= 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 gives
$$
\boxed{p_n^{\rm specialMeet}=\binom{F-1}{n}p^nq^{F-n},\qquad 0\leq n\leq F-1.}
$$
At $n=F$ the <probability> is \b[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 $\binom{F-1}{F}=0$ to include this endpoint in the same formula without a negative power of $q$.
Back to article page
