Here’s what I consider to be the most easily grasped explanation for this problem:
Assume that there are ten doors to choose from to begin with, with goats behind nine of them and a car behind only one.
You pick one of the ten doors. Your chance of selecting the door with the car behind it is 10% and the chance that the car is behind one of the other nine doors is 90%.
Then the game show host, who knows which door the car is behind and will not open that door, opens eight of the remaining nine doors to reveal goats behind them. In no way does the fact that eight of the other nine doors have now been opened to reveal goats behind them affect the 10% chance that you selected the correct door to being with. It doesn’t affect it, because YOU ALREADY KNEW THAT AT LEAST EIGHT OF THOSE NINE DOORS HAD GOATS BEHIND THEM!
You are then given the option of keeping the door you picked or of switching to the only other unopened door.
Since it is obviously 90% likely that the car is behind the sole remaining unopened door that you didn’t pick and only 10% likely that it is behind the door that you did pick, only an idiot would keep the door that he originally picked, right?
Well, the same principle applies whether there are three doors to choose from to begin with or ten or a million. The only difference is, is that the odds that you selected the door with the car behind it to begin with obviously get much worse the more doors that there are to choose from. This also means of course that the odds that the car is behind one of the doors that you didn’t select get much better.