Among Us Impostor Odds Calculator
Enter the lobby size, number of impostors, and how many games you plan to play to see your probability of being assigned the impostor role. Switch between exact, at-least, and at-most modes to answer questions like "what are the odds I get impostor at least twice in five games?" The results update instantly and include a full game-by-game breakdown chart.
How the impostor assignment works in Among Us
At the start of each game, Among Us randomly assigns the impostor role to the set number of players in the lobby. Every player has an equal chance of being selected. If a lobby has 10 players and 2 impostors, each player starts every game with a 2/10 = 20% chance of getting the impostor card. Importantly, the assignment is completely random and independent between games: getting impostor in one game does not make it more or less likely in the next. This independence is exactly what lets us use the binomial probability distribution to model your odds across a session.
Understanding the binomial probability formula
When you ask "what are the odds I am impostor exactly twice in five games?" you are asking a binomial probability question. The formula is P(X = k) = C(n, k) x p^k x (1-p)^(n-k), where n is the total number of games, k is the number of times you want to be impostor, p is the probability in a single game, and C(n, k) is the number of ways k successes can be arranged among n trials (the binomial coefficient). For the at-least mode the calculator sums all individual probabilities from k to n using the complement: P(X >= k) = 1 - P(X <= k-1). For the at-most mode it sums from 0 to k directly.
Multi-game sessions: how quickly do odds stack up?
With a 20% base chance (2 impostors, 10 players), your probability of being impostor at least once climbs quickly with session length: 20% after 1 game, 36% after 2, 49% after 3, 59% after 4, and 67% after 5 games. By 10 games it reaches 89%. This follows the complement pattern: P(at least once in n games) = 1 - (0.80)^n. The expected number of impostor games is simply n x p, so in a 10-game session with a 20% rate you should expect about 2 games as impostor on average. However, variance means the actual count will often differ by 1 or 2 games from that expectation.
Effect of lobby size and impostor count on your odds
Lobby size and impostor count are the two levers that control your base probability. Smaller lobbies dramatically increase your odds: in a 4-player lobby with 1 impostor each player has a 25% per-game chance, versus only 6.7% in a 15-player lobby. Adding impostors scales the odds proportionally. A 3-impostor game with 10 players gives each player a 30% per-game chance, 3 times the odds of a 1-impostor game at the same lobby size. The reference table above shows the full matrix of common configurations at a glance.
Quick-reference: impostor odds by lobby setup
| Players | 1 Impostor | 2 Impostors | 3 Impostors |
|---|---|---|---|
| 4 | 25.0% | 50.0% | 75.0% |
| 5 | 20.0% | 40.0% | 60.0% |
| 6 | 16.7% | 33.3% | 50.0% |
| 7 | 14.3% | 28.6% | 42.9% |
| 8 | 12.5% | 25.0% | 37.5% |
| 9 | 11.1% | 22.2% | 33.3% |
| 10 | 10.0% | 20.0% | 30.0% |
| 12 | 8.3% | 16.7% | 25.0% |
| 15 | 6.7% | 13.3% | 20.0% |
Single-game probability of being assigned impostor for common lobby configurations.
Frequently asked questions
What is the formula for impostor probability in Among Us?
For a single game, the probability is simply: P(impostor) = number of impostors / total players. For example, 2 impostors among 10 players gives a 2/10 = 20% chance. For multiple games, the binomial distribution applies: P(X = k) = C(n, k) x p^k x (1-p)^(n-k), where n is games played and k is the target impostor count.
How many games do I need to play to have a 50% chance of being impostor at least once?
It depends on the lobby setup. With 10 players and 2 impostors (20% per game), you need at least 4 games: P(at least once in 4 games) = 1 - 0.80^4 = 59%. For a lobby with only a 10% per-game chance, you need about 7 games: 1 - 0.90^7 = 52%. The general formula is n = ceil(log(0.5) / log(1-p)).
Does being impostor in one game affect my chances in the next?
No. Each game assigns roles independently. The random number generator does not keep a running tally of how often you have been impostor. This is a common misconception: if you have been crewmate five games in a row, your odds in game six are still exactly p = impostors / players, the same as always.
What does "at least X times" mode calculate?
It calculates the probability that you are chosen as impostor in X or more of your games. Mathematically it uses the complement: P(X >= k) = 1 - P(X <= k-1), where P(X <= k-1) is the sum of exact probabilities from 0 to k-1. "At least once" is the most popular version: it shows the chance you get impostor in any of your games.
Why are my odds higher in a smaller lobby?
Because the same number of impostor slots is shared among fewer players. If there is 1 impostor among 5 players, you have a 1-in-5 (20%) chance. If the lobby grows to 10 players with the same 1 impostor, your chance falls to 1-in-10 (10%). Adding more impostors scales the probability up, but the player count is the divisor.
Can I use this calculator if the lobby has a different player count each game?
This calculator assumes the same lobby size every game, because the binomial model requires a fixed per-trial probability. If your lobby size or impostor count changes game to game, you would need to treat each game as a separate Bernoulli trial and multiply individual probabilities, which is a more complex calculation.