226 Mr. H. M. Taylor on the Relative Values 



Safe check with one bishop. 



This is best found by finding the number of unsafe checks. 

 It will be seen that if the bishop be on a corner square, there 

 is one unsafe check ; for any other outside square there are 

 two unsafe checks ; and for any other square there are four. 

 The whole number therefore of unsafe checks 



= 4 + 40-2). 2 + (n-2) 2 .4 = 4(n 2 -4n + 4 + 2n-4 + l) 

 = 4(n-l) 2 . 

 Therefore the numerator of the chance-fraction 



= ^(7i-l)(2/i-l)-4(n-l) 2 = §(n-l)(n-2)(2n-3), 



and the chance of safe check 



- f (n-l)(n-2 )( 2n-3) _ 2 (n-2)(2n-3) 

 n 2 (n 2 -l) 3*" n 2 (n + l) 



If n=8, 



the chance = -J^. 



These results will be the same for a bishop restricted to squares 

 of the same colour on a board of an even number of squares. 



Simple check with one queen. 



The queen on any square of a chessboard checks all the 

 squares that a bishop on the same square would, as well as all 

 the squares that a rook on the same square would. The nume- 

 rator of the chance-fraction, therefore, will be the sum of the 

 numerators for the cases of the rook and the unrestricted bishop. 

 The chance, therefore, for the queen is the sum of the chances 

 for the bishop and rook 



= 2 2n-l _J_ = _J_ Qi-l) + 3n _2 5n-l 

 3*n(n-f-l) n + 1 n + 1' Sn 3*w(n + l) 



Ifn=8, 



the chance = -£§. 



Safe check with one queen. 

 The number of unsafe checks with one queen is seen thus : 

 if the queen be on a corner square, there are 3 unsafe checks, 

 if on any other outside square there are 5, and if on any other 

 square there are 8 unsafe checks. The total number of unsafe 

 checks, therefore, is 



4.3 + 4(n-2).5 + (n-2) 2 . 8 = 4(n-l)(2n-l). 

 The numerator of the chance-fraction, therefore, is 



§n(n-l)(2n-l) + n 2 .2(n-l)-4(27i-l)(n-l) 

 _ 20i-l)(5n-3)(n-2) 

 3 



