TITLE: Collisions between rooks taking random flights on an N by M chessboard
QUESTION [7 upvotes]: I randomly place $k$ rooks on an (arbitrarily sized) $N$ by $M$ chessboard.  Until only one rook remains, for each of $P$ time intervals we move the pieces as follows:
(1) We choose one of the $k$ rooks on the board with uniform probability.  
(2) We choose a direction for the rook, $(N, W, E, S)$, with uniform probability. 
(3) We choose a number of squares in which to move the rook along the direction chosen in [2] with uniform probability over the interval consisting of the rook's current position to the edge of the board.
(4) If the rook being moved collides with another piece while being translated in [3], just as in regular chess it will annihilate that piece and remain at the piece's former position.
NOTE - An alternative way of stating [2], [3], and [4] would be to say that the chosen rook samples all possible sets of moves, with uniform probability, and is unable to bypass other rooks without annihilating them and stopping at their former positions.
NOTE 2 - Gerhard Paseman is correct in suggesting that the original formulation for [2] and [3] will bias the rook towards shorter path lengths.  This is in part due to the choice of direction in [2] not being weighted by the resulting possible number of choices in [3], and also the over-counting of positions in [3] due to the lack of consideration that there may be a collision.  There are also problems with [2] near the board's boundaries where a direction can be chosen in which no move can take place.  Instead of [2] and [3], I'll suggest that a better method would be to number all possible position that the chosen rook from [1] can occupy (keeping the collision constraint from [4] in mind), and then use a PRNG to select the next position. 
What does the distribution look like for the number of time intervals, $P$, necessary for only a single rook to remain on the board?

REPLY [3 votes]: It occurred to me it might be of interest to see what happens if you start with a board completely clogged with rooks*, so I decided to pluck the lowest-hanging nontrivial fruit and examine the $2\times2$ case, which features 5 distinct states:  the starting state $S$ with 4 rooks, a trio state $T$, a pair of doublet states $R$ and $D$ with the rooks lined up in a row or along a diagonal, respectively, and the quitting state $Q$ with just one rook.  
In this set-up, state $S$ transitions in one step to $T$ (I'm assuming here that when you pick a rook at random, you actually have to move it).  State $T$ transitions back to itself with probability 1/3, to $R$ with probability 1/3, and to $D$ with probability 1/3.  State $D$ transitions to $R$ with probability 1, while $R$ transitions to $D$ with probability 1/2 and to $Q$ with probability 1/2.  For the expected number of steps to get to $Q$, we thus have
$$E(S) = 1+E(T)$$
$$E(T) = 1 +{1\over3}(E(T)+E(R)+E(D) $$
$$E(D) = 1 + E(R)$$
$$E(R) = 1 + {1\over2}E(D)$$
from which one finds $E(R) = 3$, $E(D) = 4$, $E(T) = 5$, and $E(S) = 6$.
It seems doubtful that the expected values will be integers in general, but it might be worth checking the $2\times3$ and $3\times3$ cases, which ought to be doable.  (The $2\times3$ case, which has 23 essentially different states, might be a good place to experiment with different conventions for the transition probabilities.)  One thing worth noting:  The states $R$ and $D$ are equiprobable when starting from $S$, but not if you create a 2-rook state from scratch by placing rooks at random.  This makes me wonder what Joseph O'Rourke's histogram would look like if you started, say with 16 rooks on a $4\times 4$ board but didn't start counting moves until you were down to the last 4 rooks.
*I wrote all this up before I read Aaron Golden's answer carefully.  His graph shows simulated results for the $8\times8$ case starting with 64 rooks, but he's allowing rooks not to move if they're on an edge of the board.