The controversy regarding this infamous problem, started on the ever so fateful day of September 9, 1990. When Marilyn vos Savant, a columnist for Parade Magazine, and the person recorded as having the highest IQ ever recorded (Guiness Book of World Records), received the following letter from a reader:

“Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, “Do you want to pick door #2?” Is it to your advantage to switch your choice of doors?”

Craig F. Whitaker

Columbia, Maryland

There is also a rather cool explanation of this problem from the movie “21”, which I’ve linked below:

Goats are funny creatures.

What would you do? Majority of people would think it would make no difference to switch because the chances of winning between door 1 and door 2 is still 50/50 or that the game show host is attempting to get you to pick a door with a goat. According to this study, when people encountered this problem for the first time they only switched 12% of the time. However, staying with your original choice will not be in your best interests, as I’ll explain below.

We can actually calculate the differences between always switching and not switching. Just to clarify, these are the rules:

The player will choose a door first, but will not open it immediately

The host, knowing where the car and goats are located, will always open a different door to the player, with a goat behind it, thereby revealing one one of the goat’s position.

After the host has opened one of the doors, he or she will then ask you if you would like to switch doors or stay with your original choice.

A Simpler Way of Looking At The Problem

The Enigmatic Table of Truth

Notice from the Enigmatic Table of Truth, that at the start, I have a total of

1/3 + 1/3 = 2/3 chance of picking a goat the first time I choose a door .

If I pick a door with a goat behind it, the host will then have to choose another door that also has a goat behind it. After this happens, the remaining door will be the one with the car behind it. This occurs with a 2/3 probability. Therefore, unless I really hate cars, I really should choose to switch.

However, 1/3 of the time, I’ll pick the door with the car behind it. Then the host will choose a door with a goat behind it, and then the remaining door will have a goat behind it as well and if I swap, I will lose (this will only 1/3 of the time).

Why do we think that it makes no difference if we switch?

We assume that switching only gives a 50% chance of winning. Which would be true if we had no extra information, however we received new information from the host, when he eliminated one of the doors with the goat. This changed the probabilities.

Basically, the fact that we have a 2/3 chance of picking a door with a goat at the outset, is actually to our benefit if we always switch. This is because if we pick a door with a goat, then the host will eliminate the remaining door, which means that the last door that is left over will have the car behind it.

Makes a bit more sense now, right?

Is your intuition a bit like, “What? That doesn’t make sense!”? Then continue reading below.

A More Rigorous Way of Looking At The Problem by Using That Magical Formula Called Baye’s Theorem

Remembering the rules stipulated earlier on, what we want to do to is figure out this problem using Baye’s Theorem.

Omg, what is Baye’s Theorem?

Baye’s Theorem is a way of calculating the probability of an event occurring, based on information of any conditions that are related to that event. This is helpful, because we can calculate whether we should switch doors, after the host opens a door with a goat behind it.

FYI: If you want to know how Baye’s Theorem is derived, you can go to the end of the article for an explanation.

Using the magical Baye’s Theorem

We can use Baye’s Theorem to calculate the probabilities. Take note of the following:

C1 = Car is behind door 1

C2 = Car is behind door 2

C3 = Car is behind door 3

H1 = Host opens door 1

H2 = Host opens door 2

H3 = Host opens door 3

The expression P(H3|C1), means the probability of the host opening door 3 (H3), given that the car is behind door 1 (C1)

Scenario: We initially choose door 1 and the host then opens door 3.

What is the probability that the host will open door 3 considering the events C1, C2 and C3?

If we originally chose door 1 (C1), and the car is behind door 1, then there is 1/2 probability of the host choosing either door 2 or door 3, therefore:

If we originally chose door 1, and the car is behind door 2 (C2), then the host can only choose door 3, therefore:

If we originally chose door 1, and the car is behind door 3 (C3), then the host cannot choose door 3, therefore:

Remember we are trying to figure out if it’s in our interest to switch doors from our original choice.

Knowing the above probabilities, what is the probability of P(C1|H3)?

All we have to do is just substitute the values in:

Substitute the values in.

P(H3) is just the sum of the probabilities of three different situations involving H3.

In other words, it is just the probability of the host choosing door 3, given that the car is behind either door 1, door 2 or door 3. i.e:

After substituting we get the following:

P(C1) = Chance of having the car behind door 1 (The doors all have a 1/3 chance of having the car behind them initially)

This means that sticking with door 1 will yield a 1/3 chance of winning the car.

Now let’s see what our chances of winning will be if we had to switch to door 2:

If we switch to door 2, we have a 2/3 probability of winning. As you can see, it’s in our interest to switch doors as we have a 66.67% chance of winning, versus 33.33% chance if we do not switch.

Conclusion:

It’s always beneficial that you switch because you will have a 66.67% versus a 33.33% chance of winning that sports car, unless you would prefer a fluffy goat of course.

Deriving The Magical Baye’s Theorem

So you want to know how to derive Baye’s Theorem? No problem! We start off with the conditional probability formula, which is formula that allows us to calculate the probability of event A occurring, given that event B occured. See below:

Conditional probability formula

A = Event A occurs

B = Event B occurs

So, P(A|B) reads like so — if event B occurs, what is the probability of event A occurring. On the right hand side of the formula above, the “P(A ∩B)” in the numerator is read as the probability of A and B both occurring. The denominator, as you might have of guessed, reads — what is the probability of B occurring.

So take the conditional probability formula and multiply P(B) on both sides to get this:

Multiply both sides by P(B), to get the above.

Now, lets switch it around, what is the probability of B occurring, if A has occurred?

Well, we just re-write the first equation by swapping the A’s and B’s around and then multiply by P(A), like so:

Note, the following:

This makes intuitive sense, the same as saying 1+2 = 2+1. (See commutative property)

Therefore, the following is true:

If P(B ∩A)=P(A ∩B), then P(B|A)P(A) = P(A|B)P(B), knowing this we can therefore substitute and get the following:

This re-arranges to what is known as Baye’s Theorem

And that’s how you derive Baye’s Theorem!

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