Red-Black Card Game Optimal Stop

This question presents a finite card-drawing game where gains and losses accumulate as you see more of the deck, and you can choose when to walk away. The candidate must decide how the remaining composition of the deck and the current running total interact to determine whether it is better to stop or continue. Because cards are drawn without replacement, the probabilities evolve in a nontrivial way as the game progresses. This style of problem is common in more theoretical quant interviews and PhD-level algorithmic trading or research roles, where one must formalize an intuitively "gambling-like" scenario into a crisp decision rule.

Solving it cleanly leans on dynamic programming, backward induction, and Markov decision process thinking: you encode the game state and derive a value function over that state space. It tests whether the candidate can frame a stopping problem as a recursion, handle symmetry and boundary conditions correctly, and argue about monotonicity and thresholds rather than brute-force enumeration. Interviewers watch for the ability to translate a narrative game into states, transitions, and value comparisons, and to reason about optimal policies under risk when you effectively hold an embedded free option to stop.

What it tests

This problem class is governed by the principle of optimal stopping in a finite, discrete-time Markov process with asymmetric payoffs and no replacement. The key structure is that at each decision point, the future is determined only by the current state (counts of each card type remaining and current score), not the path taken to reach it. The optimal strategy is to compare the immediate payoff from stopping with the expected value of continuing, which itself depends recursively on the distribution of remaining outcomes. The reason this works is that, with no replacement, the process is memoryless except for the current state, and the payoff structure allows you to always 'lock in' a non-negative outcome by stopping, which acts as a free option. The recursive value function and the use of the maximum at each step encode the trade-off between risk and reward, and the fact that the process terminates in a finite number of steps ensures the recursion bottoms out.