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 randomit 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 A1H8 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 A1H8. 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 samecolor squares individually, and got:
A1: 42/3906 (A1H8 diagonal)
H8: 42/3906 (A1H8 diagonal)
B2: 42/3906 (A1H8 diagonal); and 2/3906 (A3C1 diagonal)
G7: 42/3906 (A1H8 diagonal); and 2/3906 (F8H6 diagonal)
C3: 42/3906 (A1H8 diagonal); and 12/3906 (A5E1 diagonal)
F6: 42/3906 (A1H8 diagonal); and 12/3906 (D8H4 diagonal)
D4: 42/3906 (A1H8 diagonal); and 30/3906 (A7G1 diagonal)
E5: 42/3906 (A1H8 diagonal); and 30/3906 (B8H2 diagonal)
A3: 20/3906 (A3F8 diagonal); and 2/3906 (A3C1 diagonal)
C1: 20/3906 (C1H6 diagonal); and 2/3906 (A3C1 diagonal)
F8: 20/3906 (A3F8 diagonal); and 2/3906 (F8H6 diagonal)
H6: 20/3906 (C1H6 diagonal); and 2/3906 (F8H6 diagonal)
B4: 20/3906 (A3F8 diagonal); and 12/3906 (A5E1 diagonal)
D2: 20/3906 (C1H6 diagonal); and 12/3906 (A5E1 diagonal)
E7: 20/3906 (A3F8 diagonal); and 12/3906 (D8H4 diagonal)
G5: 20/3906 (C1H6 diagonal); and 12/3906 (D8H4 diagonal)
C5: 20/3906 (A3F8 diagonal); and 30/3906 (A7G1 diagonal)
D6: 20/3906 (A3F8 diagonal); and 30/3906 (B8H2 diagonal)
E3: 20/3906 (C1H6 diagonal); and 30/3906 (A7G1 diagonal)
F4: 20/3906 (C1H6 diagonal); and 30/3906 (B8H2 diagonal)
A5: 12/3906 (A5E1 diagonal); and 6/3906 (A5D8 diagonal)
D8: 12/3906 (D8H4 diagonal); and 6/3906 (A5D8 diagonal)
E1: 12/3906 (A5E1 diagonal); and 6/3906 (E1H4 diagonal)
H4: 12/3906 (D8H4 diagonal); and 6/3906 (E1H4 diagonal)
B6: 30/3906 (A7G1 diagonal); and 6/3906 (A5D8 diagonal)
C7: 30/3906 (B8H2 diagonal); and 6/3906 (A5D8 diagonal)
F2: 30/3906 (A7G1 diagonal); and 6/3906 (E1H4 diagonal
G3: 30/3906 (B8H2 diagonal); and 6/3906 (E1H4 diagonal)
A7: 30/3906 (A7G1 diagonal); and 6/3906
B8: 30/3906 (B8H2 diagonal)
G1: 30/3906 (A7G1 diagonal)
H2: 30/3906 (B8H2 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 20040201 02:41:36 