Chess Commons

The total number of ways to choose two distinct unit squares is $\dbinom{64}{2} = 2016$

Notice that in any 2 x 2 square, there are two distinct pairs that satisfy the parameters of the question. There are $7 \times 7 = 49$ 2 x 2 squares, so there are $49 \times 2 = 98$ distinct pairs satisfying the parameters of the problem. $\frac{98}{2016} = \boxed{\frac {7}{144} }$

There are $64$ unit squares in each chessboard.

I will separate these $64$ unit squares into $3$ groups:

Group $A$ : Unit squares which have $2$ common sides with the sides of the board, there are $4$ of them (in the corner of the board).

Group $B$ : Unit squares which have $1$ common side with the sides of the board, there are $24$ of them (they are on the $4$ sides of the board, but $EXCEPT$ from the squares of group $A$ .

Group $C$ : Unit squares which have no common sides with the sides of the board, there are $36$ of them, and they make a $6\times 6$ square in the center of the board.

Now we can easily answer this question:

With the unit squares of group $A$ , there are $3$ squares which share the common sides of each square in this group, but there are only $1$ square of these $3$ squares which satisfies the assumption of this problem, so we will have $4$ combinations of the $2$ unit squares which satisfy our assumption.

With the unit squares of group $B$ , there are $5$ squares which share the common sides of each square in this group, but there are only $2$ squares of these $5$ squares which satisfies the assumption of this problem, so we will have $6\times 2\times 4=48$ combinations of the $2$ unit squares which satisfy our assumption.

With the unit squares of group $C$ , there are $8$ squares which share the common sides of each square in this group, but there are only $4$ squares of these $8$ squares which satisfies the assumption of this problem, so we will have $6\times 6\times 4=144$ combinations of the $2$ unit squares which satisfy our assumption.

Calculate all combinations we have found, we will have $4+48+144=196$ combinations of the $2$ unit squares which satisfy our assumption.

But take a look back, all the combinations have been counted twice, so actually we have $\frac { 196 }{ 2 } =98$ combinations in total.

Finally, the probability that two distinct unit squares selected randomly from a chessboard have exactly one corner in common is:

$\frac { 98 }{ { C }_{ 2 }^{ 64 } } =\frac { 7 }{ 144 }$

Note: You can use the chessboard to prove my solution

Although my solution is quite long, but I want to show the simplest way to solve this problem, thank you!

Take one of the diagonals, it has 8 squares embedded on it, and 7 (8-1) possible pairs of squares having a common corner, now consider the parallel adjacent line of squares, which has 7 squares and 6 (7-1) possible pairs of squares having a common corner, keep going on till you reach the line with 2 squares. The total will be 7+6+5+4+3+2+1. Now the same will be true for the opposite side of the diagonal and counting that the total will be 7+2(6+5+4+3+2+1). Again, the whole process will be the same for the other other diagonal as well. Thus the net no. of possible pairs of squares having a common corner is 2(7+2(6+5+4+3+2+1) ) =98. Now the total no. of possible ways of choosing two squares is 2016. So the probability is 98/2016= 7/144.