I was reminded of this today, I think it's pretty fun.
The Monty Hall Problem is a puzzle where the scenario is as follows:
- You're a contestant in a game show.
- You have three, closed doors to choose from - behind one door is a car, behind the other two doors is a goat. You're obviously aiming for the car.
- You pick one door.
- The game show host then opens one of the other two, remaining doors doors, revealing a goat. He then asks you if you want to open the door that you have picked, or change your answer and swap to the remaining, closed door.
What would you do?
Most people say that it makes no difference, and stick to their original answer, most likely because we're hardwired to be overly confident about the choices we make. But moreover, intuitively, it just doesn't seem to make a damn bit of difference, since when there are two doors remaining, the probability, on the surface, seems to be 50/50.
But this is not the case.
At the start of the show, you pick one door. Two of the doors hide a goat, and behind one of them is the car.
So, when you first pick the door, you have a 66% chance of picking a goat door, and a 33% chance of picking the car door.
What this means is that it's two times more likely at the beginning that you picked a door that has the goat behind it, rather than the one with the car.
After you choose your door, and the game show host opens up one of the two doors that has a goat, you statistically increase your chances at getting the car by swapping doors.
Since there's a 66% chance of picking a goat door at the beginning versus a 33% chance of getting the car on the first try, it's the most likely scenario that after the host reveals one of the goat doors, the remaining door has the car behind it.
It's really obvious once you get it, but when I was younger, I went through some serious headache wrapping my head around this. It seems silly now since it seems to easy, it'd be interesting to hear if this feels counter-intuitive to people here.
@schattenjaeger