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Revisiting the Crooked Bishop. Revisiting the Crooked Bishop.[All Comments] [Add Comment or Rating]
Peter Hatch wrote on Fri, Apr 12, 2002 03:18 AM UTC:Excellent ★★★★★I think the way to find the on-board probability is to divide it into two
parts. The on-board probablity for having two paths on a (X,0) (where X
is
any even number) move would be (X,2). The probability for having just
one
path on a (X,0) move would be (X,6) (on a 8x8 board, generally (X,board
size - 2)). I think this works - moving two squares up the board can be
done on all but the last two rows, and has two paths on all but the outer
two columns.