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.
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.
It's made a lot simpler by the fact that switching doors is meaningless rather than detrimental if the host has no extra knowledge. It basically means you're still better off switching in any instance where there's a non-zero chance the host knows the answer, as there's no harm to doing it if he doesn't know, so you don't have you have to balance anything against the chance he does know.
Yup. I hadn't done the full rationale, but it does seem to always break even. If chosen randomly, remaining doors minus no-prize doors, you're still left with an even chance versus your initial pick. In the three-door scenario, it's 1/3rd vs 1/3rd, and the last third is missing, so equal chances, i.e. 50/50.
And, that's a good point, even if you don't know how he's choosing, you always switch, because if he is choosing randomly it's a coin flip, and if he isn't you're increasing (sometimes very drastically depending on number of doors) your chances, so it's an overall gain.