1. ## Chessboard

I have a question. A castle and bishop are placed on different squares of a chessboard. What is the probability that one piece threatens the other? If I remember correctly one of them moves diagonally and the other one moves only horizontally and vertically and I think that there are 64 squares on a chessboard. I don't have the slightest clue how to go about solving the problem. Could anyone help please?

2. Originally Posted by sweeetcaroline
I have a question. A castle and bishop are placed on different squares of a chessboard. What is the probability that one piece threatens the other? If I remember correctly one of them moves diagonally and the other one moves only horizontally and vertically and I think that there are 64 squares on a chessboard. I don't have the slightest clue how to go about solving the problem. Could anyone help please?
A castle (usually known as a rook) moves laterally and a bishop moves diagonally. That means that they can never both threaten each other. So the probability that one piece threatens the other is equal to the probability that the rook threatens the bishop, plus the probability that the bishop threatens the rook.

Wherever the rook is placed on the board, it threatens 14 squares (7 on its rank and 7 on its file). So if the rook and the bishop are placed on different squares, the probability that the rook threatens the bishop is 14/63, or 2/9.

For the bishop, the situation is more complicated. If it is placed on one of the 28 squares on the edge of the board, it threatens 7 squares. If it is on one of the 20 squares one rank or file away from the edge, it threatens 9 squares. On one of the 12 squares that are two ranks or files from the edge, it threatens 11 squares. Finally, if it is on one of the 4 central squares, it threatens 13 squares. So the average number of squares threatened by a bishop is $\displaystyle \frac{7\times28 + 9\times20 + 11\times12 + 13\times4}{64} = \frac{35}4$. Therefore the probability that the bishop threatens the rook is $\displaystyle \frac{35}{4\times63} = \frac5{36}$.

Adding those results together, you see that the probability that one piece threatens the other is $\displaystyle \frac{13}{36}$.