"If you have the door with car then the P(car) = 1 if you don't then
P(car) = 0. In this round, you are not selecting between two doors - if
you were then the P(car) = 1/2. But this idea fails to recognize the
fact that you had a choice in the 1st round. So P(car) is either one or
zero."
In the first round of this game the contestant had three doors to choice
from, so the probability that he will choice the car is 1/3. The host
then opens one of the doors not picked. The contestant now has two doors
to choose from (an event which will always occur). Although the
contestant has three doors to choose from from the beginning, he will not
choose the door that the host just opened. Therefore the probability
that the contestant will choose the car now is 1/2. I agree that the if
the contestant picked the car then the P(car) = 1 and P(car) = 0 for the
other door. But the contestant does not know which door's probability is
1 and which one has 0, only the host knows. So how can the 1&0 idea
affect whether the contestant should switch or not?
There really is no advantage to switch or not. It is the host's job as
part of the game and part of the suspense of the game to ask the
contestant whether he/she should switch or not after he opens the door
not choosen by the contestant that has a goat. Whether the host wants
you to win or not, he has no choice but to do his job.
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