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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 02:47 AM UTC:Excellent ★★★★★

<blockquote>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'.</blockquote> 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)