May 2nd, 2015, 10:26 am
I've probably screwed up trying to do this on the back of an envelope, but is the answer p(A wins)=4/9p(B wins)=5/9If you go through the first few tosses then see a pattern\begin{array}{|l|l|l|}\hline\textrm{probability}&\textrm{tosses}&\textrm{outcome}\cr\hline1/2&\textrm{AT}&\textrm{B is first player}\cr 1/4&\textrm{AH BT}&\textrm{ B wins }\cr 1/8 &\textrm{ AH BH AT }&\textrm{A wins}\cr1/16 &\textrm{AH BH AH BT }&\textrm{B wins}\cr1/32 &\textrm{AH BH AH BH AT }&\textrm{A wins}\cr1/&\textrm{AH BH AH BH AH BT }&\textrm{B wins}\cr\hline\end{array}