Solution:

Answer: $\frac{108}{665}$

First we note that the probability that $n$ is picked is $\frac{1}{2} \times \left(\frac{2}{3}\right)^{n}$, because this is the sequence whose terms decrease by a factor of $\frac{2}{3}$ each time and whose sum is 1 (recall that probabilities must sum to 1).

Now note that $\frac{1}{7} = .142857142857\ldots$, meaning that 2 occurs at digits $3, 9, 15, 21$, etc. We can then calculate the probability that we ever pick 2 as
$$\begin{aligned} \sum_{k=0}^{\infty} \frac{1}{2} \cdot \left(\frac{2}{3}\right)^{6k+3} &= \frac{4}{27} \sum_{k=0}^{\infty} \left(\frac{2}{3}\right)^{6k} \\ &= \frac{4}{27} \cdot \frac{1}{1-\left(\frac{2}{3}\right)^6} \\ &= \frac{4}{27} \cdot \frac{729}{729-64} \\ &= \frac{4}{27} \cdot \frac{729}{665} \\ &= \frac{108}{665} \end{aligned}$$