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I found this link about frequencies of digits appearing in π : http://www.eveandersson.com/pi/precalculated-frequencies

This made me wonder: do those frequencies ever become even on the lifespan of the currently known digits? (each of the 10 digits [0,1,...,9] appearing n times on the (10*n)'th digit)

If not, what's the closest they get to, and when?

Lyrca
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  • The word "even" is a little unclear. Do you mean "after some point only the digits $0, 2, 4, 6, 8$ occur"? Or do you mean "the frequencies of each digit become (nearly) the same"? Neither is known but it speculated that $\pi$ is a normal number. Search for "normal number". – badjohn Feb 27 '20 at 13:19
  • Sorry about that. I expected the context to have made it clear enough. I meant each of the 10 digits appearing n times on the (10n)'th digit. – Lyrca Feb 27 '20 at 14:09
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    i just wanted to be sure. As I said, it is speculated but not proven that $\pi$ is normal. So, I suggest reading up on that. However, it is not proven. It could be that after a certain point, the digit $9$ never appears. – badjohn Feb 27 '20 at 15:10

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It's unlikely. If you take $10n$ random decimal digits, the probability that you have $n$ each of $0, 1, \ldots, 9$ is $(10 n)! 10^{-10n}/(n!)^{10}$. In an infinite sequence of random digits, the expected number of positive integers $n$ for which this is true is $$ \sum_{n=1}^\infty \frac{(10 n)! 10^{-10n}}{(n!)^{10}} \approx 0.0003933209904$$ most of which ($0.00036288$) comes from the $n=1$ term.

Of course the digits of $\pi$ are not really random, but their statistical properties are (as far as we can tell) similar to a random sequence, and in this case we know that the first $10$ digits are not evenly distributed (e.g. digits $2$ and $4$ are both $1$).

Robert Israel
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