= Solution
Let $N$ be the meeting label. If $N>n$, the forward-moving object has already crossed $n$ while the reverse-moving object has not yet reached it, so the forward object wins the race to $n$. If $N<n$, the reverse object passed $n$ first. If $N=n$, the required event is exactly the special meeting from part (d).
These disjoint possibilities give
$$
\boxed{P(\text{forward object reaches }n\text{ first})
=p_n^{\rm specialMeet}+\sum_{k=n+1}^{F}p_k^{\rm meet}.}
$$
This includes the endpoints: for $n=0$ the <probability> is $1$, and for $n=F$ it is $0$, with the initial positions understood in the unwrapped convention.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-1-stopping-race.png]
{title=Example paths in a race between fixed success and failure targets}
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