Three different squares are chosen randomly on a chessboard.

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

I wonder why my last attempt at posting didn't work...
Anyway, thank goodness I saved my solution before posting! :D


I'm not sure if I got this right, but here's what I did:

The first square gets chosen at random--it doesn't matter whether it's black or white.
The probability for the next two squares being chosen on the same diagonal depends on the first square chosen. I separated the two colors and only dealt with one color, because the results for each color would be the same, and the problem doesn’t require that we specify a color.

So, if the first square was A1, the probability for the second square to be on the A1-H8 diagonal is 7/63, and for the third square, 6/62. Multiply the two, and you get 42/3906.

The same goes for all the other squares on the A1-H8. However, some squares are part of only one diagonal (A1, H8, A7, B8, etc.), while others are part of two diagonals. Therefore, I dealt with each of 32 same-color squares individually, and got:

A1: 42/3906 (A1-H8 diagonal)
H8: 42/3906 (A1-H8 diagonal)
B2: 42/3906 (A1-H8 diagonal); and 2/3906 (A3-C1 diagonal)
G7: 42/3906 (A1-H8 diagonal); and 2/3906 (F8-H6 diagonal)
C3: 42/3906 (A1-H8 diagonal); and 12/3906 (A5-E1 diagonal)
F6: 42/3906 (A1-H8 diagonal); and 12/3906 (D8-H4 diagonal)
D4: 42/3906 (A1-H8 diagonal); and 30/3906 (A7-G1 diagonal)
E5: 42/3906 (A1-H8 diagonal); and 30/3906 (B8-H2 diagonal)
A3: 20/3906 (A3-F8 diagonal); and 2/3906 (A3-C1 diagonal)
C1: 20/3906 (C1-H6 diagonal); and 2/3906 (A3-C1 diagonal)
F8: 20/3906 (A3-F8 diagonal); and 2/3906 (F8-H6 diagonal)
H6: 20/3906 (C1-H6 diagonal); and 2/3906 (F8-H6 diagonal)
B4: 20/3906 (A3-F8 diagonal); and 12/3906 (A5-E1 diagonal)
D2: 20/3906 (C1-H6 diagonal); and 12/3906 (A5-E1 diagonal)
E7: 20/3906 (A3-F8 diagonal); and 12/3906 (D8-H4 diagonal)
G5: 20/3906 (C1-H6 diagonal); and 12/3906 (D8-H4 diagonal)
C5: 20/3906 (A3-F8 diagonal); and 30/3906 (A7-G1 diagonal)
D6: 20/3906 (A3-F8 diagonal); and 30/3906 (B8-H2 diagonal)
E3: 20/3906 (C1-H6 diagonal); and 30/3906 (A7-G1 diagonal)
F4: 20/3906 (C1-H6 diagonal); and 30/3906 (B8-H2 diagonal)
A5: 12/3906 (A5-E1 diagonal); and 6/3906 (A5-D8 diagonal)
D8: 12/3906 (D8-H4 diagonal); and 6/3906 (A5-D8 diagonal)
E1: 12/3906 (A5-E1 diagonal); and 6/3906 (E1-H4 diagonal)
H4: 12/3906 (D8-H4 diagonal); and 6/3906 (E1-H4 diagonal)
B6: 30/3906 (A7-G1 diagonal); and 6/3906 (A5-D8 diagonal)
C7: 30/3906 (B8-H2 diagonal); and 6/3906 (A5-D8 diagonal)
F2: 30/3906 (A7-G1 diagonal); and 6/3906 (E1-H4 diagonal
G3: 30/3906 (B8-H2 diagonal); and 6/3906 (E1-H4 diagonal)
A7: 30/3906 (A7-G1 diagonal); and 6/3906
B8: 30/3906 (B8-H2 diagonal)
G1: 30/3906 (A7-G1 diagonal)
H2: 30/3906 (B8-H2 diagonal)


This makes a total of 58 values. To get an average, I added up all the numerators, and they gave me a total of 1176.
1176 is not divisible by 58, so instead I multiplied the denominator by 58: 3906 * 58 = 226548.

The resulting fraction is: 1176/226548 or:
14/2697

Posted by Eliza on 2004-02-01 02:41:36