Are there infinitely many primes that avoid any particular finite string of digits?

I just learned of this paper showing that for any particular decimal digit $d$, there are infinitely many prime numbers whose decimal numerals do not contain $d$.

Is anything known about the obvious generalization, i.e., that for any particular finite string of decimal digits, there are there infinitely many prime numbers whose decimal numerals do not contain that string? Or, of course, similar results for other bases?

• LOL, I should have actually thought a little bit before posting this question. :( Considering the answer & votes, I suppose it should not be deleted (however embarrassing it may be). – r.e.s. May 12 '17 at 0:14

Consider the string $d_1 \ldots d_n$ and let $P_{d_i}$ be the infinite collection of primes that don't have the digit $d_i$. Then any of $P_{d_i}$ would be an infinite list of primes that avoids the string $d_1\ldots d_n$.
• Of course every number whose decimal expansion doesn't contain any digit from a given set won't contain any string formed from those digits, but that's not enough. For example, consider the sequence $s = (9,8,7,6,5,4,3,2,1,0)$. Then the intersection of $P_9, P_8, \dotsc, P_0$ is clearly empty, but no prime less than $9876543210$ contains $s$ in its decimal expansion. – A.P. May 11 '17 at 19:25