A three-door game.

You are on a game show, and you are given a choice of three doors.  Behind exactly one of the doors, is a brand new car.  There is nothing behind the other two doors.  The game works according to certain rules.  First, you make a choice:  you are asked to PICK a door, without having it opened.  Second, soemthing is revealed to you:  the game show host opens one of the two doors which does NOT hide the car.  Third, you are allowed to rethink your choice.  You can stay with your original choice (if the host has not opened that door), or you can switch to one of the unopened doors.  Once you've made your final choice, the door you last picked is opened-and you either win a car or walk away with nothing.  The question is this: suppose that you were playing this game (or an equivalent simulation) over and over again.  Is any advantage gained by always picking any one door initially, and always switching your choice in the second phase?  Or do your odds remain the same if you always stay with whatever your first choice was?

Yay, the Monty Hall problem, one of the most trolled and beat-to-death math topics right after 0.9999...==1

Both logic and empirical studies prove that switching gives you better chances.

(in the original Monty Hall problem, the host will reveal one of the doors you HAVEN'T picked to be empty - is the OP different on purpose?)