Want a new car? Switch to the other door

By Jeffrey Ayers

Kevin Standlee/Flickr
Kevin Standlee/Flickr

Believe it or not, knowing math can make the difference between winning a new car or winning a pet goat. Don’t believe me? Let me tell you why. There is a very famous probability problem whose answer has been the cause of much debate. It’s called the Monty Hall problem, named after the very first host of the popular television show, “Let’s Make a Deal.”

The problem goes like this: In every episode of the show, a contestant is picked out of the audience to get the chance to win a grand prize, which is usually a new car, a vacation or a lot of money. They are presented with three large curtains. Behind two of them is something you don’t want, like a goat (or maybe you do want a goat; I’m not sure), but behind one of the curtains is your dream prize.

Contestants have to pick one of the curtains (curtain number one, for example). Once they have made their choice, Monty will make things more interesting by revealing what is behind one of the other curtains (curtain number three, for example). Behind curtain number three is a lesser prize such as a goat, and Monty might ask, “Do you want to keep curtain number one, or would you like to switch to curtain number two?” The Monty Hall Problem tries to determine whether you are better off sticking with your original answer or changing it. So what do you think?

I’m sure many of you will say that it doesn’t matter; you have a 50/50 shot at winning the car whether you keep your original curtain or switch but, incredibly, this is not correct. In fact, you have a two out of three chance of winning your car if you switch your curtain.

When you start out with the three curtains, you have a one in three chance of selecting the correct answer. Therefore, the other two curtains must contain the remaining two out of three odds. When Monty reveals one of the “bad” curtains, he is changing the odds in your favor. Monty is removing one of your options and in doing so, shifting the two out of three chance to the one remaining curtain. The host does not improve your chance of being right, but he does improve the chance of the other door being right.

Still not making sense? Allow me to lower the odds. Say Monty is in a bad mood and offers you a choice of 1,000 curtains. You are asked to select one and let’s say you choose curtain number 37. Now Monty does the same thing, removing all but one curtain from the equation. He keeps curtain number 763 and asks if you want to switch. I would bet that in that instance, everyone would switch. I mean, why are you going to keep your curtain that has a one in 1,000 chance of being right when Monty has systematically removed 998 other curtains. Curtain number 763 has beaten out 998 other curtains to become one of only two curtains left. Surely you’re going to like the 999/1,000 odds that have shifted to that curtain. The same idea remains even when you only have three curtains.

You probably still have one burning question, why are the odds not 50/50? It would make sense that given two curtains your odds of randomly selecting a prize are one in two, and you’re absolutely right – if you knew nothing about those curtains. But given that you know something about one of the curtains (that it’s the incorrect choice), then the odds have shifted in favor of the remaining curtain, ensuring that you will win a prize two-thirds of the time.

I hope that one day you will be fortunate enough to be on “Let’s Make a Deal,” and should you be given the option to switch curtains, please do it. The odds are in your favor.

Jeffrey Ayers is a Collegian columnist and can be reached at [email protected]