The Monty Hall Problem

Imagine you are on a game show. The host offers three doors: Door A, Door B, and Door C. Behind one of the doors is a brand new car, and behind the other two, goats. The goal, obviously, is to pick the door with the car. The host of the show knows which door contains the car, and likewise, which two doors contain the goats. Say you select Door B. The host, knowing what’s behind the door you selected, opens a different door, Door C, and behind it is a goat. With the knowledge that Door C is a goat, the host offers you two options: stay on Door B and see what’s behind it, or switch doors and see what’s behind Door A. The question, then, is as follows: does switching to Door A with the knowledge that behind Door C is a goat:
  1. increase your chances of winning the car, or
  2. decrease your chances of winning the car, or
  3. make no difference to your chances of winning the car?

The Answer

If you answered (3), that it makes no difference if you switch, you’re not alone; a lot of people pick (3). At first glance, this is somewhat logical. The probability of randomly selecting the door with the car when there are three doors is obviously 1/3. But by opening one of the doors, the host is taking a door out of the mix. Thus, it becomes a 50-50 chance of whether the door you picked is a car or a goat… right?

Actually, no. The true answer to the question is really (1), that switching doors increases your chances of winning the car. As a matter of fact, switching doors doubles your chance of winning the car, and thus, you should always switch doors.

You can set aside your skepticism now—this famous problem of probability theory, dubbed the “Monty Hall Problem”, has been formally proven over and over again by statisticians and mathematicians, students and doctorates, all around the world, and it has also been simulated innumerable times. The conclusion is abundantly and overwhelmingly clear: if you switch doors, your probability of winning the car rises to 2/3.

I can explain why.

Explanation

When you have three doors with one car, the probability of correctly selecting the door with the car by random chance alone is 1/3. Similarly, the probability of incorrectly selecting the door with the car, and instead selecting a door with a goat, is 2/3. Clearly, you are initially much more likely to choose a goat than a car at the start—twice as likely, to be exact.
In the beginning, probability of selecting the car is 1/3. The probability of selecting a goat is 2/3.
For the purposes of this discussion, let’s say that behind Door A is the car, Door B a goat, and Door C another goat.

Let’s assume the player decides to stay on the same door every time. There are now three possible scenarios:

  1. The player selects Door A, the car. The host responds by opening either Door B or Door C (it doesn’t matter which one because behind both doors are goats). The player stays on Door A and wins the car.
  2. The player selects Door B, a goat. The hosts responds by opening Door C, the other goat. The player stays on Door B and wins a goat.
  3. The player selects Door C, a goat. The hosts responds by opening Door B, the other goat. The player stays on Door C and wins a goat.

Out of the three possible scenarios, only one of them results in winning the car. Thus, we can say that the probability of winning the car is 1/3 if the player decides to stay on the same door every time, about 33%. If you think about it, this makes sense. You never changed your initial decision, so the initial probability of getting a goat is unchanged—it’s still 1/3, regardless of whether you know which of the other doors is goat. Knowledge is powerful only when you do something with it.

Now, let’s assume the player decides to switch doors every time. There are again three possible scenarios:

  1. The player selects Door A, the car. The host responds by opening either Door B or Door C (once again, both doors have goats, so it doesn’t matter which one). The player switches to the door that wasn’t opened by the host, winning a goat.
  2. The player selects Door B, a goat. The host responds by opening Door C, the other goat. The player switches to Door A, and wins the car.
  3. The player selects Door C, a goat. The host responds by opening Door B, the other goat. The player switches to Door A, and wins the car.

This time, out of three possible scenarios, two of them definitively result in winning the car. Therefore, the probability of winning the car is 2/3 if the player decides to switch every time, or about 67%. 2/3 is twice 1/3; therefore, you are twice as likely to win the car if you decide to switch doors than if you decide to stay on the door you initially selected.

Probability tree

If you're a statistics or a math geek, here's the probability tree for this situation.
Monty tree door1

Simulating the Monty Hall Problem

The Monty Hall Problem can be simulated easily in a variety of ways. You can cut a piece of paper into three doors and draw two goats and a car on them respectively. I like to use three playing cards—a face card represents that car and two number cards represent goats.

However you like to do it, it doesn’t matter; the math stays the same.
  • Create two identical T-charts and label the columns on each “successes” and “failures”.
  • Firstly, you have to choose whether or not you are going to switch or stay. Let’s say you decide to switch every single time (like a smart person would!).
  • Lay your three representative doors on a table face down, so you can’t tell which door holds which item. Shuffle the doors thoroughly.
  • Then, randomly select one of the doors, but don’t flip it over.
  • Before you flip the door you selected, flip both of the two doors you didn’t select.
  • If one of the doors is a car, that means the door you selected was a goat, so by switching you land on the door with the car; thus, you should put one tally mark in the success column of your T-chart.
  • If both doors have goats, you did not win the car since your initial selection was the door with the car but you switched to a door with a goat, so put a tally mark in the failure column of your T-chart.
  • Repeat this for many trials until you see the pattern in the results. Theoretically, 67% of the trials should be successes, and 33% of the trials should be failures.
  • Once you have completed the trials, repeat the entire simulation. But this time, instead of deciding to switch doors every time, you decide to stay.
  • If one of the unselected doors is a car, then you fail to win the car since that would mean the door you selected is a goat, and thus, you should put a tally mark in the failure column.
  • If both unselected doors have goats, then you win the car since you would a selected the door with the car to begin with; thus, you should mark a success down.
  • Theoretically, you should be successful 33% of the time, and fail 67% of the time.

Conclusion

The Monty Hall Problem is a perfect example of how sometimes, the solution to a problem is not always the first thing that comes to mind. When the Monty Hall Problem was published in 1990 in Parade magazine, “approximately 10,000 readers, including 1,000 with Ph. Ds, wrote to the magazine, most of them claiming the explanation was wrong. Even when given explanations, simulations, and formal mathematical proofs, many people to this day still do not accept that switching is the best strategy. Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation confirming the predicted result” (Wikipedia). As you attack the countless challenges that life throws at you, it is important to remember to take a step back and examine all the facts and all the logic before jumping to the first conclusion that comes to mind.
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