Problem source

I have a sliced loaf which initially contains $n$ slices of bread. Each time I finish setting a STEP question, I make myself a snack: either toast, using one slice of bread; or a sandwich, using two slices of bread. I make toast with probability $p$ and I make a sandwich with probability $q$, where $p+q=1$, unless there is only one slice left in which case I must, of course, make toast.
 
Let $s_r$ ($1 \le r \le n$) be the probability that the $r$th slice of bread is the second of two slices used to make a sandwich and let $t_r$ ($1 \le r \le n$) be the probability that the $r$th slice of bread is used to make toast. What is the value of $s_1$? 
Explain why the following equations hold:
\begin{align*}
\phantom{\hspace{2cm} (2\le r \le n-1)}   
t_r &= (s_{r-1}+ t_{r-1})\,p  
\hspace{2cm}
(2\le r \le n-1)\,;
\\
 \phantom{\hspace{1.53cm} (2\le r \le n) }           
s_r &= 1- (s_{r-1} + t_{r-1})
 \hspace{1.53cm}
( 2\le r \le n )\,.           
\end{align*}
Hence, or otherwise, show that $s_{r} = q(1-s_{r-1})$ for
$2\le r\le n-1$.
Show further that 
\[
\phantom{\hspace{2.7cm} (1\le r\le n)\,,}
s_r = \frac{q+(-q)^r}{1+q}
\hspace{2.7cm}
(1\le r\le n-1)\,,
\,
\hspace{0.14cm}
\]
and find the corresponding expression for $t_r$. Find also expressions for $s_n$ and $t_n$ in terms of $q$.

Solution source

The $1$st slice of bread can only be the first slice in a sandwich or a slice of toast. Therefore $s_1 = 0$

\begin{align*}
&& t_r &=  \underbrace{s_{r-1}}_{r-1\text{th is the end of a sandwich}} \cdot \underbrace{p}_{\text{and we make toast}} +  \underbrace{t_{r-1}}_{r-1\text{th is toast}} \cdot \underbrace{p}_{\text{and we make toast}} \\
&&&= (s_{r-1}+t_{r-1})p \\
\\
&& s_r &= 1-\mathbb{P}(\text{previous slice is not the first of a sandwich}) \\
&&&= 1-(s_{r-1} + t_{r-1}) \\
\\
\Rightarrow && s_r &= 1 - \frac{t_r}{p} \\
\Rightarrow && t_r &= p - ps_r \\
\Rightarrow && s_r &= 1 - s_{r-1} - (p-ps_{r-1}) \\
&&&= 1 -p  -(1-p)s_{r-1}  \\
&&&= q(1-s_{r-1})
\end{align*}

Therefore since $s_r + qs_{r-1} = q$ we should look for a solution of the form $s_r = A(-q)^r + B$. The particular solution will have $(1+q)B = q \Rightarrow B = \frac{q}{1+q}$, the initial condition will have

$s_1 = \frac{q}{1+q} +A(-q) = 0 \Rightarrow q = \frac{1}{1+q}$, so we must have

\begin{align*}
&& s_r &= \frac{q+(-q)^r}{1+q}\\
\Rightarrow && t_r &= p(1-s_r) \\
&&&= p \frac{1+q-q-(-q)^r}{1+q} \\
&&&= \frac{(1-q)(1-(-q)^r)}{1+q} \\
&& s_n &= 1-\frac{q+(-q)^{n-1}}{1+q} - \frac{p(1-(-q)^{n-1})}{1+q} \\
&&&= 1-\frac{1+(1-p)(-q)^{n-1}}{1+q}\\
&&&= 1-\frac{1-(-q)^n}{1+q}\\
&&&= \frac{q+(-q)^n}{1+q}\\
&& t_n &=1-s_n \\
&&&=\frac{1-(-q)^n}{1+q}
\end{align*}