This is an original research article by Yash Mantri (20/11/20)
By the time I was 10, I could solve the Rubik’s cube in under a minute. I was fascinated by the popular puzzle since the day I Iaid my hands on it. I spent that entire day trying to solve one face side and by the end of the day when I finally managed to do so, I ran around the house ecstatically to show every single person my achievement. Within a month, I learnt all the algorithms and could solve the cube easily in under a minute. That was the beginning of my cubing journey.
Recently, I watched the film “21”, a story about MIT math students who “count cards” to improve their probability of winning the card game Blackjack at casinos. Evidently, the film had a lot of mathematics related themes in it and this attributed to making the cinematic experience engaging and fast-paced. In one of the initial scenes of the movie, there was an allusion to the Monty Hall problem, a probability-based game, which piqued my interest and engendered me to take on deeper research about the problem. In this article, I explain the working of the Monty Hall Problem.
The Monty Hall problem is a brain teaser, in the form of a probability puzzle, loosely based on the American television game show Let’s Make a Deal and named after its original host, Monty Hall.
So here is the problem– Let’s say that there are three doors, and behind one of them is a car, while behind the other two are goats. If you choose the door with the car behind it, you win the car. Now, say you choose Door 1. The host Monty Hall then opens either Door 2 or Door 3, behind which is a goat. It is important to note that he knows what is behind each door, and never opens the door with the car behind it. Monty now gives you a choice: do you want to stick with Door 1, or switch to the other door. What should you do? Does it matter?
The answer is yes, and one must always choose to SWITCH. Before I get into the explanation, let us talk about what people usually decide to do.
To find this, I posed the question and conducted a poll consisting of a sample size of 110 people(who had never heard of the problem), including people from ages 13-50. The results for the same are shown below.
Looking at the pie chart, it can be observed that more people actually decide to stay with their initial choice. This could be because it is human tendency to be suspicious of being given other options and people in general do not choose to change out of paranoia or fear. And most obviously, the illusion is that both options will result in a 1/2 chance of getting the car so it does not matter to make a switch in any case. However, by probability, this is not the correct answer.
So, initially all we know is that behind 1 door is a car and behind the other 2 are goats. This gives us a probability of 1/3 for each door having a car. Now say we choose Door 1 for example. The probability of that door having a car is 1/3. If we pause this situation and see, the other two doors (Doors 2 and 3) have a combined probability of 2/3 (1/3+1/3).
Moving on, Monty Hall will now open one of the other doors which he knows has a goat in it. Let us say he opens Door 2 and shows us a goat. The probability of Door 2 having a car becomes 0 and since both the probabilities of Door 2 and Door 3 were adding up to 2/3, Door 3 will have a 2/3 chance of getting a car. Door 1 only has a probability of 1/3 that doesn’t change and that’s why it is more 33% more likely to get a car by switching.
Now you may say that eliminating Door 2 gives rise to only two remaining doors, each having a probability of 1/2 each.
Let’s look at the exact same problem with 100 doors instead of 3. You pick a random door. The probability that you have got the car now is extremely low – 1/100.
Now, instead of one door, Monty eliminates 98 doors. These are doors that he knows do not have the prize. This leaves two doors. The one you picked, and one that was left after Monty eliminated the others. When you first picked, you only had a 1/100 chance of getting the right door .It was just a random guess. Now you’re being presented with a filtered choice, curated by Monty Hall himself. It should be clear that now your odds are much better if you switch.
Another way to see the solution is to explicitly list out all the possible outcomes, and count how often you get the car if you stay versus switch. Without loss of generality, suppose your selection was door 1. Then the possible outcomes can be seen in this table:
In two out of three cases, you win the car by changing your selection after one of the doors is revealed. This is because there is a greater probability that you choose a door with a goat behind it in the first go, and then Monty is guaranteed to reveal that one of the other doors has a goat behind it. Hence, by changing your option, you double your probability of winning.
Using Bayes’ Theorem to explain the Monty Hall Problem
A few months back, I had completed a course on Data Science Math Skills by Duke University where I learnt about conditional probability and the Bayes’ Theorem. The Monty Hall is a classic problem that can be explained through Bayes’ Theorem.
Lets take the assumption that you pick door 1 and then Monty shows you the goat behind door 2.
In order to use Bayes’ Theorem we need to first assign an event to A and B.
Let event A be that the car is behind door number 1.
Let event B be that Monty opens up door 2 to show the goat.
Bayes’ Theorem Solution:
Pr(A) is easy to figure out. There is a 1/3 chance that the car is behind door 1. There are two doors left, and each has a 1/2 chance of being chosen — which gives us Pr(B|A), or the probability of event B, given A. Pr(B), in the denominator, is a little trickier to figure out. Consider that:
You choose door 1. Monty shows you a goat behind door 2.
If the car is behind door 1, Monty will not choose it. He’ll open door 2 and show a goat 1/2 of the time.
If the car is behind door 2, Monty will always open door 3, as he never reveals the car.
If the car is behind door 3, Monty will open door 2, 100% of the time.
As Monty has opened door 2, you know the car is either behind door 1 (your choice) or door 3. The probability of the car being behind door 1 is 1/3. This means that the probability of the car being behind door 3 is 1 – (1/3) = 2/3. This is why it is wiser to switch.
The Monty Hall problem is truly a fascinating problem that tends to play tricks with our brain. The important catch in the problem is that Monty knows what is behind the doors and always opens only the doors with a goat in them. If he was not aware of what lies behind the doors then it does not favour you to switch. If you are still not convinced, click on the link and try the simulation for yourself! http://www.math.ucsd.edu/~crypto/Monty/monty.html