Expected number of steps to walk through points by multiple walkers Suppose there are $m$ points.
A walker can visit the points in any order, but it will not visit a point twice.
There are $n$ walkers, and their starting points are randomly chosen.
After $k$ steps, each point is visited at least once by any of the walkers.
What is the expected number of $k$?
Some assumptions:


*

*a walker doesn't know the visited points of other walkers

*multiple walkers can be at the same position


Additional question:
What if walkers are capable of knowing the visited points of others after they have chosen their random starting points? (so the first assumption doesn't hold anymore)
 A: Einar's answer is definitely the one you should tag as an answer but I'll try and answer for your additional question.
If Walkers are capable of knowing which points have already been visited, and assuming their movement restriction expand to that knowledge (ie : they can't move to an already visited point, no matter the walker that visited it), then it is important to understand that it is the first step that fully determine $k$ (the step at which each points is visited).
First step is unique in that, it's the only step for wich walkers are allowed to be on the same point. But after that first step, each new step will see $n$ points visited, until all are.
So let's detail what happens in the first step :


*

*The first walker $n_1$ picks a point at random. His probability to pick an unvisited point is $P_1=1$ since there are only unvisited points at this stage.

*The second walker $n_2$ well he's facing a pool of $m-1$ unvisited points and $1$ visited point so his probability to pick an unvisited point is $P_2 = {m-1\over m}$.

*What's tricky for the third (and the following) walkers is that you have to account for how $n_2$ did before : either $n_2$ reached an unvisited point then $P_3 = {m-2\over m}$ or he did not and $P_3 = {m-1\over m}$ (pardon my weak notations here)


This is close to the Coupon collector's problem
From there we can reason with expectation :
Let $E_i(m)$ be the expected number of steps requiered for an unvisited point to be visited with $m$ total points and $i-1$ visited points, then $$E_i(m) = {m\over m+1-i}$$
So then, with $n$ walkers, the expected number of visited points at step $1$ is $v_1$ with :
$$v_1 = \max_{j\in\mathbb{N}}\left( j \;|\; \sum_{i=1}^j E_i(m)\le n \right)$$
To clarify the notation this is $v_1$ is the highest integer $j$ so that the sum is less than $n$
Once you got this, you can easily deduce $k$ since, as discussed at the start, once you know $v1$ there is no randomness as far as reaching every points visited, you just add $n$ each step until you're over $m$. With a more mathematical notation :
$$ k= \min_{j\in\mathbb{N}^*}\left( j-1\;|\; v_1 + (j-1)\cdot n \ge m \right)$$
I did my best to explain it but I'm not a pro at this, feel free to ask or correct me if I'm wrong.
