This is an old mathematical ‘chestnut’ that has been published quite a few times in the mathematics literature over the last 100 years.

This version of the story is true, and comes from an American tv game show.

Here is the situation:

Finalists in a tv game show are invited up onto the stage, where there are three closed doors. The host explains that behind one of the doors is the star prize – a car. Behind each of the other two doors is just a goat. Obviously the contestant wants to win the car, but does not know which door conceals the car.

The host invites the contestant to choose one of the three doors. Let us suppose that our contestant chooses door number 3. Now, our host does not initially open the door chosen by the contestant. Instead he opens one of the other doors – let us say it is door number 1. Now the door that the host opens will always reveal a goat. Remember the host knows what is behind every door!

Lucky you didn’t choose THAT door, says the host – as you can see there is just a goat there. The contestant is now asked if they want to stick with their original choice, or if they want to change their mind, and choose the other remaining door that has not yet been opened. In this case number 2. The tension in the studio rises, and the studio audience shout out suggestions. But amid all this excitement what is the best strategy for the contestant? Does it make any difference whether they change their mind or stick with the original choice?

The answer to this question is not intuitive. Basically, conditional probability theory says that if the contestant changes their mind, the odds of them winning the car will double. If they stick with their original choice, they have a 1 in 3 chance of winning the car, and if they switch they have a 2 in 3 chance of winning. And over many episodes of the tv show, the facts supported the mathematics – those people that changed their mind did indeed double their chances of winning the car.

Some people say, well there are just two doors left, and each has an equal chance of being right, so now the contestant has a 50% chance of picking the right door. Right? Well, no, actually. Now it IS true that if at this point – with just the two doors left – you brought a new contestant into the studio, who had seen nothing of what had gone before, and gave them the choice of the two doors, THEY would have a 50% chance of picking the winning door. But that is not the situation. Our contestant picked out a door from the original line-up of 3 doors, and that original choice of door had a 1 in 3 chance of being right, and it STILL has a 1 in 3 chance of being right. Whereas if they switch, they will have a 2 in 3 chance of being right.

One way of putting it is you are choosing between your original choice of door and BOTH the other doors, but with the host having conveniently shown you which of the other two doors NOT to choose – so you are really choosing BOTH the other two doors, which gives you a 2 in 3 chance of being right.

Another way of putting it is to think of a situation with 100 doors, but only 1 prize, and 99 goats. Your initial choice of door has a 1 in 100 chance of being right. The host then runs up and down opening doors all over the place, the stage is full of goats, and finally there are only two doors left. Your original choice of door, and one other. I hope you will agree that your original choice still has a 1 in 100 chance of being right, and the one remaining other door has a 99 in 100 chance of being right?

Some years ago, when the writer and journalist Marilyn vos Savant quoted this puzzle in a US newspaper article, she received over 10,000 letters mostly telling her she was wrong. One was from Robert Sachs, a professor of mathematics at George Mason University in Fairfax, Va. who said “As a professional mathematician, I’m very concerned with the general public’s lack of mathematical skills. Please help by confessing your error and, in the future, being more careful.” However a week later and Dr. Sachs wrote her another letter telling her that “after removing my foot from my mouth I’m now eating humble pie. I vowed as penance to answer all the people who wrote to castigate me. It’s been an intense professional embarrassment.”

One way of exploring this puzzle further is to do an experiment. Get a friend to set up the 3 doors – not real doors obviously, but you could use 3 pieces of cardboard on the table, one with a paper car underneath, and two with paper goats. Or just have the car, and assume the others are goats. Get the friend to set up the puzzle while you are out of the room, and then you come in and choose a door. The friend then reveals one of the other doors as not having the car behind, whereupon you decide to stick with your original choice, or to switch doors. Try the experiment 10 or 20 times, and keep a record of whether when you stick you win or lose, and when you switch whether you win or lose. The more times do you the experiment, the more the results will show 0.33333 chance of winning if you stick, and 0.66666 chance of winning if you switch.