Monte carlo methods vs Markov chains
blog.wolfram.com
blog.wolfram.com
For example, take a four-column, one-row version, with probability-of-falling like so:
* 0 0 0 0 *
1 x y y x 1
We have:x = 0 / 4 + 1 / 4 + y / 4 + x / 4
y = 0 / 4 + x / 4 + y / 4 + y / 4
giving the solutions x = 2/5 and y = 1/5. It is easier, of course, if you exploit the symmetry of the situation, as here (by writing x y y x instead of x y z w).
Of course, in a big random system there are some states that simply won't be reached by any random number generator with a finite period. Once you start looking at combinations and permutations, you can a get staggeringly large number of states. But in practice, this shouldn't matter for any problem where Monte Carlo methods make sense - if your answer is very sensitive to whether or not you have sampled a state that only crops up one in a squillion times, you shouldn't use Monte Carlo.