[SETUP] Bipartite Mafia

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[SETUP] Bipartite Mafia

Post Post #0 (ISO) » Tue Dec 04, 2018 3:28 pm

Post by Something_Smart »

3 Mafia Goons

4 Vanilla Townies


This game has no lynches or nightkills. Instead, it has a pairing mechanic:

Any player may, at any time, post
Pair:
and the name of another player (with whom they are not already paired). Those two players will attempt to pair.
If a townie pairs with a mafia member or a mafia member pairs with a townie, the pairing is successful, and it is announced as such. It's not announced which player is which.
If a townie pairs with another townie, the mafia immediately win.
Mafia members are forbidden from pairing with each other. (If they attempt to, the mafia just lose on the spot.)

Players may vote, if they wish. If one player receives a majority of votes, that player must pair with someone within a given time limit (e.g. 48 hours), and other players are forbidden from pairing during that time.

In addition, players may VOTE: Free. If Free receives a majority of votes, the game goes to a night-ish phase where the thread is locked, and the mafia must select a townie and a mafia member (who are not already paired together) to publicly pair. Obviously, it is not announced which is which. This can only be done once during the game.

The town wins if three pairings, not counting the Free pairing, are successfully made.
It's always the same. When you fire that first shot, no matter how right you feel, you have no idea who's going to die. You don't know whose children are going to scream and burn. How many hearts will be broken. How many lives shattered. How much blood will spill, until everybody does what they're always going to have to do from the very beginning... SIT DOWN AND TALK!
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Post Post #1 (ISO) » Tue Dec 04, 2018 3:45 pm

Post by Invisibility »

so tempting to throw
Invisibility is actually AWESOME!
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Post Post #2 (ISO) » Tue Dec 04, 2018 3:46 pm

Post by Invisibility »

not a bad thing set up seems interesting
Invisibility is actually AWESOME!
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Post Post #3 (ISO) » Wed Dec 05, 2018 5:12 am

Post by Sukima »

What happens if the voted player refuses to pair with someone in 48 hours or if they're away on vacation or something?
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Post Post #4 (ISO) » Wed Dec 05, 2018 10:12 am

Post by Invisibility »

people probably wouldnt vote someone V/LA without arranging stuff first
also i like randomness for not choosing
Invisibility is actually AWESOME!
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Post Post #5 (ISO) » Wed Dec 05, 2018 11:51 am

Post by Something_Smart »

I don't like random; aside from introducing provable randomness into the game it would mean that the outcome of the game could be entirely determined by chance (because if a town player is voted up and their choice is randomized, it's 50/50 odds on immediate game over).

I'd probably just say an extension can be granted for V/LA, but otherwise it works like a prod and the player gets replaced (or modkilled) if they don't decide.
It's always the same. When you fire that first shot, no matter how right you feel, you have no idea who's going to die. You don't know whose children are going to scream and burn. How many hearts will be broken. How many lives shattered. How much blood will spill, until everybody does what they're always going to have to do from the very beginning... SIT DOWN AND TALK!
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Post Post #6 (ISO) » Thu Dec 06, 2018 7:46 am

Post by Sukima »

There's no kills in this game though? I guess it'd be harder if the person in question couldn't be linked to ever, but that'd favor the team who has the AFKer.
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Post Post #7 (ISO) » Thu Dec 06, 2018 9:24 am

Post by the worst »

36 pairings d1, 10 of which instantly lose the game for town. ok that's a little fun.

Assuming:
1) any player can be a part of a limitless amount of pairs, and
2) mafia cannot use the free pair to just submit a pair that's already been made

I'm guessing the optimal town strategy is to try and pair the two scummiest players d1 then fire the three shot and try to link the two most likely scum across both pairs on d2?
who's scum? i haven't read up yet but like, it's me
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Post Post #8 (ISO) » Thu Dec 06, 2018 9:34 am

Post by the worst »

Actually if my thinking is right there's a 26/36 chance you get into day two. if you hammer two scum d1 using my strategy town has already won; so assuming both d1 and n1 pairs are {t,s} and contain different scum

I've always been pretty bad at probabilities so someone please weigh in if I'm butchering this :P

