Theorel are you sure on your numbers? I think assuming 3 scum it's about 55/45 that there's a scum. (9/12 * 8/11 of no scum). Higher if there's only 2 lower if 4. I think you are assuming fights are town v town far far too easily...
I wasn't removing myself, nor was I removing the first town player (i.e. I was just doing (10/13)^2). The second one of those is obviously terrible. The first is arguable (it's not using all information available to me/any town player, but it is still an accurate calculation for an impartial observer).
So, from an impartial observer it should be ~58%. From a town player's perspective it's ~54.5%, and from scum's perspective they already know which it is. I wouldn't say 60% is far far too easily. I mean it's only ~5% more likely.
I apologize for my bad maths.
I just noticed this post by Theorel, and I have some problems with it.
1. I wasn't removing myself, nor was I removing the first town player . The fact that you forgot to remove yourself makes me wonder if you do not automaticly see yourself as town.
2. (i.e. I was just doing (10/13)^2). How do you know for sure that there is exactly three scum?
1. He didn't remove himself because he was presenting the case to other players, who don't for sure know that he's town. I can see that, although I disagree with it.
2. 3 scum is pretty standard for a game of this size, I think. I would guess 3 scum.
I have added Theorels post below where he originally calculates this.
1. He does not present his calculations to the other players, only the result in the form: more that 50% chance they are both town. So I am not convinced he would bother to put it in our perspective.
With the phrasing "given any 2 players at random" I agree it makes sense to include himself, but when he uses this to figure out whether efhw and shraeye are both town, it makes more sense not to.
The way he did it makes me think he was focused about calculating "any 2 players at random", not "EFHW - Shraeye". Could be because he is scum and thus not really scumhunting.
2.. I agree 3 scum is pretty standard. I would guess 3 scum as well. But when calculating something I would probably consider other scum distributions as well. Ta does so here:
Theorel are you sure on your numbers? I think assuming 3 scum it's about 55/45 that there's a scum. (9/12 * 8/11 of no scum). Higher if there's only 2 lower if 4. I think you are assuming fights are town v town far far too easily...
Theorel seems to be very sure that we are dealing with three scum since he do not address other possibilities.
In this case, that's spiritbears. And although I don't think it's by any means conclusive, he does come off scummier here (IMO) for adding fuel to the fire. It seems inherently scum-like to try to alienate players from each other if both are town. Doing so helps to ensure that town will not unite against you. OTOH if efhw-shraeye are not both town, then spiritbears (if scum) was either taking a position against a team-mate, or immediately alongside a team-mate. That I think looks relatively unlikely, scum would be more likely to just try to pull attention elsewhere in that case...perhaps pushing the town v. town angle. Since, at this stage, given any 2 players at random, they are more likely both town than either scum I'm going to go ahead with this making spiritbears slightly scummier (probably taking him just past neutral. Say scumScore=26).