Advertisement
Advertisement
GAME MATHS

Impostor Odds Calculator — Among Us, over a run of games

Works out your chance of drawing Impostor in one game, at least once across a session, and exactly how many games a fair run should take.

The Among Us base game supports 4 to 15 players, and the number of Impostors is a lobby setting the host chooses. Both are inputs here because both change with every lobby you join.
The second box is the threshold for the “at least this many times” figure below. Set it to 1 and you get the plain chance of drawing Impostor at all during the run.
The tool reports how many games it takes before your chance of having been Impostor at least once reaches this level. At 50 per cent it is the median wait; at 95 it is the point where going without would be genuinely unusual.
Chance of Impostor at least once in the run
 
Chance per game
Expected times
Chance of never
Games to reach your level
 
Tip: the draw has no memory. Ten crewmate games in a row does not make you “due”, and the eleventh game is the same coin as the first. That is what the chance-of-never figure is for.
Advertisement

This Impostor odds calculator answers the question every Among Us session eventually produces: how unlucky is it, really, to go a whole evening without being Impostor? It computes the chance for a single game from the lobby settings, then extends it across a run of games using the binomial distribution, and reports the figure people actually want — the probability of having drawn Impostor at least once.

It is arithmetic only. There is no strategy here, no advice on how to play either role, and nothing about influencing the draw, because the draw is not influenceable. Arb Digital built it as a probability page that happens to use a game as its subject, and it sits deliberately apart from the general probability calculator, which handles abstract events rather than this specific selection.

What This Calculator Does

It computes five things: the chance of drawing Impostor in a single game, the expected number of times across a run, the chance of never drawing it, the chance of drawing it at least a threshold number of times, and how many games it takes before your cumulative chance of having drawn it once reaches a confidence level you set. The bars show how the at-least-once figure climbs as a session goes on.

The model is that each game draws Impostors uniformly at random from the players present, and that games are independent of each other. Both assumptions match how the selection works: nothing about a previous game carries into the next one, which is exactly why streaks feel meaningful and are not.

How to Use It

  1. Set the lobby size and Impostor count from the settings the host actually used, not from the maximum the game allows.
  2. Enter the number of games in the run you are curious about — an evening, a session, a whole week.
  3. Set the at-least threshold to test a specific claim, such as somebody drawing Impostor three times in ten games.
  4. Adjust the confidence level to see how long a genuinely unusual drought would have to be.
  5. Compare the expected count with what actually happened before concluding anything is wrong with the game.

The Maths, Checked by Brute Force

For one game, the chance a specific player is chosen is the number of Impostors divided by the number of players. That is intuitive but worth proving, because the selection picks a group rather than choosing each player independently. In a 10-player lobby with 2 Impostors there are 45 possible pairs, and 9 of them contain you, giving 9 ÷ 45 = 0.2. The same enumeration at 15 players and 3 Impostors gives exactly 0.2 as well, and at 7 players with 2 Impostors it gives 2 ÷ 7. The group-selection answer and the simple ratio agree in every case, which is the check that justifies using the ratio.

Across a run of N independent games at probability p, the count follows the binomial distribution described in the Statistics LibreTexts treatment of the binomial distribution. The chance of never drawing Impostor is (1 − p) to the power N, so the chance of at least once is 1 minus that. The chance of exactly r times is the binomial coefficient times p to the r times (1 − p) to the N minus r, and the at-least figure sums those terms from the threshold upwards.

Worked example with the defaults. Ten players, two Impostors, so p = 0.2. Over ten games the chance of never being Impostor is 0.8 raised to the tenth, which is 0.10737, so the chance of at least once is 0.89263, or 89.26 per cent. The expected number of times is 10 × 0.2 = 2. The chance of at least three times is 0.32220 — a figure this page verified by summing the binomial terms and separately by enumerating all 1,024 possible sequences of ten games, which agree to eight decimal places. The number of games for a 50 per cent chance is log(0.5) ÷ log(0.8) = 3.11, so four games.

Advertisement

Why the Numbers Feel Wrong

Two things make people distrust the draw. The first is that streaks are far more likely than intuition suggests. At p = 0.2, going eight games without Impostor has a probability of 0.8 to the eighth, about 16.8 per cent — so roughly one session in six will contain that drought, and in a group of six friends somebody is having one almost every evening. It feels rigged because somebody is always the unlucky one, and it is never the same person twice.

The second is the gambler’s fallacy. After nine crewmate games the tenth is still exactly p. The draw has no memory of what came before, so there is no accumulated debt to be paid out. The chance-of-never figure describes the run as a whole from the start; it does not describe what happens next given what has already happened. Our coin streak probability calculator works the same effect in the abstract.

