A Thoughtful Harmless Rant #2 by Daniel Wu

The Monty Hall problem is as follows:

You are in a game show, where there are three doors. Behind each door is either a goat or a ferrari. There is only one ferrari. You are told to choose any door. If you choose the door with the ferrari, you get to keep it. You choose door 1. Instead, the host opens door 2. Behind door two is a goat. You can choose another door to open. Do you choose door 1 or door 3?

Some people say that it is the same chance either way, because of common sense. Other people say that they think they’re smart say that you should choose door 3, because the probability of door one does not change, but door 3 should be one half.

My explanation is as follows:

The probability is the same. When people say that door one is only one-third chance, and door 3 is one-half chance, they are solving the problem as it was, before the conditions were changed (When the host opened door 2 and you see the goat). If you solve the equation so that conditions have been accounted for, we will have the truly right answer. The new problem is this: There are 3 doors. There is one ferrari and two goats. The door 2 does not have the car, so it has to be either door one or three, so the probability is fifty percent.