Problem for game "Forum Mafia" w/ 13 players, 10 good guys and 3 bad guys.

Quick

Junior Member
Joined
Jan 19, 2018
Messages
90
The game I am talking about is Forum Mafia.

This game has a total player base of 13 players
There are 10 good guys and 3 bad guys
Assume that 2 people leave the game per round: each round is divided up into 2 parts: a part where everyone votes someone off the game through majority vote, and a part where the bad guys pick someone to eliminate from the game.
Assume that each day, worst best case scenario happens, which means that two good guys are taken out of the game each round until one team wins.
What is the probability that the good guys need to be right in who they eliminate from the game to win the game by the time the bad guys have a majority by the end of the final round?

Bad guys win if they hold a majority of the players
Good guys win if they eliminate all bad guys

So far I have tried to multiply:
(10/13)(8/11)(6/9)(4/7)= ~20%

Is this right?
 

j-astron

Junior Member
Joined
Jan 10, 2018
Messages
181
I've played Mafia in real life, but I can't understand the question as you have posed it:

The game I am talking about is Forum Mafia.
Assume that each day, worst best case scenario happens, which means that two good guys are taken out of the game each round until one team wins.
"worst best case scenario"...what does that even mean? In what way is it the worst? In what way is it the best case scenario? For whom?

Not only that, but you said in your set up that only two people total are eliminated per round. If I take what you said above at face value, those two people are always good guys. In which case...the good guys always lose. So something is missing from your description here.

If what you meant to say was that the good guys always guess wrong and two of them are eliminated until the final round, where they can actually get it right, then they have to get it right in that final round, otherwise they lose.

What is the probability that the good guys need to be right in who they eliminate from the game to win the game by the time the bad guys have a majority by the end of the final round?
What??? You'll have to phrase this in a much less convoluted way than that. What is the question you are trying to answer? Probabilities usually refer to actual outcomes, so "probability that they need to be right" doesn't make sense to me.
 

Quick

Junior Member
Joined
Jan 19, 2018
Messages
90
Basically, I just want to know how accurate the good guys need to be to win the game.

Like what percentage do the good guys need to be right to win. That's really all I am asking.

Also, you should relax and have some tea or something.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,244
Also, you should relax and have some tea or something.
To avoid testy answers, try comprehensible questions. If your question can be asked in a single sentence, give that approach a try.
 

JeffM

Elite Member
Joined
Sep 14, 2012
Messages
3,244
...and you should be sober when formulating a question :rolleyes:
Ahh mon ami, what is the fun in that? Moreover, in my youth, back when mastodons roamed the woodlands, some of my more intriguing questions were posed while I could hardly be called sober. And sometimes the most satisfactory answers came when the girl answering was not particularly sober either. "Of course that was long ago and in a different country. Besides the mastodons are extinct," which is not a quotation from the Duchess of Malfi.
 

Quick

Junior Member
Joined
Jan 19, 2018
Messages
90
So...

Am I going to get help with my problem or are you all just going to keep making fun of me?
 

j-astron

Junior Member
Joined
Jan 10, 2018
Messages
181
So far I have tried to multiply:
(10/13)(8/11)(6/9)(4/7)= ~20%
What you are computing here is the probability that the good guys make the wrong choice in every consecutive round, until they lose. And you're doing that under the assumption that in each round, any player is equally-likely to be eliminated by the group (i.e. the group plays as though they have no information).

I.e. you are assuming P(wrong choice) = (No. of good players)/(No. of total players)

You end up with a 21.3% chance of losing, leading to a 78.7% chance of winning.

The problem with this is that the above scenario isn't the only way for the good guys to lose. For example, you could imagine the game getting down to the point where there are 2 good guys and one bad guy. A good guy makes the wrong choice and teams up with the bad guy to eliminate his team mate. The remaining good guy then gets eliminated by the remaining bad guy, and the bad guys win. This might be rare in actual gameplay, since people usually figure out who's doing the killing before it gets down to so few players. But it is at least possible, right?

