unless one of the die is seven-sided, with one face = 0?

I sorry for this, friends

We forgive you.

Does your programming include the game of forum mafia?

What is the conflict?

EDIT: Oh... Gotcha. You had me fooled.

They're a right bastard.

I'm sure he'd say the same about you.

Not at all!

Hey, just because he touches your weiner on occasion, that's no reason to talk bad about him!

HA
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

Two. After rolling twice, you'll be fed up and just assign seats arbitrarily.

I see a new (to me at least) player has expressed an interest in what seems to be the last free spot but in case they decide otherwise or you guys end up needing one last push do let me know.

You don't actually have to do that. You can easily simulate a uniform distribution (1-12) with two normal d6 dice.

e.g. multiply the result of first dice by 2. Then use the second dice to either subtract 1 from the result or leave it as it is (if the second dice rolls 1-3 you subtract 1 from the first dice's result multiplied by 2). So:

First dice determines one of those groups:

(1,2)
(3,4)
(5,6)
(7,8)
(9,10)
(11,12)

Second dice determines if it's the first or second number from each group that's chosen.

With two d6 dice rolls, you can simulate a random uniform distribution in any range up to 36.

Joe said we could boot this game to 15 players if there's enough interest, and I remember GameRager saying he could join if the signup phase lasted long enough.