Question

The Monty Hall problem is a famous problem loosely based on the game show Let's Make a Deal. You are a contestant on the game show. There are 3 doors in front of you. Behind one door is a prize, and behind the other two doors are goats. Assume the door with the prize is picked uniformly at random from the three doors.

First, you pick a door. Then, Monty Hall will open one of the other two doors that you haven't picked. Monty Hall will always open a door with a goat behind it. (He knows ahead of time where the prize and goats are.) Now you have a decision to make. You can keep the door you initially picked, or you can switch to the remaining door. (One door is now open, so those are the only two doors left.)

Which is the better strategy: stay with your current door, or switch to the other door? Or are the strategies equally good?

Here's a common (incorrect, but maybe convincing) argument: There are only 2 doors remaining since Monty Hall opened one with a goat. One of them has a prize, the other has the goat. Each door has a 1/2 probability of having the prize behind it. So, in terms of your probability of winning the prize, it doesn't matter if you switch or not.

a)  Let's use the sample space Ω = { 1 , 2 , 3 } to model this situation where the number indicates which door has the prize behind it. What is the probability measure for this sample space to model this situation?

b-c)  If you do stay with your initial door, what is the probability that you will win the prize? If you switch to the other door, what is the probability you will win the prize? Explain a bit about your answers, to convince someone who believes the probabilities are both 1/2.

Answer 1

It turns out that there are only nine different combinations of choices and outcomes. Therefore, I can just show them all to you and we calculate the percentage for each outcome.

You Pick	Prize Door	Don’t Switch	Switch
1	1	Win	Lose
1	2	Lose	Win
1	3	Lose	Win
2	1	Lose	Win
2	2	Win	Lose
2	3	Lose	Win
3	1	Lose	Win
3	2	Lose	Win
3	3	Win	Lose
		3 Wins (33%)	6 Wins (66%)

Here’s how you read the table of outcomes for the Monty Hall problem. Each row shows a different combination of initial door choice, where the prize is located, and the outcomes for when you “Don’t Switch” and “Switch.” Keep in mind that if your initial choice is incorrect, Monty will open the remaining door that does not have the prize.

The first row shows the scenario where you pick door 1 initially and the prize is behind door 1. Because neither closed door has the prize, Monty is free to open either and the result is the same. For this scenario, if you switch you lose; or, if you stick with your original choice, you win.

For the second row, you pick door 1 and the prize is behind door 2. Monty can only open door 3 because otherwise he reveals the prize behind door 2. If you switch from door 1 to door 2, you win. If you stay with door 1, you lose.

The table shows all of the potential situations. We just need to count up the number of wins for each door strategy. The final row shows the total wins and it confirms that you win twice as often when you take up Monty on his offer to switch doors.

Alternative solution:

Suppose that you choose one of the doors. Now without opening any other door, you are given a choice to switch to other two.

Clearly, there are nine cases. There are six cases in which you switch, and win in only two of them. Switch → 1/3 cases you win.

There are three cases in which you don't switch → 1 out of 3 of them you win.

So clearly, whether you switch or you don't, the probability of your winning remains the same.

Why? Because in this case, the number of choices don't change. In the original problem, we have a certain number of choices, then one of them is fixed and others are crossed out. This definitely distributes the probability unequally, 'somehow'. But I won't try to explain that before covering Part (B).

For now, all we can be sure is that if the doors are never opened, the probability distributes equally even on switching. (Which is quite intuitive.)

Part (B)

What if one of the doors was opened before you picked any of them? So, suppose Monty opened door C and showed you that it is empty, what can you say about the probability distribution for the other doors?

Without taking cases (although, that is one way to prove it, and you can check) for brevity, it is obvious that there is a 50-50 probability of a prize behind the remaining doors. Before the door was opened there was 1/3 probability of prize behind each of the door. Visualize that probability as a box. When Monty opened one of the doors, the contents of our probability-box distributed equally amongst the remaining doors.

So what changed? Why did the probability distribute equally? And that's where we realise one of the most important factors that people don't usually see when they first look over the problem: Monty Hall, our gameshow host, has to choose, too.

In the original problem, if we picked one of the doors, Monty Hall had either only one choice (if you picked the wrong door), or two choices (if you picked the correct door on the first try) of doors to open.

