The followng example is from An Introduction to Kolmogorov Complexity and Its Applications, Example 2.4.1. and is concerned with Martin-Löf-Tests for finite strings:
A string $x_1 x_2 \ldots x_n$ with many initial zeros is not very random. We can test this aspect as follows. The special test $V$ has critical regions $V_1, V_2, \ldots$. Consider $x = 0.x_1 x_2\ldots x_n$ as a rational number, and each critical region as a half-open interval $V_m = [0, 2^{-m})$ in $[0,1)$, $m = 1,2,\ldots$. Then the subsequent critical regions test the hypothesis "$x$ is random" by considering the subsequent digits in the binary expansion of $x$. We reject the hypothesis on the significane level $\epsilon = 2^{-m}$ provided $x_1 = x_2 = \ldots = x_m = 0$.
Now I have a problem with intuition. Consider some very irregular, uncompressible string $w$ of length $n$ which starts with zero, then $w \notin V_1$ and hence would be classified as not random by this test, just because its starts with zero but otherwise is very irregular (hence very "random")?
The crucial point is the level of significance of the test, if you consider $\epsilon = 2^{-1}$ then you are rejecting the null conjecture (It is a random array) in many cases. For some reason you are bound to think that many $0$ may occur. So you will reject randomness at the event the first $m$ digits are zero.
You could also think that it is not very likely that a random number will have many $1$ in its dyadic decomposition. So you would consider as evidence that it is not a random number if the first $m$ digits are all $1$.
This test is going to reject randomness at level $\epsilon = 2^{-1}$ just because its starts with $1$ but otherwise will consider it as very irregular.
As any particular region of the line is arbitrary. Your criterion of a random number will always be insufficient once random numbers may indeed belong to that region ($V_m$ for instance) and non-random numbers also belong to that region.
It is a limitation we must deal with.