>>12518933This would be true, IF you didn't know in advance that the host's decision about which door to open weren't constrained by 2 factors: 1. which door you chose initially and 2. which door the car is behind in.
The host cannot open either of those doors:
He can't open the first door you picked and he can't open the door the car is behind.
If the host could just open any door at random among those 3 after your initial choice, then assuming he opened a door with a goat after your pick, it would be a 50/50 chance between the two remaining doors.
Obviously, if he, at random, opens the door with the car, then it would 100% be that door.
But even being constrained by only 1 of those factors alone would still not be enough to create the Monty Hall situation.
For example, if the host could only open one of the doors with a goat, but he could also open YOUR door (i.e. your initial pick), then the choice between the 2 remaining doors would be 50/50.
Likewise, if the host could only open one of the doors that you didn't pick, but could also theoretically open the door with the car, then if the door he opens doesn't contain the car, the choice is likewise 50/50 between the 2 remaining doors. (Obviously, as before, if he opens the door with the car, it's 100% that door).
However, once those 2 constraints are in place simultaneously, and you know about them in advance, you can think of the show (and the problem) as a kind of cybernetic system with the host feeding you useful information in response to your initial choice.
If, however, you DON'T know about the 2 constraints in advance, then from your perspective, it will always be 50/50 after the host opens a door with a goat.
The fact that the host's choice is CONSTRAINED (and you KNOW about it) is really the key to this problem.
It's a very subtle detail and not intuitive at all.
