Let $x\in [0,1)$ be a real number. Let $x=0.a_1a_2a_3\ldots$ be the expansion of $x$ in base $b.$
Consider the average distance between two consecutive digits of $x:$
$$d(x):=lim_{N\to +\infty}\frac{\sum_{i=1}^{N}|a_{i+1}-a_i|}{N} $$
First question: Does anybody study this quantity so far?
More precisley, it is quite easy to construct a real number $x$ for which the above limit does not exist. It just suffices to choose the digits so that the quantity $$X_n:=\frac{\sum_{i=1}^{N}|a_{i+1}-a_i|}{N}$$ oscillates as $N$ increases.
Second question: If we choose a normal number $x$ then does the above limit exist?
Surely if $x$ is normal it is possible to evaluate the expectation $$\mathbb E(X_n).$$ In fact a simple calculation shows that this is equal to $$\frac{b^2-1}{3b}$$ (to this aim remember that the expectation is linear and consider the possible values assumed by $|a_{i+1}-a_i|$.)
Third question: The variance of $X_n$ seems more difficult. Do you have any idea to find it?
First we exclude the case of equality, and the case of decreasing (focus only on increasing, as the probability of decreasing by a value is equal to the probability of increasing by it overall) I'll add the case of equality at the end.
Case: $\Delta=9$
Occurrences: $0\to 9 (1)$
Case: $\Delta=8$
Occurrences: $0\to 8, 1\to 9\space(2)$
Case: $\Delta=7$
Occurrences: $0\to 7, 1\to 8, 2\to 9\space(3)$
Case: $\Delta=6$
Occurrences: $0\to 6, 1\to 7, 2\to 8, 3\to9\space(4)$
etc
Case: $\Delta=1$
Occurrences: 9
There are $10$ cases of equality, so the total cases are $$2(1+2+3+4+5+6+7+8)+10=82$$
We have $\frac{2}{82}=\frac{1}{41}$ chance of the difference being $9$
$\frac{4}{82}=\frac{2}{41}$ chance of it being $8$.
and so forth.
Overall we get:
$$\frac{2}{82}\times9+\frac{4}{82}\times8+\frac{6}{82}\times7+\frac{8}{82}\times6+\frac{10}{82}\times5+\frac{12}{82}\times4+\frac{14}{82}\times3+\frac{16}{82}\times2+\frac{18}{82}\times1+\frac{20}{82}\times0$$ $$=\frac{330}{82}\approx4.02$$