Consider an infinite random sequence $S=[i_1, i_2, ...]$ built with digits 0 to 9 (or any other alphabet). Every digit has a probability > 0 to be picked each time.
Consider a prefix of $S$, $S_k$ the finite sequence of $k$ elements from the first element to the $k$ element of $S$
$S_k$ = $[i_1,...,i_k]$
Suppose there exist some values of $k$ such that:
$S_{2k}$ = $[i_1, ...i_{2k}]$ = $[i_1, ..., i_k, i_1, ..., i_k]$ = $S_k^\frown S_k$
($^\frown$ is the concatenation operation.)
Define $S$ as birbant if $S_{2k}$ has the property to be equal to $S_k^\frown S_k$, with $k > m, m \in \mathbb{N}$ for any value of $m$.
Define $S$ as anti-birbant if does not exist any $S_{2k}$ that has the property to be equal to $S_k^\frown S_k$, with $k > m, m \in \mathbb{N}$ for any value of $m$.
Define $S$ as semi-birbant if $S_{2k}$ has the property to be equal to $S_k^\frown S_k$, with $k > m, m \in \mathbb{N}$ for any $m < C$, and does not exist any $S_{2k}$ that has the property to be equal to $S_k^\frown S_k$ for $k > m$ when $m >= C$. With $C$ costant $ > 3$.
When $S$ is semi-birbant, there are $n$ $2k$-prefixes that have the above property, $S$ is also defined as n-semi-birbant .
What is the probability that any $S$ is birbant, anti-birbant or semi-birbant ?
Is there any predominance of a particular class n-semi-birbant classes for any $S$, or is it possible to compute an expected value of n ?
Does this depend on the size of the alphabet?
Example 1 (a pipe is inserted at the end of every $S_{2k}$):
S = 1,1|,2,1,1,2|,7,2,1,5,2,1,1,2,1,1,2,7,2,1,5,2| ...
We have:
- $S_2$ equal to $S_{1}^\frown S_{1}$
- $S_6$ equal to $S_{3}^\frown S_{3}$
- $S_{22}$ equal to $S_{11}^\frown S_{11}$
If no more prefixes repeats, this sequence is semi-birbant (3-semi-birbant)
If an infinite number of repetition of prefixes occurs, this sequence is birbant
Example 2 (binary alphabet)
S = 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0...
The sequence is generated by start with 10 and double the ones on the front and the zeros at the end, then split the digits and concatenate to S and repeat.
This sequence is not-birbant.
Related: Does an infinite random sequence contain all finite sequences?
If you assume the random integers $\{I_i\}_{i=1}^{\infty}$ are i.i.d. equally likely over the 10-element set $\{0, ..., 9\}$ then $$P[S_{2k}=[S_k:S_k]]=(1/10)^k \quad \forall k \in \{1, 2, 3, ...\}$$ and so $$ \sum_{k=1}^{\infty} P[S_{2k}=[S_k:S_k]] < \infty$$ Therefore, the Borel-Cantelli theorem says that, with probability 1, there are only a finite number of integers $k$ for which $S_{2k}=[S_k:S_k]$.
So (assuming I understand your definitions properly), the probability of being Birbant is zero.
You get a similar result if the alphabet size is any positive integer $m\geq 2$ (we do not require $m=10$). For $m\geq 2$ we have $\sum_{k=1}^{\infty} (1/m)^k<\infty$.