>> 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'.
> This isn't right (I think). It can move there if e3 is empty and
> either d2 and d4 are empty or f2 and f4 are empty. So that's 0.7 * (1
> - (1 - 0.49) * (1 - 0.49) ), which works out to 0.51793, as compared
> to 0.4459. I think the generalized equation, where X is the (always
> even) number of squares moved, would be 0.7^(X/2 - 1) * (1 - (1 -
> 0.7^(X/2))^2)
(We're talking about the probability of the zFF being able to make a four
step
move, for example from e1 to e5.)
My verbal description is saying that the choice between the two paths
is made only once, and therefore the two-path probability correction should
be made only once in the calculation; this gives me a simpler formula
for doing the calc by hand. Upon review I am even more convinced that
this is correct, but in order to feel perfectly secure I must find
your error.
You are saying 'if e3 empty and ((d2 empty and d4 empty) or (f2 empty and
f4 empty))'. The verbal description is clearly correct, although it
makes things more complicated when you extend to 4 step and 6 step moves.
The probability that d2 empty and d4 empty is 0.49; the probability that
p or q is (1 - ((1 - p) * (1 - q))). Ouch, that's convincing.
Wouldn't another fair way of stating it be
'(d2 empty and e3 empty and d4 empty) or (f2 and e3 and f4)'?
But that gives me a completely different number, even higher.
Aha! '(d2 and e3 and d4) and (f2 and e3 and f4)' is incorrect because
in effect it applies the two-path correction to e3, but e3 non-empty
blocks both paths!
But then by the same token, your 'e3 and ((d2 and d4) or (f2 and f4))'
must apply the two-path correction twice!!
I'm right, you're wrong. Nyaah, nyaah! (If I were a licensed
mathematician
I would be able to say Q.E.D., but since I'm not I can only say nyaah
nyaah.)
That was difficult. My head hurts.