TITLE: Drunkards walk on a sphere.
QUESTION [8 upvotes]: I simulated the following situation on my pc. Two persons A and B are initially at opposite ends of a sphere of radius r. Both being drunk, can take exactly a step of 1 unit(you can define the unit, i kept it at 1m) either along a latitude at their current location, or a longitude. A and B are said to meet, if the arc length distance between A and B becomes less than equal to 1km.
Note: the direction of possible motion of each man is fixed w.r.t thr axis of the globe. Either latitude or longitude. Assume such a coordinate system exists before hand(just like the 2d analog on a plane, moving in x or y only and not absolutely randomly).
The simulation returned results, which i could not comprehend fully. The average time to meet, was about 270 years for a sphere of radius 100km!. Can someone shed some light on how i can proceed with proving this result. I want the expected time of meeting given the radius and step length, given that each move requires 1 sec. I tried considering a spehrical cap of arc length √n, after n steps, in analogy with the 2d model. But then,i cant calculate the expected time. If possible please help or suggest some related articles.

REPLY [3 votes]: 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.