Game One: You are given a choice of three doors. You pick number one. The host opens one of the other two doors, having been given instructions that, if you pick the car, the host is to open one of the other doors, and if you pick a goat, the host opens the other door with a goat. Stalemate. It is a predetermined outcome.
Game Two: The prior game’s outcome stands. The new choice you have is do you keep door number one, or do you switch?
How do you have a 2/3 chance of winning if you switch?
In: 107
In addition to what others have said, it becomes easier to think of this as a single game if you imagine the Monty Hall problem with 100 doors instead of just three.
You choose Door 1.
Host opens all doors except 1 and 55.
The likelihood that you chose the one correct door out of 100 is unlikely. This likelihood doesn’t change even if a bunch of dud doors get opened. When you chose door number one, it will forever be a 1 in 100 choice. If you swap to door 55, it becomes a 1 in 2 choice.
The higher the number of doors, the more obvious this becomes. You can have one billion doors but the host will still open all but your first choice and one other one. It would be nearly impossible to pick the correct number out of a billion, but far, far more possible if the host narrows things down to just two numbers.
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