RIP Monty Hall: Explaining the Monty Hall Problem

Monty Hall Problem

Update: This article was originally posted in May 2010. We’ve just learned that Canadian-American game show host Monty Hall died last Saturday at the age of 96. Read more about the puzzle that was named after him below.

It’s a fact: a lot of people get confused over probability problems. For example, most geek reading this site will probably realize that if a coin is tossed ninety-nine times and comes up heads every time, then in probability terms, a hundredth toss still has an equal shot of coming up heads or tails. Those who think otherwise are either misreading the question (looking at the probability of the hundred-head streak, not the individual toss) or are sensible enough to take a practical approach (if a real coin comes up heads ninety-nine straight times, it’s probably weighted.)

But other probability issues cause much more headscratching. One of the most famous is the so-called Monty Hall problem which is written in several different ways but follows these principles:

  • There are three doors with a car and two goats placed behind them at random. The game show host knows which is placed where.
  • You must start off by choosing a door.
  • The game show host opens one of the two doors which you did not choose, revealing a goat. (He or she will always open a door that will reveal a goat. He will never open a door which will reveal the car.)
  • The host then offers you the chance to change your original pick.

The question is whether it is better to change or stick with your original choice. The answer — which can be and regularly has been demonstrated by running the scenario over and over — is that you are more likely to win if you change. But many, if not most people simply can’t process this and insist that it cannot make any difference whether or not you switch and that your chances of winning are the same either way.

What fascinates me is that the real puzzle isn’t what the correct answer is, but rather how to explain it in a way that people can understand why their initial belief is incorrect. As an exercise in both linguistic and numerical persuasion, I suspect it may be that different explanations will be more effective for different people. Here’s a few of my attempts:

Explanation one:

(This is the most basic explanation, but even when people read and accept it, they sometimes still can’t overcome their original “logic.”)

There are three equal possibilities. For example, if you originally pick A:

1) It’s in A (sticking wins)
2) It’s in B, so C is revealed (switching wins)
3) It’s in C, so B is revealed (switching wins)

So switching is more likely to win.

Explanation two:

One way to understand it is to realize that because the host opens one of the doors you don’t pick, you can treat both of the two doors you don’t pick as effectively being one singular option.

So say that the doors are A, B and C and you pick A. There’s a 1 in 3 chance that the car’s behind door A and a 2/3 chance that it’s behind either door B or C.

What you then have to take into account is that if the car isn’t behind A, it doesn’t matter whether it’s behind B or C: If it’s behind B, the host reveals the goat behind C. If it’s behind C, the host reveals the goat behind B. Either way, it’s always been more likely that the car is behind B/C than A. Your first guess will always probably be wrong, which is why you are better off switching once you no longer need to worry about having to choose between B and C.

Explanation three:

Even though it’s a mathematical problem, it can be better understood by looking at the situation in a completely literal and practical manner. What is physically behind the doors never changes. That’s why you can’t apply mathematical “logic” after the reveal and call it a 50-50 chance. The prize goes behind one door at the start. Either it’s behind the door you choose first, or it isn’t. What happens with the reveal doesn’t physically change that by making it more or less likely.

Thinking it’s a 50-50 chance is taking a logical approach to a *different* situation. If you were to put a prize behind one of two doors, it would indeed give a 50-50 chance of being behind either door. But that hasn’t happened here. The prize is already in position *before* you come to the point where two boxes are left.

To say the same thing a different way: Probability relates to random events, not to states. The random event in this situation is the placing of the car and goats. Selecting a door to open, whether that be by the contestant or the host, has no bearing on this event.

Explanation four:

Regardless of the wording he or she uses, the two questions the host asks you are actually very different. The original question is to guess the correct door, which is difficult (1 in 3 chance). The question after the reveal shouldn’t be thought of as a repetition “What door is correct?” but rather “Was your original guess wrong?” And as we’ve just established, the chances are it was.

So, are you any clearer now? Or did you understand it already? And if so, have you got a clearer way to explain it?

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