By Abid Iqbal Choon
This week’s article is going to be a little bit relaxed (it is the exam plus winter season after all) instead of the usual, heavy, technical stuff(hopefully!) and is about a fun statistical riddle that makes people doubt their initial intuition. Imagine a game where there are three doors, where if you choose the door with a car behind it, you win it. The other two doors will have goats behind them. Sounds easy right? The odds of winning a car is 1/3. But here is the twist, after you picked a door, say door number 1, the host reveals another door which has a goat behind it, say door number 3, leaving an unopened door number 2 and offers you an opportunity to switch into door number 2. This problem is called the “Monte Hall Problem” and the name comes from a TV game show, Let’s Make A Deal hosted by Monte Hall.
Intuitively, it might seem that the host is just introducing another 50:50 game by doing so and therefore, you might think that whether we switch or not does not matter when in fact, by probability, if you switch doors, you will always have a higher chance (2/3) of winning the car. Sounds peculiar, right?
Do not worry if it sounds weird and counter-intuitive at first. Vos Savant (she is an American who is known for having the highest IQ according to the Guinness Books of Records)  at that time wrote an answer to a reader’s letter which asked about this problem in the “Ask Marilyn” column in Parade magazine in 1990. Even PhD holders at that time wrote out to Vos Savant, claiming that her explanation was wrong. The answer she wrote was as follows:
“The contestant should switch to the other door. Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance.”
The first way to convince yourself that switching is indeed better and gives a higher chance of picking into the car is to try it yourself with a simulator. You can try it out from the link in the related links section below. 
Obviously one would need to run it over and over with a big sample size to let the probability approach 2/3 when switching. However, what if you wanted a different way of understanding the reason as to why is it always better to switch?
Firstly, we need to understand that initially, the probability of us choosing the “correct” door is only 1/3 and we are more likely to choose the wrong door (because it has a probability of 2/3). [In other words, behind the two doors considered together, the probability of the car being behind either one of them is 2/3 if we picked the “wrong” door] Also, as we know in statistics, the sum of the probabilities for individual doors containing the prize must give us 1 as illustrated in the equation below.
Probability of choosing the “correct” door + the “wrong” door = 1/3 + 2/3 = 1
Here comes the part where you really need to pay attention to understand the solution, the fact that the host will always choose to reveal a “wrong” door, which makes the remaining other door (for example, the blue door in the diagram below if we picked the red door) has the whole probability of 2/3 of containing the car. Why? Because the total probability must add up to 1. (Of course, this assumes that we picked the “wrong” door initially)
So maybe you are thinking, “Okay, but what if I actually chose the correct door in the first place?” I explained earlier the fact that we are always more likely to choose the “wrong” door initially (2/3 chance). So knowing that we will pick the “wrong” door initially most of the time assures us that switching will be the best choice statistically to win the prize.
Actually, the discussion up to this point has an underlying assumption, which is the host will never open the door you chose initially if you picked the “wrong” door and will only reveal the other “wrong” door. Of course, if that is the case, you should obviously switch. (because you knew you picked the “wrong” door, unless you are really, really dumb, or does not care about the winning the prize)