Problem source

I have a Penny Black stamp which I want to sell to my friend Jim,
but we cannot agree a price. So I put the stamp under one of two cups,
jumble them up, and let Jim guess which one it is under. If he guesses
correctly, I add a third cup, jumble them up, and let Jim guess correctly,
adding another cup each time. The price he pays for the stamp is $\pounds N,$
where $N$ is the number of cups present when Jim fails to guess correctly.
Find $\mathrm{P}(N=k)$. Show that $\mathrm{E}(N)=\mathrm{e}$ and
calculate $\mathrm{Var}(N).$

Solution source

\begin{align*}
&& \mathbb{P}(N = k) &= \mathbb{P}(\text{guesses }k-1\text{ correctly then 1 wrong})\\
&&&= \frac12 \cdot \frac{1}{3} \cdots \frac{1}{k-1} \frac{k-1}{k} \\
&&&= \frac{k-1}{k!} \\
&&\mathbb{E}(N) &= \sum_{k=2}^\infty k \cdot \mathbb{P}(N=k) \\
&&&= \sum_{k=2}^{\infty} \frac{k(k-1)}{k!} \\
&&&= \sum_{k=0}^{\infty} \frac{1}{k!} = e \\
&& \textrm{Var}(N) &= \mathbb{E}(N^2) - \mathbb{E}(N)^2 \\
&& \mathbb{E}(N^2) &= \sum_{k=2}^{\infty} k^2 \mathbb{P}(N=k)  \\
&&&= \sum_{k=2}^{\infty} \frac{k^2(k-1)}{k!}  \\
&&&= \sum_{k=0}^{\infty} \frac{k+2}{k!} \\
&&&= \sum_{k=0}^{\infty} \frac{1}{k!} + 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 3e \\
\Rightarrow && \textrm{Var}(N) &= 3e-e^2
\end{align*}