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Let's take for example $\pi$. The numbe $\pi$ is irrational, hence the decimal pattern is non-periodic. However, could it happen, that by just observing a huge but finite amount of digits of $\pi$, it would "look periodic"?

It is not known, but it is conjectured that $\pi$ is a normal number, i.e. any sequence of digits of length $n$ occurs with density $10^{-n}$. So let's assume this is true, and then ask what is the probability that for some sufficiently large $n$, $\pi$ looks like

$$\pi\approx 3.\underbrace{x_1x_2\cdots x_n}_{\text{first period}}\overbrace{x_1x_2\cdots x_n}^{\text{second period}}\underbrace{\cdots}_{\llap{\text{other}}\,\rlap{\text{digits}}}\;.$$

Of course, this question is not really restricted to $\pi$ or to digits of numbers, but can be asked in the context of normal sequences (joriki pointed out in the answer that for a specific number the probabilit is always one or zero).


Remark. I know that general periodic number can have the form $$0.x_1x_2\cdots x_n\overline{y_1y_2\cdots y_m}$$ with an initial non-periodic sequence $x_1x_2\cdots x_n$. But since in a normal number any finite sequence occures, it is certain that any sequence will also repeat an arbitrary amount of times. The question is specifically about whether such a period can start with the first digit.

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  • $\begingroup$ The title "Will a normal number look like a periodic number?" was scarily imprecise (even misleading, I believe). Modified. $\endgroup$ – Did Aug 29 '18 at 8:29
  • $\begingroup$ @Did Why do you think so? I am not specifically asking for $\pi$, this is just a catch (and I try a little bit of poetry in my titles as well, maybe inappropriate). For $\pi$ the probability is zero or one (see the answer of joriki). However, I ask for a probability in the title now if this is better. $\endgroup$ – M. Winter Aug 29 '18 at 8:31
  • $\begingroup$ This maybe so but the former title is not asking what you mean to ask, about pi or about any other number. The new version you adopted does not either. Note that 0.1231234756178... is (in general) not periodic thus it solves the question in the body but not the question in the title. $\endgroup$ – Did Aug 29 '18 at 8:32
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One way to answer your question would be: “Since $\pi$ is a fixed number and not a random variable, the probability is either $0$ or $1$; we just don't know what it is.” I suspect that that's not the interpretation you intended for your question, so I'll flesh out the concept of “probabilities” in this context in a different way, by replacing $\pi$ by a number whose decimal digits are randomly uniformly and independently chosen.

Then the probability is not $0$, since the probability for the period $n=1$ to work is $\frac1{10}$.

The probability is also not $1$, since it's bounded above by the sum of the marginal probabilities for all $n$:

$$ \sum_{n=1}^\infty10^{-n}=\frac1{10}\cdot\frac1{1-\frac1{10}}=\frac19\;. $$

Determining the exact probability seems hard, since one would have to account for all cases with multiple periods. This Java code determines the probability for a period to appear up to finite $n$, with the following results:

\begin{array}{r|l} 1&0.1\\ 2&0.109\\ 3&0.1099\\ 4&0.1099891\\ 5&0.109998001\\ 6&0.1099988911\\ 7&0.109998980101\\ 8&0.109998989001028\\ 9&0.10999898989102999\\ 10&0.1099989899800301008 \end{array}

So the probability is apparently very close to $0.11$, which is the estimate you get by subtracting out the cases where a sequence with a longer period starts with two identical digits:

$$ \frac1{10}+\frac9{10}\sum_{n=2}^\infty10^{-n}=\frac1{10}+\frac9{10}\cdot\frac1{100}\cdot\frac1{1-\frac1{10}}=\frac1{10}+\frac1{100}=0.11\;. $$

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  • $\begingroup$ Of course, I ask this for a truely random normal sequence ;). Are you okay with me editing the question to specifically ask for the exact probability if it can be computed nicely? I ask because simply changing the question after an answer is not okay. $\endgroup$ – M. Winter Aug 29 '18 at 8:28
  • $\begingroup$ @M.Winter: Thanks for asking. Yes, I think that would be good, as it makes for a more interesting question :-) $\endgroup$ – joriki Aug 29 '18 at 8:42

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