Let $e$ be the expected number of tosses. It is clear that $e$ is finite.
Start tossing. If we get a tail immediately (probability $\frac{1}{2}$) then the expected number is $e+1$. If we get a head then a tail (probability $\frac{1}{4}$), then the expected number is $e+2$. Continue $\dots$. If we get $4$ heads then a tail, the expected number is $e+5$. Finally, if our first $5$ tosses are heads, then the expected number is $5$. Thus
$$e=\frac{1}{2}(e+1)+\frac{1}{4}(e+2)+\frac{1}{8}(e+3)+\frac{1}{16}(e+4)+\frac{1}{32}(e+5)+\frac{1}{32}(5).$$
Solve this linear equation for $e$. We get $e=62$.
For $i=1,\ldots,r,\;$ define event $B_i = \text{"First tail occurs on $i^{th}$ toss"}$. Since the process ends after $r$ successive heads then, given initially that $A=H$, exactly one of these $r$ events must occur.
Note that, given $A=H$, if $B_r$ occurs then $E$ also occurs, which is to say $P(E\mid B_r\cap A=H) = 1$. Also, given $A=H$, if any of the other $B_i$ events occur then it is like we are starting over but, instead, initially given $A=T$. Therefore, conditioning on whether or not $B_r$ occurs,
\begin{eqnarray*}
P(E\mid A=H) &=& P(E\mid B_r\cap A=H)P(B_r\mid A=H) + P(E\mid B_r^c\cap A=H)P(B_r^c\mid A=H) \\
&& \\
&=& p^{r-1} + P(E\mid A=T)(1-p^{r-1}). \qquad\qquad\qquad\qquad\qquad(1) \\
\end{eqnarray*}
$\\$
Next, we can find $P(E\mid A=T)$ by way of $P(E^c\mid A=T)$ and that is found in a similar way to $P(E\mid A=H)$.
For $i=1,\ldots,s,\;$ define event $C_i = \text{"First head occurs on $i^{th}$ toss"}$. Then,
\begin{eqnarray*}
P(E\mid A=T) &=& 1 - P(E^c\mid A=T) \\
&& \\
&=& 1 - [P(E^c\mid C_r\cap A=T)P(C_r\mid A=T) + P(E^c\mid C_r^c\cap A=T)P(C_r^c\mid A=T)] \\
&& \\
&=& 1 - [1\cdot (1-p)^{s-1} + P(E^c\mid A=H)(1-(1-p)^{s-1})] \\
&& \\
&=& 1 - [(1-p)^{s-1} + (1 - P(E\mid A=H))(1-(1-p)^{s-1})] \\
&& \\
&=& (1-(1-p)^{s-1}) P(E\mid A=H). \qquad\qquad\qquad\qquad\qquad(2) \\
&& \\
\end{eqnarray*}
We now proceed to find $P(E)$. Substitute $(2)$ into $(1)$:
\begin{eqnarray*}
P(E\mid A=H) &=& p^{r-1} + (1-p^{r-1}) (1-(1-p)^{s-1}) P(E\vert A=H) \\
&& \\
\therefore P(E\mid A=H) &\times& (p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}) \;\; = \;\; p^{r-1} \\
&& \\
P(E\mid A=H) &=& \dfrac{p^{r-1}}{p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}} \qquad\qquad\qquad(3) \\
\end{eqnarray*}
Substitute this into $(2)$:
\begin{eqnarray*}
P(E\mid A=T) &=& \dfrac{p^{r-1} - p^{r-1} (1-p)^{s-1}}{p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}} \qquad\qquad\qquad(4) \\
\end{eqnarray*}
$\\$
Finally, using results $(3)$ and $(4)$,
\begin{eqnarray*}
P(E) &=& P(E\mid A=H)P(A=H) + P(E\mid A=T)P(A=T) \\
&& \\
&=& \dfrac{p^r + p^{r-1}(1-p) - p^{r-1} (1-p)^s}{p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}} \\
&& \\
&=& \dfrac{p^{r-1} - p^{r-1} (1-p)^s}{p^{r-1} - p^{r-1}(1-p)^{s-1} + (1-p)^{s-1}}.
\end{eqnarray*}
Best Answer
"But I am confused with the term "initial run" in my question. What does mean it ?"
In e.g. $HHTHTT$ the initial run is $HH$, in e.g. $TTTHHT$ the initial run is $TTT$.
I hope that these examples are enough to make clear what is meant by "initial run".
Problems of this sort can often be solved without looking too much at distributions.
Let $\mu$ denote the expectation of the length of the first run.
Let $\mu_H$ denote the expectation of the length of the first run under condition that the first toss results in heads.
Let $\mu_T$ denote the expectation of the length of the first run under condition that the first toss results in tails.
Defining $q:=1-p$ we find:$$\mu_H=p(1+\mu_H)+q1=1+p\mu_H\text{ and }\mu_T=q(1+\mu_T)+p1=1+q\mu_T$$leading to:$$\mu_H=q^{-1}\text{ and }\mu_T=p^{-1}$$ Then finally we find:$$\mu=p\mu_H+q\mu_T=pq^{-1}+qp^{-1}$$Of course you can make it an expression in $p$ only by substituting $q=1-p$.