How this instrument works
Being the impostor in Among Us is a per-game coin flip weighted by how many impostor slots exist relative to the lobby size — with 2 impostors in an 8-player game, your chance of being one of them in any single game is 2/8, or 25%. Across many games, whether you land the impostor role a specific number of times follows the binomial distribution, the standard probability model for counting successes across a fixed number of independent trials with the same success probability each time.
The binomial formula multiplies three things: the number of ways to choose which games you were impostor in (a combinations count), the probability of being impostor raised to the power of how many times that happened, and the probability of NOT being impostor raised to the power of the remaining games. Games where the impostor role is assigned independently each round — no memory of past assignments — are exactly the scenario this model was built for.
This matters for a common gut-check: after being the impostor five games in a row, many players assume the game is 'due' to assign it to someone else, or suspect something's rigged. The binomial distribution shows streaks like that are not actually unlikely — with independent per-game odds, unusual-feeling streaks happen more often than intuition suggests, especially across dozens of games.
- Enter Games played — the total number of matches you're analyzing.
- Enter Players per game and Impostors per game to set your per-round probability of being an impostor.
- Enter Times you want to be the impostor — the exact count you're checking the probability of.
- Read Probability (%) for the chance of landing exactly that count of impostor rounds across all games played.
- Lower Times as impostor to 0 to see the probability of never once being assigned the role.
Worked example — impostor 3 times in 10 games
A standard 8-player lobby with 2 impostors per game gives p = 2/8 = 0.25. Over 10 games, the probability of being impostor in exactly 3 of them is C(10,3) x 0.25^3 x 0.75^7 = 120 x 0.015625 x 0.13348 = 0.2503, so the instrument reports a probability of 25.03%.
That's higher than many players expect for landing a specific count exactly — the expected number of impostor games here is 10 x 0.25 = 2.5, so getting exactly 3 is close to the most likely single outcome, not an unusual one, even though it feels like a notable run in the moment.
Questions
How is the probability of being the impostor a certain number of times calculated?
It uses the binomial probability formula: P(k) = C(n,k) x p^k x (1-p)^(n-k), where n is the total games played, k is the exact number of times you want to have been impostor, and p is your per-game probability (impostors per game divided by players per game). C(n,k) counts how many different orderings of games could produce exactly k impostor rounds. This assumes the impostor role is assigned independently and with equal probability to every player each game, which matches how Among Us actually assigns roles.
Why does being impostor several games in a row feel more common than the math suggests it should be?
It doesn't happen more often than the math says — streaks just feel more notable when you experience them than when you don't. With independent per-game probability, runs of 2-3 impostor games in a row are expected to occur reasonably often over dozens of games; a coin doesn't 'remember' it landed heads last time, and neither does the impostor assignment. If a streak feels unusually long compared to what this calculator predicts as a very low-probability count, it's worth checking your inputs (accurate lobby size and impostor count) rather than assuming bias.
What's the difference between 'exactly k times' and 'at least k times'?
This instrument computes the probability of exactly the count you enter — being impostor in precisely 3 of 10 games, for instance, not 3 or more. 'At least k times' would be the sum of the exact probabilities for k, k+1, k+2, all the way up to n games, which is always a larger number than the single-value probability shown here. If you want a cumulative 'at least' figure, sum this calculator's output across each value of k from your target up to the total games played.
How many impostors and players does Among Us actually support?
A standard Among Us lobby holds 4-15 players, with the host able to set 1, 2, or 3 impostors depending on lobby size (larger lobbies typically support more impostors). This instrument accepts the same ranges, so you can model any lobby configuration you actually play, not just the game's own default of 1 impostor in a 10-player lobby.
Does this account for skill, voting patterns, or anything besides random assignment?
No — this is a pure statistical model of role assignment probability, assuming every player has an equal, independent chance of being made impostor each game. It has nothing to do with how good you are at deceiving other players once you're impostor, how often you're voted out, or any other gameplay skill factor. It only answers: given the lobby size and impostor count, how likely is a specific count of impostor assignments across a run of games.