But this raises a different problem with intuition:
If Monty doesn't know where the car is, then if 999,998 doors were opened showing goats, leaving two doors, the odds that the car is behind your door or behind the remaining door is 1:1 ... this defies many people's intuition.
The difference between the two cases is that, if Monty knows where the car is, then his opening 999,998 doors with goats behind them is exactly what we expect, whereas if he doesn't know where the car is, then his opening 999,998 doors with goats behind them is an extraordinarily unlikely event. But if that does happen despite being extraordinarily unlikely, then there's still a 50% chance that the car is behind your door.
1) I will probably lose when Monty opens a car door.
2) If I don't, I am really gambling between whether I made a 1-in-a-million pick or Monty did (in the choice of which door to leave shut), which obviously has even odds.
Interestingly, by compressing this problem back down to the 3-door version, it makes it pretty obvious why that's the case (and aligns with people's intuition about the original problem). Also interesting that in this case, even if the intuition is wrong (that 'obviously' they must have picked the car), the outcome (sticking with the chosen door) is an optimal strategy.
If Monty doesn't know where the car is, then if 999,998 doors were opened showing goats, leaving two doors, the odds that the car is behind your door or behind the remaining door is 1:1 ... this defies many people's intuition.
The difference between the two cases is that, if Monty knows where the car is, then his opening 999,998 doors with goats behind them is exactly what we expect, whereas if he doesn't know where the car is, then his opening 999,998 doors with goats behind them is an extraordinarily unlikely event. But if that does happen despite being extraordinarily unlikely, then there's still a 50% chance that the car is behind your door.