Round Table Adjacent Seats Problem If there are 10 seats in a round table and 7 people are already sitting in some random pattern(edit: uniform distribution, thank you for the correction) what is the probability that the next two people entering the room will get to sit in adjacent seats?

I am very challenged in trying to start to think about the problem. I would sincerely appreciate any thoughts and I will definitely indicate my progress thereafter.
We assign seats at random, by putting place mats labelled $1,2,3,\dots, A, B, C$ at random in front of the $10$ chairs. The place mats labelled $A,B,C$ indicate where any person apart from $1$ to $7$ may sit.  A pair of lovers now enters the room, and they wish to sit together. We find the probability that they can do so without asking anybody to move. 

The only thing that matters is the placement of $A, B, C$. And since the table is round, all that matters is the placement of $B$ and $C$ relative to $A$. So we have $9$ empty spaces, and have to choose $2$ of them. There are $\binom{9}{2}$ choices, all equally likely. 

We count the bad choices. For a choice to be bad, it must involve choosing $2$ non-adjacent seats from the $7$ not next to $A$. There are $\binom{7}{2}$ ways to choose $2$ seats, of which $6$ give an adjacent pair. So there are $15$ bad choices, out of the $\binom{9}{2}$ choices. Thus the probability there will be an adjacent pair is $\frac{21}{36}$.
Consider 10 chairs arranged in a circle. 
The number of ways of leaving three seats empty is C(10,7) = 240.

Of these, the number of ways in which atleast 2 of the empty seats are adjacent is 10.8 = 80.
(Think of this situation as having 10 fixed spaces. There are 10 ways to choose a pair of adjacent spaces, considering the first and last also as adjacent-one for each space being the left space of the chosen pair. There are 8 ways to choose one of the remaining 8 spaces)

Hence, the required probability is 80/240 = 1/3.