Wimbledon's final 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.
 A: 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.
A: A more systematic approach is using Markov chains with difference equations. You have two absorbent states: $T_2, J_2$ ( I use $\{T, J\}$ for players and $T_i, J_j$ for $i$ games ahead. You start with a '0' state/noone ahead, and the probability of winning the game for either gamer is $h_{0,T_2}, h_{0,T_2}$:
$$
h_{0, T_2} = ph_{T_1, T_2} + (1-p)h_{J_1,T_2}\\
h_{0, J_2} = ph_{T_1, J_2} + (1-p)h_{J_1,J_2}\\
h_{T_1, T_2} = p \times 1 + (1-p)h_{0,J_2}\\
h_{J_1, J_2} = (1-p)\times1 + (1-p)h_{0,J_2}
$$
Obvisouly $h_{T_2,T_2} = 1=h_{J_2,J_2}, h_{T_2, J_2}=0=h_{J_2, T_2}$.
From here, you need to construct 2 more difference equations.
EDIT, for 2), consider $H$=Tom wins, $T$=Jack wins. You have the following probabilities ($S$: game finishes in 4 or less trials):
$$
P(S) = P(HH)+P(TT)+ \\
+ P(HTHH) + P(THTT) + P(HTTT)+P(THHH)
$$
A: A possible approach is via computing series. Say that Tom wins after $2n+2$ games: then he either won the last two games, which happens with probability $p^{n+2}(1-p)^n$, or he won the last three games, which again happens with probability $p^{n+2}(1-p)^n$ but can only happen if $n>0$. Since there are no other possibilities, it is enough to compute the value of the series above and add up the result.
