Updated: Jun 12, 2020
3 doors, 2 lead to death and 1 to your safety. Can you pick the right one to ensure your survival?
Let's go back a few decades, 45 years to be exact, and look back on a problem which astounded not just the media and the public but even Nobel prize winners. This was a question shrouded in mystery and criticism, attracting the attention of not just mathematicians but even psychologists. But first, what is the Monty Hall problem?
Imagine this: You're walking along a dark and decrepit corridor, the light, eerie whistle of the wind fills your ears as you arrive at a room with 3 doors. A hooded figure stands at the center of the room, beckoning you to choose a door. He explains, "2 of these doors will lead to your death and only one is the pathway to safety. Choose one of these doors and I will tell you which of the other 2 doors is the wrong one. Then, you will have a choice: choose a different door, or stick with the one you have picked." 
To put it in simpler terms, say you choose door 1 and the hooded figure tells you that door 3 is the wrong door. Would you rather stick with door 1 or change to door 2?
At this point, you're probably thinking that this decision is down to a 50-50 chance. Since there are only 2 doors, the probability is 1 / 2, right?
But, that's actually a very common thought process which is also why this problem is known as a veridical paradox.
- It is a paradox in which the outcome appears to be ridiculous at first glance but is found to be true on further study -
So, what is the solution to this problem? Well, the answer would be to change the door that you have picked. There is, in fact, a 2 / 3 or 67% chance of getting the right door, which leads to the question, why is this so?
The easiest way to visualise this is in the form of a tree diagram:
This tree diagram shows every possible outcome. From this, we can see that out of the 3 times that we change doors, 2 outcomes lead to safety. Hence, it can be concluded that the probability of choosing the right door if we change to be 2 / 3, much more than the initial estimated probability of 1 / 2.
There are numerous ways of solving this problem, be it intuitively or mathematically. An intuitive way would be to realise that to get the right door after switching, you would have had to pick the wrong door in the first place. And what is the probability of that? 2 / 3 !
How about a slightly more complex and mathematical approach to this problem? We shall briefly state and use 2 theorems in probability: Bayes' Theorem and Law of Total Probability.
First, let us assume that you have picked door 1 and the hooded figure, let's call him A, points to door 3.
Then, what we need to find is P( S | D3 ) where S is the event that you switch choices and obtain the right door while D3 denotes the event that A points to door 3.
So, using Bayes' Theorem, we get:
Now, we know that P( S ) is equal to 1 / 3 since there is an equal probability for each door to be the right door if there is no extra information.
P( D3 | S ) is the probability that the figure points to door 3, given that you switched to the right door. This means that you must have originally chosen the wrong door, hence, the other 2 doors are the correct and wrong door. Since A is only allowed to choose the wrong door, he has no choice but to point to it, so P( D3 | S ) = 1.
For P( D3 ), we have to use the Law of Total Probability. This law allows us to calculate the probability of an event occurring by using the conditional probabilities of another related event. So, let's take things slowly and break it down part by part.
P( D3 | 1 ) is the probability of door 3 being pointed to, given that door 1 is the correct one. Since we assumed that you picked door 1, this means A can choose door 2 or 3 randomly. Hence, it is equal to 1 / 2.
P( D3 | 2 ), on the other hand, is equal to 1 simply because if door 2 was the right door, A would have no choice but to point to door 3.
P( D3 | 3 ) is just 0 because according to the rules, A is not allowed to point at the right door- in this case, door 3.
By substituting these values in, we get P( D3 ) = 1 / 2
Hence, by using the original formula, we get P( S | D3 ):
Now that you have seen multiple ways of tackling this problem, hopefully this will encourage you to think out of the box when faced with a challenge, even when the answer seems obvious (like in this case). So, before we come to an end, how about trying a variation of the Monty Hall problem as a brain teaser?
In this case, the hooded figure offers you the option to switch, only if your initial choice is the winning door. Do you stay or switch? Leave your answers in the comments section below : )