January 7th, 2021, 03:34 PM

It is kind of what I said, at least conceptually. The thing is, this paper actually assumes the town still has a chance of winning when there's 50% mafia and 50% town. It simply assumes there is a 50/50 chance of a mafia getting lynched at that phase. If there is only 1 town and 1 mafia, the paper assumes the winner is determined by a coinflip rather than simply handing mafia the tie-winner.

That's why the probability of a mafia win goes up for odd numbers rather than even numbers. In this paper, the even numbers allow for 3 mafia 3 town, which is better odds than 3 mafia 4 town. If we assume mafia automatically wins ties, then even numbers become worse again like you'd expect.

p;edit this paper is really goofy. When there is a single mafia and 100s of town, the probability that the single mafia wins a game with 2n players becomes asymptotically close to an equation of n with pi in it. 2n = root(2/n*pi) (roughly). For example - for 2000 players, the probability of mafia victory is extremely close to root(2/(1000*pi))

It is kind of what I said, at least conceptually. The thing is, this paper actually assumes the town still has a chance of winning when there's 50% mafia and 50% town. It simply assumes there is a 50/50 chance of a mafia getting lynched at that phase. If there is only 1 town and 1 mafia, the paper assumes the winner is determined by a coinflip rather than simply handing mafia the tie-winner.

That's why the probability of a mafia win goes up for odd numbers rather than even numbers. In this paper, the even numbers allow for 3 mafia 3 town, which is better odds than 3 mafia 4 town. If we assume mafia automatically wins ties, then even numbers become worse again like you'd expect.

p;edit this paper is really goofy. When there is a single mafia and 100s of town, the probability that the single mafia wins a game with 2n players becomes asymptotically close to an equation of n with pi in it. 2n = root(2/n*pi) (roughly). For example - for 2000 players, the probability of mafia victory is extremely close to root(2/(1000*pi))

January 7th, 2021, 03:10 PM

[QUOTE=yzb25;914737]It's based on the fact the probability of mislynching technically goes up if there are more towns to mislynch from without additional lynches, which happens if there are say, 5 towns and 3 scums rather than 4 towns and 3 scums. (5/8 mislynch chance vs 4/7 mislynch chance supposedly)

But it's a very innaccurate way of modelling the outcome of games, because I don't know many ppl who would prefer to play as town from a position with 5 towns and 3 scums rather than 4 towns and 3 scums. This perverse result is caused by simply assuming every person has an equal chance of getting lynched irrespective of game-state. A simple way to improve the model may be to assign each person a randomly selected list of preferred lynches, then keep running AV until a majority winner is found (and assume scums are unlikely to "prefer" to lynch their own, especially at or near LYLO). I'd argue that process comes somewhat closer to representing how a "simple" game is more likely to play out, but obviously that has a lot of flaws too. It would probably make games appear more scum-sided than they really are.[/QUOTE]

hmm, this looks more serious than I assumed. Maybe they didn't do what I said. I'll give it a read then get back to you ^^.

Originally Posted by

**yzb25**
It's based on the fact the probability of mislynching technically goes up if there are more towns to mislynch from without additional lynches, which happens if there are say, 5 towns and 3 scums rather than 4 towns and 3 scums. (5/8 mislynch chance vs 4/7 mislynch chance supposedly)

But it's a very innaccurate way of modelling the outcome of games, because I don't know many ppl who would prefer to play as town from a position with 5 towns and 3 scums rather than 4 towns and 3 scums. This perverse result is caused by simply assuming every person has an equal chance of getting lynched irrespective of game-state. A simple way to improve the model may be to assign each person a randomly selected list of preferred lynches, then keep running AV until a majority winner is found (and assume scums are unlikely to "prefer" to lynch their own, especially at or near LYLO). I'd argue that process comes somewhat closer to representing how a "simple" game is more likely to play out, but obviously that has a lot of flaws too. It would probably make games appear more scum-sided than they really are.

hmm, this looks more serious than I assumed. Maybe they didn't do what I said. I'll give it a read then get back to you ^^.

January 7th, 2021, 03:08 PM

[QUOTE=Marshmallow Marshall;914429]Trying to quantify Mafia is very hard. The social element adds a ton of instability to any calculations you'd try to make.

I do not understand that statement at all. It's very counter-intuitive, and the logic behind it is beyond me, if it exists.[/QUOTE]

It's based on the fact the probability of mislynching technically goes up if there are more towns to mislynch from without additional lynches, which happens if there are say, 5 towns and 3 scums rather than 4 towns and 3 scums. (5/8 mislynch chance vs 4/7 mislynch chance supposedly)

But it's a very innaccurate way of modelling the outcome of games, because I don't know many ppl who would prefer to play as town from a position with 5 towns and 3 scums rather than 4 towns and 3 scums. This perverse result is caused by simply assuming every person has an equal chance of getting lynched irrespective of game-state. A simple way to improve the model may be to assign each person a randomly selected list of preferred lynches, then keep running AV until a majority winner is found (and assume scums are unlikely to "prefer" to lynch their own, especially at or near LYLO). I'd argue that process comes somewhat closer to representing how a "simple" game is more likely to play out, but obviously that has a lot of flaws too. It would probably make games appear more scum-sided than they really are.

p;edit many people who [I]wouldn't[/I] prefer to play from a position with 5 towns and 3 scums

Originally Posted by

**Marshmallow Marshall**
Trying to quantify Mafia is very hard. The social element adds a ton of instability to any calculations you'd try to make.

I do not understand that statement at all. It's very counter-intuitive, and the logic behind it is beyond me, if it exists.

It's based on the fact the probability of mislynching technically goes up if there are more towns to mislynch from without additional lynches, which happens if there are say, 5 towns and 3 scums rather than 4 towns and 3 scums. (5/8 mislynch chance vs 4/7 mislynch chance supposedly)

But it's a very innaccurate way of modelling the outcome of games, because I don't know many ppl who would prefer to play as town from a position with 5 towns and 3 scums rather than 4 towns and 3 scums. This perverse result is caused by simply assuming every person has an equal chance of getting lynched irrespective of game-state. A simple way to improve the model may be to assign each person a randomly selected list of preferred lynches, then keep running AV until a majority winner is found (and assume scums are unlikely to "prefer" to lynch their own, especially at or near LYLO). I'd argue that process comes somewhat closer to representing how a "simple" game is more likely to play out, but obviously that has a lot of flaws too. It would probably make games appear more scum-sided than they really are.

p;edit many people who *wouldn't* prefer to play from a position with 5 towns and 3 scums