If two queens are randomly placed on distinct squares of an ordinary chessboard, what is the probability that they attack each other?
The first queen can be placed anywhere on the board with equal likelihood. The placement of this first queen then determines the number of places where the second queen's placement would result in a mutual attack and therefore the probability of such attack.
Consider a chessboard labeled thus:
4 8 8 8 . . . .
. 4 8 8 . . . .
. . 4 8 . . . .
. . . 4 . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
The probabilities need only be considered for the placement of the initial queen in one of the ten numbered positions. The numbers refer to how many total positions on the chessboard that that one position represents in terms of attacked squares.
The positions on the first row attack 21 squares each.
The numbered positions on the second row attack 23 squares each.
The numbered positions on the third row attack 25 squares each.
Each of the central four positions attacks 27 squares.
For simplicity, the weights of the probabilities need only be proportional:
total conditional weighted
(weight) probability probability
1 2 2 2 . . . . 7 21/63 147/63
. 1 2 2 . . . . 5 23/63 115/63
. . 1 2 . . . . 3 25/63 75/63
. . . 1 . . . . 1 27/63 27/63
. . . . . . . .
. . . . . . . .
. . . . . . . .
. . . . . . . .
----- ---------
16 (364/63)/16 = 13/36
The final answer then is 13/36.
Edited on August 16, 2013, 1:48 pm
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Posted by Charlie
on 2013-08-16 13:47:12 |
solution