Fair Value of Number Game

This quant interview question revolves around expected value under optimal stopping, where you repeatedly sample from a known discrete distribution at a fixed cost. It sits at the intersection of probability, decision theory, and dynamic programming, and is a classic example of how quant traders think about paying for information to obtain better payoffs. In quant prep, it is a quintessential expected value puzzle with a strategic twist.

It trains your intuition for optimal stopping, stationarity, and memoryless structure, as well as comfort with sums, expectations, and fixed-point reasoning under repeated independent draws. It pushes you to formalize tradeoffs between immediate payoff and the option to continue, and to quantify the marginal value of another sample relative to its cost.

It matters for quant interviews because market making, options trading, and algorithmic execution all involve repeated decisions under uncertainty with costs to waiting or re-quoting. Interviewers use such problems to check whether you can turn an informal "keep trying until it's good enough" idea into a precise stopping rule with a well-defined fair value. It reveals not just raw probability skills, but also whether you can frame trading-style problems rigorously, which is central to high-level quant work.

What it tests

Optimal stopping problems with independent, identically distributed draws and a fixed cost per attempt are governed by threshold strategies: at each step, compare the immediate reward to the expected value of continuing, and stop if the reward meets or exceeds this value. The underlying structure is that the process is memoryless and stationary, so the optimal action at each round depends only on the current observation, not the history. The threshold emerges because the opportunity to redraw is always available at the same cost, so the tradeoff is between taking a sure payoff now or paying again for another independent sample. The principle holds because, with a known distribution and cost, the expected gain from continuing can be computed and compared directly to any observed value. This reduces the problem to finding the fixed point where the value of stopping equals the value of continuing, which is why the solution is always a threshold.