Suppose $x_1,x_2,...,x_m\in S_n$ are uniformly randomly chosen ($m=O(1)$, or at least $m\ll n!$) and that they generate the entirety of $S_n$. (The probability that two random generators (i.e., $m=2$) generate $S_n$ is already about $3/4$ for large $n$, [0] so heuristically you'd expect most combinations for even $n=3$ to work anyway.) Let $a_0=I$ begin at the identity permutation and $a_{i+1}=a_ix_k$, where $k$ is chosen uniformly at random. For a fixed, uniformly randomly chosen $p\in S_n$, what is the expected value of the first $i$ such that $a_i=p$? In other words, what is the expected length of a random walk ending on $p$ on the (directed) Cayley graph of $S_n$ generated by $\{x_i\}$?
I'm looking for crude heuristics, not necessarily rigorous ones—I'm just struggling to visualize the problem. I suppose the difficulty in visualization relates to the unintuitive fact that permutations tend to have a rather small order and thus don't "mix" well. I can find some related results: for example, for two randomly chosen $g,h\in S_n$ which generate $S_n$ (or in fact $A_n$), along with their inverses $g^{-1},h^{-1}$, the average shortest word to an arbitrary permutation is $O(n^2 (\log n)^c)$ for some absolute constant $c$. [1] [2] contains some results and a conjecture on the worst-case diameter with $O(1)$ generators. But I can't seem to find estimates for a random walk.
[0] https://math.stackexchange.com/a/453068/677124
[1] https://www.sciencedirect.com/science/article/pii/S0021869314004797
[2] https://pages.uoregon.edu/kantor/PAPERS/STOCdiameter.pdf
