I'm fucking losing it, lads.
I'm aware that it is a 33% and 66% for the <total> probability when either staying or switching in the Monty Hall problem. But my main issue is that I don't understand why 3 doors are even being taken into consideration to calculate the probability of staying or switching. Monty will always choose a door with a goat that isn't the one you've picked. To me, that means the deciding choice will always boil down between two doors (Goat or Car). Picking any door prior to this seems more like statistical fluff, because you would never have two doors with two goats when making the winning choice. If I assign 'staying' the variable A and 'switching' B, then mapping those randomy to either door A or B, then I always get a 50% probability. So can anyone explain to me why we even take the 3 doors into consideration to come to the 2/3 1/3 conclusion?
I'm aware that it is a 33% and 66% for the <total> probability when either staying or switching in the Monty Hall problem. But my main issue is that I don't understand why 3 doors are even being taken into consideration to calculate the probability of staying or switching. Monty will always choose a door with a goat that isn't the one you've picked. To me, that means the deciding choice will always boil down between two doors (Goat or Car). Picking any door prior to this seems more like statistical fluff, because you would never have two doors with two goats when making the winning choice. If I assign 'staying' the variable A and 'switching' B, then mapping those randomy to either door A or B, then I always get a 50% probability. So can anyone explain to me why we even take the 3 doors into consideration to come to the 2/3 1/3 conclusion?
