How does meeting probability on a finite $d$-dimensional grid depend on the dimension? At the beginning, we put two people $A$ and $B$  randomly on a finite $d$-dimensional closed grid.
Along each dimension there are $2^n$ positions $p_k$, so $p_{2^n}=p_0$. Think of a circle if $d=1$ or a torus if $d=2$ and so on...
Let's say $m$ is the minimal number of steps to reach each other. $m$ is unknown to $A$ and $B$.
Now both start to move step by step along a randomly chosen direction on the grid.
If we wait for less than $m/2$ steps, it is not be possible for them to meet at all.
If we wait for infinitely many steps, I would expect them to meet somewhen for sure.

How does the meeting probability as a function of steps on a finite $d$-dimensional grid depend on the dimension?

 A: The random walk is a Markov chain on the torus graph $(\mathbb{Z}/k\mathbb{Z})^d$. In the general case of a random walk on a graph, let $N$ be the total number of nodes. Let $v$ be the probability distribution (an $N$-dimensional vector) of the initial position of the walker. Let $P$ be the $N \times N$ matrix whose entry $P_{ij}$ is the probability of moving from point $j$ to $i$ if points $i$ and $j$ are adjacent, and $0$ otherwise (we do not require symmetry or $0$s on the diagonal). The probability of the walker being at point $i$ after $m$ steps is the $i$th component of the vector $P^mv$.
Set $v_a$ and $v_b$ to be the probability distribution according to which the initial position of $A$ and $B$ are chosen. For example if $A$ is known to start from position $i$ then $v_a = e_i$ where $e_i$ is the standard basis vector with $1$ at the $i$th position. The probability $p_a(m, i)$ of finding $A$ at the position $i$ at the $m$th step is equal to the $i$th entry of $P^m v_a$, i.e., $$p_a(m, i) = \langle P^mv_a, e_i \rangle$$ and similarly $p_b(m, i) = \langle P^mv_b, e_i \rangle$. Using the independence of the two random walks, the probability of $A$ and $B$ meeting at the $i$th position at the $m$th step is equal to $p_a(m, i)p_b(m, i)$. The total probability of $A$ and $B$ meeting at the $m$th step is then
$$\sum_i p_a(m, i)p_b(m, i) = \sum_i \langle P^mv_a, e_i \rangle \langle P^mv_b, e_i \rangle = \langle P^m v_a, P^m v_b \rangle.$$
If the initial positions of $A$ and $B$ are randomly chosen with every position equally probable, then $$v_a = v_b = (\underbrace{\frac{1}{N}, \ldots, \frac{1}{N}}_{N \text{ times}}) =: v$$ for all $i$. Each grid point has the same number of neighbours ($2d$ if only counting axes-parallel neighbours, $3^d-1$ if we include diagonally adjacent neighbours, and add $1$ if we allow the walker to remain stationary), and each non-zero entry of $P$ has the same value. This means that $P^mv_a = P^mv_b = v$ for every $m$, i.e. the probability $p(m, i)$ of finding $A$ at the point $i$ at the $m$th step is simply $1/N$ and does not depend on $m$. Therefore the probability of finding both $A$ and $B$ at point $i$ at the $m$th step is equal to $\displaystyle p(m, i)^2 =\frac{1}{N^2} = \frac{1}{k^{2d}}$. The total probability of $A$ and $B$ meeting at some point on the grid at the $m$th step is therefore $1/N$.
Interestingly enough we did not need the particular structure of the torus graph; we just needed that the graph was regular and that the initial and transition probabilities were uniform. So this result still holds if we slightly relax our definition of adjacency, or if the grid has a different number of points in different dimensions.
