Analysis of board game die rolls violates intuition My friends and I occasionally play a board game called Feudality.  Recently some of us analyzed the dice-rolling aspect of the game and found that some assumptions we'd been making based on our intuition were not in fact true.  What follows is a simplification of the rules in question.
A player's board consists of an 8x8 grid of cells.  On one's turn, one rolls a pair of eight-sided dice, color-coded so that one die indicates the horizontal axis and the other die the vertical axis.  The rolled coordinate is "activated," as are all of the adjacent cells.  If both dice show the same number, the process is repeated until doubles are not rolled.  For example, if (2,2) was rolled twice, followed by (4,2), then the board's cells would be activated in the following numbers:
2 2 3 1 1 0 0 0
2 2 3 1 1 0 0 0
2 2 3 1 1 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0

The question is: For each cell, what is the average number of activations per turn?
My intuition tells me that the non-corner cells on and adjacent to the main diagnonal should have the highest average activation rate, since they can be activated more than once per turn, and all other cells can be activated at most once per turn.  Seems entirely obvious, but when I run simulations of the game, I find that:


*

*The corner spaces have the lowest activation rate (expected)

*The other edge spaces have the next-lowest activation rate (expected)

*All other cells have essentially the same (highest) activation rate, whether on or off the diagonal (unexpected!)


In my most recent simulation of ten million games, I find that the activations per turn for the corner cells is about 0.071, for the other edge cells it's about 0.107, and for every other cell it's about 0.161.
I've been trying to work out the exact rate.  It's pretty simple for the cells away from the diagonal, and quite involved for those near the diagonal.  But in any case, the simulations show that if there any difference, it must be miniscule, and not worth basing any strategy around.
Can anyone help my intuition by showing me where it's gone wrong?
The actual Feudality game is a bit more complicated.  A player's board is 7x7, and rerolls also happen if either die is an 8.  My friends and I also expected that cells near the right and bottom edges would be activated more often due to the rerolls, but this is also not the case, according to our simulations.  I can almost see how edge rolls might not be activated any more often than other cells since a die showing an 8 activates fewer cells than other rolls, which might compensate for the increased frequency.  But as in my simplified version of the game above, I can't understand why the diagonal cells are not activated more often than the others.
 A: This was really interesting.
For each cell I defined the terms:


*

*$NRNA$ (Non-Repeating-Non-Activator): a roll that doesn't activate the cell and doesn't trigger an additional roll, i.e. it's not a double.

*$NAR$ (Non-Activating-Repeater): a roll that doesn't activate the cell but does trigger another roll.

*$AR$ (Activating-Repeater): a roll that activates the cell and triggers another roll.

*$NRA$ (Non-Repeating-Activator): a roll that activates the cell but doesn't trigger another roll


For a particular cell, the possible ways it can be triggered once are:
$$NAR^n, NRA$$
(meaning some number $n$ of non-activating-repeaters followed by a non-repeating-activator) or
$$NAR^{n_1}, AR, NAR^{n_2}, NRNA$$
for some $n_1$, $n_2$.
To find $\Pr(1)$ for this cell we must sum over all possibilities:
\begin{align}\Pr(1) &= \sum_{n=0}^\infty \Pr(NAR)^n \Pr(NRA) + \sum_{n_1 = 0}^\infty\sum_{n_2 = 0}^\infty \Pr(NAR)^{n_1}\Pr(AR)\Pr(NAR)^{n_2}\Pr(NRNA) \\ &=\Pr(NRA)\frac{1}{1-\Pr(NAR)} + \Pr(AR)\Pr(NRNA)\frac{1}{(1-\Pr(NAR))^2}\end{align}
Similarly the ways it can be triggered twice are:
$$NAR^{n_1}, AR, NAR^{n_2}, NRA$$
$$NAR^{n_1}, AR, NAR^{n_2}, AR, NAR^{n_3}, NRNA$$
which gives $$\Pr(2)=\Pr(AR)\Pr(NRA)\frac{1}{(1-\Pr(NAR))^2} + \Pr(AR)^2\Pr(NRNA)\frac{1}{(1-\Pr(NAR))^3}$$
Proceeding in this way, the general $\Pr(n)$ is given by:
$$\Pr(n)=\frac{\Pr(AR)^{n-1}\Pr(NRA)}{(1-\Pr(NAR))^n} + \frac{\Pr(AR)^n\Pr(NRNA)}{(1-\Pr(NAR))^{n+1}}$$
This is all getting a bit cumbersome, so setting \begin{align}a &= \Pr(AR) \\ b &= \Pr(NRA) \\ c &= \frac{1}{1-\Pr(NAR)} \\ d &= \Pr(NRNA)\end{align}
the expected number of activations per round for a specific cell is $$\sum_{n=1}^\infty n\Pr(n)$$
which, after some manipulation, turns out to be $$\frac{(b+adc)c}{(1-ac)^2}$$
For the non-diagonal corner cells there are $52$-$NRNA$, $8$-$NAR$, $0$-$AR$, $4$-$NRA$ so you can work out $a = 0$, $b = \frac{1}{16}$, $c = \frac{8}{7}$, $d = \frac{52}{64}$ and get the expected number of activations as $\frac{1}{14} = 0.07142\ldots$ in line with your experimentation.
For a central diagonal cell there are $49$-$NRNA$, $5$-$NAR$, $3$-$AR$, $6$-$NRA$ which gives an expectation of $0.15975\ldots$
I'll leave the other cells up to you if you want to take it further, just plug in the values.
