You stand before three closed doors. The doors are evenly spaced and appear identical, aside from being numbered from 1 to 3. One of the doors conceals a car, while each of the other two doors conceals a goat. The host of this game, Monty Hall, asks you to select a door. If you select the car door, you get to keep the car. You select Door 1, hoping to win the car. But wait. Hall opens Door 2 to reveal a goat. Hall, who knows where the car is, always reveals a goat. (Also assume that, when a contestant chooses the car door, Hall chooses which of the two goat doors to open with equal probability.) You chose Door 1, but are now given the option to switch your guess to Door 3. What’s the probability of winning if you switch?
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What I never understand about this problem is that there was only ever 2 choices, a 50/50 chance. The presenter knows which 2 doors are wrong and takes 1 away after the contestant chooses. One door was always going to be irrelevant to the contestant , he just doesn’t know which one. I’m not not arguing with the logic – I just think the original premise is a bit deceptive.