Math free explanation of Monty Hall problem.
Key to understanding of why by switching initial selection leads to increased chance of win is to realize how balance is kept during the game.
Each time the decision has to be made two doors are closed. Situation is same for all variants of the game. 3 or 1000 doors.
- There is one closed door with prize and one closed empty door
That's the natural consequence of game rules. Please think about it until you are pretty sure you understand why.
- Role of the host is to keep this balance during the game by selecting second door WHICH WILL NOT BE OPENED
This is small shift in thinking about role of the host. Host has its own agenda. Just for a while think about the host as someone who must choose second door which will not be opened. Door(s) opened by the host are not interesting. We need to understand how and why second door was chosen.
- Every time guest selects empty door, host MUST select the door with prize.
- Every time guest selects door with prize, host selects one of the remaining empty doors.
All other empty doors are opened. Since there are more empty doors and only one door with prize:
- Guest selects empty door more often, consequently host MUST select door with prize more often.
- Guest selects door with prize rarely, consequently host rarely selects empty door.
And we are done!
If guest understands that his guess was likely to be incorrect, he can be sure to the same extent that the other closed door MUST contains the prize.
That is the simple mechanics of why changing the decision leads to winning more often.
