Probability of a random chair being empty and having a seat to the right which is not empty is $\frac{m(n-m)}{n(n-1)}$ Let there be a round table with $n$ chairs. $m$ people choose their chair (max 1 person per chair) and let $m < n$.
I am pretty sure that the probability of a random chair being empty and having a seat to the right which is not empty is $\frac{m(n-m)}{n(n-1)}$
How can I show that this is indeed the right formula for calculating the probability of the described event?
I have tried for several hours but I can't  figure it out, maybe  someone of you finds an easy solution?

Your expression is correct. The probability that the chair is empty is $\frac{n-m}{n}$.  Given that it is empty, the probability the chair to the right of it  is occupied is $\frac{m}{n-1}$. Multiply.
