Rooks on a 8 by 8 checker board. Probability problem

Question

Two rooks are placed on an 8 by 8 chessboard such that they're in different squares. What is the probability they can't attack each other? Rooks can attack each other if and only if they're in the same row or column. I know that there are questions like this but they give the general numbers not the probability I am not sure how to start the questions and I have looked at this questions for hours. Any help would be welcome.

Answer 1

Place one rook down. There will always be $14$ remaining slots that another rook can attack from, giving a probability of...

Answer 2

Place the first rook. No matter where it is placed, it attacks $2\cdot 7 = 14$ squares of the chessboard. Because both pieces are rooks, if one attacks the other, they both attack each other.

This means that $\dfrac{63-14}{63} = \dfrac{7}{9}$ places on the chessboard are safe for the other rook such that they won't attack eachother.

Answer 3

Making this very explicit (almost pedantic) as I know that introductory probability is often challenging for students. knowing when to condition, how to count, when double counting is occurring, etc...

Call the rooks, $R1$ and $R2$

Fix the position of $R1$ on the chessboard.

Place $R2$ in a random open spot. $R1$ can only attack $R2$ if they are in the same row and column and vice versa. Because $R1$ takes up a space in the $8 + 7 = 15$ (to avoid double counting) possible unsafe spots for $R2$, there are $14$ unsafe spots out of $63$ spots (because $R1$ is already taking up a spot) for $R2$ to be placed. Therefore, given $R1s$ position,

$P(\text{$R2$ $safe$} | R1) = \frac{63 - 14}{63} = \frac{49}{63} = \frac{7}{9}$

Next,

$P(\text{$R2$ $safe$}) = \sum_{\text{all positions for $R1$}} P(\text{$R2$ $safe$}, $R1$) = \sum_{\text{all positions for $R1$}} P(\text{$R2$ $safe$} | R1)P(R1) = 64\cdot\frac{7}{9} \cdot \frac{1}{64}$

Therefore: $P(\text{$R2$ $safe$}) = \frac{7}{9}$

$R1$ is safe exactly when $R2$ is safe, so by symmetry, the probability the two rooks can't attack each other is $\frac{7}{9}$

Note: if you want to be totally sure the formula works regardless of order, you can assume order is random and then condition on which rook is placed down first. $2 \cdot \frac{1}{2} = 1$ — again, no change.