[jibal]

"But it's not at all clear or intuitive that jumping from 3 to 1,000,000 doors should lead the host to open 999,998 other doors rather than 1 other door."

It SHOULD be clear, because you have two givens: 1) Monty never reveals the car. 2) He opens all the doors except 1.

[pyhtel]

"2) He opens all the doors except 1"

How is this a given exactly? In the original problem he only opens 1 other door. Now that also happens to be all doors except 1, but from just the 3 door problem that seems more coincidental than a fundamental part to the question

[jibal]

It's a given by the person who mentioned 999,998 doors. I think you're missing the point, but I won't pursue this further.

[kqr]

Sorry, but you are the one missing the point. The person who mention 999,998 doors didn't give any reasoning for why that would be the logical extension of the problem.

Obviously, you and I know it is, but the person grappling with the Monty Hall problem is right in not being convinced of that just because someone says it is!

[will_pseudonym]

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.

Monty Hall is asking you a simple question, whether or not you should switch, and so in my example of 1,000,000 whether or not you open 999,998 doors, or 1 door, you will always have worse odds to win if you don't switch to another door. Removing 999,998 doors just takes the proposition to an extreme.

Another component to utilize one's intuition using the 999,998 example, would be to imagine the game being played 3 times in a row. What are the odds that not switching will help you? So basically, not switching is disregarding everything Monty Hall is doing. You are either behind a door or you are not. You don't switch. If that is how you play the game, your chance of choosing right when not switching is 1/1,000,000 each game, or 1/10^18 for it to happen 3 times in a row. Now, consider what Monty is doing. He's removing every chair but two, yours and another. If the odds of you winning are 1/1,000,000 if you don't switch, What are the odds of doing _the opposite_? Since there are only two options, the probability of winning if you switch is 1-1/1,000,000, or 999,999/1,000,000, as the sum of the probabilities of all possible events has to add up to 1.

The "999,998" chairs removed example is an attempt at making the dichotomy between "stay" and "switch" more extreme, so that you would feel it in your gut rather than trying to mentally account for the moving pieces.

I'm always interested in improving my ability to explain these kinds of phenomena, and I appreciate the pointing out of why the dots don't get connected for some with the example.

[em500]

> 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_...).

[will_pseudonym]

pyhtel, I gave it a go at explaining the rationale in this comment I made below: https://news.ycombinator.com/item?id=24643272