I have this question, by my intuition the probability would be 0, since we do not know what the next irrational number decimal digit will be in base 10.
Does anybody have a more concrete idea, or is there already a solution to this ?
EDIT: Why I think it is not $1/10$ is because we could change the base from base 10 to let's say base 100 and then the probability would be $1/100$. That is also why I think it is zero.
EDIT 2 : What I wanted to ask was, that if we take the decimal digits of an irrational number. Let's say we take those digits in base 10. We do not know what will the next decimal digit be, as they go to infinity. But we could say, that the next digit can be any digit from 0 to 9. But we can change the base of these digits to base 2, where the next digit would have 1/2 probability to be 0 or 1. But that is a contradiction since 1/2 is not the same as 1/10 or if we pick a base 100, it is not the same as 1/100
EDIT 3: So we also have the probability $1/100$ to reconstruct the digits in base 100 back to base 10, since the numbers are always generating at random order since the number is irrational.
First... you seem to be confusing the concept of an irrational number with the concept of a normal number.
It is not true that given a specific irrational number that the digits are equally distributed. Consider Liouville's Constant written in base 10, $0.11010010001000010000010000001\dots$ This number has no $2$'s, $3$'s, $4$'s, etc... appearing in its decimal expansion anywhere.
Now, you are probably talking about choosing an irrational number at random (which you never clearly specified). We can in this case refer to the $n$'th digit in base $b$ of such an irrational number and it will be a particular digit with probability $\frac{1}{b}$ and the $n+1$'st digit will be independent of the first and also be that digit also with probability $\frac{1}{b}$.
Let us define random variables. Let $X$ be our randomly selected irrational number. Let $X_n^{(b)}$ be the $n$'th digit right of the decimal of $X$ as represented in base $b$.
We do have that $\Pr\left(X_n^{(b)}=X_{n+1}^{(b)}\right)=\frac{1}{b}$ for every $b$.
Your complaint that $\Pr\left(X_n^{(p)}=X_{n+1}^{(p)}\right)=\frac{1}{p}\neq \frac{1}{b}$ is completely irrelevant. $X_n^{(p)}$ is not the same as $X_n^{(b)}$... There is no contradiction here. There is no reason to expect $\Pr\left(X_n^{(b)}=X_{n+1}^{(b)}\right)$ to equal $\Pr\left(X_n^{(p)}=X_{n+1}^{(p)}\right)$. When you change from one base to another... things change. Yes, some of the fundamental properties of the number remain the same, however the frequency of the digits does not need to stay the same at all...