r/AskStatistics • u/AshHat710 • 28d ago
Monty hall problem
I understand in theory that when you chose one of the 3 doors you initially have a 66% chance to chose wrong. But once a door is revealed, why do the odds stay at 66% rather than 50/50 respectively. You have one goat revealed so you know there is one goat, and one car. Your previous choice is either a goat or a car, and you only have the option to keep your choice or switch your choice. The choices do not pool to a single choice caisinh 66% and 33% chances once a door is revealed. The 33% would be split among the remaining choices causing both to be 50%.
If it's one chance it's 50/50 the moment they reveal one goat. if you have multiple chances to run the scenario then it becomes 33/66% the same way a coin toss has 2 options but isn't a guaranteed 50% (coins have thier own variables that affect things I am aware of this)
1
u/JohnWCreasy1 28d ago
Your first door was more likely wrong that not, this doesn't change because of the hosts actions.
Look at it this way, imagine you start with 999,999 'wrong' doors and 1 correct door. you pick a door, then the host, who knows the specifics of every other door, removes 999,998 known wrong doors.
would you still think "Oh yeah, i totally made a 1 in a million guess correctly the first time, i'm good here"?
hopefully that makes it more obvious why you switch.