The Monty Hall Statistics Problem

By Caroline Wells

What would you do if you were given the chance to choose among three doors, one of which holding a car behind it while the other two present a goat? If you were given the chance to edit your choice (either choose to stay with the previously chosen door or switch to a different one), would you?

This is the context presented in the famous Monty Hall statistics problem: contestants in the game show Let’s Make a Deal are given the choice among three doors and then the chance to alter their decisions after Monty Hall (the host) opens one of the two remaining doors revealing a goat. This game may appear to be based upon pure luck; however, there is a correct answer. So, what is the answer and why?

We can use concepts of probability, the predicting of random events, to help solve this. Let’s start with a scenario in which a contestant decides to stay with the original choice. The probability of winning the car would be ⅓, because only 1 door out of the 3 presented holds a car. Likewise, the probability of losing would be ⅔.

Things become more complicated if the contestant chooses to switch. If a contestant originally chooses a door holding a goat, Monty Hall presents the other goat-holding door, and the contestant decides to switch to the other remaining door, he or she would win the car, because the car is the only door left.

With this being said, the probability of winning when switching doors equals the probability of losing when sticking with the original decision, or the original choice bearing a goat-holding door. We now know that in order to win in a world where the contestant is switching their choice, the contestant must originally choose wrong in order to switch to the correct door.

Switching when originally incorrect results in a win, and the only way to lose when switching is when the original choice is correct, the probability of which being ⅓.

Let’s tie it all together. Without switching, the probability of choosing the one correct door out of three is ⅓. When the option to change comes into play, the contestant will switch to the door opposite (without knowing) of the original choice. So, if the original choice was a goat-bearing door, the contestant would switch to the winning car-holding door, and if the original choice was the car-holding door, the contestant would switch to the unrevealed goat-bearing door. With this in mind, because the probability of originally choosing incorrectly is more likely, it is thus more likely to choose the correct door when switching. So, because the probability of winning when switching doors is ⅔ while the probability of winning when staying with the original choice is only ⅓, switching may be the better option.

The world of statistics is extremely fascinating, and the use of probability in the Monty Hall problem shows its power to predict the outcome of and the logic behind seemingly random events.