> The rationale for opening 999,998 doors in my example versus the 1 door left, is that in both examples, Monty Hall opens every door except the one you're on, and 1 other door. It happens that in the normal Monty Hall, if you open every door except the one you're on, and 1 other door, you have only opened one door.
And the rationale for opening 1 other door in the million door example is that in both examples the host is opening 1 other door. The normal Monty Hall problem is usually formulated such that the host opens 1 other door, not that he opens all other doors. As you noted, the two formulations are equivalent in with 3 doors, but with more than 3 doors, they're not. I just don't see why it's "intuitive" that if the number of doors is increased, the natural extension of the game is that the host opens all other doors that don't have the prize. In fact I'd argue the opposite.
Imagine an actual Monty Hall game with 4 doors. The contestant opens 1 door with a goat, and the host might open (a) 1 other door with a goat or (b) 2 other doors with a goat. Both are valid, reasonable, but different extensions of the game. In both versions, the best strategy for the contestant is to switch[1], because in both versions the host is giving her extra information. But in version (b) he's giving her much more information than in version (a). Of course it's much easier to intuit in version (b) that switching is better, but it's not clear to me why version (b) rather than (a) is the natural 4-door analog to the 3-door Monte Hall game. If a 4 door version were played in real life, it's far more likely IMO that version (a) would be played. In (a) the host gives a little bit of extra info where the prize is, without giving away the solution. And in this case you need a much better model to see why this is the case in stead of relying on intuition and analogy.
[1] A usually unstated assumption in most formulations is that the host must open another door with a goat if the contestant initially chooses a wrong door. In an actual TV show the host will likely have discretion whether he opens another door at all, to increase suspension and not become predictable in repeated games. In this case the problem becomes much more difficult as you need to model the strategy of the host. All of thise and pretty much any other solutions and explanations on pretty much every forum is already extensively documented in Wikipedia (https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...).
And the rationale for opening 1 other door in the million door example is that in both examples the host is opening 1 other door. The normal Monty Hall problem is usually formulated such that the host opens 1 other door, not that he opens all other doors. As you noted, the two formulations are equivalent in with 3 doors, but with more than 3 doors, they're not. I just don't see why it's "intuitive" that if the number of doors is increased, the natural extension of the game is that the host opens all other doors that don't have the prize. In fact I'd argue the opposite.
Imagine an actual Monty Hall game with 4 doors. The contestant opens 1 door with a goat, and the host might open (a) 1 other door with a goat or (b) 2 other doors with a goat. Both are valid, reasonable, but different extensions of the game. In both versions, the best strategy for the contestant is to switch[1], because in both versions the host is giving her extra information. But in version (b) he's giving her much more information than in version (a). Of course it's much easier to intuit in version (b) that switching is better, but it's not clear to me why version (b) rather than (a) is the natural 4-door analog to the 3-door Monte Hall game. If a 4 door version were played in real life, it's far more likely IMO that version (a) would be played. In (a) the host gives a little bit of extra info where the prize is, without giving away the solution. And in this case you need a much better model to see why this is the case in stead of relying on intuition and analogy.
[1] A usually unstated assumption in most formulations is that the host must open another door with a goat if the contestant initially chooses a wrong door. In an actual TV show the host will likely have discretion whether he opens another door at all, to increase suspension and not become predictable in repeated games. In this case the problem becomes much more difficult as you need to model the strategy of the host. All of thise and pretty much any other solutions and explanations on pretty much every forum is already extensively documented in Wikipedia (https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...).