# Introduction

I'm sure many of you have seen streams or played games yourselves where people suddenly start "speed running" by voting people out without any substantial evidence, basically at random, continuing to do so until the game ends. While this is not how the game is meant to be played, it got me thinking, who would such a scenario favor? Would this lead on average to more impostor or more crewmate victories? Let's find out!

# Probability model

We are going to look at a standard game of 10 people, with 2 impostors and 8 crewmates. Starting from the beginning of the game, people keep calling meetings, where each time one person (who they agree on) is voted out. Since there are no kills and no tasks are completed, there are no clues whatsoever, so people are voting entirely randomly. This is modelled by picking one player at random from the people still alive each meeting, and killing them. Our goal is to find the probability of the impostors winning the game if it proceeds as described.

The way we combinatorically handle this problem is to count all possible voting orders, and then count those that result in an impostor win. Their quotient will be the probability of the impostors winning. The problem we face immediately is that the length of games varies depending on whether people voted out are crewmates or impostors—a game can be as short as 2 votes (if both impostors are are found immediately) or as long as 8 votes (if 1 impostor and 1 crewmate is left alive in the end). This is a problem, because these games are not the same in terms of probability: a certain vote sequence of 8 votes is much less probable than any sequence of 2 votes, solely on account of it being longer ("more votes have to happen exactly that way"). To combat this, we take into account all possible orderings of the 10 players, which is as if the game would go on until there is only one person who then votes out themselves. The first player to be voted out can be selected from all 10 players, the second from the remaining 9, and so on. This is in total 10*9*8*7*6*5*4*3*2*1 = **10! = 3,628,800 possible orderings**.

*Note: these ordering also distinguish between individual players, not just whether or not they are impostors.*

These orderings are now all the same probability, so they can be used for calculation of the final win rate as described above. We just have to find those orderings which mean an impostor victory. Keep in mind, we are only interested in the first part of the 10-long sequences which actually describe a game. What happens afterwards is not relevant. (Imagine after crewmates win, they start voting among themselves: it will not change the fact that they won.)

# Impostor wins

In a standard 10-people, 2-impostor game of Among us, impostors can win on two distinct occasions (and only then):

- after 6 deaths, when there are 4 players left, if both impostors are alive, or
- after 8 deaths, when there are 2 players left, if one impostor is alive.

*Note: impostors cannot win before 6 deaths (votes, in this case), and they also cannot win after exactly 7 deaths, since for that they would need 2 impostors alive, in which case they would already have won in the previous round.*

For the first way of impostor wins, neither impostor can be voted out in the first 6 votes. In other words, both impostors have to be voted in the last 4 of the 10 (theoretical) votes. There are 4C2 = 6 possible place combinations for 2 impostors in the last 4 votes (you can check this for yourself), where nCk denotes the *n choose k."* This, however, does not yet take into account all the possible inner orderings of both impostors and crewmates. There are 8! = 40,320 ways to order the crewmates among themselves, and also 2! = 2 ways to order the impostors among themselves (either one comes first or the other). This gives a total number for the orderings: 4C2*2!*8! = 483,840. Thus, the probability of impostors winning by voting out 6 innocent people in a row is 4C2*2!*8!/10! = 2/15 = 13.33%.

The second way is a little trickier. Here we have to make sure exactly one impostor dies in the first 6 votes (if none die, we would get the previous case; if both die, the crewmates win), AND the second impostor does not die in the following two votes, but only in the last two (theoretical) votes. This, using the logic before, means 6C1*2C1*2!*8! = 967,680 possible orderings, and by division by 10! a final probability of 4/15 = 26.67%.

This means impostors actually have twice the chance to win with only one of them remaining than both of them staying alive, and **the total probability of an impostor win after voting randomly is exactly 40%.**

# Interpreting the result

The math shows that going purely with random votes, crewmates have a considerable, 3:2 edge over impostors. Of course, this is not the way the game is meant to be played—impostors can also kill crewmates or win by sabotage, and crewmates can win by tasks and provide useful info on who the impostor might be, so that votes are not entirely random.

It still, nevertheless, puts the whole game into perspective: should crewmates fail to maintain a winrate close to 60% in the long run, theoretically they would improve by following this strategy. Even more so, since as long as they vote in unison, impostors cannot prevent it with their votes, only by calling sabotages to prevent the emergency meetings themselves. In a way it also sheds light on the fact then the game's winrate *should* be at least 60% for the crewmates, otherwise this would be the mathematically superior strategy, which, admittedly, defeats the whole purpose of the game and is ultimately no different from coin flipping.

# Varying the number of players

An interesting point is to examine how the impostor win rate changes if there are fewer players (either impostors or crewmates). These cases may be relevant for situations where either there are fewer than 10 players playing, or when the random voting begins after a number of people have already been killed. In case there is only one impostor, the picture is simplified significantly—the impostor can only win if all ejected players are crewmates, or, in other words, if they are among the last two players. This probability will come out to be 2/N, where N is the number of players.

The exact probabilities for the impostor win for different impostor and crewmate numbers are shown in the table below:

Impostors / crewmates | 1 | 2 |
---|---|---|

2 | 66.67% | – |

3 | 50.00% | 80.00% |

4 | 40.00% | 66.67% |

5 | 33.33% | 57.14% |

6 | 28.57% | 50.00% |

7 | 25.00% | 44.44% |

8 | 22.22% | 40.00% |

9 | 20.00% | – |

As expected, the impostor chances increase by decreasing the crewmate numbers or increasing the impostor number. For example, with an 8 crewmate/2 impostor setup, if one crewmate is killed, and the "random voting" begins after this (say, nobody is suspicious), the impostors' chance to win goes up to 44.44% from 40%. If they manage to do a double kill instead, they are already at a 50% chance.

An interesting observation from the table is that for each case, **the impostor win probability is exactly twice the ratio of impostors among all players: 2*I/N**, where I is the number of impostors and N is the total number of players, which provides an easy way to estimate the success rate of this strategy at any point in a game.

Source: https://www.reddit.com/r/AmongUs/comments/kupv9w/impostor_win_probability_if_every_round_soneome/