= 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 $Y\geqslant T_2$.
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 \b[$\mathbb E[Y]\geqslant6$], 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.
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