The Monty Hall Problem


Sometimes, intuition can be proven wrong with mathematics. This is the case with what’s known as the Monty Hall Problem, from the old TV game show, “Let’s Make a Deal.”

In this game, there are three doors. There is a new car (or other nice prize) behind one of the doors, and joke prizes like goats behind the other two doors. The contestant on the show starts out by choosing one of the doors. Monty Hall, the show’s host, then opens one of the two doors that wasn’t chosen; he would always open a door with a joke prize behind it. The contestant then had the opportunity to either switch to the other unopened door, or stick with their initial choice. The question is, does the contestant have a better chance of winning the grand prize if she sticks with her initial choice, or if she switches to the other unopened door? Or is there a 50-50 chance of winning regardless of whether she sticks or switches?

Before reading any further, try out the game below. First, choose one the of the three doors. After the goat is revealed, click the same door if you'd like to stay, or click the other available door if you'd like to switch. You'll then find out if you won or not!

After Monty opens one of the losing doors, it seems to most people that the chance of winning is 50%. At this point, there are two choices, one with a car and one with a goat, so common sense/intuition seems to say that there is a 50-50 chance, regardless which of these two doors is chosen.

However, if you play the game 100 times or 1,000 times (you can run a simulation of 10, 100, or 1,000 games here), you see that if you stay with your initial choice, you win about 33% of the time, and if you switch to the other choice, you win about 66% of the time. Clearly, you have a better chance of winning if you switch. But why?

 Let’s take a closer look. 

One way to think about it is, if you begin by correctly choosing the door with the car, and you stick with this choice, you have a 1/3 chance of winning, since you chose one out of three doors. This means that if you switch to the other door, you must have a 2/3 chance of winning, since the probabilities must always add up to 1.

Here’s another explanation, in case the above reasoning didn’t click for you. Let’s assume the car is behind door A, and goats are behind doors B and C. The contestant’s initial choices are the following:

  • The contestant begins by choosing door A, the door with the car.
  • The contestant begins by choosing door B, a door with a goat.
  • The contestant begins by choosing door C, a door with a goat.

After this initial choice, an unchosen door is opened to reveal a goat. For each of the three initial choices, determine whether the contestant wins if she switches. For option 1 above, if she switches, she loses the car. For option 2, if she switches, she wins the car. And for option 3, if she switches, she wins the car. In other words, if her initial guess is wrong and she switches, she will always win the car. Thus, by switching, she wins 2/3 of the time, and by sticking with the initial guess, she wins only 1/3 of the time.

For a third (excellent) explanation, check out this video. And if you ever find yourself in this situation, you now know the logical choice to make!


From Visually.

Learn math with my online courses

My online math courses 

Want to learn more about math? I have step-by-step courses for that. :)