Problem source

\begin{questionparts}
\item A bag of sweets contains one red sweet and $n$ blue sweets. 
I take a sweet from the bag, note  its colour, 
return it  to the bag, then shake the bag. 
I repeat this until the  sweet I take is the red one.
Find an expression for the probability that I take
 the red sweet on the $r$th attempt. 
What value of $n$ maximises this probability? 
\item Instead, I take sweets from the bag, without 
replacing them in the bag, until I take the red sweet.
Find an expression for the probability that I take
 the red sweet on the $r$th attempt. 
What value of $n$ maximises this probability? 
\end{questionparts}

Solution source

\begin{questionparts}
\item This is the probability of having the sequence $\underbrace{BB\cdots B}_{r-1 \text{ times}}R$ which has probability $\displaystyle \left ( \frac{n}{n+1} \right)^{r-1}\frac{1}{n+1}$.

Maximising this, is equivalent to maximising $\log$ of it, ie

\begin{align*}
&& y &= (r-1) \ln n - r \ln (n+1) \\
\Rightarrow && \frac{\d y}{\d n} &= \frac{r-1}{n} - \frac{r}{n+1} \\
&&&= \frac{(r-1)(n+1)-rn}{n(n+1)} \\
&&&= \frac{r-n-1}{n(n+1)}
\end{align*}

Therefore this is maximised when $n = r-1$

\item 
\end{questionparts}