 Solution

Let $T_2$ be the time of the second occurrence of $3$, whether or not it is adjacent to the first. On every digit sequence, completion of a consecutive pair requires at least two occurrences of $3$, so $Y⩾T_2$.

The waiting time for the first $3$ has mean $3$. Once that digit has appeared, the fresh independent digits give another waiting time with the same mean for the next $3$. Thus $\mathbb{E}[T_2]=3+3=6$, the mean of the corresponding negative binomial distribution. Monotonicity of expectation now proves $\mathbb{E}[Y]⩾6$, without computing the waiting time for the consecutive pair itself. The argument remains valid even before establishing that $Y$ has finite mean, since these waiting times are nonnegative.