While fooling even PhDs, Monty Hall Problem is a good exercise in using Probability Trees – the ones math tuition students are all too familiar with.
“It’s either you do or you don’t – it’s a 50/50,” is probably one the most common Probability misconceptions that people often say. If you’re guilty of thinking along the same lines, you just might want this refresher on the O Level Math Topic of Probability.
For instance, if you try to go into your O Level Maths Exam without any practice, guidance or math tuition, it’s probably fair to say you wouldn’t have much chance of getting an A1. You could be said to have two possibilities – getting an A1 or not – but you wouldn’t have equal probabilities between them.
On the other hand, if you had gone through one of TSV’s O Level Fast Track Programmes, while you’d still have the same two possibilities, you’d have a much higher probability of securing that distinction, rather than equal probabilities.
A key part of the O Level and H2 Math syllabus is known as Probability & Statistics – which includes learning how to calculate probabilities of multiple events. Sometimes, this may be less intuitive than it seems at first – as we shall soon see.
Let’s take a step out of the air-conditioned comfort of our math tuition classrooms, and step right on up to the stage of a 1960s American game show, to take a look at one of the most popular math paradoxes of all time – The Monty Hall Problem.
Using the example of the Monty Hall Problem, we shall cover one of the essential tools to breaking down probability problems – the probability tree diagram. The Monty Hall Problem goes a little like this:
On the stage before you are three doors. Behind one of the doors is a car, and behind the others, a goat. You, the player, seek to choose the door with the car and win.
Once you make your first choice, the game show host opens one of the two doors not chosen, revealing a goat. You are then given your final chance to choose the winning door – do you switch your choice, or not?
Though certainly a strange situation to find yourself in, this sequence of events is what Steve Selvin, professor emeritus at the University of California, Berkeley, thought up in 1975. This was based off of the 1960s Let’s Make a Deal game show host Monty Hall, building excitement in his audience by similarly opening a wrong door.
To most people, intuitively, left with two seemingly identical doors, the chance of either door hiding the car would be 50% – a probability of ½. Yet, when it comes to the Monty Hall Problem, taking up the option to switch doors increases your chances to ⅔. To understand why, let’s go over the problem in detail, breaking down the key assumptions.
When initially presented with three doors, the chance of any door hiding the car is as intuitive – ⅓. However, when the game show host opens a door, his choice of door to open is actually limited – he cannot open the door chosen by the player, nor the door hiding the car.
Additionally, we make two assumptions – firstly, we assume that if left with two unchosen doors that both hide goats, the game show host shows no bias to which door he opens. Second, the game show host is also required to abide strictly as per the problem – he must always choose to reveal a goat, behind an unchosen door.
Now, we shall utilise a familiar tool of the secondary school math syllabus – the probability tree diagram. For easier reference, let’s define door A as the door the player initially chooses, and give the other two doors the arbitrary names B and C.
As we can see, there are four possibilities as follows:
There is an equal number of winning possibilities by switching or not switching. However, remember how we started this article? Possibilities do not equate probabilities.
Using the probability tree diagram, we can see that given the assumptions, the sum of probabilities for winning by not switching is ⅓, whereas the sum of probabilities for winning by switching is ⅔.
In statistics, the term Conditional Probability refers to the probability that an event occurs, given that another event is already known to have occurred. As per the JC H2 math syllabus, conditional probability can be presented as follows:
Probability of event A occurring,given that event B has occurred
Since the door names for Doors B and C are arbitrary, we can simplify the problem by narrowing the scenario to one where the host has revealed Door C, and the player wins by switching only when Door B has the car.
Hence, the probability of winning by switching, given that Door C was revealed is as follows:
But what if Door B was revealed by the host instead? Well, since the door names for Doors B and C are arbitrary, it should not matter – that is to say, if we were to calculate conditional probability given that Door B revealed, it would give us the same results.
At TheSequoiaVision, students are exposed to a wide variety of question types to achieve mastery. Similarly, we will now look at variations of the Monty Hall Problem.
Let’s say instead of three doors, a hundred doors are placed before the player.
Initially, choosing a door at random will only result in a 1-in-100 probability of choosing the door hiding the prize car. Following the same assumptions we applied for three doors, when the game show host then opens 98 other doors with goats behind them, switching their choice of door would give them a 99-in-100 probability of selecting the door with the car behind it. A significant increase in their odds, wouldn’t you agree?
So, now that we’ve covered the maths behind the game, why aren’t we all going on game shows to make use of what we learn in math tuition classes? Well, unfortunately in reality, the two assumptions of the problem do not hold true.
Starting off with the second assumption, given what we now know of the increased probability of winning by switching doors, it is easy to understand why the game show host may not always reveal a goat behind an unchosen door – or even offer the option to switch doors in the first place. After all, most game shows have to earn their profits, one way or another.
More interestingly, if the first assumption does not hold true, the odds are no longer ⅔ to ⅓ in favour of switching. However, even if the host always favours the same door when the two unopened doors are goats, in the worst-case scenario where the door favoured is revealed to hide a goat, the odds actually only become 50-50. In other words, not switching the original choice of door never actually becomes the better choice.
Luckily for math students, while probability and statistics are a critical component of the Secondary School and JC math syllabuses today, such counterintuitive paradoxes are a specific and esoteric portion of mathematics that are few and far between. If anything, by definition, the nature of these paradoxes are meant to be confusing.
In truth, getting a good grasp of math principles – be it primary school math, secondary school math, IB math or even JC H2 maths – doesn’t have to be as confusing as the paradox we just covered. It is for this very reason that it is so important to have the proper support available for learning maths.
At TheSequoiaVision, it is our passionate belief that it is an integral part of our role to make math fun and accessible for everyone. Whether you’re in Primary School, Secondary School, or preparing for your JC H2 math exams, TheSequoiaVision has the perfect support system for you – find out more about our values, plentiful resources, expert consultations, and flexible class schedules on the TSV math tuition website here.
At TheSequoiaVision, we’re passionate about making math fun and accessible for everyone! Whether you’re in Primary School, Secondary School,, or preparing for your JC H2 math, we’ve got the perfect support system for you. With teachers who are math experts with strong teaching experience, be prepared to excel with us through our proven track record!
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