The Monty Hall Problem: An Introduction

The Monty Hall problem is a probability puzzle-based on a game show called "Let's Make a Deal" hosted by Monty Hall. Here's how the game goes: 1. There are three doors, behind one of which is a car (the prize), while the other two doors hide goats. The player doesn't know where the car is, and the probability of the car being behind any of the doors is 1/3. 2. The player chooses one of the three doors (say, Door A). 3. The host, Monty Hall, who knows what's behind each door, opens one of the other two doors (say, Door B), which has a goat. Monty Hall will never open the door with the car. 4. You now have a choice to stick with your initial choice (Door A) or switch to the other unopened door (Door C). The question is what's the best strategy - sticking with your initial choice or switching? From a probability standpoint, it's actually to your advantage to switch. Initially, the car has a 1/3 chance of being behind the door you initially chose, and a 2/3 chance of being behind one of the other two doors. Even after Monty reveals a goat behind one of the doors you didn't pick, the probability doesn't change. The combined chance that the car is behind one of the other two doors is still 2/3. Therefore, switching equates to selecting the 2/3 probability. If I were in the situation, I would switch my choice. The reason is simple- it's based on probability. At the start, the odds were twice as high that the car was behind one of the two doors I didn't choose. Monty's action of opening one of the non-selected doors that has a goat doesn't change those initial probabilities, but it does give me more information. By switching, I get the higher probability.