riddle of the day

A seemingly simple decision—how to distribute one hundred balls into two boxes—hides a strategy that defies instinct. It's worth stopping before handing out.

White balls in two boxes

Strategist
Chance and uncertainty

Riddle statement

You have 50 white balls and 50 black ones.

You must distribute them into two boxes, with the only condition that none of them remain empty. Then one of the two boxes is chosen at random and, from that box, a ball at random.

How should you distribute the balls to maximize the probability of getting a white one?

Show solution

Solution

Answer: Place 1 single white ball in one box, and in the other the remaining 99 (49 white and 50 black).

With this distribution:

$\mathbb{P}(\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\text{blanca}) = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{49}{99} = \frac{148}{198} \approx 0{,}7475.$

To see that it is the maximum possible, observe that in any optimal distribution the "small" box should not contain black balls: they would only reduce its fraction of white ones. If that box has white $x$ and no black ones, the total probability is:

$p(x) = \frac{1}{2} + \frac{1}{2} \cdot \frac{50 - x}{100 - x}, \quad 1 \le x \le 50.$

The function $p(x)$ decreases with $x$, so that the maximum is reached at $x = 1$.

Conclusion: the maximum probability is exactly $\dfrac{148}{198} \approx 74{,}75\%$.