Problem source

A biased coin, with a probability $p$ of coming up heads and a probability $q=1-p$ of coming up tails, is tossed repeatedly. Let $A$ be the event that the first run of $r$ successive heads occurs before the first run of $s$ successive tails. If $H$ is the even that on the first toss the coin comes up heads and $T$ is the event that it comes up tails, show that 
\begin{alignat*}{1}
\mathrm{P}(A|H) & =p^{\alpha}+(1-p^{\alpha})\mathrm{P}(A|T),\\
\mathrm{P}(A|T) & =(1-q^{\beta})\mathrm{P}(A|H),
\end{alignat*}
where $\alpha$ and $\beta$ are to be determined. Use these two equations to find $\mathrm{P}(A|H),$ $\mathrm{P}(A|T),$ and hence $\mathrm{P}(A).$

Solution source

\begin{align*}
&& \P(A|H) &= \P(\text{achieve }r\text{ heads immediately}) + \P(\text{don't and then achieve it from having flipped a tail}) \\
&&&= p^{r-1} + (1-p^{r-1}) \cdot \P(A|T) \\
&& \P(A|T) &= (1-q^{s-1})\P(A|H) \\
\\
&&\P(A|H) &= p^{r-1}+(1-p^{r-1})(1-q^{s-1})\P(A|H) \\
\Rightarrow && \P(A|H) &= \frac{p^{r-1}}{1-(1-p^{r-1})(1-q^{s-1})} \\
&& \P(A|T) &=  \frac{(1-q^{s-1})p^{r-1}}{1-(1-p^{r-1})(1-q^{s-1})} \\
&& \P(A) &= \frac{(2-q^{s-1})p^{r-1}}{2(1-(1-p^{r-1})(1-q^{s-1}))} 
\end{align*}