$The probability $A$ wins is $\frac{1}{2}$ by symmetry.$
$Alternatively, $A$ can win with $B$ winning exactly $k$ games in the first $n+k$ games, where $k$ can be $0,1,2,\ldots,n$ (this happens with probability $\binom{n+k}{n}\cdot \left(\frac{1}{2}\right)^{n+k}$; note $B$ can't win...