Vtyjkyuk

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

$$piapprox 3.underbracex_1x_2cdots x_n_textfirst periodoverbracex_1x_2cdots x_n^textsecond periodunderbracecdots_llaptextother,rlaptextdigits;.$$

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 $$0.x_1x_2cdots x_noverliney_1y_2cdots y_m$$ with an initial non-periodic sequence $x_1x_2cdots x_n$. 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.

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 $frac110$.

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

$$
sum_n=1^infty10^-n=frac110cdotfrac11-frac110=frac19;.
$$

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:

beginarrayr
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
endarray

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:

$$
frac110+frac910sum_n=2^infty10^-n=frac110+frac910cdotfrac1100cdotfrac11-frac110=frac110+frac1100=0.11;.
$$