There is also a counting bias. People remember the games they were Impostor far better than the ones they were not, both for themselves and for the friend who “always gets it”. Compare the expected count with the actual before deciding anything is off, and remember that in a group of ten players, seeing one player draw Impostor three times in ten games is not surprising at all — it happens about 32 per cent of the time for any given player.

What Changes These Numbers

Lobby settings, first and foremost. The Impostor count is chosen by the host, and Innersloth states the game is played with 4 to 15 players, so p can range widely between one lobby and the next. A 15-player lobby with one Impostor gives 6.7 per cent a game; a 5-player lobby with two gives 40 per cent. If a session mixed lobby sizes, no single p describes it, and the honest approach is to run the calculation separately for each configuration.

Game version matters too. Among Us has added roles over time, and the exact set of roles and settings available depends on the release and the platform you are on. This calculator models only the Impostor-versus-Crewmate draw from lobby size and Impostor count, which is the part that has stayed structurally the same. If a mod, a custom server or a future update changes how selection works, the arithmetic here stops describing it, so check the settings screen rather than assuming.

Finally, players leaving and joining between games changes the lobby size, and therefore p, without anybody noticing. A session that started at twelve players and drifted to seven has been running at meaningfully different odds throughout. Our binomial distribution calculator handles a run at a fixed probability in the general case, and the expected value calculator is useful when you want to weigh outcomes rather than count them.

Want free tools that verify their own maths?

Arb Digital builds calculators that check the formula against a brute-force enumeration and say so on the page. Browse the library, or tell us what your community keeps arguing about.

Browse Free Tools Talk To Arb Digital

Common Mistakes to Avoid

  • Thinking you are due. Games are independent. Nine crewmate rounds change the tenth by nothing at all.
  • Using the maximum lobby size instead of the actual one. Players drift in and out, and the probability moves with them.
  • Confusing the expected count with a guarantee. An expectation of two means two on average across many runs, not two in every run.
  • Treating a streak as evidence. Long droughts are common at these probabilities, and in any group somebody is always having one.
  • Averaging across mixed lobbies. If the settings changed during the session, a single probability does not describe it. Calculate each configuration separately.

Related Free Tools From Arb Digital

Work a run at a fixed probability with the binomial distribution calculator, handle abstract events with the probability calculator, explore streaks with the coin streak probability calculator, weigh outcomes with the expected value calculator, translate between odds formats with the odds probability converter, and handle conditional questions with the conditional probability calculator. Everything else is in the free online tools hub.

Frequently Asked Questions

What are the odds of being Impostor in Among Us?

In a single game it is the number of Impostors divided by the number of players in the lobby. Ten players with two Impostors gives one in five, or 20 per cent. Because both figures are lobby settings chosen by the host, there is no single answer that applies to every game.

Does the group selection change the simple ratio?

No, and this page checked it by enumeration. In a ten-player lobby with two Impostors there are 45 possible pairs and nine of them include you, which is exactly 0.2. The same holds at fifteen players with three Impostors and at every other configuration tested.

How likely is it to go a whole session without being Impostor?

More likely than it feels. At a 20 per cent chance per game, eight games without Impostor happens about 16.8 per cent of the time, so roughly one session in six. In a group of friends somebody is having that run almost every evening, which is why the draw feels rigged when it is not.

Am I more likely to be Impostor after a long streak of crewmate games?

No. Each game is an independent draw with no memory of the last one, so the chance is unchanged whatever came before. The cumulative figures on this page describe a run from its beginning, not what happens next given what has already happened.

How many games until I am probably Impostor at least once?

Take the logarithm of one minus your confidence level, divide by the logarithm of one minus the per-game chance, and round up. At 20 per cent per game, a 50 per cent chance takes four games and a 95 per cent chance takes fourteen.

Which version of the game does this model?

The standard Impostor and Crewmate draw from lobby size and Impostor count, which is the structure the base game has used across releases and which supports 4 to 15 players. Added roles and platform differences do not change that draw, but mods, custom servers and future updates could, so check the settings screen you are actually playing on.

Can I improve my chances of being picked?

No. The selection is a random draw from the players in the lobby, and nothing about how you play, when you join or which colour you pick affects it. The only thing that changes your odds is the lobby settings, which is why they are the only inputs here.

Why does my friend seem to get Impostor constantly?

Partly memory and partly arithmetic. People recall Impostor games far more vividly than crewmate ones, and drawing it three times in ten games at a 20 per cent chance happens about 32 per cent of the time for any given player. Compare the expected count with the actual before concluding anything is unusual.

This page is a probability calculation about a game and nothing more. It offers no strategy advice, describes no way to influence the draw, and models only the standard Impostor selection from lobby size and Impostor count. Lobby settings, game updates, platform differences and mods can all change how a lobby behaves, so read the settings screen rather than assuming these numbers apply.

Advertisement
Advertisement

Take it further