Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 12 / f / Solution

First take $1\le n\le F-1$. The forward object reaches $n$ at its $n$th move, while the reverse object reaches that label at its $(F-n)$th move. If the forward object wins after $k$ reverse moves, then $0\le k\le F-n-1$. The final choice is forward; among the preceding $n+k-1$ choices, exactly $k$ are reverse. By the negative binomial stopping argument, this case has probability $\left(\genfrac{}{}{0ex}{}{n-1+k}{k}\right)p^n{q}^k$.

Sum these mutually exclusive cases and equate the result to part (e), using parts (b) and (d). This proves the success-failure stopping identity here:The right side splits the same race into a special meeting at $n$ and all meetings beyond $n$; the left side splits it according to the number of reverse moves at the winning time.

At $n=F$, both sides are zero with an empty sum and $\left(\genfrac{}{}{0ex}{}{F-1}{F}\right)=0$. At $n=0$, the event has probability one; the right side equals ${\sum }_{k=0}^F\left(\genfrac{}{}{0ex}{}{F}{k}\right)p^k{q}^{F-k}=1$. The printed left side includes $\left(\genfrac{}{}{0ex}{}{-1}{0}\right)$ and is outside the usual nonnegative-upper-index combinatorial definition. With the explicit generalized convention $\left(\genfrac{}{}{0ex}{}{-1}{0}\right)=1$, its $k=0$ term is one and all $k\ge 1$ terms vanish, so the identity also holds at that endpoint. Without this convention the displayed sum is undefined there; the zero-target race must instead be stated separately. All identities extend to $p=0,1$ by their polynomial forms.