In this scenario I don't have a twelve sided dice handy, so I'm using two dice to randomly determine a seat.
In this scenario I would take offence if somebody thought I would confuse a random selection method from 12 places with a skewed selection method from 11 places, so please factor that into your calculations.
Assuming that regardless of the number of dice you use, you get a number in range [1,12] with a uniform probability, then:
In essence, you need to roll until you hit an empty seat (the fact that the person replaces someone in a seat that's already taken doesn't matter).
The average (mean, not median) number of times, it takes to roll an empty seat is the inverse of probabililty (proof below for those interested). So when there are k empty seats, it takes an average of 12/k times to roll for one.
Therefore the answer is:
Sum(n: 12 to 1)(12/n) = 12/12 + 12/11 + 12/10 + ... + 12/2 + 12/1 ≈ 35.74
Now for the proof. If you google (or visit your library) you can find quite a few of them. Here is my favourite that uses recursion.
There is an event with a probability p. We need to find on average how many times we need to roll to get p.
Assume that the average is x.
Now, there is a probability p of getting it on first try. And a probability of not getting it on first try is (1-p).
If we don't get it on the first try, then we used up one try but, since x is the answer, then on average it takes x further tries to get it
Taking that into account, we got a simple equation to solve.
x = (p * 1) + ((1-p) * (1+x))
Which we can solve to get
x = 1/p