I came across a video (not sure if I'm allowed to post it because it had some kid was swearing due to the nature of the game's lobbies) where the guy claimed that you can apply Monty Hall to Among Us.
So basically he picks some random guy in his head to be imposter, lets the round take place and then he switches to a different guy - also in his head - after a body is found. He claims that this increases the probability of him finding the imposters by pure statistics. Obviously this works better with more people on board with him.
So I dug into it and I tried to apply Monty Hall to guess the imposter but to the game I couldn't.
The main difference between the 100 doors Monty Hall problem and Among Us is that in the 100 doors game, the game master will eliminate 98 doors for you and you'll be left with one door plus the one you initially chose. Choosing to switch increases your chances of picking the correct door.
However in Among Us, there is no way you're going to be left with one crewmember and an imposter. You'll usually have 1 or 2 dead bodies which is NOT equivalent to the game master eliminating all the n-2 doors.
Is there a way to apply Monty Hall recursively while eliminating n-x doors rather than n-2?
EDIT: For those who don't know how Among Us is played. It's usually played with the following settings. 2 imposters and 8 crewmembers on a map. The map has various tasks for the crewmembers to do. If they finish all tasks they win. If the imposters kill them to the point of where their number is greater to or equal to the number of remaining crewmembers, the imposters win. Each round the imposters usually kill 1-2 crewmembers (it could be more if the crewmembers are that unlucky/bad). Once a body is discovered by a crewmember, the crewmember will report it and all players (imposters and crewmembers) will be given a chance to discuss their whereabouts and people will start accusing each other with one objective, to find the imposters and vote on ejecting them. The players get to eject only ONE person per round. It could be an imposter or it could be a crewmember. It's only revealed after the game is concluded (or it can be revealed right after the ejection if they create it with the alternate setting).
Let us look at a far simplified game with two other crewmates and one imposter. (four players including ourselves with us as player $4$)
Let us assume that we are not an imposter and are not the first victim. Without loss of generality, let us have picked player $1$.
The following six scenarios are equally plausible:
It should be plain to see that your initial guess is only right $\frac{1}{3}$ of the time. If the person you initially guessed happens to have been the first murder victim... then obviously you should switch. Not switching guarantees you picked the wrong person and switching only gives you $50$-$50$ odds. Now... given that your initial pick was not the first murder victim, we see that the first four cases are equally likely and not switching versus switching each give the same chance of winning.
The whole point to take away from this is that in the original Monty Hall problem the host would never willingly reveal the door you initially selected before giving you the chance to switch. The "desire to switch" rather than staying with your original choice which would purportedly increase your chances of winning all effectively got absorbed by the last cases listed where our initially selected player was first killed.
Long story short, no this is not the Monty Hall problem and in this scenario, without additional information (player behaviors, locations, etc...) there is not enough information to justify switching unless the player is dead in which case it is obvious that you need to switch. This strategy does not provide any benefit whatsoever.