 Jaewoo Song

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“Monty Hall problem” is a very famous mathematical problem related to conditional probabilities.

Many people, including myself, have been confused about it and of course, this is not easy to understand intuitively.

This post is based on a lecture in Statistics 110 from Harvard University which I’ve been taking recently.

First, let me introduce what the Monty Hall problem is.

“Monty Hall” is a show host who hosted the entertainment show called “Let’s Make a Deal”.

Monty Hall problem came from this show and as you can see, the name itself is from his name. Assume that you participate in a game show where you can have a car or a goat depending on the door you choose to open.

First, Monty suggests you to select one door.

Once you choose one, then Monty opens a door which you didn’t choose and has a goat behind to show you that this door is not the answer.

Then Monty gives you an additional chance to change the door.

In this situation, which would be more advantageous to get a car, keeping the selected door before or selecting the other one?

OK, I think that there can be various opinions on that, such as “Is the probability of picking a car still $1/3$? Then what does it matter if I change the door?” or “The number of doors left are two, so the probability becomes $1/2$. Changing doesn’t matter.” etc.

Also this can be a little bit psychological, since a lot of people feel frustrations or anxieties when they change their mind especially in crucial cases like this.

So many people just conclude that if the door currently selected is correct, then changing would be a big mistake, so it is reasonable not to change.

And it is very interesting that this conclusion is fully understandable considering the human mind.

So what is the answer?

The answer is that changing the door is more advantageous.

There are several ways to prove it, but I will talk about the details in two ways just I learned in the lecture $6$ from Statistics 110.

1. Solving with a tree diagram

Assume that the door first we choose is $1$ and the door with a car $2$.

We all know the answer but in the actual situation we don’t know what the answer is, so the probability of each door having a car is $1/3$ each.

This is the initial state of the game and let’s draw a tree diagram with that. Although we chose the door $1$, but still each probability is $1/3$ since we didn’t check any door.

So the overall conditions are not different from the beginning.

Now this is the important part, Monty should open a door with a goat among the two doors which we didn’t select.

So Monty doesn’t have a lot of choices since the number of options is either $1$ or $2$. This is what happens next.

If we picked the door $1$ and the door with a car is $1$, Monty have two options $2$ or $3$.

So the probability of $1$ being the answer is $1/3$ and each probability that Monty opens $2$ or $3$ is $1/2$, so eventually by the multiplication rule, each result becomes $1/6$.

On the other hands, if $2$ or $3$ has a car, then Monty has only one choice.

So the each probability becomes $1/3$ after the multiplication.

But as we know, the actual answer is $2$, so Monty is never gonna open the door $2$.

So actually, we should exclude the cases that the $2nd$ door is opened. Since we reduced the possible cases into two, then all the possibilities should be also normalized.

This is a fundamental prerequisite in conditional probabilities, which when the condition is changed, we should make the sum of all probabilities become $1$ again.

So as we can see, if the answer is $2$ and the first selected door is $1$, changing the door is a more profitable choice.

And the normalization part is very important because it means that due to Monty’s choice, the condition has changed.

That’s why it seems that keeping previous option is not a bad idea, but actually it is.

This is a very confusing part of conditional probabilities.

Let’s see another explanation on this problem solving.

2. Solving with the Law of Total Probability

Let’s assume that the condition is same as before.

We chose the door $1$ first and the answer is actually $2$.

We have to make a few definitions about cases.

• $S$: the case that we get a car by changing the door.
• $D_j$: the case that the $j$th door has a car. ($j \in \{1,2,3\}$)

Then we can calculate the probability $P(S)$ as follows.

$P(S) = P(S \cap D_1) + P(S \cap D_2) + P(S \cap D_3)$

$= P(S \mid D_1)P(D_1) + P(S \mid D_2)P(D_2) + P(S \mid D_3)P(D_3)$

Then let’s think about those values more thoroughly.

First, $P(D_j)$ becomes $1/3$ equally, since this is a prior probability with no extra conditions.

Now, each conditional probability can be inducted with simple thoughts.

• $P(S \mid D_1)$: If there is a car behind the door $1$, then we can never get it after changing the option. So this becomes obviously $0$.
• $P(S \mid D_2)$: If the door $2$ has a car, then Monty should open $3$. Then by changing the option, we can always win.
• $P(S \mid D_3)$: It is same as $P(S \mid D_2)$ which is totally symmetric.

So the calculation can be like below.

$P(S) = P(S \mid D_1)P(D_1) + P(S \mid D_2)P(D_2) + P(S \mid D_3)P(D_3)$

$= 0 \times 1/3 + 1 \times 1/3 + 1 \times 1/3 = 2/3$.

Therefore, changing our mind can have a larger probability of winning.

Why do these things happen?

I think many people still don’t understand why above solutions are correct and why these solutions are quite different from our intuitions.

I think this dilemma come from the confusion between the conditional probability and the prior probability.

Let’s check several things about it.

• We should consider conditional probabilities, not just probabilities.

A lot of people have difficulties in separating the probability and the conditional probability.

First, when we start the game, there are no extra conditions at all.

So obviously the probability that the door we first choose is the answer is $1/3$ equally.

This is the original probability we have considered so far naively.

But how about after Monty opens a door?

Then this is no longer the same situation with the previous state, so we cannot conclude the probability stays the same as $1/3$.

In other words, the conditions have changed!, so we should calculate the probability with this in mind.

We should focus on the conditional probability, not the other one.

That’s why the probability of choosing a car is no longer $1/3$ and is also not $1/2$ since this value came from another simple assumption that there are just two options total, which is not different from another naively calculated probability.

• Actually the key is not your choice, but Monty's.

More intuitively, Monty’s choice gives us a hint actually.

As I mentioned above, Monty doesn’t have many choices since he should exclude the door we chose obviously, and also must not select a door with a car.

If we choose a correct answer in the first place (which is $1/3$), then Monty has two choices.

Whatever Monty opens, we’re gonna lose a car by changing our minds and in this case keeping the door we chose is more advantageous.

But if we choose a wrong answer (which is $2/3$), then Monty has no choice but to open another door with a goat.

So only one left has to have a car inevitably.

To put it simply, even if we don’t think about complex calculations of conditional probabilities, we can intuitively find out that Monty tells us changing the door is more profitable.

This is a very interesting problem worth of trying, which tell us the importance of understanding conditional probabilities properly.

Additionally, in most of real situations, we have to conditional probabilities rather than prior probabilities since there are a lot of conditions included in real life.

In order not to get confused, we should keep this in mind always.

Lecture 6: Monty Hall, Simpson's Paradox | Statistics 110 . (2013, Apr 29). https://www.youtube.com/watch?v=fDcjhAKuhqQ&list=EC2SOU6wwxB0uwwH80KTQ6ht66KWxbzTIo.