A bag contains $b$ balls in total, $r$ of which are red, while the rest are white. In a game a player removes balls one at a time from the bag (without replacement). He may remove as many balls as he wants, but if he removes even a single red ball he loses, and gets no reward at all. If he does not remove a single red ball, he is awarded 1 pound for each white ball removed.
He should aim to remove $n$ white balls on the first $n$ draws in order to maximize his expected total reward. Find the value of $n$ (in terms of $b$ and $r$).
The winning probability is $$ \frac{b-r\choose n}{b\choose n}$$ and the expected win is therefore $$ E_n=n\cdot\frac{b-r\choose n}{b\choose n}=\frac{(b-r)!}{b!}\frac{n\cdot(b-n)!}{(b-r-n)!}$$ Then $$\frac{E_n}{E_{n-1}}=\frac{n(b-n)}{(n-1)(b-r-n)}.$$ Whenever this factor is $>1$ it says that $n$ is better than $n-1$. This should help you find the optimal $n$.