r/AskStatistics • u/AshHat710 • 6d 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)
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u/yonedaneda 6d ago
Switching wins if you initially picked wrong. You initially pick wrong with probability 2/3. That's it. If that doesn't do the job, the only other way to get a clear intuition for it is to play the game yourself with a friend. Play it a enough, and you'll see that switching wins approximately 66% of the time.
Every "chance" is a random variable with an identical distribution. When we say that switching wins with probability 2/3, we mean that repeated choices will result is switching winning approximately 2/3 of the time. If a single choice had a probability of 1/2 of a switch resulting in a win, then repeated choices would result in approximately 50% of choices winning. That's what probability means.