The Monty Hall Problem is a famous brain teaser based on a probability puzzle that often leads to counterintuitive conclusions. The puzzle was first published and solved by Steve Selvin in 1975 in The American Statistician.

Here’s the content of the puzzle:

There are three doors. Behind one door is a car (the prize), and behind the other two doors are goats. All three doors are closed, and the player doesn’t know what’s behind each door.

• First, the player chooses one of the three doors (they choose but don’t open it).
• Then, the host, who knows where the car is, opens one of the remaining doors to reveal a goat. (The host will always open a door with a goat, never the door with the car.)
• Next, the player is given two choices:
  • Stick with their original choice.
  • Switch to the other unopened door.

The Question: If the player opens the door with the car behind it, he/she wins the car. Should the player switch doors or stick with their original choice to maximize their chances of winning the car?

Answer: The player should switch to the other unopened door (choice 2) because the odds of winning will increase to 66% instead of sticking with the original choice (33% chance of winning).

Actually, you can search Google for “Monty Hall problem” and you’ll get countless results with solutions and explanations of why choice 2 is better than choice 1, and about prior probability and posterior probability… So I think it would be redundant for me to explain it again.

However, I want to point out a shorter, simpler, and easier-to-visualize explanation that most people can easily understand:

💯

Imagine if the game had 10 doors instead of 3, with only 1 door hiding a car and the other 9 hiding goats. You choose 1 door, then the host opens 8 doors with goats behind them, leaving 1 door closed. Now there are only 2 closed doors: the door you initially chose and the door the host didn’t open. Would you change your choice? :))