## [EV] Soulbind except the mafia cant change the pairs

This forum is for discussion of individual Open Setups, including theoretical balance.
Awoo
Goon

Joined: September 01, 2017
Pronoun: He
Soulbind except the mafia pick what the pairings are pregame and its LYLO. I hope this will be the first EV thread / post I have made that is actually accurate.

Pregame the mafia put everyone into pairs except for one person who is alone.
If one half of a pair is lynched they both die and the next day starts without a night phase.
If the person who is alone (referred to as the loner) is lynched, the game goes to night and the mafia kill one member of a pair. The member of the pair that was not targeted remains alive as the new loner.

tl;dr: BNL's soulbind but the mafia pick pairs pregame and cant change them.

EV (2 town 1 mafia) = 1/2 - trivial

EV (3 town 2 mafia) = 1/3. Proof:

Define the probability the mafia will make each type of pairing pre-game.

P{ (MM)(TT)(T) } = a
P{ (TM)(TM)(T) } = b
P{ (TM)(TT)(M) } = 1 - a - b

where 0 <= a, b <= 1. The values of a and b are public knowledge.

Town can either choose to lynch a pair or lynch the loner.

EV (lynch a pair) = a/2 + b( X ) + ( X ) (1-b-a)/2
EV (lynch loner) = (1-b-a)/2

Let A and B be events.

A: We lynched a pair and it flipped (MT).
B: The loner is mafia.

P(A) = b + (1-a-b)/2
P(B) = 1-a-b
P(A|B) = 1/2

Where X = max(P(B|A), 1 - P(B|A))).

Since the town will just lynch the more likely of {pair, loner} to flip mafia. Which means X is has a minimum at 1/2 when the conditional probability becomes 1/2.

We want to minimize X, so let us find a formula for this.

P(B|A) = P(A|B)P(B) / P(A)
P(B|A) = ((1-a-b)/2) / (((1-a-b)/2) + b )
which simplifies to
P(B|A) = (1-a-b)/(1-a+b).
We want to have this equal to 1/2 to minimize town's EV.
1/2 = (1-a-b)/(1-a+b)
which yields the solution
b = (1-a)/3.

Let us put these results back into the original definitions:
Note that (1-a-((1-a)/3)/2 = (1-a)/3

P{ (MM)(TT)(T) } = a
P{ (TM)(TM)(T) } = (1-a)/3
P{ (TM)(TT)(M) } = 1 - a - (1-a)/3

EV (lynch a pair) = a/2 + ((1-a)/3)(1/2) + (1/2)((1-a)/ 3) = a/2 + ((1-a)/3)
EV (lynch loner) = (1-a)/3

Now the mafia have control over a, and they want to minimize town's EV. choose a = 0 has the result

EV(lynch a pair) = 1/3
EV(lynch a loner) = 1/3

and it follows that the strategy that gives this result as

P{ (MM)(TT)(T) } = 0
P{ (TM)(TM)(T) } = 1/3
P{ (TM)(TT)(M) } = 1- (1/3) = 2/3

, and it also follows that the worst play the mafia can make (short of claiming mafia) is to have a = 1, which yields a town EV of 1/2.

Hypothesis:

[1] in this setup, with N mafia and N+1 town, the town EV is 1/(N+1).
[2] As N increases, the probability that the loner is mafia approaches 1.

Awoo
Goon

Joined: September 01, 2017
Pronoun: He
EV(4-1) = 2/3. Mafia places himself as the loner 1/3 of the time to achieve this result.

Every setup of this type has an EV graph that looks kind of like this: (ignore values outside of [0, 1])

EV(3-2)

EV(4-1)
Where X is probability that the loner is mafia, Y is town's EV. g(x) is the conditional probability that the loner is mafia given that a TT lynch occurred yesterday. f(x) is the greater of the EV of (lynch loner, lynch pair) in 2-1 lylo. p(x) is the EV for lynching a pair d1. q(x) is the EV for lynching the loner d1.

