Player A and B both have equal chance of winning any given point regardless of who's serving, so there is a 0.5 probability for either of them to win any point and therefore the game. My question is if Player A wins the first point and goes 15-0 up (as per the tennis scoring system) what are their chances of winning the game? Also what about if Player A is 30-15 up, or 40-15 up?
Thanks for any help, James
It’s easy but tedious to build the (sparse) transition matrix $P$ of an absorbing Markov chain that represents a tennis game, with each state corresponding to a possible score and fixed probabilities $p$ of player A winning a point and $q=1-p$ of player B winning a point. For example, if the first state corresponds to 0-0, then the first row will have $p$ in the column that corresponds to 15-0 and $q$ in the column that corresponds to 0-15. There will be, if I’m not mistaken, 20 states. The last two states should be winning and losing so that this matrix is in canonical form.
Once you have the transition matrix $P$, compute its fundamental matrix $N=(I-Q)^{-1}$, where $Q$ is the upper-left $18\times18$ submatrix consisting of the non-absorbing states (i.e., everything but win and lose). This is not a pleasant calculation to do by hand, so I suggest using an online matrix calculator of some sort (e.g., Wolfram|Alpha can do this for you). Once you have $N$, you can easily answer your questions and many others. For example, the chance of A’s winning if the score is 15-0 is given by the product of the row of $N$ that corresponds to this score and the “win” column of $P$. (By my calculations, with $p=\frac12$, this is ${21\over32}$).
