Drunkards walk on a sphere.
For now, I will give a reformulation of this problem in terms that should make it easier to attack with analytic methods (at least to get results on the asymptotic behavior) and greatly simplify simulations. In a second time, I will maybe also attempt to solve it, but I don't guarantee any kind of success.
My reformulation of this problem is based on the following two remarks:
- On spheres, it is always easier to work with angles than with distances. Therefore, we will always work on a sphere of radius $r=1$, define an angle $\epsilon$ corresponding to one unit, and say that $A$ and $B$ meet if the angle between them is less than or equal to some angle $\beta$.
- If possible, it is better to have only one thing moving around. Therefore we will change our point of view and fix ourselves in the reference system of one of the two people, say person $B$. We will take this point to be the north pole (for the spherical coordinates we'll find ourselves in). In light of the first observation, the only variable of interest to us will be the latitude $A$ finds itself on.
For the mathematical details:
First we have to fix our coordinate system. Let $\phi$ be the longitude (this variable will be useful for calculations, but irrelevant in the end), and $\theta$ be the latitude. We put the north pole $B$ at latitude $\theta = 0$, and the south pole is at $\theta = \pi$. We'll denote positions by coordinates $(\theta,\phi)$.
Notice that by spherical symmetry, $A$ and $B$ bot taking a step is the same as $B$ staying put and $A$ taking two steps. Let's start by seeing what happens when $A$ takes one step. Let's say that $A$ starts in position $(\theta,\phi)$. We are only interested in the probability of $A$ landing at latitude $\theta'$. Notice that $P[A\text{ lands at }\theta'\le\theta_0]$ is given by the ration of the circumference of the circle at angle $\epsilon$ from $A$ below the meridian at $\theta_0$. To find this, I refer you to this answer by @Aretino, giving $$P[A\text{ lands at }\theta'\le\theta_0] = 1 - \frac{1}{\pi}\arccos\left(\frac{\cos\theta' - \cos\epsilon\cos\theta}{\sin\epsilon\sin\theta}\right)$$ whenever the term in the brackets is in $[-1,1]$, and $0$ or $1$ else (depending on $\theta'$). The distribution function $f_\theta(\theta')$ giving the probability to land at $\theta'$ after one step starting at $\theta$ can then be found as usual by differentiating this probability: \begin{align} f_\theta(\theta') = & \frac{\partial}{\partial\theta'}\left(1 - \frac{1}{\pi}\arccos\left(\frac{\cos\theta' - \cos\epsilon\cos\theta}{\sin\epsilon\sin\theta}\right)\right)\\ = & \frac{\sin\theta'}{\sqrt{1 - \left(\frac{\cos\theta' - \cos\epsilon\cos\theta}{\sin\epsilon\sin\theta}\right)^2}}\\ = & \frac{\sin\epsilon\sin\theta\sin\theta'}{\sqrt{\cos\theta'(2\cos\epsilon\cos\theta - \cos\theta') + \sin^2\theta - \cos^2\epsilon}}, \end{align} and $0$ outside the domain of definition of the original function. The probability of landing at $\theta'$ after two steps starting at $\theta$ is therefore given by $$F_\theta(\theta') = \int_0^\pi f_\theta(\theta'')f_{\theta''}(\theta')d\theta''.$$ I have some doubts this can be done analytically, but maybe some approximation can give something useful.
Given this data, we can have a hope to be able to do something at least to find asymptotic bounds for when $\epsilon\to0$, and if not, it will at least greatly simplify simulations, as we are reduced to simulating a walk on a line (parametrized by $\theta$) with a non-uniform probability to move to nearby points.
As stated in Daniel's answer, it is easier to think about the problem if we say only one of the two people is moving. Suppose person $B$ stays at the North pole and person $A$ starts at the South pole, and each time step, $A$ moves 1m either longitudinally or latitudinally.
If $A$ moves latitudinally, then $A$'s distance from the North pole remains the same. Therefore, we ignore these latitudinal movements and say instead that $A$ has a 50% chance of not moving and a 50% chance of taking a 1 meter step along a great circle containing the North and South pole. We parameterize this circle by $x = 0$ at the South pole, $x = 1$ one meter in one direction from the South pole, $x = -1$ one meter in the opposite direction, and so on.
If the radius of the sphere is 100000m, and $A$ needs to be 1000m from the North pole to see $B$, then $A$ will see $B$ if the distance of $A$ from the South Pole is $$ \pi 100000m - 1000m \approx 313159m$$
If $f(x)$ is the expected number of remaining time steps when $A$ is at position $x$ then $f$ satisfies $$ f(x) = \frac{f(x-1) + 2f(x) + f(x+1)}{4} + 1 $$ And $$f(313159) = f(-313159) = 0$$ We see that this is satisfied by $$ f(x) = 2(313159)^2 - 2x^2 $$
So $$ f(0) = 2(313159)^2 = 1.9 \times 10^{11} $$ So if the time steps are half seconds (since two movements happen each second - one for each person), the expected time should be 3108 years.