[st1x7]

I see this argument a lot and for some reason it doesn't help me with the intuition at all. If you (wrongly) get caught up on the fact that the remaining door and your pick have the same initial probabilities of being a car, then you'll still think that switching doesn't make a difference even in the million-door case.

Here's what works for me:

- the switching strategy always gives you the opposite of your initial choice

- the initial probabilities are 2/3 goat and 1/3 car so by switching you get 2/3 car and 1/3 goat

[debbiedowner]

When Monty opens doors he uses 2 pieces of information: the door you picked and the correct door.

After he opens 999,998 doors he has given you quite a bit of information. There is a 1/1000000 chance though that he has given you no information (you picked the correct door)

But you're right that thinking about it in partitions also makes sense. You try to pick a partition size 1 that contains the prize, while Monty picks the partition size 999,999, if you agree with his partition and it has the prize you get it

[cgriswald]

I think the point a lot of people miss is that the trick to understanding the 3-door question and the 1000000-door question is the same trick. If you don't grok the trick, the 1000000-door example might make it easier to grok, but there's a fair chance it won't.

To muddy the waters further, it's not always understood that in the 1000000-door case, 999998 other doors are opened (as evidenced by discussion elsewhere in these threads). Sometimes people think it's still just one door. I suspect this is because the original problem is usually stated as "...Monty Hall then opens one of the doors you didn't pick" and because people suggesting the 1000000-door often just say "...what if there were one million doors?"