[EV] Nomination Mafia (Generalized)

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BulletNLynchproof
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Post Post #0  (ISO)  » Sat Dec 29, 2018 1:54 am

M Mafia
T Townies (M+T is odd)

Nightless. Before every even Day, the mafia nominate N people. Only those nominated people can be voted for that Day.
Lynching is compulsive.

Note that the original Nomination Mafia has N = 3 and (M,T) = (2,5) or (3,8).

What I've noticed is that choosing who to nominate isn't all WIFOM when N is low compared to M+T.
Consider the extreme cases: when N = 1, the Mafia will nominate a Townie every even Day. It isn't an issue that the Mafia confirm a Townie as Town are forced to lynch the Townie anyway, hence giving no extra information to the town on future Days. In fact, it is equivalent to Mountainous, with odd Days being mountainous Days, and even Days being Mountainous Nights.
On the other extreme, when N = T+M, this game clearly becomes a simple Nightless.

I'll do some examples where M = 1 and T = 4. There is no Day 1 strategy other than random, so I'll show the Day 2+ strategy assuming town mislynches on Day 1. The Mafia has two strategies on Day 2: self-nominate (A), or don't self-nominate (B). Town has two strategies on Day 3: Lynch one of Day 2's nominees (X), or don't lynch Day 2's nominees (Y). Also, let the probability that the Mafia choose strategy A be p, and that of Town choosing strategy X be q.

Suppose N = 3. Then the table of town EVs according to Mafia's and Town's strategies is as follows:

AB
X2/30
Y1/31


The Nash equilibrium is when p = 3/4 and q = 1/2, which leads to an overall EV of 1/2. There's actually a simpler way to get to this result: town just choose the two scummiest people without looking at who is nominated, then lynching one of those on Day 2 who is nominated, lynching the other on Day 3, making this equivalent to Nightless. This also explains p = 0.75, which is really choosing one player uniformly randomly to be not nominated, giving away as little information as possible.

Now instead suppose N = 2:
AB
X10
Y1/21/2


Here I believe the EV is 1/2 as well, but the results are weird: p = 1/2 and q = 0 (!) I believe are the strategies. This actually surprised me, because I thought that scum could afford to play WIFOM here to include themselves less than usual. I guess this is because there's already a 1/2 chance that scum gets lynched by putting themselves in the pool.

However, I believe when M+T is suffciently large, it benefits scum to put town in the pool more than average: suppose scum always puts 2 town in the pool, then the game becomes mountainous except a bulletproof IC; this probably has a lower EV for town than uniformly randomising the nomination (=Nightless), and there may be an even better strategy somewhere between the two. However, calculations with more than one nomination becomes very complicated because the nominations may not be independent of each other. Hence, I would like to call on the EV calculation experts (callforjudgement and mith) for extensions of this.
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callforjudgement
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Post Post #1  (ISO)  » Sat Dec 29, 2018 9:48 am

This is a really interesting problem, but the number of potential scum strategies is too high to even list them all by hand, even in something small like a 2:7 (the issue is that nominating the same player twice, and nominating a separate batch of players, are distinct scum strategies). So I think the correct approach here is to write a script to automatically generate a description of everything the players can potentially do, and then run it through an existing WIFOM solver like Gambit. (This approach doesn't work for setups in general, but does work in a setup like Nomination Mafia where any asymmetry between townies is public knowledge, as it means that there are only two sides to the game, town and scum.)
scum · scam · seam · team · term · tern · torn · town


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