I'm trying to understand the Monty Hall problem. When the contestant switches the door the probability of them winning a car is higher than if they continued with their first choice. Why would the probability change from 1/3 go up to 2/3 if both doors had the same chances of being opened?
Update: There are many posts about this subject in SO but none of then answer my question. Let me re define my question.
I know that there are many models out there that shows that if you change the doors you have 2/3 of probability (higher than 1/3). After Graham Kemp answer I will update a scenario that I have in my mind.
You choose door1 (1/3) and the host opens the door 2 (the goat). By the current solution the door 3 has 2/3 of probability to be the one with the car. Now let's think a little change. After the host open the door 2 he will give you a coin with numbers 1 and 3 and you should torn it. Now is not a choice, you should stay with the number you get in the coin. In this new scenario based on the previous model the door 3 has 2/3 of success probability than door 1. With clearly makes no sense. In this new scenario why door 3 is not 1/2 of probability? I am pretty sure that a model with 10M attempts will show 50% for each door. Is it correct?
You select a door at random, knowing nothing about the placement of the car, other than that is presumably behind any particular door with equal probability (1/3).
Monty opens one of the other two doors, knowing the location of the car, he always chooses to reveal a goat.
You are asked if you want to switch to the remaining door.
If the car was behind your door, you cannot win if you switch. It was there with probability 1/3.
If the car was not behind your door, you surely win if you switch. It wasn't there with probability 2/3.
Therefore the probability that you win if you switch is 1 minus the probability that the car was behind the door which you selected. IE: 2/3