Have Numbers for a, b, c three doors, one door behind have a car, there was a lucky audience, he can be specified to open a door, for example a, this time, the host random opened a door of the other two doors, such as b, found no car behind, this time, the lucky viewers have a chance to replace the choice, he is replaced, please select the probability of open the door to get the car c high or adhere to a choice, opened a door to get the high probability of the car?
If the host is not random open a door of the other two doors, but deliberately opened a car door, excuse me this time lucky audience and how to choose?
CodePudding user response:
Default is a third rule out is 50% after a bai