Generally, when the loner is lynched and the game continues, the mafia can manipulate how often the new loner will be mafia using the nightkill (the value of x), which is crucial to mafia's objective of lowering town's EV. However, if mafia put themselves into pairs of (MM) pregame, they diminish their ability to do this with no other reward. Therefore I will be assuming mafia does not pair themselves into (MM) pairs going forward, even though I'm unable to come up with a general proof for it.

When a pair is lynched, the town forces a static X value going into the next day. I believe that based on the previous day's X value and the pair's flip (MT) or (TT), the town can determine a strategy for the next day (lynch pair or lynch loner) using conditional probability: the loner is mafia given that the flip was (MT) or (TT).

callforjudgement
Mafia Scum

Joined: September 01, 2011
There's one obvious advantage to MM pairing: town are less likely to actually lynch scum, as fewer possible lynch choices will lead to a scum death. It does, however, indeed hurt scum if the loner is lynched or (obviously) if scum is somehow lynched anyway.
scum · scam · seam · team · term · tern · torn · town

Awoo
Goon

Joined: September 01, 2017
Pronoun: He
I did prove that making an MM pairing is strictly suboptimal in (3-2). Do you think it would change in other situations? Because I feel like a lot of mafia's power to reduce EV in this comes from the ability to manipulate how often the loner is mafia.

Maybe if there are 3 mafia it's different because you have 1 free mafia if 2 pair up. But you can still get into some clunky endgames where the 2 are glued together and I still don't think it's worth it, though I will probably investigate that later.

Awoo
Goon

Joined: September 01, 2017
Pronoun: He
Furthermore, town is already has incentive to lynch pairs over the loner because then mafia can't manipulate the pairs overnight. I think proving this for a general case would be very hard since it would need to account for 4 different variables. Still working on a general case/recursive formula.

Awoo
Goon

Joined: September 01, 2017
Pronoun: He
Okay, I found the recursive formula. It is a bit of a monster to look at, but it always simplifies into a function that consists of 2 straight lines, starting at (0, 1), decreasing until it reaches a minimum, then increasing over the rest of the domain we are interested in.

For reasons currently beyond my comprehension, that monster expression "EV for lynching a pair" ALWAYS simplifies to 1 - x for every setup I have tried. The somewhat less monstrous expression "EV for lynching a loner" always simplifies to ax + b, where a and b are between 0 and 1.

And assuming the "EV for lynching a pair" expression really does always simplify to 1 - x, then it would be true that the lowest EV mafia can achieve for town is always equal to 1 minus the probability that they make the loner mafia with optimal choice.

Here are some functions I have found already for your amusement. Maybe you can see a pattern? The EV for each is just 1 minus the minimum x value.

EV(2, 1)(x) = max(1-x, x), min @ x = 1/2
EV(4, 1)(x) = max(1-x, 1/2 x + 1/2) min @ x = 1/3
EV(6, 1)(x) = max(1-x, 1/3 x + 2/3), min @ x = 1/4
EV(3, 2)(x) = max(1-x, 1/2 x) min @ x = 2/3
EV(5, 2)(x) = max(1-x, 1/3 x + 1/3) min @ x = 1/2
EV(7, 2)(x) = max(1-x, 1/4 x + 1/2), min @ x = 2/5

Sadly, my function started producing nonsense results when I tried to evaluate EV(4, 3). I believe the cause for this is that my assumption that mafia should never make an MM pair only holds for m = 2, or at the very least does not hold for m = 3.

mith
Godfather

Joined: March 27, 2002
Location: Plano, TX
Pronoun: He
Yeah, I wouldn't assume this would hold for m > 2.

The pattern for EV at m = 1 is just t/(t+2), and for m = 2 it's (t-1)/(t+3). I would guess that pattern continues for a general formula of (t-m+1)/(t+m+1), but maybe MM pairings change that.

(And remember that nightless is (t-m)/(t+m), so this is obviously very similar but a bit worse for town.)

[ + ]