Problem source

A biased coin has probability $p$ of showing a head
and probability $q$ of showing a tail, where $p\ne0$, $q\ne0$
and $p\ne q$. When the coin is tossed repeatedly, runs occur. 
A \textit{straight run} of length $n$ is a sequence of $n$ consecutive
heads or $n$ consecutive tails. An \textit{alternating run} of length 
$n$ is a sequence of length $n$ alternating between heads and tails. An 
alternating run can start with either a head or a tail.
Let $S$ be the length of the longest straight run beginning with the
first toss and let $A$ be the  length of the longest 
alternating  run beginning with the
first toss.
\begin{questionparts}
\item Explain why  $\P(A=1)=p^2+q^2$ and find $\P(S=1)$. Show that
 $\P(S=1)<\P(A=1)$.
\item Show that
$\P(S=2)= \P(A=2)$
and determine the relationship between
$\P(S=3)$ and $ \P(A=3)$.
 \item Show that, for $n>1$, $\P(S=2n)>\P(A=2n)$ and determine
the corresponding relationship between $\P(S=2n+1)$ and $\P(A=2n+1)$.
[You are advised \textit{not} to  use $p+q=1$ in this part.] 
\end{questionparts}

Solution source

\begin{questionparts}
\item The only way $A = 1$ is if we get $HH$ or $TT$ which has probability $p^2+q^2$. The only way we get $S=1$ is if we have $HT$ to $TH$, ie $2pq$.

Since $(p-q)^2 = p^2 + q^2 - 2pq >0$ we must have $\mathbb{P}(A=1) > \mathbb{P}(S=1)$.

\item $\,$ \begin{align*}
\mathbb{P}(S=2) &= p^2q + q^2p \\
\mathbb{P}(A=2) &= pq^2 + qp^2 = \mathbb{P}(S=2) \\
\\
\mathbb{P}(S=3) &= p^3q + q^3p = pq(p^2+q^2) \\
\mathbb{P}(A=3) &= pqp^2 + qpq^2 = pq(p^2+q^2) = \mathbb{P}(S=3)
\end{align*}
\item For $n > 1$ we must have 
\begin{align*}
&& \mathbb{P}(S = 2n) &= p^{2n}q + q^{2n}p \\
&& \mathbb{P}(A=2n) &= (pq)^{n}q + (qp)^{n}p \\
&&&= p^nq^{n+1} + q^np^{n+1} \\
&& \mathbb{P}(S = 2n)  &> \mathbb{P}(A = 2n) \\
\Leftrightarrow &&  p^{2n}q + q^{2n}p & > p^nq^{n+1} + q^np^{n+1}\\
\Leftrightarrow &&  0 & < p^{2n}q+q^{2n}p - p^nq^{n+1} -q^np^{n+1}\\
&&&= (p^n-q^n)(qp^n - pq^n)
\end{align*}
which is clearly true.

\begin{align*}
&& \mathbb{P}(S=2n+1) &= p^{2n+1}q + q^{2n+1}p \\
&& \mathbb{P}(A=2n+1) &= (pq)^{n}p^2 + (qp)^{n}q^2 \\
&&&= p^{n+2}q^n + q^{n+2}p^n
\end{align*}

The same factoring logic shows that $\mathbb{P}(S = 2n+1) > \mathbb{P}(A=2n+1)$
\end{questionparts}