Problem source

In the basic version of Horizons (H1) the player has a maximum of $n$ turns, where $n \ge 1$. At each turn, she has a probability $p$ of success, where $0 < p < 1$. If her first success is at the $r$th turn, where $1 \le r \le n$, she collects $r$ pounds and then withdraws from the game. Otherwise, her winnings are nil.
Show that  in H1, her expected winnings are
$$
p^{-1}\left[1+nq^{n+1}-(n+1)q^n\right]\quad\hbox{pounds},
$$
where $q=1-p$.
The rules of H2 are the same as those of H1, except that $n$ is randomly selected from a Poisson distribution with parameter $\lambda$. If $n=0$ her winnings are nil. Otherwise she plays H1 with the selected $n$.
Show that in H2, her expected winnings are
$$ 
{1 \over p}{\left(1-{\e^{-{\lambda}p}}\right)}
    -{{\lambda}q}{\e^{-{\lambda}p}}
\quad\hbox{pounds}.
$$

Solution source

\begin{align*}
&& \E[H1] &= \sum_{r=1}^n r \cdot \mathbb{P}(\text{first success on }r\text{th turn}) \\
&&&= \sum_{r=1}^n r \cdot q^{r-1}p \\
&&&= p\sum_{r=1}^n r q^{r-1} \\
\\
&& \frac{1-x^{n+1}}{1-x} &= \sum_{r=0}^n x^r \\
\Rightarrow && \sum_{r=1}^n r x^{r-1} &= \frac{-(n+1)x^n(1-x) +(1-x^{n+1})}{(1-x)^2} \\
&&&= \frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2} \\
\\
&& \E[H1] &= p\sum_{r=1}^n r q^{r-1} \\
&&&= p\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} \\
&&&= p^{-1}(1-(n+1)q^{n} + nq^{n+1})
\end{align*}

Not that if $n =0$ , the formula for $\E[H1] = 0$.

So

\begin{align*}
&& \E[H2] &= \E[\E[H1|n=N]] \\
&&&= p^{-1}\E \left [ 1-(N+1)q^{N} + Nq^{N+1}\right] \\
&&&=  p^{-1}\E \left [ 1-((1-q)N+1)q^{N} \right] \\
&&&= p^{-1}\left (1 - p\E[Nq^N] - G_{Po(\lambda)}(q) \right) \\
&&&= p^{-1}(1-e^{-\lambda(1-q)}) - \E[Nq^N] \\
&&&= p^{-1}(1-e^{-\lambda(1-q)}) -  q\lambda e^{-\lambda(1-q)} \\
&&&= p^{-1}(1-e^{-\lambda p}) - q\lambda e^{-\lambda p}
\end{align*}