The easiest explanation to the Monty Hall problem

The last one you are going to need to look at

Anup Sebastian
Feb 13 · 5 min read

The Monty Hall problem is a one of a kind math problem that has puzzled everyone from Average Joes to world renowned mathematicians. It has a really simple premise, that is easily understood, yet solutions and explanations to it have been reasons for some of the greatest debates on the internet, only to be surpassed by the “Dress” and “Laurel or Yanny”. The biggest reason for this is most ways of approaching the solution to this problem can be counter intuitive. This explanation should put away your questions once and for all, and also allow you to explain it very easily to others.

The problem

It gets it’s name from Monty Hall, the host of the game show “Let’s make a Deal”. It was originally posed by Steve Selvin an American Mathematician in 1975, but was popularized by the show — hence the name.

Okay, let’s now have a look at the problem.

Say, there are three doors as shown below. Behind two of them are goats, and behind one is a car, and if you pick the door with the car, you win.

Simple enough, but here’s the catch. After you pick a door, Monty always opens one of the doors with a goat behind it, and asks you if you want to switch to the other door. Do you switch or do you stick to your original choice?

This is what confuses people. Your first instinct is to think, well, it doesn’t matter, now there are 2 doors and there’s a 50% chance that the car is behind either of them. But you would be wrong. The solution is to always switch. In fact, you double your odds of winning by just switching.

Confused? Yeah, everyone usually is at this point. So now let’s break it down step by step visually.

Say you picked the first door, the chance that it is behind your door is 1/3 and the chance that it is behind one of the other doors is 2/3. Agreed?

In other words, the chance that the car is behind your door is 1/3, and the chance that is is behind the ones you did not pick is 2/3. In our image above 1/3 of the odds are on the left and 2/3 are on the right.

Now Monty opens one of the doors you did not pick — one of the doors from the right side of the image. It is always a goat behind.

The odds still have not changed, left side (original choice) has 1/3 odds and right side (all other choices together) have 2/3 odds, except now that one door is open, the total odds on the right side is on that one door.

If you did not get it still, lets try with more doors, say 10.

You pick a random door, and in our diagram lets put your choice on the left and the rest on the right. There is a 1/10 chance that you are right, and 9/10 chance that you are wrong. Or in other words, 1/10 chance that there is a car behind your door, and 9/10 chance that it is behind any one of the other doors.

Now Monty opens 8 doors from the ones that you did not originally choose, all of them reveal a goat. Do you switch or stay? Right now, you are probably screaming, of course switch. After the doors are opened, there is a 9/10 chance that the remaining door on the right has the car behind it.

Just think of the act of Monty opening all the other doors as allowing you to pick all the 9 doors ion the right at the same time instead of just one.

That’s it. It’s that simple. It is important to remember that this is a game of probability so you could still lose. The solution is about strategy. What you should do to maximize odds or do to win the most times.

Let’s take it further

Let’s say Monty decides not to take it so easy on you and opens only one door from the right. What do you do? Switch to one of the ones on the right or stay?

For this we have to do a little math. Probability of winning on the right side is 0.9(Total Odds with unselected doors). To calculate the probability of winning for unopened door on the right, we do

Slightly higher than your odds on the original door of 0.1, so you should still switch, granted your chances of winning are much lower. You can calculate the your chances of winning for any number of doors with any combination of opened doors using the same procedure.

So, that’s the Monty Hall problem in a nutshell, in as simple of an explanation as I can put it. This is a different way of looking at it compared to the conventional way on most other websites. Feel free to solidify your understanding by looking at other explanations to this problem. I hope that was useful to you.

Images from:

Laurel-Yanny dress:

Doors and Goats:

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Anup Sebastian

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Currently studying Data Science

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