Explanation of the Monty Hall Problem

106 115
The Monty Hall Problem, named after the host of the classic television game show Let’s Make a Deal, is a result from probability that seems paradoxical at first. There are many ways to justify the solution to the problem, and one of these is an exercise in conditional probability.

Statement of the Problem

Suppose that you are on a game show in which you are shown three closed doors. Behind one of the doors is a fabulous prize, such as a new car.

The other two doors contain undesirable prizes such as goats. The host allows you to choose any of the three doors. The door remains closed. The host of the show, who knows what is behind each of the doors, opens a door that has a goat behind it.

The host offers you a choice. You can stay with the door that you originally picked, or you can switch to the other unopened door. After your decision has been made all doors will be opened and you be awarded what lies behind the door that you chose.

The key question in the Monty Hall Problem is: should switch doors or stay with the one that you initially picked?

The Answer

While it may seem like it wouldn’t matter what you do, it turns out that one of the strategies is better than the other. If you switch doors, you will have a 2/3 probability of winning the car, whereas if you do not switch you will only have a 1/3 probability of winning the car. This may seem quite strange, so we will explain why this is the answer.


There are many explanations to the Monty Hall Problem.

One of these involves conditional probability. Suppose that you initially pick door # 1. There is a 1/3 probability that the car is behind this door. Given that door one was selected, the game show host will open either door #2 or door #3, each with probability of 1/2. The probabilities are multiplied together giving the following:
  • Probability of the car being behind door #1 and door #2 being opened: 1/3 x 1/2 = 1/6.
  • Probability of the car being behind door #1 and door #3 being opened: 1/3 x 1/2 = 1/6.
In both of these situations, switching doors will result in your losing the car. This means that switching for these two scenarios results in a 1/6 + 1/6 = 1/3 probability of losing.
Still suppose that your initial pick was door #1, but this time the car is behind door #2. The probability of this being the location of the car is 1/3. The host has to open door #3, as he cannot reveal the car. This has probability of 1. A similar argument applies for your initially picking door #3.
  • Probability of the car being behind door #2 and door #3 being opened: 1/3 x 1 = 1/3.
  • Probability of the car being behind door #3 and door #2 being opened: 1/3 x 1 = 1/3.
In both of these cases, switching doors from door #1 to the other unopened door results in winning the car. This has probability of 1/3 + 1/3 = 2/3. Therefore switching doors is the better strategy.

Alternate Explanation

Sometimes the above explanation does not make much sense. An alternative way to think about this is to consider 100 doors. Only one has a car behind it, the other 99 doors have goats. You pick a door, the host opens all of the doors but your door and one other one. What is more likely, that you initially picked the correct door with the car, or that you picked incorrectly and the door with the car behind it is the other unopened door?


The Monty Hall Problem has a history of stirring up trouble. Due to its counterintuitive nature, there has been much scrutiny of the problem, and many journal articles deal with subtleties. This is a great illustration of how strange the subject of probability can be at times.
Subscribe to our newsletter
Sign up here to get the latest news, updates and special offers delivered directly to your inbox.
You can unsubscribe at any time

Leave A Reply

Your email address will not be published.