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


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

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


1 Answer 1


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 $n$.

  • $\begingroup$ Given this analysis, could you now answer the questions? Specifically, (1) should the algorithm look at overlapping windows or not (and why?) and (2) what will happen when $3$ is increased (and, implicitly, why exactly might $3$ have been chosen for this test)? $\endgroup$
    – whuber
    May 28, 2015 at 21:51
  • $\begingroup$ Answering (1) objectively would (should?) involve knowing whether or not there exists a solution in the overlapping case. I don't know about that. I don't think the video is specific about this. As far as (2) goes, that's what my last paragraph answers. And I don't know why 3 was chosen for this test...perhaps because it was small, greater than 1, and not 2. $\endgroup$
    – Taylor
    May 28, 2015 at 23:52
  • $\begingroup$ Taylor, thank you for reply bud i still did not get the point. My math background is not as strong as yours. I went lost when you start talking about summing indicators (just fancy name for saying if I've already saw the sequence?) and binomial distribution. $\endgroup$ May 29, 2015 at 0:35
  • $\begingroup$ So to be more direct: don't look at overlapping sequences, and also make sure the total length is divisible by 3 so you don't have any leftovers/incomplete seqs of length 3. Question 2: there is no strong reason not to look at sequences of length 4,5,... etc. Questuon 3: no not really $\endgroup$
    – Taylor
    May 29, 2015 at 16:46
  • $\begingroup$ 1. when there is very long sequence then the remainder wont have any significant impact on result or will it? The question might also stand: how long the input sequence must be so the possible remainder wont affect the result? Or what is the tolerable deviaton among histogram members? I thing there is not a clear answer for this and it is some deal in statistics or am I wrong? $\endgroup$ Jun 3, 2015 at 21:22

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