Is it, though? It seems apparent that, after the first guess, the host opens all but the last two doors, which just so happens to be 1 door.
To check the math:
Start with the $NUM_DOORS open doors. Now open all but the last two. So that’s $NUM_DOORS-2, which is 3-2, which equals 1 open door.
1. You pick a door.
2. You get the offer "Do you want to keep that door, or choose both [all] of the other doors? In either case, you'll keep anything that isn't a goat."
3. Nobody opens any doors.
Should you keep your one door, or switch to the two doors?
Let's demonstrate with a slightly different construction: You're no longer playing with monty, but with a demon. This demon wants you to lose, but also picked a very bad game for themselves. You pick a door, then the demon opens all-but-one of the remaining doors. Then, you can pick any door, open or closed, and you get what's in it.
If the demon opens doors at random, nearly all the time (with 100 doors) you'll see the car and be able to pick it directly. In this situation, switching between the closed doors doesn't really matter, but you'll usually know exactly which door to pick, because you can see the car.
So instead, the demon only opens doors that don't have a vehicle behind them. You only ever see goats. At this point, he's not opening doors at random. If he were, you'd see the car 98% of the time, but you never do. At this point, since he's using additional information, it is in your best interest to switch.
The fact that Monty Hall opens the doors deterministically (not randomly) is KEY.
In the original problem, Monty ALWAYS opens a door with a goat. In using 50 doors, Monty would ALWAYS open doors containing goats, and not the car. It's not random.
Knowing it's not random, it should be very intuitive.
I think it has to do with the difference between "probable outcome in reality" and "probably outcome based on personally known information".
Lets say when you get down to door #1s and #49, Monty brings in someone new, with no information and says pick a door. For that new person, standing right next to you, doors #1 and #49 have a 50-50% chance, while for you they are a 2% vs 98% chance.
How can door #1 simultaneously have a 2% chance for you and a 50% chance for Bob? The answer is that the chance is not a single fixed property of the door itself- which is hard to wrap ones head around.
And for that matter, Monty Hall himself knows one of the doors is 100% and the other is 0%.
This ambiguity is resolved by something called do calculus - https://arxiv.org/pdf/1305.5506.pdf
Your personal, information limited calculation of the chance a car is behind door #1 has no impact on if there is a car behind door #1. Reality is binary and constant. There was always a car there, or there always wasn't.
Most people correctly intuit that of course the real probability that the car is behind door #1 cant change with reveled information. It isn't a quantum car. They just get caught up on the fact that predictive chance is a attribute of the model, not the real door.
The car isn't moving, as you say, but that intervention by the host lets us trade one door for both of the other doors.
The best way I’ve heard it explained to help people get it through intuition is by changing the number of doors and goats. Say there are 100 doors, and they all have goats except one, which has a car. You pick door 1. Monty then proceeds to open doors 2 through 48, skips door 49, and then opens the remaining doors. After all that, he stops and asks you, would you like to switch?