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Originally Posted by Lenny2
Hmmmm this is an interesting explanation
http://en.wikipedia.org/wiki/Monty_Hall_problem
It's very counter-intuitive, but if looked at the right way can be understood.
The advantage to switching comes from the fact that the game show host doesn't open a door at random, he ALWAYS opens a non-winning door.
That is, if the player picks a door with nothing the game host picks the other door with nothing.
And if the player picks the door with a car, the game host randomly picks either of the two doors with nothing
This means there are three possible scenarios, each with equal probability (1/3):
The player picks empty door number 1. The game host picks the other empty door. Switching will win the car.
The player picks empty door number 2. The game host picks the other empty door. Switching will win the car.
The player picks the car. The game host picks either of the two empty doors. Switching will lose.
Kind of hard to get your mind around that one, but I'll be damned if it isn't right.
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i may be an idiot but i think there is a fourth option:
The player picks empty door number 1. The game host picks the other empty door. Switching will win the car.
The player picks empty door number 2. The game host picks the other empty door. Switching will win the car.
The player picks the car. The game host picks one of the two empty doors. Switching will lose.
The player picks the car. The game host picks one of the two empty doors. Switching will lose.
that would make it 50/50
its like the are counting the empty door collectively.