JoeSapphire: 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.
ZFR: 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
Neat! I got it done in 24 rolls (meaning 48 dice were rolled but that's a technicality)
Looks like I beat maths.
JoeSapphire: While we wait to find out who the last sign-up is I have a problebility for you:
If I have 12 seats that need to be filled, and every time someone rocks up to sit in a seat I roll two dice to randomly determine a seat for that person. If someone is sat in a seat that is occupied, the occupator is displaced, and I roll again to determine them a new seat. What is the average number of times that I will roll a dice to seat twelve people?
Lifthrasil: Two. After rolling twice, you'll be fed up and just assign seats arbitrarily.
It looks like we have enough players. I haven't heard from MindCollapse but I'll try and put something together so that GameRager and dedoporno can both get involved and hopefully begin the game over the weekend.