10/36 = d1 loss
3/36 = d1 pair contains two scum = effective d1 win
23/36 = proceed to d2, which assuming {t,s}+{t,s} is like a 3/4 chancd of town win

d1 loss = 27.78%
d1 technical win = 8.33%
d2 win = 47.91%
d2 loss = 15.97%

town ev using this strategy looks like 56.24% which is interesting, first impressions were it felt pretty townsided. Uncharacteristically I think I'd rather be scum in this setup. :giggle:
who's scum? i haven't read up yet but like, it's me
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Post Post #9 (ISO) » Thu Dec 06, 2018 9:35 am

Post by the worst »

just spotted the "not including the free pairing"
Huh that's difficult. Let me think
who's scum? i haven't read up yet but like, it's me
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Post Post #10 (ISO) » Thu Dec 06, 2018 9:46 am

Post by the worst »

same strategy but on d3 I think you pair the same person from the d1 pairing with the other person in the n1 pairing (allowing: d1 has a chance of containing two scum and should contain a higher confidence scumbutt so should always be relied on before the n1 pairing which is a manipulative pairing anyway)

I've left the house now but I think this only marginally swings the win % back to scum
who's scum? i haven't read up yet but like, it's me
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Post Post #11 (ISO) » Thu Dec 06, 2018 12:01 pm

Post by callforjudgement »

Town can guarantee a 50% win rate by using the free pair immediately, then choosing the scummier of the two players in it and voting them three times. Whichever player gets voted three times will lose for their faction; they have a 50% chance of being scum, so that's 50% EV.

This seems like it might miss the point of the setup, though, and I'm not convinced that it isn't possible to do better.
Last edited by callforjudgement on Thu Dec 06, 2018 12:03 pm, edited 1 time in total.
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Post Post #12 (ISO) » Thu Dec 06, 2018 12:02 pm

Post by RadiantCowbells »

This is an awfully townsided setup
2019 stats: Town WR 76.7%, overall WR 81.667%, 1 scum defeat involving a major mod error in lylo vs 8 scum wins.
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Post Post #13 (ISO) » Thu Dec 06, 2018 12:04 pm

Post by RadiantCowbells »

Actually it just occurred to me that scummy scum can force pair with towny town to leave the game down to the nullreads, that makes it better
2019 stats: Town WR 76.7%, overall WR 81.667%, 1 scum defeat involving a major mod error in lylo vs 8 scum wins.
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Post Post #14 (ISO) » Thu Dec 06, 2018 12:37 pm

Post by Something_Smart »

In post 7, the worst wrote:Assuming:
1) any player can be a part of a limitless amount of pairs, and
2) mafia cannot use the free pair to just submit a pair that's already been made
Both are correct.

It's cool to hear some other people's thoughts about strategy. I came up with some town strategy ideas but didn't share them immediately so I could see what other people came up with. The worst's idea of picking two scummy people to pair because if you pick two scum you win on the spot is interesting.

CFJ, I'm pretty sure it is possible to do better than 50%; after considering a few different strategies, the best EV I got was 55%. (And actually, to get 50%, you don't even need to repeatedly vote out the scummier player of the pair; you can just order every town player to pair with them :P Remember that a majority isn't necessary for a pairing to occur, only sufficient.) My idea was that you get the free pair first, then select someone and force that person to pair with one of the already paired players. Then you force that player and the one they didn't choose to collectively pick another player and both pair with that player.

If the first selected player is scum, it's a guaranteed town win because the scum are forced to either deviate from the plan, outing themselves, or give town the three pairings they need to win. The odds of that are 2/5.
If the first selected player is town (3/5 chance), the odds are 1/2 that they pair with the scum in the initial pairing and 1/2 again that they select another scum to double-pair with, making town's total EV 2/5 + (3/5)*(1/2)*(1/2) = 11/20.

Of course, it will probably be very different in practice because scum are definitely not going to allow scum to be picked for the second pairing 40% of the time. I don't know if it would be townsided in practice, given that just one townie being wrong is enough to cost town the game.
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Post Post #15 (ISO) » Thu Dec 06, 2018 2:26 pm

Post by callforjudgement »

It's probably townsided in theory, but not in practice (you need a disciplined town to pull the optimal strategies off).

