[{"id":"_meta","name":"x squared = lie","author":"BigBrainBen10","bootlegger":["There will ALWAYS be 1 to 3 mathmatician(s). One of the extras is poisoned. if there are 2, split them into drunkness + special and poison + special. if there are 3, add a regular mathmatician to the case of 2 mathmaticians. either way, tell every mathmatician what they detect at the start of the game, even if there's 1.","For the Sweetheart's ability, replace \"drunk\" with \"poison\".","The Acrobat's effect is instead: \"at the start of the game, three players are shown to you; one of them is THE drunk. Each* night, choose three players; you will know how many of them are drunk or THE drunk.\""]},{"id":"empath"},{"id":"mathematician"},{"id":"fortuneteller"},{"id":"innkeeper"},{"id":"monk"},{"id":"acrobat"},{"id":"courtier"},{"id":"philosopher"},{"id":"soldier"},{"id":"amnesiac"},{"id":"fool"},{"id":"virgin"},{"id":"magician"},{"id":"drunk"},{"id":"sweetheart"},{"id":"heretic"},{"id":"puzzlemaster"},{"id":"poisoner"},{"id":"cerenovus"},{"id":"assassin"},{"id":"xaan"},{"id":"scarletwoman"},{"id":"nodashii"},{"id":"imp"},{"id":"vortox"},{"id":"lleech"},{"id":"bootlegger"}]