 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 giveThis 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.