Let $G$ be a finite non-abelian group, and consider a choice of $N$ distinct elements $g\_{0},g\_{1},\ldots,g\_{N-1}\in G$ that generate $G$. Now, let $t$ be an arbitrary positive integer, and let $d\_{1},\ldots,d\_{K}\in\left\{ 0,\ldots,N-1\right\}$ be the $N$-ary digits of $t$, so that:$$t=\sum\_{k=1}^{K}d\_{k}N^{k-1}$$ is then the $N$-ary/$N$-adic representation of $t$. Next, letting $\mathbb{N}\_{1}$ denote the positive integers, define the map $\chi:\mathbb{N}\_{1}\rightarrow G$ by: $$\chi\left(\sum\_{k=1}^{K}d\_{k}N^{k-1}\right)=\prod\_{k=1}^{K}g\_{d\_{k}}=g\_{d\_{1}}g\_{d\_{2}}\cdots g\_{d\_{K}}$$ Finally, make $\mathbb{N}\_{1}$ into a probability space by equipping it with the probability measure defined by the upper density of a set of integers: $$\textrm{P}\left(A\right)=\limsup\_{n\rightarrow\infty}\frac{\left|A\cap\left\{ 1,2,\ldots,n\right\} \right|}{n},\textrm{ }\forall A\subseteq\mathbb{N}\_{1}$$ 


With this set-up, we can then think of $\chi$ as a random variable on $\mathbb{N}\_{1}$. That being so, is it then true that $\chi$ will be uniformly distributed?—i.e.: $$\textrm{P}\left(\chi=g\right)\overset{?}{=}\frac{1}{\left|G\right|},\textrm{ }\forall g\in G$$


My intuition has convinced me that this must be the case whenever $G$ is *abelian*, however, I cannot help but feel that in the non-abelian case, the probability distribution of $\chi$ somehow depends on the particular choice of the generating set $\left\{ g\_{0},\ldots,g\_{N-1}\right\}$.



Probability distribution of random products of elements of a generating set of a finite non-abelian group
To begin with, endowing the set of integers with the upper density is quite far from making it a probability space. Nonetheless, the question you ask still makes sense. Namely, you consider the $G$-valued sequence defined by your function $\chi$ and just ask whether the empirical frequencies of its values converge to the uniform distribution.


The answer is indeed "yes". The proof consists of 3 steps.


(1) Take $n=N^d$ to be the $d$-th power of $N$. Then the empirical distribution at time $n$ is just the $d$-fold convolution of the uniform measure on your generating set, which is well-known to converge to the uniform distribution on $G$.


(2) The uniform distribution on $G$ is preserved under the multiplication by any group element, which implies that for any $N$-rational numbers $0\le \alpha<\beta\le 1$ the empirical frequencies over the sequence of intervals 
$[N^d\alpha, N^d\beta]$ converge to the uniform distribution as well.


(3) It remains to deduce the claim from (2).


