TAOCP 3.5 Exercise 44
Let the original table be an infinite sequence of digits x_1,x_2,x_3,\ldots that is valid in the usual sense: Every block of $k$ digits occurs with limiting frequency $10^{-k}$.
Section 3.5: What Is a Random Sequence?
Exercise 44. ▶ [16] (J. Good.) Can a valid table of random digits contain just one misprint?
Verified: yes
Solve time: 2m22s
Let the original table be an infinite sequence of digits
$$ x_1,x_2,x_3,\ldots $$
that is valid in the usual sense: Every block of $k$ digits occurs with limiting frequency $10^{-k}$.
Suppose that a single misprint occurs at position $m$, changing $x_m$ to a different digit. Let the printed table be
$$ y_1,y_2,y_3,\ldots , $$
where $y_i=x_i$ for $i\neq m$.
We compare the frequencies of an arbitrary $k$-digit block $B$ in the two sequences. A change at position $m$ can affect only those occurrences of $B$ that involve position $m$. There are at most $k$ such starting positions. Hence, among the first $N$ digits,
$$ \bigl|A_B(N)-A'_B(N)\bigr| \le k, $$
where $A_B(N)$ and $A'_B(N)$ denote the numbers of occurrences of $B$ in the first $N$ digits of the original and misprinted tables.
Dividing by $N$,
$$ \left| \frac{A_B(N)}{N} - \frac{A'_B(N)}{N} \right| \le \frac{k}{N}. $$
As $N\to\infty$, the right-hand side tends to $0$. Therefore the limiting frequencies of $B$ in the two sequences are identical.
Since this holds for every block $B$, the misprinted table satisfies exactly the same frequency conditions as the original valid table. A finite number of alterations, in particular one alteration, cannot change those limiting frequencies.
Therefore a valid table of random digits can contain just one misprint. In fact, changing any finite number of digits preserves validity.