If two bishops are placed randomly on a chessboard, what is the probability that they will be able to attack each other? Also, we exclude the option that two bishops can inhabit the same square on the chessboard. 
I've done this problem before with two rooks and it was fine, but I'm unsure how to sort it out with bishops. I wrote a short program to compute the probability by simulation and I got that the  bishops will attack each other just less than 14% of the time. Is this correct? I'd like to understand this problem mathematically - could anyone please help?
 A: I am assuming the bishops are indistinguishable, then the total number of ways you can put them is $\binom{8\times 8}{2},$ now attacking each other means that they are in the same diagonal, from $2$ to $7$ enumerate diagonals from any corner, then the total ways you can do this is $$2\left (2\sum _{n=2}^7\binom{n}{2}+\binom{8}{2}\right ),$$ where we multiply $2$ because there are $2$ possible orientations.The probability, which is the quotient, is $0.13889.$
A: This is a pedestrian solution, the one of a chess player. (No graph, no higher combinatorics.)
The squares of the chess board are in bijection with the set $J^2$, where $J=\{1,2,3,4,5,6,7,8\}$. (We use this convention, instead of the usual one in chess, using letters for the first identifier, then number for the second one.)
There is a total of $64\cdot 63$ possibilities to place two bishops on the board (where the first one has a clown hat to make it different while counting).
Let us count now the favorable cases. We place the first bishop on the board on the field $(j,k)$. We assume first $1\le j\le k\le 4$, using the symmetry of the board.


*

*If $j$ is $1$, than we have $7$ possibilities for the second bishop.

*If $j$ is $2$, than we have $9$ possibilities for the second bishop.

*If $j$ is $3$, than we have $11$ possibilities for the second bishop.

*If $j$ is $4$, than we have $13$ possibilities for the second bishop.


Let us count 


*

*the cases with $\min(j,k) = 1$, these are the boundary cases, $8^2-6^2$. We eliminate them from the board. A $6\times 6$ square remains. 

*the cases with $\min(j,k) = 2$, these are the cases at the boundary of the $6\times 6$ square, we count $6^2-4^2$ cases.

*the cases with $\min(j,k) = 3$, there are $4^2-2^2$ cases.

*the cases with $\min(j,k) = 4$, we count the final $2^2-0^2$ cases.


This gives a total of 
$$
7(8^2-6^2) +
9(6^2-4^2) +
11(4^2-2^2) +
13(2^2-0^2) 
=
7\cdot 8^2+ 2\cdot 6^2+ 2\cdot 4^2+2\cdot 2^2
=
560
$$
good cases. The wanted probability is thus:
$$
\frac{560}{64\cdot 63}
=
\frac {5}{36}
=
0.1388888\dots\ .
$$
A: This is basically the same answer dan_fulea gave, with a picture. On each square of the chessboard, write down the number of squares that a bishop on that square would attack.

Note that there are $4$ $13$’s, $12$ $11$’s, $20$ $9$’s, and $28$ $7$’s. A randomly-placed bishop then has a $4\over64$ probability of attacking $13\over63$ of all other locations, a $12\over64$ probability of attacking $11\over63$ of the other locations, a $20\over64$ probability of attacking $9\over63$ of the other locations, and a $28\over64$ probability of attacking $7\over63$ of the remaining locations.
Thus the probability of two randomly-placed bishops attacking each other is $${4\over64}\cdot{13\over63} + {12\over64}\cdot{11\over63} + {20\over64}\cdot{9\over63} + {28\over64}\cdot{7\over63}= {560\over64\cdot63}={5\over36}=0.13\bar8.$$
