The ballot problem

Setup. There are $\binom{a+b}a$ equally likely sequences of $a$ pluses and $b$ minuses. $A$ is strictly ahead throughout iff every partial sum is positive. Then the first vote must be $+1$.

Sequences starting with $+1$. There are $\binom{a+b-1}{a-1}$ of them. Among them, call a sequence bad if some partial sum equals $0$.

Reflection. Given a bad sequence, let $t$ be the first time its partial sum is $0$, and flip the signs of the first $t$ entries. The result starts with $-1$, and has the same total $a-b$. Conversely, every sequence starting with $-1$ and with total $a-b>0$ has a first time $t$ where its partial sum is $0$; flipping the first $t$ entries gives a bad sequence starting with $+1$. These maps are inverse to each other. Hence the number of bad sequences equals the number of sequences starting with $-1$ with $a$ pluses and $b$ minuses:
$$\binom{a+b-1}{a}.$$

Counting. The number of good sequences is
$$\binom{a+b-1}{a-1}-\binom{a+b-1}{a}=\binom{a+b}a\cdot\frac{a-b}{a+b}.$$

Probability. Dividing by $\binom{a+b}a$ gives $\dfrac{a-b}{a+b}$.

Check. For $a=5$, $b=3$ a direct count over all $56$ sequences gives probability $\frac14$.

Lesson. The reflection principle turns "paths that touch a line" into "paths ending elsewhere".