Will You Ride Home In A Goat or Car?
In today’s post, I will explain about a probability-based puzzle which caused quite a flutter among mathematicians and statisticians in 1990. It is called the Monty Hall Problem.
Though the problem was initially posed in a letter by Steve Selvin, a statistician to the American Statistician (a peer-reviewed journal mostly covering statistics), it became famous as a question from a reader’s letter quoted in Marilyn vos Savant’s “Ask Marilyn” column in Parade Magazine in 1990
The Problem
The problem is called the Monty Hall Problem and is loosely based on the American television game show Let’s Make A Deal and named after its original host, Monty Hall. I will state the problem below as was stated in the letter sent to the “Ask Marilyn” column:
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say №1, and the host, who knows what’s behind the doors, opens another door, say №3, which has a goat. He then says to you, “Do you want to pick door №2?” Is it to your advantage to switch your choice?
As stated in the letter sent to the “Ask Marilyn” column
Based on whether you stay on the door or choose to switch, you can go home either riding a goat or a car.
The basic question here is: What should you do to maximize your chances of winning a car?
a. Stay with your initial choice of door
b. Switch to the other unopened door
c. It doesn’t make a difference. The chances are 50–50
(You might want to give some time and think about it before reading further.)
What Should You Do?
So, if you thought it doesn’t make a difference and went with option C, you might be happy to know that this was the same answer given by many people, including Ph.D’s when they first came across this puzzle. But, that is not the correct answer!!!
Vos Savant’s reply was that the contestant should switch to the other door. A contestant who switches doors has a 2/3(66%) chance of winning a car while contestants who chose to stick with their original choice had only a 1/3rd(33%) chance of winning a car
Many people who read Savant’s answer refused to accept it. Even many PhD’s wrote to the magazine claiming she was wrong. Even Paul Erdos, one of the world’s most prolific and celebrated mathematicians refused to believe until he was shown a computer simulation which confirmed that Savant’s answer was right.
Reasoning 1
One way to approach this is by using the method of elimination or, to put it in the words of Sherlock Holmes:
Suppose, you pick one of the doors with a goat by random chance. What is the probability of it happening? Since there are 2 goats, it is 2/3(66%)
Now remaining are 2 doors: One with the car and one with another goat. The host knows what is behind each door. He opens the door with the goat and shows it to you.
Out of the 2 doors with goats, one has been chosen by you and one has been opened by the host. So, what remains is the door with the car which is unopened. So, it is evident you should switch. The probability of this is 2/3
Reasoning 2
For those of you who are unable to digest the above explanation, allow me to explain on a case by case basis. Let us run through all the possible cases and evaluate them:
There are 3 doors for you to choose from and 3 doors behind which the car could be. So, there are a total of 9 possibilities.
Suppose I choose Door 1 and:
- If the car is in Door 1, in order to win, I should not switch
- If the car is in Door 2, in order to win, I should switch
- If the car is in Door 3, in order to win, I should switch
Similarly, we can do for Door 2 & Door 3, which I have tabulated below:
As you can see, it is evident that when you Switch Doors, you win a car 66% of the time.
This does not mean that you will always win if you switch, just that on switching, you get a higher chance of winning the car.
Why Does our Brain Scream 50–50?
Assume you choose Door №1. That is a completely random event(it is equiprobable that you would choose any of the 3 doors). After that, the host opens one of the doors out of the remaining 2.
Now, you might think, there are 2 closed doors. I have to choose any 1(i.e either stay or switch to the other door). It is a 50–50 chance. So, what difference does it make whether I stay or switch doors?
The difference is that, the act of the host choosing the door is not random. He does not do it randomly. The host already knows what is behind each door. He selectively opens a door such that he does not reveal the prize.
When he asks you to stay or switch, it is not a 50–50 chance as many people think it is. Our mental assumptions do not match the actual process of how the game works
There is a reference to the Monty Hall problem in the English Movie “21” starring Kevin Spacey that was released in 2008. You can watch it here below(watch from 1:03)
For another similar puzzler on probability, you can read my post on the Birthday Paradox in my blog.
Thanks for reading!!!
