Single Comment

I'm going to weigh in on Peter's side.  The probability of getting to (0,4)
should be 0.51793.

There are two ways to get there.  One requires (1,1) and (1,3) open.  The
other requires (-1,1) and (-1,3) open.  Both require (0,2) open.  

The probability of (1,1) and (1,3) open is 0.49.  (Same for the other
route.)  So the probability of this part of each route being closed is
0.51.  So the probability of both being closed is 0.51^2 = .2601, and the
probability of at least one of them being open is 1-.2601 = .7399.

Of course, in either case (0,2) must be open, so the final probability of
getting to (0,4) is .51793.

This generalizes to the probability of getting to (0,2n) being 
(1-(1-p^n)^2)*p^(n-1), where p is the magic number.  This is equivalent to
Peter's original formula, although a more compact writing of the formula
is (2-p^n)*p^(2n-1) or 2*p^(2n-1) - p^(3n-1).

If you need help convincing yourself your original statement

> Would 0.91 times 0.7 times 0.7 be correct? Yes, this is the answer
> to 'it can move there if either d2 or f2 is empty AND e3 is empty AND
> the corresponding square (d4 if d2, or f4 if f2) is empty'. 

is wrong, try thinking about it this way: suppose both d2 and f2 are
empty.  Then the crooked bishop may pass through either d4 or f4 on its
way to e5.  And the probability of being blocked on BOTH squares is 0.09,
not 0.3.  So the difference between your calculation and the answer Peter
and I get lies entirely in the cases where d2 and f2 are both empty.  In
these cases, it doesn't matter which square (d2 or f2) it passed through
to get to e3, so no binding decision about which path was taken has been
made yet.  In fact, no such decision has to be made until the first d-file
or f-file blockade is encountered.  Your method of calculation forces the
decision to be made at the very first step, even when both paths are
open.

Make any sense?