Idea and Explanation

You're the contestant on a game show. Before you are 3 doors - there are goats behind 2 of the doors and the grand prize, a brand new car, is behind the third door. You start by choosing a door. The host then opens one of the other doors with a goat. He then turns against you and asks: "Do you want to remain on the same door or pick the other one?". Is there a greater chance that the car is behind the door you did not pick?

The answer to that question is "yes" - but why?
Well, let's first take a look at the doors to choose from. Behind them are 2 goats and 1 car. The goal is to win the car. We can all agree about that there is 2/3 risk of choosing a goat in the first pick and only 1/3 of getting the car.



If the doors are open we can see the two goats and the car:



So, we have now established that you have a greater risk of choosing a goat in the first pick. What happens next? The host will now reveal one door with a goat behind. Let's think about the different possible scenarios. If we picked the car in the first pick (33% chance) the host can choose between 2 doors. If we instead picked a goat at first (66% risk) the host will have to reveal the only remaining door with a goat behind - ergo, the remaining door would be the car.
Said in another way - if we pick a goat the remaing door would be the car. Since there's a bigger possibility of choosing a goat (66%) there's a bigger chance of winning if we change our pick to the remaining door!

By Lewinzki