I'm a little intoxicated so this may be way over doing it or just not make sense, but...
Let's make the number even bigger, so we don't even need to deal in probabilities -- a sextillion (10^21) doors. That's more doors than there are grains of sand on the planet. Say you're given a sextillion doors, and you pick one lonesome door, call it door A. What are the odds that door A is the right one? Essentially zero, right? I'm going to say straight up zero because now is not the time for you to be a dreamer.
Ok so now that you've made your futile pick, poor Monty has to go and open every other door, all sextillion of them, except for yours and one other (let's call it door B) which he leaves closed. Every door he opens contains a goat.
So now out of this vast universe of doors, there are (10^21-2) doors opened with goats bleating away, and only two doors left unopened: door A, aka yours; and door B, the one that Monty mysteriously left unopened. One of them must contain the car.
(I'm guessing at this point it will be obvious for many people that door B contains the car, but I'll elaborate further in case it isn't clear yet.)
So we're left with your door A and Monty's door B. At this point there are only two scenarios that matter. Either you picked the right door originally, in which case staying put is the winning move. OR, you picked the wrong door originally, in which case switching is the right move. And we already know you picked the wrong door (remember, no dreamers allowed here), so you should for sure switch to this by-now-extremely-conspicuous-looking door B.
Door B isn't just some random door; the fact that it was chosen as the only door to not be opened tells you a tremendous amount of information about it. Think about the game from Monty's perspective. Either you pick the correct door initially and he gets to make door B a random goat door, or you picked the wrong door and he has no choice but to make door B the door with the car behind it. Since we know you picked the wrong door, we know what Monty's gonna have to do.
Let's make the number even bigger, so we don't even need to deal in probabilities -- a sextillion (10^21) doors. That's more doors than there are grains of sand on the planet. Say you're given a sextillion doors, and you pick one lonesome door, call it door A. What are the odds that door A is the right one? Essentially zero, right? I'm going to say straight up zero because now is not the time for you to be a dreamer.
Ok so now that you've made your futile pick, poor Monty has to go and open every other door, all sextillion of them, except for yours and one other (let's call it door B) which he leaves closed. Every door he opens contains a goat.
So now out of this vast universe of doors, there are (10^21-2) doors opened with goats bleating away, and only two doors left unopened: door A, aka yours; and door B, the one that Monty mysteriously left unopened. One of them must contain the car.
(I'm guessing at this point it will be obvious for many people that door B contains the car, but I'll elaborate further in case it isn't clear yet.)
So we're left with your door A and Monty's door B. At this point there are only two scenarios that matter. Either you picked the right door originally, in which case staying put is the winning move. OR, you picked the wrong door originally, in which case switching is the right move. And we already know you picked the wrong door (remember, no dreamers allowed here), so you should for sure switch to this by-now-extremely-conspicuous-looking door B.
Door B isn't just some random door; the fact that it was chosen as the only door to not be opened tells you a tremendous amount of information about it. Think about the game from Monty's perspective. Either you pick the correct door initially and he gets to make door B a random goat door, or you picked the wrong door and he has no choice but to make door B the door with the car behind it. Since we know you picked the wrong door, we know what Monty's gonna have to do.