Maximizing discrete probability I'm stuck with the following problem:
Let's assume we have two buckets: bucket one contains $k$ white spheres and $l$ red spheres. Bucket two contains $n-k$ white spheres and $n-l$ red spheres (n a fixed constant).
The probability to pick a white sphere from any of the two buckets (assuming uniform  probability) is given by:

$$\mathbb{P}(\ White\  sphere)=\frac12\cdot\frac{k}{k+l}+\frac12\cdot\frac{n-k}{2n-(k+l)}$$

The question reads as follows: how to choose k and l such that the above proability is maximal?
The given hint is: "first consider $k+l=m$ fix"
The solution states: $k=1$ and $l=0$. I've tried applying Lagrange's multipliers without any apparent success... can someone help me out?
Thank you in advance
 A: We apply the hint: Fix $m=k+l$. Then the equation becomes
$$\frac{1}{2}\left(\frac{k}{m}+\frac{n-k}{2n-m}\right)=\frac{1}{2}\left(\frac{2nk-mk+nm-mk}{2nm-m^2}\right)=\frac{1}{2}\left(\frac{k(2n-2m)+nm}{2nm-m^2}\right)$$
This value increases as $k$ increases since all other values are fixed. Since $m=k+l$, for a given $m$, the maximum value of $k$ is $m$.
Letting $k=m$ in the above equation, we get 
$$\frac{1}{2}\left(\frac{m(2n-2m)+nm}{2nm-m^2}\right)=\frac{1}{2}\left(\frac{2n-2m+n}{2n-m}\right)=\frac{1}{2}\left(\frac{3n-2m}{2n-m}\right)$$
This value increases as $m$ decreases. Hence the maximum value occurs when $m$ is at a minimum, that is, at $m=1$. In conclusion this gives us that the maximum occurs when $k=m=1$ and $l=0$.
A: If $k+l<n$, then the second denominator is greater than the first.  Any balls in the first bucket have a higher probability of being chosen than balls in the second bucket.  So, subject to $k+l$ being constant, you want all of the $k+l$ balls to be white - that is, $l=0$.
Now that $l=0$, how do you maximize the probability?
There is another solution if $k+l>n$.
