Probability involving chess board if 2 cells are chosen at random on a chess board what is the probability that they will have a common side i tried solving the question by considering different cases for the cells on: 1\. corner 2\. edge other than corner 3\. cell in middle but i guess the cases might be repeating so please help

We have $64$ cells:

* $36$ cells have $4$ neighbors each
* $24$ cells have $3$ neighbors each
* $ 4$ cells have $2$ neighbors each



The total number of ways to choose $2$ out of $64$ cells is $\binom{64}{2}=2016$.

The number of ways to choose $2$ adjacent cells is $\frac{36\cdot4+24\cdot3+4\cdot2}{2!}=112$.

So the probability of choosing $2$ adjacent cells is $\frac{112}{2016}=\frac{1}{18}$.