I thought there should be some way to win if you accidentally pick a townie to make the first non-free pairing. I found the strategy you listed a bit hard to follow, so with letters:

Free pair: {A, B}
Town vote for a scummy player C
C pairs with their choice of A and B; let's say A (the choice of B is symmetrical)
B and C collectively choose a player D; B and C both pair with D

And the EV calculation seems correct (I thought it was wrong but miscounted, so here's the full calculation):
  • 40%: C is scum; no scum strategy avoids a town win from here
  • 60%: C is town:
    • 50% (30% of total): A is also town, scum win (town C paired with town A)
    • 50% (30% of total): A is scum:
      • 50% (15% of total): D is town (scum win; town C paired with town D)
      • 50% (15% of total): D is scum (town win; three successful pairings B-D, C-D, C-A)
Total = 55% to town, 45% to scum.
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Post Post #16 (ISO) » Mon Dec 17, 2018 7:10 am

Post by mith »

I have at least convinced myself that you can't do better by holding off on the free pair until you get two successful pairings. First pair (say A-B) will lose 2/7 of the time, win 1/7 (basically force A to pair with B, and if A refuses then they are both scum and you just win by continuing to pair A with whoever), and continue 4/7. If you try to pair A (or B) again with C, you will lose 3/10 (1/2 * 3/5), and 2/7 + 4/7*3/10 is already 16/35, which is greater than 45%. If you try to pair C with D, again probability of immediate loss is 3/10 (3/5 * 2/4).

If you get the free pair after a successful A-B, you need to win 57/80 of those games to hit 55%. Mafia has two options:

1. If Mafia gives a new pairing C-D, you win 3/4. You try to pair one of A/B with one of C/D - if you succeed, then you win with the other pair (also guaranteed to have one scum).

2. However, if Mafia chooses from A-B to pair with C, town has three strategies I see: you can try pairing D with C and with B/A (the one not chosen for the free pair), and you always win if A was town but only win half if A was scum (you need to choose D scum). You can also try pairing A with D, and then one of E-G with one of B-D. If A is scum, A-D will work, and then you have a 2/3 for picking from E-G. If B+C are scum, you have a 1/4 of picking D scum, and then win if you do. And finally, you can try A-D and A-E, just assuming A is scum. You win if A is scum and lose if A is town. I doubt that town can force an equilibrium high enough, but I haven't worked out what the equilibrium is yet.
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Post Post #17 (ISO) » Wed Dec 19, 2018 5:43 am

Post by Sukima »

Mith, I can't understand your math entirely on this one, but are you assuming that the free pair counts towards the 3 necessary pairs to win?
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Post Post #18 (ISO) » Wed Dec 19, 2018 6:10 am

Post by mith »

No.

The first paragraph is just showing that you will lose more than 45% of the time if you
don't
take the free pair before trying to vote on two pairings. (EV for that strategy is even worse - Mafia can always pick a free pair such that town can't guarantee success on the third pairing.)

The other cases (free pair after one success) are broken down as follows (with some room for choosing e.g. B instead of A in some cases):

1. Pair A-B, Free Pair C-D, Pair A-C, Pair B-D.
2. Pair A-B, Free Pair A-C,
a. Pair C-D, Pair B-D
b. Pair A-D, Pair B-E
c. Pair A-D, Pair A-E

tl;dr - Asking for the free pair first is probably the way to go (but my intuition could be wrong about the equilibrium between Mafia's choice in 2 - whether they pick the scum or town from A-B - and Town strategies a-c), and I don't see any way of doing better than 55% yet.
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Post Post #19 (ISO) » Wed Dec 19, 2018 6:56 am

Post by mith »

Yeah, looks like the best town can do there is 2/3 (picking a 2/3 of the time and c 1/3; b isn't helpful). That only gets you to an EV of 1/7 + 4/7*2/3 = 11/21 (52.38%).
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Post Post #20 (ISO) » Wed Dec 19, 2018 7:14 am

Post by mith »

Oh, actually, b is a little better than I thought. On the A-D pairing, A could refuse to pair (if they are both scum), and town just wins here. There is a 1/2 chance of this happening, and a 2/3 chance of winning if A is scum and D is town (two of E-G are scum), so 5/6 overall for the A scum case, not 2/3. (Still doesn't improve the equilibrium though; the optimal a-b mixed strategy is 7/13 and 6/13, with an EV of 17/26, just a hair under 2/3.)

I can't improve on 55%. After Free Pair A-B and Pair A-C, town has the same three strategies a-c, with a giving 55% and the other two giving 50% (basically, you are taking the scum choice out of their hands and forcing an A 3/5 C 2/5 split, which makes strategy a. optimal). The other option is Free Pair A-B, Pair C-D, which has a 1/10 chance of winning immediately, 3/10 chance of losing immediately, and 6/10 chance of continuing with a 3/4 chance of winning from there (case 1 above), for a total of 1/10 + 9/20 = 11/20 = 55% again.
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