Parade, a popular American magazine, has a section called ‘Ask Marilyn’. Readers send in questions on all kinds of topics, and Marilyn vos Savant, occasionally billed as “The World’s Most Intelligent Person”, answers them.

In 1990, she received the following question from one Craig F. Whitaker:

Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what’s behind the doors, opens another door, say number 3, which has a goat. He says to you, “Do you want to pick door number 2?” Is it to your advantage to switch your choice of doors?

Well, what do you think?

Before revealing Marilyn’s answer and discussing the problem a bit more, let’s clarify one aspect of the question. The answer depends on how you think the game show host behaves. Does he always show contestants what’s behind another door? Perhaps he does this only if they originally chose the door with the car. But let’s suppose not. Let’s suppose that he does always open another door and offer the contestant the chance to switch their choice. Also, since the host knows what’s behind the doors, let’s suppose he never opens the door with the car behind it (because that would spoil the fun rather).

OK, with these extra suppositions, would you switch?

Marilyn’s answer was that the contestant should switch. To be more precise, she argued that, if you switch, you will have two chances in three of winning the car (that is, a probability of 2/3 of winning), but if you stick with your original choice, you’ll only have one chance in three of winning (that is, a probability of 1/3). In other words, if you switch, you still might not win, but you’ve improved your chances.

Maybe that’s what you thought. But maybe not. Many people think along the following lines. After the host has shown the goat behind door number 3, there are two doors left that the car might be behind. You still don’t know which, so there’s one chance in two that the car is behind door 1, your original choice, and one chance in two that it’s behind door 2, the choice you’ll make if you switch. So it doesn’t make any difference if you switch. You can if you like, but there’s no disadvantage in sticking with your original choice.

If that’s what you thought, you’d be in good company. The original Parade article generated a huge correspondence, including letters from professors who advised Marilyn vos Savant to stick to what she knew and not mess about with questions of probability. Since then, there have been articles and correspondence in the press in many different countries, heated discussions on Internet mailing lists for statistical experts, several articles in learned journals, and even a couple of books that address the problem. The car and the goats continue to crop up again and again in the media. For instance, after the economist John Kay described the problem in his column in the Financial Times in August 2005, a correspondence about it ran for a fortnight.

OK, why do your chances of winning increase to two out of three if you switch? There are several ways to explain this. I’ll go through a couple of them. However, I’m aware from all the previous discussions that none of these ways might convince you! Different people are convinced by different arguments, and some never are.

One way to think of the problem is in terms of three different scenarios. There are three ‘prizes’: the car, goat 1 and goat 2. The possible scenarios are as follows.

1. The contestant originally picks goat 1. The host shows goat 2. The contestant will win the car by switching.
2. The contestant originally picks goat 2. The host shows goat 1. The contestant will win the car by switching.
3. The contestant originally picks the car. The host shows either one of the goats. The contestant will lose by switching.
These three scenarios are all equally likely (because the contestant has no idea where the car is, originally, and is making the original choice at random). In two of the scenarios, the contestant wins by switching. In the third, the contestant loses by switching. So the contestant has two chances in three of winning if they switch, and only one chance in three if they don’t switch.

Looking at it this way also shows what’s wrong with the argument that there are two possibilities after the host has opened the door — car behind door 1 and car behind door 2 — and that these are equally likely, so that it doesn’t matter whether you switch or not. There are indeed two possibilities, but they aren’t equally likely.

Convinced? Some people are persuaded by this; others can’t see why it makes sense to treat these three scenarios as equally likely.

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