If the host opens the door with the car, the game would be over anyway. It is clearly implied that the host always opens a door with a goat.
"the host opens another door that he knows has a goat behind it, say No. 3,"
Using the knowledge of where the car is to avoid showing it, is more specific than just having the knowledge, and possibly still opening at random.
I think the source of the paradox lies somewhere in the biases (the Endowment effect or Status Quo bias) as discussed in the Wikipedia article about the problem - not with how the question was stated and especially not with the part about how the host selects a door.
If you model this game where the host chooses randomly, and might open the car door (when that hapoens the player instantly loses) , then there is no point switching.
I think the fact this does matter is the source of many people's confusion.
I don't think I've ever seen this rule in any variants.
Since we're making up rules: Could the player not still have the option to switch, and thus win the game?
I'm not convinced that I am wrong after your explanation.
Either way, my point is that focusing on the problem statement as the "main thing that seems to cause the paradox" allows the reader to stop there and disregard a far more interesting discussion about biases.
Why not? Your chances of choosing the correct door were 1 in 3 from the start. That doesn't change with the fact that the host opens the doors at random.
Otherwise, as a host, I would only offer the players to change if they picked the car. (And if they heard about this paradox before, they might even change their choice)
Given the description from the article:
> Suppose you're on a game show, > and you're given the choice of > three doors: Behind one door is > a car; behind the others, goats. > You pick a door, say No. 1, and > the host, who knows what's behind > the doors, opens another door, > say No. 3, which has a goat. He > then says to you, "Do you want > to pick door No. 2?" Is it to > your advantage to switch your > choice?
I'd say, no, it is not to my advantage because I'd think the host would only ask me to switch if I had taken the "right" choice and want to make me lose. Unless I knew that the host always ask if one wants to change, in which case the paradox indeed apply.
If the host does not have to open a door for the candidate (which is quite likely, he's playing his own strategy of making the show exciting, not predictable), then the usual solution (always switch) is not correct.
All this, and probably any variation and every solution of the problem that comes up in this thread (and probably every other Monty Hall forum thread in the post 3 decades) are discussed extensively in the Wikipedia entry (e.g. https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...).
Maybe every time a new Monty Hall thread comes a mod should just stick the Wikipedia entry at the top. I've never seen anything in any of the following discussion that hasn't been covered in Wikipedia already.
I can’t argue with what you’ve observed, but my own experience is not that there’s a misunderstanding about the question, but that people simply don’t think properly about the question. Nobody I’ve ever seen confused about the answer was confused by the question, and many numerically literate people I know have struggled with it.
The strange thing is that "choosing to stay" somehow doesn't behave like a new 50/50 choice, but choosing to change does.
Intuition implies that given a random choice between a, b and c - chances are 1:3. And given a new choice a or b, chances are 50/50. But you only get to act on new information (make a new choice) if you change.
I must admit I'm still confused on this, and wonder if it bears out in simulation (that always change is better than random stay/change when offered to choose again
Ed: paultopia has a nice simulation and explanation below:
https://paul-gowder.com/montyhall/
Significantly - staying means the player remains at 1:3, but since we now know the chosen door, or the alternate, holds the car - if staying is 1:3 wins, changing must account for the remaining 2:3 - since there are only two options. And random choice is also better than staying (it's 50/50) - but worse than changing.
This is not the correct interpretation. In fact, the choice of getting the car if you switch is not 50/50, up from 1/3. It's 2/3. The choice of winning if you don't switch is 1/3.
The reason for this is that the second choice is not independent of the first choice, so you shouldn't model them as different choices.
Essentially the problem could be reformulated as: you choose one door at random. Then the host asks you if you would rather keep that door, or choose to get what is behind both of the other doors. The door opening is just a red herring.