White balls in two boxes

1. And it is verified that p(x) decreases with x, so the maximum occurs at x=1.
2. With that distribution: $\mathbb{P}(\text{blanca})=\frac12\cdot1+\frac12\cdot\frac{49}{99}=\frac{148}{198}\approx0.7475$.
3. Place 1 single white ball in one box, and the other 99 (49 white and 50 black) in the other.

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

$$ \mathbb{P}(\text{blanca})=\frac12\cdot1+\frac12\cdot\frac{49}{99} =\frac{148}{198}\approx0.7475. $$

To justify optimality, in an optimal solution the “small” box should not contain black balls (they only worsen its white fraction).
If that box has white $x$ and 0 black ones, the probability is:

$$ p(x)=\frac12+\frac12\cdot\frac{50-x}{100-x},\quad 1\le x\le50. $$

And it is verified that $p(x)$ decreases with $x$, so the maximum occurs in $x=1$.
Conclusion: the exact maximum is $\frac12+\frac12\cdot\frac{49}{99}\approx74.75\%$.