Splitting Balls Between Two Urns

Key idea. One urn can be made a sure win with a single white ball. The other urn then holds everything else, and its white fraction stays just below $\tfrac12$.

Suppose urn 1 holds $w$ white and $b$ black balls and urn 2 holds the remaining $50-w$ white and $50-b$ black balls. Assume both urns are non-empty. If one urn is empty, the draw comes from the other, which holds all $100$ balls, and the survival probability is $\tfrac12$. Let $p_1$ and $p_2$ be the white fractions of the two urns. Then the survival probability is

Achievability. Put one white ball in urn 1 and the other $99$ balls ($49$ white, $50$ black) in urn 2. This gives

Optimality. Label the urns so that $p_1\ge p_2$. The overall white fraction $\tfrac12$ is a weighted average of $p_1$ and $p_2$, so $p_2\le\tfrac12\le p_1$. There are two cases.

• If $p_2=\tfrac12$, then $p_1=\tfrac12$ as well, and the survival probability is $\tfrac12$.

• If $p_2<\tfrac12$, let urn 2 contain $x=50-w$ white and $y=50-b$ black balls, so that $x<y$.

  • Since $x$ and $y$ are integers, $y\ge x+1$.
  • Also $w>b\ge 0$, so $w\ge1$ and hence $x\le 49$.

  • Therefore

    $$p_2=\frac{x}{x+y}\le\frac{x}{2x+1}\le\frac{49}{99},$$

    because $x\mapsto x/(2x+1)$ is increasing.

  • Combining this with $p_1\le1$ gives $\tfrac12(p_1+p_2)\le\tfrac12\bigl(1+\tfrac{49}{99}\bigr)=\tfrac{74}{99}$.

In every case the survival probability is at most $\tfrac{74}{99}$, and the arrangement above attains it. Hence