TITLE: Wimbledon's final
QUESTION [7 upvotes]: Tom and Jack are playing the final of Wimbledon and they are 6:6 at the last set. They play to the bitter end until one of them is ahead by two games. For Tom the probability to win the next games is $p$, and for Jack $1-p$. Every games is independent from the others.

Find the probability that the match end 9 to 7 for one of them.

For $A=($Tom wins 9 to 7$)$ and $B=($Jack wins 9 to 7$)$, we have
$\rightarrow \mathbb{P}(A\cup B)=2p(1-p)[p^2+(1-p)^2]$

Find the probability that it needs more than 4 games to end the match.

For $X=($# games to the end$)$, we have
$\rightarrow \mathbb{P}(X>4)=1-\mathbb{P}(X\leq 4)=1-\mathbb{P}(X\leq 4|A\cup B)=1-\frac{\mathbb{P}(X\leq 4 \cap A)+\mathbb{P}(X\leq4\cap B)}{\mathbb{P}(A)+\mathbb{P}(B)}=1-\frac{2[p^2+(1-p)^2]}{[p^2-(1-p)^2]}$

Find the probability that Tom wins.


Hoping 1) and 2) are right, do you have any ideas for point 3)? Thanks in advance.

REPLY [6 votes]: I think the simplest way is as follows. Almost surely, they eventually finish, and this must be after an even number of games $2k$.
If Tom wins after $2k$ extra games, than means they won one game each in all of the first $k-1$ pairs, then Tom won twice, which has probability $q_kp^2$ for some $q_k>0$; similarly Jack wins after $2k$ games with probability $q_k(1-p)^2$ for the same $q_k$. This means that, conditional on the number of games being $2k$, the probability of Tom winning is $$\frac{P(T\text{ wins after }2k)}{P(\text{game ends after }2k)}=\frac{q_kp^2}{q_kp^2+q_k(1-p)^2}=\frac{p^2}{p^2+(1-p)^2},$$ which does not depend on $k$. So this must also be the unconditional probability that Tom wins.