hohum wrote:
So your notion is the town was lucky enough to hit on TWO scum, both fake claiming PR to avoid lynch? Do you have any idea what the odds of that are?
It's also possible that just one of them is a power role.
As for the odds...
I think we can safely assume that there are either 3 or 4 scum and at least 2 PRs. (Has anyone seen a 12 person normal that didn't fit those criteria?) I'll also assume we don't have a retarded vanilla who false claimed.
That leaves a range of 0-7 vanilla townies. Now, assuming random selection of two targets, what are the odds that neither is a vanilla townie?
Let 2nv = 2 non-vanilla and t be the number of townies
P(t=x|2nv) = P(t=x)*P(2nv|t=x)/P(2nv)
P(2nv|t=x) is easy to solve for. It's simply (12-x/12)(11-x/11).
[mrow]x[col]P(2nv|x)[col]P(x|2nv)0[col]1[col].24
1[col].83[col].20
2[col].68[col].16
3[col].54[col].13
4[col].42[col].10
5[col].31[col].07
6[col].23[col].06
7[col].15[col].04
Now, in order to get to a final conclusion, I assumed that all of the values of t are equally likely. I just divided all the likelihoods by 4.16 (the total of the probabilities) to give us the probability of each situation.
The number of PR's = 12 - x - scum. I would say we've got at least a 1/4 chance of having 4 power roles or less. Probably much greater than that given the greater power probability of a reasonable number of power roles.