Wimbledon's final

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.

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Probability

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