Among Us Impostor Odds Calculator (chance to be impostor & suspect odds)

Understanding the Among Us impostor odds calculator — concept and principle

In the social deduction game Among Us, a few players are secretly impostors. This calculator works out the basic odds: your chance of being an impostor, the chance that any other player is one once you know you are a crewmate, the chance that your suspects include an impostor, and how often you can expect the impostor role over a series of games.

How it is calculated

P(you are impostor) = k ÷ n

P(a given other player is impostor | you are crewmate) = k ÷ (n − 1)

P(at least one impostor among m suspects) = 1 − C(n−1−k, m) ÷ C(n−1, m)

P(all m suspects are impostors) = C(k, m) ÷ C(n−1, m)

P(impostor at least once in G games) = 1 − (1 − k/n)^G

Expected impostor games = G × k ÷ n; average games per impostor turn = n ÷ k

• n players including you, k impostors, m suspects, G games; C(a, b) is the number of combinations.

The principle

If roles are dealt at random, every player has the same chance of being an impostor. Once you see you are a crewmate, you have gained information: the impostors must be among the other n − 1 players, so each of them becomes slightly more likely. Choosing a group of suspects is like drawing without replacement, which follows the hypergeometric distribution: the easiest way is to count the ways your picks could all be innocent and subtract from 1. Separate games are independent, so the chance of never being impostor is (1 − k/n) multiplied G times.

Worked example

Default inputs: 10 players, 2 impostors, 2 suspects, 10 games.

1. You are impostor: 2 ÷ 10 = 20%; crewmate 80%.
2. A given other player: 2 ÷ 9 = 22.2%.
3. At least one impostor in your 2 picks: 1 − C(7,2) ÷ C(9,2) = 1 − 21 ÷ 36 = 41.7%.
4. Both picks impostors: C(2,2) ÷ C(9,2) = 1 ÷ 36 = 2.8%.
5. Impostor at least once in 10 games: 1 − 0.8¹⁰ = 89.3%.
6. Expected impostor games = 10 × 0.2 = 2; average 5 games per impostor turn.

How to read the result

These are starting odds before any evidence. Votes, alibis, task progress and who reported a body all update the odds during play, so use the numbers as a baseline. Even with an 80% chance per game of being a crewmate, ten games without the impostor role would happen only about 11% of the time, so long streaks are not unusual but not common either.

Common mistakes

• Thinking the chances of each player are independent; if one suspect is cleared, the others become more likely.
• Expecting impostor turns to arrive evenly; random streaks are normal.
• Using k ÷ n for other players after you already know you are a crewmate.

Frequently asked questions

What are my odds of being impostor with 3 impostors in 15 players?

3 ÷ 15 = 20%.

Why does knowing I am a crewmate change the odds for others?

The impostors must be among the remaining players, so each has a slightly higher chance.

How many games until I am likely to be impostor?

On average n ÷ k games, but chance varies widely game to game.

Sources: Standard combinatorial probability (equally likely outcomes; hypergeometric and binomial distributions)