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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 $r$).

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    $\begingroup$ Please consider updating your question with some information about what you have tried or where you are getting stuck. You will find people are much more willing to help if you do! $\endgroup$ Commented May 3, 2014 at 13:31

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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 $n$.

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