Probability can be a tricky devil when it comes to intuition. It can fool you just like how you see what isn’t there in an optical illusion, or how you can hear two different words from the same sound. In this case, what seems like such a simple probability question leaves half the world with their brains exploding.

The Monty Hall Problem is among the most famous of perplexing probability puzzles, with people still struggling to understand the answer. It is said that Marilyn vos Savant, who correctly answered the question in her Parade magazine column, received thousands of incredulous letters from readers, many with PhDs, explaining how they thought she was wrong! So, if you have not heard of the problem, enjoy the puzzle and see if you can grasp the solution.

#### The Problem

The Problem goes like this. Steve Selvin wrote a letter to The American Statistician in 1975, basing it off the show “Let’s Make a Deal”, which was hosted by Monty Hall.

Suppose you are on a game show, and you are given a choice of 3 doors to open. Behind one door is a gleaming Ferrari, but behind the other two are old goats. Your objective is to win the gleaming Ferrari. After you choose one door, the host opens another door, which contains a goat. The host then gives you a choice of either staying with your original choice, or switching to the remaining unopened door. Is there a better chance of claiming the Ferrari by remaining with your original choice or switching to the other door, or does it not matter?

Now, when presented with this problem, almost everybody intuitively thinks of the same answer. In short, there are two doors left: one with a goat behind it and one with a Ferrari behind it. Surely there is only a 50% chance of a Ferrari being behind each door, so it doesn’t matter if you switch or stay? Well, that would be the case if Monty had either just presented two doors (with one Ferrari and one goat) or had opened one of the three doors to reveal a goat before we made our first choice. Most people think that Monty throwing open a door after we choose our first choice only narrows the odds to a 50/50 on whether we should switch or stay.

However, Monty showing us the goat after we made our first choice makes all the difference. The answer is that we are twice as like to win that golden Ferrari by switching our choice instead of remaining. In other words, there is a $\frac{2}{3}$ chance of winning by switching, with a $\frac{1}{3}$ chance of winning by staying. And this is why.

#### Understanding the Solution

Imagine you’re Monty Hall. After your guest chooses a door, you must throw open another door revealing a goat. You know which door hides the Ferrari and which ones hide the goats. You also know that you want to make it as difficult as possible for your guest.

Let’s assume the guest switches and see what happens. There are 2 scenarios after the guest makes his/her first choice of a door.

Scenario 1: Uh oh. The guest managed to guess correctly on his/her first try. It doesn’t matter which of the other doors I open as they both reveal goats. I will open this one and hope they switch. Yes! They switch and end up with a goat.

Scenario 2: Good shift! My Ferrari is safe for now, as they chose a goat door. Ahh, but what is this? I must reveal another door? Well it is certainly not going to be my Ferrari, so it will have to be the other goat. Surely he keeps his first choice? What!? He is switching for no reason to the door which reveals the Ferrari? He must be cheating.

Scenario 1 happens when we, the guest now, choose the door with a Ferrari behind at first. In this scenario, switching proves fatal as we give up our winning door for a losing one. However, there is only $\frac{1}{3}$ chance of choosing the Ferrari first time. On the other hand, Scenario 2 happens when we choose a goat door first. Monty is forced to reveal the second goat door and we win the Ferrari by switching. This happens $\frac{2}{3}$ of the time as there is a $\frac{2}{3}$ chance of choosing a goat first. Therefore, if we switch, we have a $\frac{2}{3}$ chance of winning, so because our only two options are switching or staying, we only have $1-\frac{2}{3} = \frac{1}{3}$ chance of winning if we stay.

If you do not get this, and there’s no shame in not doing so, keep on thinking about all of the possibilities. Maybe even try this experiment yourself. The question and the solution baffled many clever people in the 20th century, so keep on trying to understand it!