In fencing, you win a round if you are the first to reach $15$ points. Suppose that when $A$ plays against $B$, at any point during the round, $A$ scores the next point with probability $p$ and $B$ scores the next point with probability $q=1-p$. (However, they never can both score a point at the same time.) Suppose that in this round, $A$ already has $14-k$ points, and $B$ has $14-\ell$ (where $0\le k,\ell\le 14$). By how much will the probability that $A$ wins the round increase if $A$ scores the next point?
Problem
Source: Kürschák 2015, problem 1
Tags: probability, combinatorics
shinichiman
08.10.2016 08:27
The answer is $\binom{k+l}{k}q^{l+1}p^k$. $A$ needs $k+1$ points to win and $B$ can only have at most $\ell$ points. Note that the probability for $A$ to win when after that round, $B$ has $a \le 14$ points is $\binom{k+a}{k}p^{k+1}q^a$. This is because we have a total of $\binom{k+a}{k}$ ways to put $a$ points of $B$ in $p$ squares $$\square \quad p \quad \square \quad p \quad \square \quad \cdots \quad \square \quad p$$Thus, if $A,B$ start with $14-k,14- \ell$ respectively then the probability that $A$ wins the round is $\sum_{i=0}^{\ell} \binom{k+i}{k}p^{k+1}q^{i}$. If $A,B$ start with $15-k,14- \ell$ respectively then the probability that $A$ wins the round is $\sum_{i=0}^{\ell} \binom{k-1+i}{k-1}p^kq^i$. Thus, the difference in probability is $$p^k \sum_{i=0}^{\ell} \left [ \binom{k-1+i}{k-1}q^i - \binom{k+i}{k}pq^i \right ]=p^k \cdot A.$$Since $p=1-q$ and $\binom{n}{k+1}= \binom{n-1}{k+1}+\binom{n-1}{k}$ so $$\begin{aligned} A & = \sum_{i=0}^{\ell} q^i \left [ \binom{k-1+i}{k-1}- \binom{k+i}{k} \right] q^i +\sum_{i=0}^{\ell} \binom{k+i}{k}q^{i+1}, \\ & = -\sum_{i=1}^{\ell} \binom{k-1+i}{k} q^i + \sum_{i=0}^{\ell}\binom{k+i}{k}q^{i+1}, \\ & = \binom{k+l}{k}q^{l+1}. \end{aligned}$$Thus, the winning probability of $A$ increases by $\binom{k+l}{k}q^{l+1}p^k$.