Optimal Stopping Rule: Marbles

We have a bag with 100 marbles: 99 white and 1 black. We draw marbles one by one without replacement. Each white marble drawn earns us $1, but if we draw the black marble, we lose everything. We may stop at any time and keep our current earnings. What is the optimal stopping strategy, and what is its expected value?

Answer

The thing is that, to find the optimal strategy, we must determine the optimal stopping rule. Let $$x$$ be the wealth we have before the next round. We need to find the smallest such $$x$$ for which the expected value of the next round’s payoff is $\le x$ , or equivalently, the expected profit of continuing is $\le 0$ . That means, once $$x$$ is reached, the next draw has negative EV, so we should stop at $$x$$ . This critical value $$x$$ turns out to be 50, so the optimal strategy is to stop as soon as you have drawn 50 white marbles.

Why $$x=50$$ is the threshold: If there are $$100-x$$ marbles left, with $$(99 - x)$$ white and 1 black, drawing again yields an expected increment

\frac{99 - x}{100 - x} \times (+1) \;+\;\frac{1}{100 - x}\times(-x) \;=\;\frac{99 - 2x}{100 - x}.

This quantity becomes nonpositive at $x \ge 50$ , so continuing is unfavorable once you have 50 dollars.

Now, given this strategy, the final payoff is either $50 or $0. The probability of actually earning $50 is the probability that no black marble appears among the first 50 draws, which is $\frac{50}{100} = \frac12$ . The reason why it is $\frac{1}{2}$ is because we are looking for a specific marble out of 50 draws from a "deck" of 100 (you know this from the card questions).

Therefore, the expected value is

\left(\frac12\right) \times 50 \;+\; \left(\frac12\right)\times 0 \;=\; 25.

So the optimal strategy yields an expected value of $\boxed{25}$ .