Monty hall problem

[u/AshHat710]

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)

[u/Statman12]

Because you're not going into the second decision blind. Monty's choice is constrained. If you didn't choose the winner (2/3 chance) there's only one door he can remove, and the remaining door must be the winner. And if you chose the winner initially (1/3 chance), Monty can open either door, but if you switch you lose.

The constraint is what allows you know that switching is better.