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?
re: Corrected solution
