I’m so stupid that I had to create a simulation of Monty Hall’s problem to understand it.
And now I’m making it available so you can understand it too.
Trying to explain in my own words, when you choose 1 door, your door has a 1/3 chance of being with the car, and the other 2 doors combined have a 2/3 chance of being with the car.
When 1 of the doors you didn’t choose is revealed as not having the car, the doors you didn’t choose still have a 2/3 chance of having the car, but now you are sure which one doesn’t have the car, thus making the only door available, with a 2/3 chance of having the car, while yours still has a 1/3 chance.
Map Link: