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THE MONTY HALL PROBLEM

royston

Administrator
Staff member
This problem amused me at work when we all had a big row about it - try it out on you colleagues

I have taken this solution from the Grand Illusions Site but you can find more formal solutions all over the Web

This 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, the 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. The door that the host opens will always reveal a goat. Remember the host knows what is behind every door!

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 studio audience shout suggestions. 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, the theory says that if the contestant changes their mind, the odds of them winning the car double. And over many episodes of the tv show, the facts supported the theory - those people that changed their mind had double the chance of winning the car.

Why should this be so? After all, the contestant doesn't know which door the car is behind, and so the chance of the car being behind any one particular door is one third, isn't it? So surely the chance of winning the car if they stick with their original choice is one third, and the chance of winning the car if they change their mind is also one third? How can the odds double?

The answer goes like this. When the contestant makes their first choice, the chance of them being correct is indeed one in three, or one third. And if, after the host has done his patter and opened another door, they stick with their original choice, then their chance of being correct is unaltered, it is one third. Indeed, how could this possibly change?

We get some pretty abusive e-mails here at Grand Illusions telling us we are talking rubbish. But before sounding off at us, take note - this is an old mathematical puzzle, and has been published in probability textbooks for over 100 years. However, as we said, it is counter-intuitive, and the maths that most people do in school does not cover this.

This is not an example of simple probability (suppose there are two doors, therefore there is a 1 in 2 chance of the car being behind either of the doors). This is an example of conditional probability: what is the chance of something happening, given that something else already has.

OK, so how does this puzzle work? Here is one way of explaining it. Let us assume that our contestant has chosen door 3. And let us assume that our contest never changes their mind. There are three equally likely possibilities-

Door 1 Door 2 Door 3
car goat goat lose
goat car goat lose
goat goat car win

In one case out of three, the contestant will win the car. The green background shows the door that the host opened. Only in the last case did he have a choice, when he could have opened either door 1 or door 2.

Now let us consider what happens if the contestant always changes their mind. Again they initially pick door 3.

Door 1 Door 2 Door 3
car goat goat win
goat car goat win
goat goat car lose

Their chance of being right initially is still only 1 in 3. But now in both the first two cases the host opens a door revealing a goat, and the contestant changes their mind - and 2 times out of 3 they will be right.

So, if the contestant sticks with their original choice, they will win the car 1 time in 3, and if they switch doors, they will win the car 2 times in 3.

Some of you will still be shaking your heads at this, and saying we are wrong. So here is another way of thinking about it. Imagine there are 100 doors, with a car behind only 1 of them. You choose a door. Your chance of being right is 1 in 100, right? Now the host opens 98 of the remaining 99 doors, in each case revealing a goat. You can now either stick with your original choice, or you can switch to the one remaining door that is closed. We say if you stick with your original choice, you still have a 1 in 100 chance of being right. And if you switch, you have a 99 in 100 chance of being right.

When Marilyn vos Savant quoted this puzzle in the US a few years ago, 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."

Well, we said it was counter-intuitive. Even professional mathematicians get it wrong.
 

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