Well, I can reassure you that was not my angle of approach. The cherry-picking part I introduced to illustrate how improbable it is to get a correlation out of pure randomness.

That's not entirely fair - I think it is a matter of starting point.

A major problem in getting your question answered is that your terminology is sloppy, in fact truly sloppy.

You failed to make a definition. The terms "random" and "truly random" are neither used nor defined in probability texts. And after reading more of your posts it is not clear to me what you mean.

Let me give a simple concrete QM example:
Given an entangled pair from state √½(|00⟩ + |11⟩), we let A measure one of the pair at angle 0º, i.e. with measurement operator/observable ##Z =\begin{pmatrix}1&0\\0&-1 \end {pmatrix}##.
We let B measure the other at 30°, i.e. with observable ##½Z + √¾X =\begin{pmatrix}½&√¾\\√¾&-½\end{pmatrix}##.

The joint probability density of (A,B) is (1,1) with prob ⅜, (1,-1) with prob ⅛, (-1,1) with prob ⅛, (-1,-1) with prob ⅜. (1 & -1 are eigenvalues of the observables)
We see A and B agree with prob = ¾ = cos²30º, as usual.
The correlation coefficient is ½.
The marginal density of A is 1 with prob ½, -1 with prob ½. Same for B. A and B are not independent.

All of this is justified by repeated trials in the lab.

Can you ask your question from the above formulation?

I don't know what you mean by fragment. Do you have a link?
If a1,a2, ... ,al with l=20 is a binary sequence, what is a fragment of 10 bits? Is it a subset of size 10? Is it a contiguous subset like a7,a8, ... ,a16? Or what?

I now think you were trying to define a binary normal sequence, but failed.

"A sequence of bits is random if there exists no Program shorter than it which can produce the same sequence." ~ Kolmogorov
So obviously it is impossible to exhibit a Kolmogorov random sequence.

Neither normality or K-random imply one another. But all of this should be in the Probability section of PF. And none of this is relevant to QM.