So if we picked one of the wrong doors, we limited Monty's choices and forced - yes, forced - him to open the other wrong door. We had 2/3 chance of choosing a wrong door. That implies we had 2/3 chance of forcing Monty to open the wrong door. We had 2/3 chance of determining that the correct door is left. (Although not rigorous, can you see that we have 2/3 chance of winning if we switch? Bravo! But we'll return to calculations in Part (D))

In the case of Part (B), there was no limitation. Monty had two choices and could choose any of them. So the probability distributed equally.

Part (A) is the exact converse of Part (B). Part (A) shows how Monty's opening of doors affects us! When he opened the doors, and distributed the contents of our 'probability-box', he consciously changed his choice. And so he gave more preference to one of the doors. This preference distributed our probability unequally.

Part (C)

Many people say that the fact that Monty knew the correct door is all there is to this solution. In fact, that's what I just wrote. But many other people also argue that Maths can't distinguish consciousness from randomness. Of course, this is silly, but if you were a layman and you looked at the maths (I'll show in Part (D)), that would seem confusing.

The trick that your intuition has to grasp is that Maths is about results. We don't care if Monty knew the correct door or otherwise. If his choice would have been random, we would have gotten cases where he would have opened the correct doors and ended the game.

And those cases would be useless to us. So we assume that those cases don't exist. Saying that, "Monty's careful and planned choice changed the probability distribution" is a ‘fancy’ way of saying that the cases where he mistakenly opened the correct doors just don't exist at all. Of course, in real life consciousness is how we guarantee ourselves that those cases won't exist. But our Maths isn't really interested in how you make those cases non-existent. It just cares they are.

Makes sense? Can you determine the probability of your losing the gameif the doors were opened at random?

Part (D)

Hooray! We are here. This is just the mathematical part. For the laypeople, I'll make cases and show how this system makes sense. But for the sake of keeping this answer, here's a rigorous mathematical way to the whole problem:

Let the doors be X,Y,ZX,Y,Z.

Let CxCx be the event that the car is behind the door XX, and CyCy be for YY ... and so on.

Let HxHx be the event that the host opens door XX ... and so on.

Supposing you choose door XX, the possibility that you win the prize if you then switch your choice is given by:

P(HzP(Hz ^ Cy)+P(HyCy)+P(Hy ^ Cz)Cz)

=P(Cy).P(Hz|Cy)+P(Cz).P(Hy|Cz)=P(Cy).P(Hz|Cy)+P(Cz).P(Hy|Cz)

=(13.1)+(13.1)=23=(13.1)+(13.1)=23

Tada!

What's fascinating is that what we've been talking about in this answer is clearly shown in the Maths. The crux of the question as we've determined is that Monty Hall would never open the correct door. That directly translates to P(Hz|Cy)P(Hz|Cy) (Probability that Monty Hall opens door Z, given that prize is behind door Y) and P(Hy|Cz)P(Hy|Cz) being equal to 11.

What if that were not true? What if he could open random door? The probability then would be:

=P(Cy).P(Hz|Cy)+P(Cz).P(Hy|Cz)=(13⋅12)+(13⋅12)=13=P(Cy).P(Hz|Cy)+P(Cz).P(Hy|Cz)=(13⋅12)+(13⋅12)=13

Which is equal to the original probability of winning without switching. So if Monty were to choose random doors, switching would have no effect. And that's what we had concluded!

Or using our trusty cases, here's the best I could do on a mobile app. (Gah, I hate writing answers on mobile.)

(So now that you see what features this problem had, do you think it is intuitive now?)

Just because it's bugging me, I will add the 'thousand/million/whatever doors' way of thinking. It's actually fun! So consider the same problem but instead of three there are a thousand doors. The host makes you pick and then opens 998 doors. Would you switch to the remaining?

Imagine yourself standing like a fool while Monty goes around the humungous stage opening doors for an hour. And then comes back and you see that he opened the doors numbered from 1 to 556 and then left 557 closed and then opened 558 to 999 (Door number 1000 was your choice). “What trickery does this evil man apply to not open 557?” you think. And as you rub your chin, you say, “Aha! He's a bloody fool! That must be the door with prize!”