If the host picks randomly, then on some occasions he reveals the car and the game cannot proceed to conclusion. In other words the door picked by the host reveals a goat, by definition. So what the host knows is irrelevant except to permit the game to proceed. This leads to the contestant facing consistent odds.
My understanding pertains to the Monty Hall problem as described by Wikipedia, in which game it always makes sense to take the host's offer to switch.
For those who have not yet intuitively grasped why it pays to always switch, the key to the Monty Hall problem is knowing that the host will only reveal a goat, and that when he does an unwanted option has been removed by the host, improving your chances on a re-pick.
To see why this is so, imagine the game offers you many doors (say 100) instead of 3. You pick a door. Then the host shows you what's behind all doors but 2, your pick and one remaining door (you see 98 goats). With so many doors revealed (all goats) it's easy to see you should take the opportunity to switch: Your first pick's chance of getting the car was 1-in-100; very unlikely! After 98 doors are revealed (cast out) your switch pick will be 1-in-2; pretty good.
Returning to the 3-door scenario instead of many doors: a change in odds from 1-in-3 to 1-in-2 is much less significant than going from 1-in-many to 1-in-2, but your switch still improves your chances.
That's the essence of the puzzle.
NoPicklez•7mo ago
Extrapolating it to 100 doors has been the only way I could understand the problem and why its beneficial to change.
Even the Wikipedia article extrapolates this out to 1 million, but assumes all the time that the host knows which door the car is behind and will avoid picking the door with the prize in it when narrowing it down to 2 doors for the contestant.
I think where the author is coming from, is what happens to the probabilities if the host doesn't know which door the car is behind and in the scenario of 1 million doors, there's a risk that the door with the prize is thrown away at random when narrowing it down to the 2 doors. Where the remaining choice to keep your original door or switch could result in both doors not having the prize at all.
However by changing that the author has missed the original intent of the Monty hall problem, because they have removed the fact that one of the remaining doors has to have a car in it. If the host didn't know which door the car was behind, there is a chance the contestant is no better off switching. Which isn't the intent of the problem.
If I have understood correctly, it is important to the monty hall problem that the host knows which door has the car in it and that door is always behind one of the doors left at the end.
NoPicklez•7mo ago
Having written another comment, overall my anecdotal understanding is that the whole Monty Hall problem is based on the host knowing which door and will avoid removing any doors that don't have the car. That's the premise of the problem and isn't counterintuitive, but completely part of the intent of the problem.
If you remove the host knowing what is behind each door, then its no longer the monty hall problem.
_wire_•7mo ago
If the host picks randomly, then on some occasions he reveals the car and the game cannot proceed to conclusion. In other words the door picked by the host reveals a goat, by definition. So what the host knows is irrelevant except to permit the game to proceed. This leads to the contestant facing consistent odds.
My understanding pertains to the Monty Hall problem as described by Wikipedia, in which game it always makes sense to take the host's offer to switch.
For those who have not yet intuitively grasped why it pays to always switch, the key to the Monty Hall problem is knowing that the host will only reveal a goat, and that when he does an unwanted option has been removed by the host, improving your chances on a re-pick.
To see why this is so, imagine the game offers you many doors (say 100) instead of 3. You pick a door. Then the host shows you what's behind all doors but 2, your pick and one remaining door (you see 98 goats). With so many doors revealed (all goats) it's easy to see you should take the opportunity to switch: Your first pick's chance of getting the car was 1-in-100; very unlikely! After 98 doors are revealed (cast out) your switch pick will be 1-in-2; pretty good.
Returning to the 3-door scenario instead of many doors: a change in odds from 1-in-3 to 1-in-2 is much less significant than going from 1-in-many to 1-in-2, but your switch still improves your chances.
That's the essence of the puzzle.
NoPicklez•7mo ago
Even the Wikipedia article extrapolates this out to 1 million, but assumes all the time that the host knows which door the car is behind and will avoid picking the door with the prize in it when narrowing it down to 2 doors for the contestant.
I think where the author is coming from, is what happens to the probabilities if the host doesn't know which door the car is behind and in the scenario of 1 million doors, there's a risk that the door with the prize is thrown away at random when narrowing it down to the 2 doors. Where the remaining choice to keep your original door or switch could result in both doors not having the prize at all.
However by changing that the author has missed the original intent of the Monty hall problem, because they have removed the fact that one of the remaining doors has to have a car in it. If the host didn't know which door the car was behind, there is a chance the contestant is no better off switching. Which isn't the intent of the problem.
If I have understood correctly, it is important to the monty hall problem that the host knows which door has the car in it and that door is always behind one of the doors left at the end.