So I would say that no, what you have above is not the correct answer.
 

j-astron

Junior Member
Joined
Jan 10, 2018
Messages
181
Basically, I just want to know how accurate the good guys need to be to win the game.

Like what percentage do the good guys need to be right to win. That's really all I am asking.
I think that this game is pretty tricky to model with pencil and paper. There are a lot of outcomes to keep track of. What I was going to try to do was to assume a certain rate of correct guesses by the good guys, as you suggested, and then see how many rounds it took for the good guys to win under that assumed success rate. Then I thought back to actually playing mafia, and how, especially in the early rounds, all the accusations of who is a mafia member are completely arbitrary and pointless, because nobody has any more information than anyone else about who is who. That's what always frustrated me about this game as a party game IRL.

So I decided to go with the idea that the group was equally-likely to pick any player to eliminate, in every round. This approximation becomes worse as the game progresses, and more information is gained. But I wasn't sure what else to do.

Even then, it was too laborious to compute outcomes by hand, so I wrote a program to simulate 100,000 games and to look at the percentage of them that are won by the good guys, as a function of number of rounds. I pretty consistently get results like this:

(No. of rounds to win, Percentage of games)
(3, 0.44)
(4, 1.80)
(5, 5.10)
(6, 13.61)
(They lose, 79.05)

(No. of rounds to win, Percentage of games)
(3, 0.50)
(4, 1.83)
(5, 5.20)
(6, 13.56)
(They lose, 78.92)

So, assuming I did this right , then the assumption that any player is equally-likely to be eliminated in each round heavily favours the bad guys, since there are so many more good guys to choose from.
This also says that the good guys can't win in more than six rounds. I think that makes sense, since a win in 7 rounds would imply 3 "right choice" rounds and 4 "wrong choice" rounds. But four wrong choice rounds eliminate 8 good guys total, leaving them with a minority. So they can't win in 7 or more.

The code (in Python):

Code:
Ngames = 100000

# List recording the number of rounds it took the good guys to
# win in each game. Will be set to 42 if they lost
rounds = [] 
            
for game in range(Ngames):

    # Initial conditions:
    N_good = 10
    N_bad = 3

    N_players = N_good + N_bad

    # These are relative to the good guys
    win = False 
    lose = False

    N_rounds = 0

    while not (win or lose):
        # Select someone to be eliminated at random by choosing their
        # player number (ranges from 0 to 12 in round 1)
        elim = np.random.randint(0, N_players) 
                                              
        # If we eliminated a good guy (0 to 9 in round 1)
        if elim < N_good: 
            N_good -= 1
        # If we eliminated a bad guy (10 to 12 in round 1)
        else:
            N_bad -= 1

        # Now the mafia eliminates a guy:
        N_good -= 1

        N_players = N_good + N_bad
        N_rounds += 1
        #print(N_rounds, N_good, N_bad, N_players)

        # check for game-ending conditions:
        if N_bad == 0:
            win = True
            rounds.append(N_rounds)
        if N_bad > N_good:
            lose = True
            rounds.append(42)

counts = np.bincount(np.array(rounds))
probs = np.float64(counts)/Ngames
 

j-astron

Junior Member
Joined
Jan 10, 2018
Messages
181
I updated the above code to record both the number of rounds required for the good guys to win, and the number of rounds required for them to lose. That way we can get a sense of the probability mass function for each outcome, which is attached in a plot below.

This asymmetry makes me understand why the game mafia often has a good-guy player who is the "detective". He knows who the bad guys are. He can't reveal them, but he can try to steer the selection towards them. If he's too obvious about it, he gets eliminated.

mafia_pmf.jpg
 

Quick

Junior Member
Joined
Jan 19, 2018
Messages
90
So if Town is ~79% accurate with their reads, will they break even?
 
Top