Gerrymandering urns (redux)

Assume that we decide to put $50-r$ balls into urn $A$ and $50+r$ balls into urn $B$ where $r$ is between $0$ and $49$ inclusive. Furthermore we decide to put $s$ red balls into urn $A$ and $50-s$ red balls into urn $B$. The number of blue balls in urn $A$ will then be $50-r-s$, and the number of blue balls in urn $B$ will then be $r+s$. Since all four numbers of balls have to be nonnegative we obtain the conditions $$s\geq0, \quad s\leq 50-r,\quad s\leq 50,\quad s\geq -r\ .$$ As $r\geq0$ these conditions are fulfilled iff $0\leq s\leq 50-r$.

The probability $P$ to draw a red ball computes to $$P={1\over2}\ {s\over 50-r}+{1\over2}\ {50-s\over 50+r}\ .$$ Keep $r$ fixed for the moment. Increasing $s$ by one produces a $$\Delta P={1\over2}\Bigl({1\over 50-r}-{1\over 50+r}\Bigr)>0\ .$$ In order to maximize $P$ we should therefore give $s$ the maximal allowed value $50-r$. The resulting value of $P$ is then $$P={1\over2}\ {50-r\over 50-r}+{1\over2}\ {50-(50-r)\over 50+r}={1\over2}+{1\over2}\Bigl(1-{50\over 50+r}\Bigr)\ .$$ This is maximal when $r$ is maximal, i.e., if $r=49$, and this leads to $s=1$.

Therefore the optimal way to distribute the balls is putting one red ball into urn $A$ and all the other balls into urn $B$. The resulting probability $P$ of drawing a red ball is then about ${3\over4}$.

It's absolutely possible to solve this using calculus; in fact that's how I would solve it. With a problem like this, it's important to make use of the symmetries to simplify the calculations. Your formulation of the problem is asymmetric both with respect to the two urns and with respect to the two colours. You can see from your plot that there's a lot more symmetry in the problem than is apparent from your equations.

To make use of the symmetry, let $r$ be the number of red balls beyond the average of $25$ in the first urn, let $b$ be the number of blue balls beyond the average of $25$ in the first urn, and calculate the difference between the probability of drawing a red ball and the average probability $1/2$:

$$ \begin{align} & \frac12\left(\frac{25+x}{25+x+25+y}+\frac{25-x}{25-x+25-y}\right)-\frac12 \\ =&\frac12\left(\frac{(25+x)(50-x-y)+(25-x)(50+x+y)}{(50-x-y)(50+x+y)}-1\right) \\ =&\frac12\left(\frac{50^2-2x(x+y)}{50^2-(x+y)^2}-1\right) \\ =&\frac12\frac{(y-x)(y+x)}{50^2-(y+x)^2}\;, \end{align} $$

Now switch to independent variables $s=y+x$ and $d=y-x$ to obtain

$$ \frac12\frac{sd}{50^2-s^2}\;. $$

It's now almost immediate that the only stationary point is at $s=d=0$, and this is clearly not a maximum. Thus the maximum of the continuous problem must be at the boundary. Since the result is positive iff $s$ and $d$ have the same sign and in that case increases monotonically with the magnitude of either, the maximum of the discrete problem must also be at the boundary. If we arbitrarily choose the common sign of $s$ and $d$ to be positive, the maximal value of $d$ given $s$ is $50-s$, so the maximum along the boundary is the maximum of

$$ \frac12\frac{s(50-s)}{50^2-s^2}=\frac12\frac s{50+s}\;, $$

and this is clearly maximized for maximal $s$. The maximal value of $s$ consistent with the constraints is $49$, so the solution is $s=49$, $d=1$, or $x=24$, $y=25$.