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.
Yeah, I think I was wrong. I was thinking about it before I left for work, but didn't have time to edit. It's weird to think about, and still hard to completely grasp, but I do believe intent matters. It's just like how something crazy improbable can happen, and it doesn't affect the next result. You can flip 99 coins, and have them all turn up heads...you're already in the crazy universe, but the next flip is still a 50/50 chance.
So, you could have ten doors, and the host could randomly remove eight losers...the remaining two doors are still equally likely, you just got a really unlikely outcome...which I suppose means logically if not mathematically, you actually have more likelihood of the car being behind the already chosen door.
So, if the host chooses randomly, and you know that, I suppose you're actually better sticking with your choice. The more doors he reveals (again, if randomly, and you know it's random) without revealing the car, the more likely the car isn't in the doors that were unchosen.
Ugh. This gets more and more confusing, because you do have to know he's not choosing randomly, to be able to reach the "answer." If you don't know how he's choosing, it gets into more mind games than logic or math.
So I suppose I was a little hasty. I was still right that all those people on Twitter were retards and didn't even get the basics, but there were more assumptions you have to make than I took into account, initially. You have to know how the host is removing doors to be able to calculate whether or not you switch.