remaining three boxes

Let V4 be the value with four boxes left, V3 the value with three boxes left. We have V3 = (1/3)×100 + (2/3)×V2. Continuing the recursion: V2 = (1/2)×100 + (1/2)×V1, and V1 = 100.

Working it out: V2 = 100, V3 = 100, V4 = 100.

But since you have to pay the game fee again every time you continue opening another box, you can set up a recurrence for it. Solving for the point where the player is just barely willing to start the game gives X = 40.

Follow-up: Say you've already paid 40 yuan in the first round and didn't find the 100 yuan. There are three boxes left now. Do you keep going or stop?

Answer: Keep going.

Reason: the 40 yuan you already paid is a sunk cost — it shouldn't factor into the decision. What you're facing now is a new decision: the expected value of the remaining three boxes is still higher than the cost of continuing, so continuing is the optimal strategy.

Problem 2: Rolling a die three times

There's a fair six-sided die.

Rules:

• You can roll up to 3 times
• After each roll, you can choose to stop
• Once you stop, your payoff is whatever number you rolled last
• If you continue, you have to accept the next result
• You only pay once for the whole game

Question: What's the most you'd pay to play this game?

This is fundamentally an optimal stopping problem.

Let V1 be the expected value with only the last roll left, V2 the expected value with two rolls left, and V3 the initial state.

First: V1 = (1+2+3+4+5+6)/6 = 3.5.

Next, for two rolls left: if the first roll is 4, 5, or 6, stop immediately; if it's 1, 2, or 3, keep rolling. So V2 = (1/6)(3.5+3.5+3.5+4+5+6) = 25.5/6 = 4.25.

Finally, for three rolls (the initial state), the threshold becomes 5: stop on 5 or 6, keep rolling on 1 through 4. So V3 = (1/6)(4.25+4.25+4.25+4.25+5+6) = 4.6667.

Final answer: the most you'd pay to play this game is about 4.67 yuan.

Both of these are pretty common expected-value questions at WorldQuant.