The Monty Hall Problem: Understanding the Paradox

If you're a fan of game shows or math puzzles, you might have heard of the Monty Hall problem. It's a classic probability puzzle that has confused and confounded people for decades. In this blog post, we'll take a deep dive into the Monty Hall problem, what it is, how it works, and why it's so confusing. We'll explain it in easy words, with headings and subheadings to make it easier to understand.

What is the Monty Hall problem?

The Monty Hall problem is a probability puzzle based on a game show called Let's Make a Deal, which aired in the 1960s and 1970s. In the show, the host, Monty Hall, would present a contestant with three doors. Behind one of the doors was a valuable prize, like a car, and behind the other two doors were goats. The contestant would choose one of the doors, and then Monty would open one of the other doors to reveal a goat. The contestant would then be given the option to switch their choice to the remaining door or stick with their original choice. The question is, should the contestant switch or stay with their original choice?

The problem has become famous because the answer is not what most people expect. In fact, the correct answer is so counterintuitive that even mathematicians and statisticians have been known to argue about it.

The basic setup of the Monty Hall problem can be summarized in the following steps:

1. A contestant is presented with three doors, behind one of which is a valuable prize and behind the other two are goats.

2. The contestant chooses one of the doors.

3. The host, who knows what is behind each door, opens one of the other two doors to reveal a goat.

4. The contestant is then given the option to stick with their original choice or switch to the remaining door.

Why is the Monty Hall problem so confusing?

The Monty Hall problem is confusing because it goes against our intuition. Most people assume that there is a 50/50 chance of the prize being behind either of the two remaining doors, and that switching or staying with the original choice would not make a difference. However, this assumption is incorrect.

To understand why, let's break down the probabilities involved in the Monty Hall problem.

Probability analysis of the Monty Hall problem

The Monty Hall problem can be solved using conditional probability. Conditional probability is the probability of an event occurring given that another event has already occurred. In the case of the Monty Hall problem, the conditional probabilities are as follows:

1. The probability of the prize being behind Door 1 (the contestant's original choice) is 1/3.

2. The probability of the prize being behind Door 2 (the remaining door after one of the doors has been opened) is 2/3.

Let's take a closer look at how we arrive at these probabilities.

Step 1: The probability of the prize being behind each door is 1/3.

When the contestant first chooses a door, there is a 1/3 chance that they have chosen the door with the prize behind it. This is because there are three doors and only one prize, so the probability of choosing the correct door is 1/3.

Step 2: The probability of the prize being behind the remaining door is 2/3.

After the host opens one of the other two doors to reveal a goat, there are only two doors left - the one the contestant chose and the other remaining one. At this point, the probability of the prize being behind either door is no longer 1/3 each, as many people assume. Instead, the probability is now 1/2 for the remaining door that was not chosen by the contestant, and 1/2 for the door that the contestant originally chose. 

To understand why this is the case, we can consider what would happen if the contestant always stuck with their original choice. In this scenario, the contestant would win the prize if and only if they initially chose the door with the prize behind it. Therefore, the probability of winning the prize in this scenario is 1/3. 

Now consider what would happen if the contestant always switched to the remaining door after one of the other doors was opened to reveal a goat. In this scenario, the contestant would win the prize if and only if they initially chose a door with a goat behind it. This is because the host will always reveal a door with a goat behind it, and therefore the remaining door must be the one with the prize behind it. Since there are two doors with goats behind them, the probability of winning the prize in this scenario is 2/3. 

So, by switching to the remaining door, the contestant has a higher probability of winning the prize. This is why the Monty Hall problem is so confusing - it goes against our intuition that the probability of the prize being behind each door should be 1/2 after one of the other doors has been opened.

Why do people struggle with the Monty Hall problem?

There are a few reasons why people struggle with the Monty Hall problem. One is that it's counterintuitive - our brains are wired to think in certain ways, and the Monty Hall problem goes against these instincts. Another reason is that people often make assumptions about the problem without fully understanding the probabilities involved. For example, many people assume that the probability of the prize being behind each door is 1/2 after one of the other doors has been opened, when in fact it is 2/3 for the remaining door.

How can we approach the Monty Hall problem?

One way to approach the Monty Hall problem is to use probability theory and conditional probability, as we did above. Another way is to use simulation. We can simulate the Monty Hall problem by writing a computer program that randomly assigns the prize behind one of the three doors, then randomly selects a door for the contestant, and finally randomly opens one of the other two doors to reveal a goat. We can then run the simulation many times and see how often the contestant wins the prize by switching or staying with their original choice. The results will show that the contestant has a higher probability of winning the prize by switching to the remaining door.

Conclusion

The Monty Hall problem is a classic probability puzzle that has confounded and confused people for decades. It goes against our intuition, but it can be solved using probability theory and conditional probability. The key takeaway is that by switching to the remaining door after one of the other doors has been opened, the contestant has a higher probability of winning the prize. Understanding the Monty Hall problem can help us to think more critically about probability and avoid common misconceptions.