You participate in a game show where you must choose between 3 doors. Behind 2 doors is a goat, but behind 1 door is a car. You pick a door, say door no. 1, but before you open the door, the host opens another door, say door no. 3, which has the car behind it. "Oops," says the host.
You lose.
The door the car is behind is fixed, no amount of door opening will change its actual position, but your knowledge is probabilistic and the "smart" choice leverages that knowledge to maximise your chances with your knowledge at that time.
You have to consider the whole probability tree at any given time, so if you're saying the host has no knowledge of the car's location the branch points for switching are that 1/3 times you picked the car first time and switching is automatically losing, or that 2/3 times you choose a goat, then 1/2 of those times (i.e. 1/3 of the total possibilities) the host accidentally reveals the car and you lose immediately, or 1/2 he reveals a goat and switching will win. Bringing you back down to 1/3 total chance of winning if he only accidentally reveals a goat and you switch. Just because it didn't happen doesn't mean it wasn't part of the probability space and without the host having additional knowledge, to your knowledge you have no way knowing if you're on the now equally likely "chose right the first time and switching loses" or "chose wrong the first time and the host got lucky" branch, so switching is meaningless
But if the host knows which door has the car and can't reveal it, then 2/3 times you chose wrong first and 1/1 times he uses that knowledge to essentially show you by omission which door the car is definitely behind and gives you the chance to open it, or 1/3 times you chose right and he reveals none of that knowledge by randomly picking one of the two wrong doors and gives you a chance to open a door that the car is definitely not behind. That's where the inversion from 1/3 to 2/3 if you switch happens.
You're right. I understand my error now. Thanks for explaining.
I think, for me, the easiest way to visualize this is this.
1 )Instead of 3 doors, you have, lets say 10.
You first pick a door. You have a 1/10 chance of been right the first time. This chance doesn't change at all.
Host eliminates 8 of the remaining doors with no car, leaving just 1. We are back the "50/50" situation, except its not truly 50 /50. Your initial chance of 1/10 has not changed.
Ask yourself this - Do you stick with your initial chance of 1/10? He already removed 8 bad choices for you. Remember the host knows where the car is always. Given the new information, you should definitely positively switch.