Does every irrational number contain "$666$" in its decimal expansion?

Question

Let $x\in\mathbb{R}$. Say $x$ is satanic if $x$ contains "$666$" somewhere in its decimal expansions. Let $S$ be the set of such numbers. Is $$ \mathbb{R} \setminus \mathbb{Q} \subseteq S $$ More generally, for any finite string of digits, say $y = y_1y_2y_3\ldots y_{n-1}y_n$, does $x\in \mathbb{R}\setminus\mathbb{Q}$ necessarily contain $y$ in its decimal expansion.

It seems intuitive that, at least, for any $y$, $y$ should be contained in $\textit{almost every }$ irrational number, since the irrational numbers have an infinite decimal expansion. But, I could envision there exists some irrational number that follows some pattern such that it doesn't contain some $y$.

Answer 1

No. There are even transcendental numbers all of whose digits are $0$ and $1$.

Answer 2

Almost all numbers are satanic, in the sense that the Lebesgue measure of the numbers that do not contain 666, or any other particular finite sequence, is zero.

Also in terms of Baire category almost all numbers are satanic, since any finite sequence can be extended to be satanic.

On the other hand, in terms of cardinality there are as many non-satanic numbers as satanic numbers, because there are $2^{\aleph_0}$ of each: just consider numbers whose decimal expansion contains nothing but 0s and 1s.

Answer 3

For a concrete example, take $$x=0.101001000100001000001\ldots1\underbrace{00\dots0}_{n\text{ zeros}}1\underbrace{00\dots0}_{n+1\text{ zeros}}1\dots.$$ This decimal expansion isn't periodic (since it contains arbitrarily long runs of zeros), and so $x$ is irrational. You can do something similar to avoid any string of digits you want, not just $666$.