You participate in a game show where you must choose between 3 doors. Behind 2 doors is a goat, but behind 1 door is a car. You pick a door, say door no. 1, but before you open the door, the host opens another door, say door no. 3, which has the car behind it. "Oops," says the host.
You lose.
Wow, people in that thread are so stubbornly and arrogantly wrong, it's insane. Normies having opinions slightly outside of normie-norm is even worse than normies being normal normies.
A lot of people are leaving out the point that the host KNOWS which door has the car behind it and is presenting you a much higher-quality choice by opening all the other doors. If the host doesn't know himself, it doesn't make sense and you get an outcome like the above.
Well, yes, a host would never choose randomly, for that reason (and that would be hilarious), but...even if he did choose randomly, but "randomly" chose a non-prize...it does still make sense to switch. Wild, but true.
It's not the knowledge that changes the odds.
No, if he chose randomly and didn't reveal the car by pure chance then both doors remain equally likely to have the car behind them and switching doesn't matter. Most teachers are bad at explaining/don't understand the monty hall problem themselves and they take shortcuts when explaining it so that's an understandable confusion though.
The point is that the host choosing to reveal a door has more knowledge than you and he divulges new information by elimination when he opens a door that lets your make a more favourably informed choice in the second round.
But lot of people try to explain the system where the probability of your first choice is just frozen in time to before the outcome is revealed, and you get to chose a second time at 1/2 odds simply because there happen to be only two doors left. The leads to the nonsense outcome where if host knows nothing and accidentally reveals the car at random, then you definitely shouldn't chose to switch doors because there's still a 1/3 chance you win a car if you stick, but a 0/2 chance if you switch. When in reality the choice is meaningless.
No, actually both explanations you’ve given are wrong as I understand it. Here’s the solution as I know it:
When you picked, you had a 1/3 chance to pick “right.” Since probabilities sum to 1, that means there is a 2/3 chance that the prize is behind a door you did not pick. One of those doors is then eliminated and you are shown that door did not contain the prize. It doesn’t matter whether that door was eliminated at random or purposefully—what matters is that you know the prize was not behind that door, and that out of the selection categories remaining, [door you originally picked, 1/3rd chance of having a prize] and [all doors you did not pick, 2/3rd chance of having a prize] you get to repick your category. If it helps, you may think of repicking as opening every door in the latter category, since that’s effectively what will happen.
If you switch, you have a 2/3rd chance of hitting the right door, because that’s the necessary inverse of your original 1/3rd chance, now that the other possibility of a dud door has been removed.