I don’t know how many of you are familiar with the monty hall problem, but i am guessing since you are reading this post you must already be familiar with it. Nonetheless i will also give you a small introduction, so that you may know exactly what it is about.
Introduction
The Monty Hall problem is a probability puzzle, that is loosely based on the American television game show Let’s Make a Deal and named after its original host, Monty Hall.The main idea behind it can be summarized in the follwing scenario:
Suppose you are in a game show and you are given the choice of three doors. Behind the first door there is a car, behind the second door there is a goat and behind the third door there is also a goat. This is by the way the reason why it is also known as the goat problem. If you choose the door with the car behind it you win the car but if you choose the other two doors you lose. Now let’s say you make a choice and you chose door number 2( the numbers are arbitrarily chosen). The host, who also knows behind which door the car is hidden, in order to spice things up a bit, reveals one of the doors that lose and presents you with the following dilema. You can either stick to your initial choise or choose the other door that is left. So what do you do?
Intuitive answer
Most people usually say that it doesn’t matter what you do. You have 50% chance of winning and 50% chance of losing whether you stick to your initial choice or not. At first it sounds pretty logical doesn’t it? I mean you have a door that wins and a door that loses. So the probability is 50-50 right?. Well not exactly. What we are forgetting is that when we made our initial choice we could choose from three doors and that two of them were hiding a goat behind them. This is a very valuable piece of information that at the same time is the solution to our dilema. If you are not convinced read along as i explain WHY.
Solution
First of all let’s use some very basic maths to model our problem. When we make our very first choice there are two goats and one car, that is two of the doors lose and one of them wins, or if you prefer we have 2/3 = 66% chance of making the wrong choice and 33% of making the right one. Now like i said earlier this is a very important piece of information that is telling us that we are most likely to make the wrong choice in our first guess. Let me repeat this one more time we are most likely to make the wrong choice in our first guess. Now while keeping this in mind we proceed to the part where the host reveals one of the goats and presents us with the dilema: Stick to your initial choice or change?
At this point is where you smile and say i abandon my initial choice and choose the other door that is left and hopefully by doing so you will win a car with a probability of 66%. Why is that you say?
Well having probably picked the wrong door in our initial guess (two out of three times you will pick the wrong door), and since the door with the other goat is revealed from the host, it only makes sence to go for the door that is left, because in that case, two out of three times it will be the door that has the car.
Generalization
Now if you are still not convinced and you think there is a scenario where sticking to your initial choice would make more sense than to chose the other door, we can go through all the cases and see what happens.
We actually have goats and cars and we want to combine them in triplets. There are eight different combinations which are the following.
| Door 1 | Door 2 | Door 3 |
|---|---|---|
| goat | goat | goat |
| goat | goat | car |
| goat | car | goat |
| goat | car | car |
| car | goat | goat |
| car | goat | car |
| car | car | goat |
| car | car | car |
However we are only interested in the cases where we have two goats and one car, which are the following:
| Door 1 | Door 2 | Door 3 |
|---|---|---|
| goat | goat | car |
| goat | car | goat |
| car | goat | goat |
Now let’s examine scenario number one. For every scenario there are three different cases since we have three doors. What i mean by that is that when the initial choice is made you can chose one of the three doors.
Scenario number 1
In scenario number 1 if we chose door number 1 then the host will reveal the goat behind door number 2. So by choosing door number three as our final choice we win
If we choose door number 2 then the host will reveal the goat behind door number 1. So by choosing door number three at the end we win once again.
If now we choose door number 3 then the host will reveal the goat behind door number 1 or door number 2, doesn’t really matter which one since both of them lose. Now by changing we will end up with the goat and therefore lose…
So all in all during scenario number 1 if we choose NOT to stick with our initial choice and choose the only available door left thenwe win two out of three times
Scenario number 2
In scenario number 2 if we chose door number 1 then the host will reveal the goat behind door number 3. So by choosing door number 2 as our final choice we win
If we choose door number 2 then the host will reveal the goat behind door number 1 or door number 3, doesn’t really matter which one since both of them lose. Now by changing we will end up with the goat and therefore lose…
If now we choose door number 3 then the host will reveal the goat behind door number 1. So by choosing door number 2 at the end we win once again.
So all in all during scenario number 2 if we choose NOT to stick with our initial choice and choose the only available door left thenwe win two out of three times
Scenario number 3
In scenario number 3 if we chose door number 1 then the host will reveal the goat behind door number 2 or door number 3, doesn’t really matter which one since both of them lose. Now by changing we will end up with the goat and therefore lose…
If we choose door number 2 then the host will reveal the goat behind door number 3. So by choosing door number 1 at the end we win.
If now we choose door number 3 then the host will reveal the goat behind door number 2. So by choosing door number 1 at the end we win once again.
So all in all during scenario number 3 if we choose NOT to stick with our initial choice and choose the only available door left thenwe win two out of three times
As you can see in all possible scenarios the best course of action is to abandon your initial choice and choose the only available door left. Since if you do that you have an overall chance of winning equal to 66%.