The Monty Hall Problem

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What You Need:

  • A printout of the problem below
  • Some people to debate it (Note: Don’t look at the answer until you’re done with the debate!)

What You Do:

Read this scenario:

  1. You’re on the game show “Let’s Make A Deal” looking at 3 doors. Behind one is a brand-new car. Behind the others are gag gifts. You get to choose one door: if it’s the car, you win!
  2. You choose Door #1. But before he unveils its contents, Monty opens Door #2 and reveals a gag gift. You may have picked the right door!
  3. Monty asks if you’d like to make a deal. If you want, you can switch to Door #3.
  4. What would switching do: improve your odds, worsen your odds, or leave them the same?
  5. Debate!

The answer:

  1. When you choose Door #1, your odds of winning are clearly 1/3.
  2. Therefore, the odds that the prize is behind Door #2 or Door #3 is 2/3.
  3. When Monty reveals that there is nothing behind Door #2, it doesn’t change the original probabilities. There is still a 2/3 chance that the prize is behind door two or three. (Knowing the contents of door two doesn’t change the odds once you’ve started playing.) Therefore, door three now has a 2/3 odds of winning.
  4. You should switch!

This is a very hard concept to absorb. Our intuition tells us that once Door #2 is eliminated, Door #1 and Door #3 each have a 1/2 probability of winning. But because the odds are set for good with your first pick, the winning odds of your original door will always stay at 1/3.

I worked with a NASA scientist who wouldn’t believe this answer. So we sat down and simulated the “game” a hundred times, where I was the host and he changed doors every time. Sure enough, he won about 2/3 of the games. If you need to be convinced, try it!

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