King and Rook

Dimitri places a Black King and a White Rook on an empty chessboard. If the probability that Dimitri places the King and the Rook on the chessboard such that the King is NOT in check (that is, the Rook is not attacking the Black King), can be expressed as m n \frac{m}{n} , in which m m and n n are coprime positive integers, find n m n-m .

Details and Assumptions: e.g. If the Rook is on f 2 f2 , the f 1 f1 , f 3 f3 , f 4 f4 , f 5 f5 , f 6 f6 , f 7 f7 , f 8 f8 , g 2 g2 , h 2 h2 , e 2 e2 , d 2 d2 , c 2 c2 , b 2 b2 and a 2 a2 squares are under attack.

The King and the Rook cannot be placed in the same square.


This is the fourth problem of the set Look after the King!


The answer is 2.

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2 solutions

Firstly, note that Dimitri can place the Rook in anyone of the 64 64 squares. As the King and the Rook can't be placed in the same square, Dimitri has 63 63 possible squares to place the Black King. Therefore, there are a total of 64 63 64\cdot 63 possible positions.

Secondly, note that, wherever the Rook is placed, will be attacking its whole row and column. This means that the King can't be placed on the same column or row of the Rook. As each row and column is formed by 8 8 squares, and we are counting twice the square on which the Rook is, there are a total of 63 14 = 49 63-14=49 squares on which Dimitri can place the King. Therefore, in each of the 64 64 possible positions of the Rook, the King can be placed on anyone of the 49 49 remaining squares, so the total positions are 49 64 49\cdot 64 .

Thus, the probability is 49 64 63 64 = 7 9 \frac { 49\cdot 64 }{ 63\cdot 64 } =\frac { 7 }{ 9 } . Hence, m = 7 m=7 and n = 9 n=9 , so n m = 2 n-m=2 .

Or, (bit simpler):

Given that the King has been placed in any of the squares, there are only 14 places out of the 63 squares left (King is occupying one) that the Rook can be placed and not attack the King.

m ÷ n = 14 ÷ 63 = 7 ÷ 9

n - m = 9 - 7 = 2

Michael Moneke - 5 years, 1 month ago
K T
Nov 15, 2018

After placing the rook, the king can go to 63 other squares, of which 49 are 'safe', no matter where the rook is. He is in check if he is on the same row or file as the rook, which is in 14 out of 63 cases. Now 49÷63=7÷9, and 9-7=2.

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