Single Comment

Revisiting the Crooked Bishop. Revisiting the Crooked Bishop.[All Comments] [Add Comment or Rating]

Peter Hatch wrote on Sun, Apr 14, 2002 06:15 AM UTC:

I still think I'm right, but I'm not as sure anymore...

I think I'm not fully applying the two-path for the f4/d4 - getting the
full two path bonus twice would be 0.91 * 0.7 * 0.91 = 0.57967, just
getting it once is 0.91 * 0.7 * 0.7 = 0.4459, and my method is in my first
comment as 0.51793.  So it is between the values of one and two two-path
bonuses.

Mostly I think it is right because the verbal description seems to be
right, and you agreed with my math on turning that description into a
formula.  I think that formula is properly figuring out how much of a
two-path bonus to give.

Does this seem reasonable to you?