Two squares are chosen at random from a chess board . What is the probability that the two squares chosen have exactly one corner in common ? 
According to my reasoning , if we consider one of the longest diagonals consisting of $8$ squares , we get $7$ such pairs that have one corner in common. If we then move back towards the next diagonal consisting of $7$ squares we get $6$ such pairs. Then continuing in this manner ,we come to the smallest diagonal consisting of two squares in which one pair is formed. This gives us
$7+6+5+4+3+2+1= 28$ such pairs for one half of the chess board . We can continue in a similar manner for the other half and we will get the same answer. Hence the total number of such squares is $(2)(28)=56$ squares.
According to me the probability should be $\dfrac{56}{\binom{64}{2}}= 0.02$ but the actual answer is $0.04$. Can you please help me to find my mistake ?
Thank you !
Best Answer
First, look at the diagonals going from top-left to bottom-right. There is one big diagonal which is $8$ squares long, giving us $7$ pairs. Then, there are $2$ diagonals which are $7$ squares long on either side of the big diagonal, each giving us $6$ pairs. On the other sides of those diagonals, there are $2$ diagonals which are $6$ squares long, each giving us $5$ pairs. This pattern continues until we get $2$ diagonals which are $2$ squares long, each giving us $1$ pair. Thus, the number of pairs for all of the top-left to bottom-right diagonals is:
$$7+2*6+2*5+...+2*1=49$$
Now, look at the diagonals going from top-right to bottom-left. The analysis is exactly the same, except the diagonals are just going in a different direction. Thus, these diagonals give us another $49$ pairs.
Therefore, the total number of pairs is $98$ and the answer is:
$$\frac{98}{\binom{64}{2}}=\frac{7}{144}$$