I think what's confusing, is that if one doesn't switch, the first selection is still 1:3 - since you randomly picked 1 door out of 3 (the fact that one is revealed does not change that).
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:
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.
> 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.
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.
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.