Problems

In certain forms of Tennis two players \(A\) and \(B\) serve alternate games. Player \(A\) has probability \(p\low_{A}\) of winning a game in which she serves and \(p\low_{B}\) of winning a game in which player \(B\) serves. Player \(B\) has probability \(q\low_{B}=1-p\low_{B}\) of winning a game in which she serves and probability \(q\low_{A}=1-p\low_{A}\) of winning a game in which player \(A\) serves. In Shortened Tennis the first player to lead by 2 games wins the match. Find the probability \(P_{\text{short}}\) that \(A\) wins a Shortened Tennis match in which she serves first and show that it is the same as if \(B\) serves first. In Standard Tennis the first player to lead by 2 or more games after 4 or more games have been played wins the match. Show that the probability that the match is decided in 4 games is \[ p^{2}_Ap_{B}^{2}+q_{A}^{2}q_{B}^{2}+2(p\low_{A}p\low_{B}+q\low_{A}q\low_{B})(p\low_{A}q\low_{B}+q\low_{A}p\low_{B}). \] If \(p\low_{A}=p\low_{B}=p\) and \(q\low_{A}=q\low_{B}=q,\) find the probability \(P_{\text{stan}}\) that \(A\) wins a Standard Tennis match in which she serves first. Show that \[ P_{\text{stan}}-P_{\text{short}}=\frac{p^{2}q^{2}(p-q)}{p^{2}+q^{2}}. \]