The number of rabbits at the (i+1)'th step is equal to the number of rabbits at the i'th step minus that one rabbit which is /maybe/ being eaten. And the "maybe" I think can be quantified as the probability the Troll has to find a rabbit at the i'th step. So:

R_(i+1) = R_i -1 * P(Troll finds a rabbit)

And now we have to calculate the probability of the event "Troll finds a rabbit" (i'll call it event A). There are 2 cases: * the Troll falls into a black hole (btw, i'm gonna assume the blackholes teleport to any cell with uniform probability); * the Troll does not fall into a black hole;

I'm going to call these two events B and not(B).

So:

P(A) = P(A given B) + P(A given not(B))

Well, P(A given B) is the probability that Troll is being teleported to a cell with a rabbit in it, that is number of rabbits over number of cells: R_i / (nm)

P(A given not(B)) is the probability that Troll will step on a rabbit in a nearby cell which is: rabbits' density over 8: R_i / (8nm)

So:

R_(i+1) = R_i - ( R_i/nm + R_i/8nm ) = R_i * ( 1 - 9/8nm ) = ... iterating ... = R_0 * ( 1 - 9/8nm )^i

R_0 being the initial number of rabbits. I'm not sure all this is correct, but that formula confirms the following intuitions: * The number of rabbits is monotonically decreasing at every step; * The bigger the grid, the slower the decrease.

edit: The closed formula would be:

int numOfRabbits(int numOfSteps) {

    if (numOfSteps == 0) return R_0;

    else return R_0 * (1 - 9/8nm)^(numOfSteps-1);

}

It's a silly question unless it's for a quantitative role.

As a first go I'd assume that the topology was that of a torus, so as to not deal with the boundary. Now, as the exact definition of a "round" wasn't specified, I'll just assume the troll can walk forever.

Now, it's been proven that a simple symmetric random walk on a 2D lattice has probability 1 of reaching any given point as the number of steps goes to infinity. So, ignoring the teleporting, the closed form answer for the number of rabbits eaten is mnR.

With teleporting, I'd guess the Markov chain would mix better. So, the closed form answer would again be mnR.

E(number_of_rabbits) = rabbits_in_last_turn - prob(land on rabbit) - sum_i_to_infinity(prob(land_on_blackhole)^i*prob(land on rabbit))