Researching the frequency stability property I have seen this short, wonderful video about the frequency stability property (also there is another video from this guy under his Kickstarter project ArtOfTheProblem; they are both great, and I highly recommend them).
It basically said that when you have two sequences of 0s and 1s you can determine which of those two sequences was truly generated randomly (using a coin flip procedure) and which one was generated by a human trying to act randomly. The trick to discern the two sequences from one another starts with counting non-overlapping segments of length 3. For any sequence you pick, the proportion of times it occurs should be the same as all the other sequences. 
Consider following example
1010001101

Three questions:
1st: Which 3 numbers should we observe? Or in another words, should the window of size 3 overlap? Is this the correct histogram?
2 x 101
1 x 010
1 x 100
1 x 000
1 x 001
1 x 011
1 x 110

Or if the window should not overlap then what to do with last 1:
1x 101
1x 000
1x 110
1??

2nd: Why should be observer 3 subsequent numbers? Does it have something to do that the string consist of two numbers (0 and 1)? What would happen if we observe 4,5,6 ... subsequent numbers?
3rd: Do you know some other channel on youtube like this one?
EDIT:
I've done experiments using (non)overlapping window, results are here
 A: I take it that he means the sequences don't overlap. If you look at a sequence of length $n=3$, and the sequences are equally likely, then the probability of any sequence is $\frac{1}{2^3}=\frac{1}{2^n}$, since there are $2^3=2^n$ possible sequences. 
Now say you observe length $n$ sequences $m=10$ times. Then that's a total of $n\cdot m = 30$ numbers. But instead of thinking about $30$ numbers, think about $10=m$ indicator functions that are based on the $10$ sequences of three you saw. Define $1_1, \ldots 1_m$ as indicator functions that are just $1$ if your sequence that you pick is observed, and $0$ otherwise. The sum of these indicators follows a binomial distribution with parameters $m$ and $p = 2^{-n}$. Let's call it $Y$.
So $Y = \sum_{i=1}^m 1_i$. By the law of large numbers $\bar{1} = \frac{Y}{m} \to \frac{m 2^{-3}}{m} = 2^{-3}$. This is true for any arbitrary sequence. 
$\bar{1}$ is the proportion of occurences of any sequence you pick. If you looked at sequences of a difference length, replace $n$. 
