How this calculator works
A lobby with ten players and one impostor gives each player a one in ten chance of being the impostor in that game. The roles are drawn again next game, so the chance does not change from a previous result. Over a fixed number of games, the count of impostor turns follows a binomial distribution, which is built from the chance of each individual sequence of roles. The page sums the cases that match your question: exactly X, at least X or at most X.
Formula
P(X = k) = C(n, k) x p^k x (1 - p)^(n - k), where n is the number of games, k is the number of games in the role, p is the role chance per game, and C(n, k) is the binomial coefficient n! / (k! (n - k)!). For impostor, p = impostors / players. For crewmate, p = 1 - impostors / players. P(X >= k) is the sum of P(X = j) for j from k to n, and P(X <= k) is the sum for j from 0 to k. Mean = n x p. Variance = n x p x (1 - p). Standard deviation = sqrt(variance).
Worked example
Ten games with one impostor among ten players gives p = 0.1. The chance of being impostor exactly once is C(10, 1) x 0.1 x 0.9^9 = 0.3874, or 38.74 percent. The chance of at least once is 1 - 0.9^10 = 0.6513, or 65.13 percent. The expected count is 10 x 0.1 = 1 game, the variance is 0.9 and the standard deviation is about 0.949 games.
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Frequently asked questions
Does the calculator use the game version role limits?
It uses the lobby counts you choose. In the 2026 game versions, Among Us allows 4 to 15 players and 1 to 3 impostors in the standard Classic mode. Hide and Seek has no impostors, and some custom modes change the rules, so this model applies to a normal lobby with the counts you enter.
Why is the crewmate chance not one minus the impostor chance?
Because the question is about a count, not a single game. The chance of being crewmate exactly k times in n games is the chance of being impostor in the other n - k games, which is a different binomial tail. For a single game the two chances do add to one, but for a count they do not.
Are the games really independent?
The game draws roles again each round, and this calculator assumes the draw does not depend on previous rounds. If a host changes lobby size, impostor count or game mode between rounds, the chance changes and no single binomial model covers the session. Use one calculation per fixed setting.
What does the chart show?
It shows the probability of every possible number of games in the chosen role, from zero to the number of games played. The bars peak around the expected count and spread out as the number of games grows, which is why a run of bad luck is normal rather than proof of a rigged lobby.
Can I use this for another game with hidden roles?
Yes, if the game assigns a fixed number of special roles independently each round. Enter the number of special roles and the number of players, and interpret X as the count of rounds in that role. Games with teams, rotating roles or a role queue need a different model.