Problem source

Three pirates are sharing out the contents of a treasure chest containing $n$ gold coins and $2$ lead coins. The first pirate takes out coins one at a time until he takes out one of the lead coins. The second pirate then takes out coins one at a time until she draws the second lead coin.  The third pirate takes out all the gold coins remaining in the chest. 
Find:
\begin{questionparts}
\item the probability that the first pirate will have some gold coins;
\item the probability that the second pirate will have some gold coins;
\item the probability that all three pirates will have some gold coins. 
\end{questionparts}

Solution source

\begin{questionparts}
\item The first pirate will have some gold coins as long as the very first coin drawn is a gold coin, ie $\frac{n}{n+2}$.

\item The probability the second pirate will have some gold coins is the probability the two lead coins are separate. There are $(n+2)!$ ways to arrange the coins, and there are $(n+1)! \cdot 2$ ways to arrange the coins where they are together, therefore the probability is:

\[ 1 - \frac{2(n+1)!}{(n+2)!} = 1 - \frac{2}{n+2} = \frac{n}{n+2} \]

\item For all three pirates to have some gold coins we need the lead coins to be separate and not first or last. If we line up all $n$ gold coins, there are $n-1$ gaps between them we could place the $2$ lead coins in, therefore $\binom{n-1}{2}$ ways to place the lead coins with the restriction. Without the restriction there are $\binom{n+2}{2}$ ways to choose where to put the coins, therefore

\begin{align*}
&& P &= \frac{\binom{n-1}{2}}{\binom{n+2}{2}} \\
&&&= \frac{(n-1)(n-2)}{(n+2)(n+1)}
\end{align*}

[Notice this is clearly $0$ if there are only $1$ or $2$ gold coins]
\end{questionparts}