Now that you mention it I could do with some ham. Especially since Kryps' cheese sandwich never came.

As long as it is country ham. None of that honey glazed nonsense.

Honey glazed.... what? People put honey on ham? This sounds pompous as hell and not even that tasty. What's next? Chocolate?

Yes.

Try it first. If done correctly, it's lovely.

Nah, I'd assume pineapple.

Yes, I agree with Krypsyn's math. But if we approach this probabilistically, we'll have a 50% chance of hitting scum, no matter whom we lynch. So, in half the cases we'll sacrifice one town to kill one scum. In the other half of the cases we will kill two townies.
So, after thinking about it, I think it is important for town to have all the information that is available. Scum knows each other's partner anyhow and therefore also know, who the four townies are, that aren't paired with scum.

Time to even the playing field by giving town more info: my lover is drealmer7

CSPVG is my Lover and I guess Krypsyn and JMich are Lovers.

As for ham, I'm not a big fan of the honey glaze either, but I do like a brown sugar glaze.

This

hurts

my

head.

Permatownbucket has 0% of being scum even when scum.

Yep.

So we have:

Hijack + Cristi
JMich + Krypsyn
trent + CSPVG
drealmer + me

and we're still in RVS, basically. At least I have nothing solid yet. Just one suspicion, but that will take additional watching before I can reveal it (otherwise the additional watching would be pointless).

I did the math as well, and while I still can't quite follow Krypsyn's process, I do agree that randomly choosing someone other than you or your pair gives you a 5/9 chance of killing scum. I do have a slightly higher chance for a townie's partner to be scum though, since it seems that for any given townie, 30 out of the 90 combinations result in his partner being scum, so 1/3.

Now to see who to vote for, even though this game does look like it will be totally random.

Well, it certainly isn't a Monty Hall problem in my mind, after thinking a little on it, so you aren't 50% likely to be Mafia (as I asserted in the QT).

The reason this isn't a Monty Hall problem is that the reveal of a town player as town is not predicated by any choice by that player. Monty Hall works the way it does because the contestant first picks a door, then the host reveals a door based on that choice. In this case there is no choice made by the player to affect how who the moderator revealed as town This doesn't even count that, in 25% of cases, the revelation wasn't town at all, which also breaks the Monty Hall problem model.

In this case, it would be as if the host in the Monty Hall problem, before of waiting for the contestant to choose, just opened up one of the 3 Doors to reveal a goat. The remaining two would still be a 50/50 chance to contain the car, not some 33/67 chance, because the player's choice couldn't, since it was never made, have ever narrowed the host's door choices at all.

-------------------------

Anyway. The rest of my math is just based on probability distribution when there is a given. Knowing that one is town affects the distribution in the way I described.

The basic assumption is that everyone has an equal probability to be town, but the chances that the town player's Pair has Mafia is reduced because the chance that player is town is 100%. However, since a Pair can only ever contain 1 Mafia, each roll for another pair effective reduces the pool available to be scum by 2, except in the case that the player's partner is Mafia, in which case the pool is only reduced by 1 player. It is this difference in pool size that makes the town player's partner slightly less likely to be town, individually, than any other player.

You guys are killin' me.

Meh. It isn't a Monty Hall problem. The Monty Hall problem requires a certain behavior from the host; it must be an intelligent choice based on actions made by the player. In this case the reveal of the player as town was completely random, and not predicated on any action made by the player before hand.