A rook and a bishop of a standard chess set and having different colors are randomly placed on a standard chessboard.
Determine the probability that one is attacking the other.
The bishop's position can be considered, in terms of the number of squares under attack from it, as in one of ten positions:
1 2 3 4 * * * *
* 5 6 7 * * * *
* * 8 9 * * * *
* * *10 * * * *
* * * * * * * *
* * * * * * * *
* * * * * * * *
* * * * * * * *
Position 1 can attack any of 7 positions.
Position 5 can attack any of 9 positions.
Position 8 can attack any of 11 positions.
Position 10 can attack any of 13 positions.
On the full board there are four of each of the above types of bishop positions.
Position 2 can attack any of 7 positions
Position 6 can attack any of 9 positions
Position 9 can attack any of 11 positions
Position 3 can attack any of 7 positions
Position 7 can attack any of 9 positions
Position 4 can attack any of 7 positions
There are eight of each of these types.
Taking a weighted average:
number weight value
7 28 196
9 20 180
11 12 132
13 4 52
 
64 560
So the average square the bishop occupies attacks 560/64 = 8.75 = 35/4 squares out of the 63 other squares on the board making the probability that the rook will be attacked by the bishop 35/252 = 5/36.
The rook attacks the bishop if they are on the same rank or file. They can't be both, as we're assuming they were chosen not to be coincident on the board. The rook always attacks 14 squares out of the 63 others, so the probability that the rook attacks the bishop is 2/9.
They can't be mutually attacking, so the mutually exclusive probabilities are added: 5/36 + 2/9 = 13/36 is the probability that one is attacking the other.

Posted by Charlie
on 20140514 12:56:47 