A queen on an English chessboard is able to attack in the same row, column and diagonal. The probability that 2 randomly placed queens on a 8 by 8 chessboard will be able to attack each other can be expressed as ab, where a and b are co-prime integers. What is a+b? Clarification: The 2 queens will not be placed on the same square.
mrcausality
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You must mean "a/b", otherwise we aren't talking about probabilities. If we call P(n) the probability on an nxn board, I get the general formula for even n:
P(2n) = ∑ {k=0 ... n-1} [3(2n-1) + 2k] / (4n^2-1) *4[2(n-k) - 1]/(4n^2)
This was obtained by decomposing the probability into a sum over conditional probabilities where the 1st queen is placed on the outer edge square (i.e. anywhere on the perimeter), the first inner edge square, and so on.
As an example, for P(4):
P(4) = 9/15*12/16 + 11/15*4/16 = 19/30
Here (a,b) = (19,30) and a+b=49.
For P(8):
P(8) = 21/63*28/64 + 23/63*20/64 + 25/63*12/64 + 27/63*4/64
= 1456/4032 = 13/36
Thus (a,b) = (13,36) and a+b=49---identical to P(4).
By the way, the summand can be summed to give the following:
P(2n) = (10n-1)/((3n)*(2n+1))
which can be verified to give the same result.