 Solution

For the forward-moving object to have been waiting at the meeting position, the last move must be a reverse move. The preceding $F-1$ choices must then contain exactly $n$ forward moves and $F-n-1$ reverse moves. Choosing their order givesAt $n=F$ the probability is zero, since the reverse-moving object was already at that unwrapped position initially. At $n=0$ the displayed formula is $q^F$, consistent with the forward object waiting at its initial position throughout. Equivalently, use the binomial convention $\left(\genfrac{}{}{0ex}{}{F-1}{F}\right)=0$ to include this endpoint in the same formula without a negative power of $q$.