That explanation never worked for me, because you can turn it around to the situation where Monty does not know where the car is. Say there are 1000 doors, and you pick door 429. On his way to open door 429, Monty stumbles, falls, and accidentally knocks open every door except door 128. If by some coincidence all opened doors happened to contain goats, you will have nothing to gain from switching. Very counter-intuitive, but just as true as the original problem.
A possible intuition here is that Universes where your first pick was the door with the car, which initially were just 1 in a 1000 compared to Universes in which you picked a goat, will suddenly become massively overrepresented. After all, in these types of Universe Monty's Fall couldn't possibly have shown a car, whereas most of the other Universes will not survive to the next "round".
Of course, if this happened in real life, Bayesian thinking would increase the likelihood of hypotheses such as, for example, "The door containing the car has a better lock" to such an extent that I would switch.
You can't really turn it around, because Monty knowing and using his knowledge of where the car is to reveal only goats is what makes switching advantageous.
In the case of the clumsy Monty of your example, it goes like this:
1. There is a 1/1000 chance door 429 has the car.
2a. If it has the car, then when Monty accidentally opens 998 doors no car will be revealed. This does not change the chances that 429 has that car, which remain 1/1000.
2b. If 429 does NOT have the car, then 998/999 times that Monty accidentally opens 998 doors, he will reveal a car, which presumably ends that game. There is only a 1/999 chance that he will not reveal the car and the game proceeds.
3. Thus, there are two cases where the game reaches the point of two remaining doors, with 998 revealed, the car is behind one of the two, and you have a chance to switch.
3a. Your door has the car, which happens 1/1000 games.
3b. Your door does not have the car, which happens 999/1000 x 1/999 games, or 1/1000 games.
In other words, if the clumsy Monty version is played repeatedly, 998 out of 1000 games end without even getting too the point you get a chance to switch, and 2 get to where you get the chance. In those two, one has the car in your door, one not. There is no advantage to switching.
In the case of the systematic Monty who knows where everything is and ALWAYS opens 998 goats, it goes like this:
1. There is a 1/1000 chance your door, 429, has the car.
2a. If it had the car, Monty opens 998 doors that do not have the car, leaving one door besides your yours.
2b. If your door did not have the car, it is one of the 999, and Monty systematically opens the 998 of those 999 that do not have the car.
3. You always reach the choice stage. You can either get there via 2a, which always results in the car being behind your door, or via 2b, which always results in the car being behind the other door.
3a. You get there via 2a in 1/1000 games.
3b. You get there via 2b in 999/1000 games.
If you do not switch, you only if and only if you got there via 2a, so you only win 1/1000 games. If you always switch, you win if and only if you get there via 2b, so you win 999/1000 games.
In the original problem, Monty always opens a door to reveal a goat. This strongly implies that he knows which doors have goats.
Your alternative is different. If the doors opened are truly selected randomly, then the odds are high that he would reveal a car, especially in the 1000-door version. What are the odds that Monty could choose 998 doors randomly out of 1000 and NOT pick the one with the car?
If Monty doesn't know where the car is, then you're arguing with a completely different version of the game that could result in him opening the door with the car, so the strategy is going to be different.
Changing the scenario is the exactly the point of why I am unsure increasing the number of doors gives true intuition. In the original problem with the original scenario, people intuitively think switching doesn't matter, when it provably does. With 1000 doors and the Monty Fall scenario, again the intuition is wrong. So are we gaining true intuition for the Monty Hall scenario by expanding the setup, or is it just some subconscious Bayesian thinking accidentally steering us toward the right answer.
A possible intuition here is that Universes where your first pick was the door with the car, which initially were just 1 in a 1000 compared to Universes in which you picked a goat, will suddenly become massively overrepresented. After all, in these types of Universe Monty's Fall couldn't possibly have shown a car, whereas most of the other Universes will not survive to the next "round".
Of course, if this happened in real life, Bayesian thinking would increase the likelihood of hypotheses such as, for example, "The door containing the car has a better lock" to such an extent that I would switch.