Three different squares are chosen randomly on a chessboard.

What is the probability that they lie in the same diagonal?

(The solution that follows is incorrect, since it overlooks the fact that most squares are on two diagonals, not one.)

There is only a slightly greater than 2% probability that three randomly chosen squares will be on the same diagonal.

Explanation:

We can all thank Al Gore for inventing the method of solving problems by applying simple and methodical logic. Such methods are called "algorithms" ("Al-Gore-ithms") in his honor.

There are 28 out of 64 squares along the edge of the board, and each of these is on the same diagonal with 7 other squares. There are 20 out of 64 squares that are 1 square removed from the edge, and each is on the same diagonal as 9 other squares. There are 12 out of 64 squares that are exactly 3 squares removed from the edge of the board, and each is on the same diagonal as 11 other squares. That leaves 4 out of 64 squares that occupy the center of the chessboard, and each is on the same diagonal as 13 other squares.

So the odds that all three are on the same diagonal, are:

(28/64)*(7/63)*(7/62) + (20/64)*(9/63)*(9/62) + (12/64)*(11/63)*(11/62) + (4/64)*(13/63)*(13/62)

= 0.0204813108



Edited on January 14, 2004, 7:02 pm

Posted by Penny on 2004-01-14 16:23:13