50:50 chance on the other 4. 1 in 3 chance out of us 3. I'd say its better to go for a 2:1 split here
I dont know what the rules one diting here are but assuming theyre the same as mafia ill repost to clarify. I mean we shud take 2 of myself/weeman/tart and 1 other imo as the chances are there's only one spy in the original 3 and 2 out of the other 4 so on probabilities its the best way
I'm going to object based on the fact that I know I'm not a spy and so one of the other two must be. But I realise that from your perspective we're all pretty much equally as likely to be the spy.
There's an 89% chance that there is a spy in this group and I think we can do better than that (assuming Blunder/Tart are 1/3 each).
Is there any way we can gurantee either 1) a passed mission or 2) catch a guaranteed spy if the missions fails? I'm thinking there might be but I'm shit at maths.
Just going to drop in and say that there's a reason why you can say no to groups proposed by the leaders.
One fail. Mission fails. 2-1 to spies now. Weeman Tart Number 11 Blake Ritchie Blunder Teacups 4 required for the fourth mission. NOTE: Two fails needed for the spies to fail this mission. Blake leader.
I would argue to go again... but its based off some logic that makes sense to me but may not to the rest of you. One of myself, weeman and Tart are spies we know. Weeman picked the first group and picked me. If he's a spy then he wouldnt pick me beczause its too much of a risk (I know its wifom here but getting 2/3 of your spies on the first mission is crazy and uneccesary) If he's resistance then he's picked me at random so it's a 50% chance of getting a hit... but its first dependent on weeman being innoncent