Monty Hall Problem

I think most of the people who read this sentence don’t understand Monty Hall problem and while searching, they came across with this article. Also, I didn’t understand the problem until two days ago. I had to pass this topic that I took in probability course 3 years ago without understanding because I had to look another topics and courses. Last week, it entered my mind and I thought “Maybe, All of the mathematicians think wrong. I would become one of the most famous mathematician in the world by proving this solution is wrong.”

Anyone don’t know about the problem may watch “Blackjack 21” movie scene in that problem is well explained. (There is a turkish subtitle because my native language is Turkish. I didn’t want to add one more video.)

After watching the video, people generally think ” Are you kidding me? Isn’t the chance of both doors the same?” Strictly speaking, movie couldn’t explain clearly. I thought and searched to find the answer. Unhappily, I couldn’t find the fault of mathematicians so, changing the door increases your chance but I understood what the main reason is behind the problem. There is no mystic and facinating calculations in the problem. This problem has been the subject of movies and TV series, and there are many articles and videos on the internet that cannot be counted.

The essence of the problem

I saw things trying to find the truth in such a ridiculous way that confused me. I will not mention them. I focused on the essence of the topic with a question. “Why is not the chance of doors same and 1/2 after the game show host open a door?” For the answer to be 1/2, the game show host should have the freedom to choose the door. Let me explain: There are three doors. A,B and C. The contestant chose “Door A” and There is a goat behind the door.  So, can game show host open “Door A” to show the goat as he know there is a goat behind it ? No because the game show host ask to contestant “Do you want to change your door?” If he open “Door A”, contestant will not have a door to stay. It is the difference; the choice of contestant protect a goats behind “Door A” not to reveal. It limits the number of doors that the game show host can open. In fact, the game show host has only one option even though there are two goats because he cannot choose “Door A” if there is a goat behind it. The game show host decreases the risk of doors that contestant didn’t choose. So, changing the door increases the chances of winning. Having 3 doors in competition would confuse your mind. If there are 4 doors(A, B, C and D) and you choose “Door A” and after that the game host open “Door B”, you can choose Door C or D to switch and the chances of winning of these two doors are the same. I hope you understand. I will show it with Bayes theorem.  If you see there are some unclear points, I can improve the article.

Monty Hall Proof with Bayes Theorem

I will not explain Bayes theorem because it takes too long. You can search on internet. There are a lot of resources about it. It may not help if you read it down without knowing the Bayes theorem. I can explain expressions shortly:

P(CarA | B); Probability of the car being behind Door A in case the  host has opened Door B

P(B | CarA); Probability of the host opening Door B if the car is behind Door A

P(CarA); Probability of the car being behind Door A

We can express formula like that:

Now, think the game show host opened a door which there is a goat behind, after contestant had chosen a door. For example, the contestant chose Door A. The host opened Door B and asked whether want to change or not the Door A with Door C. The probability of winning if he change the Door write like that:

The numbers were written in the order of the above formula. What this formula means is that when the contestant selects Door A and the host opens Door B, the probability of car being behind the Door C is 2/3. So, it is twice of the probability of car being behind the Door A. I want you to focus on the blue rectangle. In the rectangle, probability of car being behind the Door C and probability of the host opening Door B if the car is behind Door C are multiplied. The car is likely to be at Door A, B or C, so 1/3 but if car is behind the Door C, why is the probability of opening Door B by the host 1? If car is behind the Door C, there is also a goat behind the Door A which was chosen by contestant. Can it be chosen? No, it cannot be! That is the difference. He have to choose Door B so, it increases chance of Door C.

Let’s say the host doesn’t care what the contestant chose. He can open any door. Let’s say the contestant had chose Door A and the host opened Door A or he said that he also will open a door that there is a goat behind and it can be Door A and the contestant have to choose a new door if it can happen. The host doesn’t limit himself. Let’s say he opened Door B. In this situation, Should the contestant change the door? The formula of this expression look like that:

No! He can stay or not. The result is always same. The probability of winning is same and 1/2. Probability of choosing a door by the host is 1/2 if the car is behind the Door C because he know there is a goat behind Door A and he can choose also this door. In the same way, probability of car being behind Door A is 1/2.  Both probabilities are same. I hope it’s clear. The person seen in the top of the article is Monty Hall. His program name is “Let’s Make a Deal.” The problem became famous with him and it is called with his name.

Yazının Türkçesine buradan ulaşabilirsiniz.

Leave a Reply

Your email address will not be published. Required fields are marked *