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
Well, I'd say the coin is biased or not. If it isn't biased, the probability of each outcome (H or T) is 1/2 (disregarding the possibility of the coin landing on its side and remaining in that position). In this case it doesn't matter what you choose.
If the coin is biased, it'll be biased towards H or T. Considering the outcome of the first ten flips, the probability of the coin being biased towards T is smaller than it being biased towards H. Choosing H will be the better choice in this case.
In short: choose H.