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.
_wire_•2h 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.