Solution
= Solution
Ignore rolls showing $2,3,4,5$ and consider the first roll among $\{1,6\}$. By symmetry, the event $A$ that a 1 occurs before a 6 has probability $1/2$. The event that the very first roll is 1 has probability $1/6$ and is contained in $A$. Hence
$$
\boxed{
\mathbb P(\text{first roll is }1\mid A)
=\frac{1/6}{1/2}=\frac13}.
$$