Three different squares are chosen randomly on a chessboard.

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

(In reply to

Corrected solution by Penny)

Penny, you wrote:

*"So the true odds are:
*

(4/64)*(7/63)

+ (8/64)*(6/63)

+ (8/64)*(2/63)*(1/62) + (8/64)*(5/63)*(4/62)

+ (8/64)*(3/63)*(2/62) + (8/64)*(4/63)*(3/62)

+ (4/64)*(2/63)*(1/62) + (4/64)*(7/63)*(6/62)

+ (8/64)*(3/63)*(2/62) + (8/64)*(6/63)*(5/62)

+ (8/64)*(4/63)*(3/62) + (8/64)*(5/63)*(4/62)

+ (4/64)*(4/63)*(3/62) + (4/64)*(7/63)*(6/62)

+ (8/64)*(5/63)*(4/62) + (8/64)*(6/63)*(5/62)

+ (4/64)*(6/63)*(5/62) + (4/64)*(7/63)*(6/62)

= 0.0076804916"

I don't understand how you are coming up with these probabilities, but they appear incorrect.

______________________

BTW, this Al-Gore-ithm thing... did this occur before or after Al invented the internet?