Time for some balance theory from me.
Based on the fact that oEJo flipped scum I am perfectly amenable with the idea of symmetry that armlx proposed ages ago.
Here's my own contribution:
We have 4 scum, divided up 2 and 2 into the two scum teams. How do I know this? Simple, three player scum teams would be physically impossible if the town would have any chance at all to win. Here's why.
-3 scum plus 3 scum equals 6/16 players as scum, each with a kill at Night.
-Assume one successful lynch of scum and two successful kills of townies each cycle, with one scum group being eliminated completely before another is hit. That gives us:
-After D1 = 15 (5 scum)
-After N1 = 13 (5 scum)
-After D2 = 12 (4 scum)
-After N2 = 10 (4 scum)
-After D3 = 9 (3 scum)
See the problem yet? In this situation the town gets only 1 mislynch, and only gets that once one scum faction is entirely gone. To illustrate this point, let's look at the same situation, assuming the town lynches scum of alternating factions on successive Days. That would give us:
-After D1 = 15 (5 scum)
-After N1 = 13 (5 scum)
-After D2 = 12 (4 scum)
-After N2 = 10 (4 scum)
-After D3 = 9 (3 scum)
-After N3 = 7 (3 scum)
-After D4 = 6 (2 scum)
-After N4 = 4 (2 scum)
-After D5 = 3 (1 scum)
-After N5 = 2 (1 scum)
That's a scum win, with the only thing up in the air being which faction is the final winner.
Now if you factor in the fact that the town needs to successfully determine which weapon to use on each scum, the town tends more towards a tendancy to leave each scum alive for a day through a failed lynch. Let's look at this situation now, assuming that one half of the town lynches fail the first time. We'll factor in the same deal for the scum as well, even though their percentage to hit successfully is signifigantly higher (we assume one faction is eliminated entirely as in the first example):
-After D1 = 16 (6 scum)
-After N1 = 14 (6 scum)
-After D2 = 13 (5 scum)
-After N2 = 13 (5 scum)
-After D3 = 13 (5 scum)
-After N3 = 11 (5 scum)
-After D4 = 10 (4 scum)
-After N4 = 10 (4 scum)
-After D5 = 10 (4 scum)
-After N5 = 8 (4 scum)
-After D6 = 7 (3 scum)
At this point the miss chance goes away for both parties because they know that each other are the only remaining group. And wouldn't you know, it's LyLo. If we assume the alternating situation like in my second example, then the town loses in this fashion:
-After N6 = 7 (3 scum)
-After D7 = 7 (3 scum)
-After N7 = 7 (3 scum)
-After D8 = 6 (2 scum)
-After N8 = 4 (2 scum)
-After D9 = 4 (2 scum)
-After N9 = 4 (2 scum)
-After D10 = 3 (1 scum)
-After N10 = 2 (1 scum)
And there's the same situation, a scum win, with the only variable being which faction wins the prize.
No matter how you look at it, the town has, at most, one mislynch, and then only late in the game. This would be unreasonable odds for the town to expect to win with. Cut the scum teams down to two players each and you solve all of the problems, giving the town a fighting chance of winning with good play.
What does all of this mean, you ask? It means that if armlx is right about the neighbor scum ratio, we have another scum guarenteed among 4 players.
That said, I'd like to see how this case of yours pans out armlx. I'm not quite sold on it yet though. I'll have to reread your original case. Until then,
vote: ribwich
vote: gun
My point about the alteration of the lynch type at the end of D1 is right back on the table, since the way it panned out, we would have lynched scum D1 if not for that.
Because, no matter how you dress it up, that's what the world is. A community of idiots doing a series of things until the world explodes and we all die.