Solution
= Solution
Now the event is to hit $20$ before $1$, starting at $3$. Up to the first of those two <hitting times>, translate the interval by one unit; its endpoints become $0,19$ and the starting point becomes $2$. The <gambler's ruin> <hitting probability> is therefore
$$
\boxed{\mathbb P_3(T_{20}<T_1)=\frac{3-1}{20-1}=\frac2{19}.}
$$
Stopping at $1$ is essential: using the original absorbing endpoint $0$ would answer a different event.