The Monty Hall math problem


I was watching Brooklyn 99 Season 4 Episode 8 around the 5 minute mark

The problem goes “There are 3 doors behind one of which is a car. You pick a door and the host, who knows where the car is, opens a different door showing nothing behind it. He asks if you want to change your answer.

Apparently the math dictates that you have better chances if you change your decision. Why? 2 doors 50/50 chance, no?

One character (Kevin) says it’s 2/3 if you switch 1/3 if you don’t. What? How? Please help.

In: Mathematics

The key detail: the host can’t reveal the prize door, and the outcomes of the choices don’t ever change. By being forced to only reveal a non-prize door, the host is disclosing information about the other choices.

But it’s not two doors, 50/50. You chose one of three so you have a 1/3 chance of having the right one and there’s a 2/3 chance of it being one of the other two. Now the host opens one of the other doors THAT THEY KNOW WON’T REVEAL THE PRIZE. That hasn’t affected the basic 1/3 to 2/3 split, nothing has changed as the host knows which door they can safely open. So that leaves the other door with the whole of the 2/3 chance.

The host never reveals a door which has the prize behind it. You already know from the start that two doors don’t have prizes, so when you pick one you already know at least one of the doors you don’t pick has no prize.

Which has a better chance of having the prize, 1 door or 2 doors? Obviously 2 doors. One of those 2 doors is always going to not have the prize but the host eliminated it by showing you. So by switching you in essence pick the 2 doors option.

The Monty Hall problems come down to this, is it better to choose one door, or two doors? If you decide to stay with your original pick, you are basically saying that you want to keep your one door instead of the two other ones. When you change doors, you are picking both the other ones.

A lot of people get stumped by the door reveal, but that door reveal is actually 100% inconsequential. The prize can only be behind a single door, which means that most doors will be empty. In the two doors that you did not pick, one door *must* be empty. That does not improve or worsen your odds in any way, because it is expected that the door will be empty. Furthermore, the host knows not to open a door that has the prize, so the door will be empty.

Since this step is inconsequential, you can skip it. So, Monty says pick a single door. Then once you pick a single door, he skip opening an empty door and says “okay, do you want to keep your single door, or switch to the other two doors?” Which do you choose, one door or two?

Another way to help this problem is add the amount of doors. Let’s say that there was 100 doors, and you have to pick one. You pick one, and then Monty says “do you want to switch to the other 99.” Keep in mind that with those 99 doors, at least 98 of them will be empty. If you see a bunch of empty doors, that should not be a surprise. Which do you chose?

I think the easiest way to wrap your head around this is to increase the number of doors. Instead of 3, lets say there’s 100 doors. You choose a door, then the host opens 98 wrong answer doors. There’s only two doors left, but its of course not 50 50. Your original door had a 1 in 100 chance of being correct, so choosing the only other door left has a 99 in 100 chance.

Say there are three doors, A, B, C. One of them has the prize, and you pick one, hoping you picked the one with the prize. Simple enough.

To make it even more simple, we can say that door A is the one with the prize in it. Now the game starts. You pick a door at random. 1/3 chance of A, 1/3 of B, 1/3 of C, right? So at this first pick, you have a 1/3 chance of picking correctly (door A).

Now Monty Hall opens one of the other doors. The door he opens **won’t** have the prize behind it, or the game would be over.

If you picked A, he could reveal either B or C. Switching now would lose you the prize.
If you picked B, he would reveal C. Switching now (to door A) would win you the prize.
If you picked C, he would reveal B. Switching now (to door A) would win you the prize.

So after the reveal, that single “50/50” choice to switch doors is actually based on the original 1/3 choice. Switching afterwards will get you the prize 2/3 of the time, because you would’ve only picked correctly 1/3 of the time in the first pick.

There are a ton of resources on the web or youtube that explains this which will go through the steps of reasoning this out. Much better than a wall of text.

Monty Hall illustrates a very important point about the nature of new information and how it updates your prior probabilities. For example:

You fall down some stairs and hurt your right foot and believe there is a 25% chance it is broken. You go to a clinic and they examine your left foot and say “left foot isn’t broken”. Does this make it more or less likely that your right foot is broken?

Now in the same situation, the clinic tests your right foot and say “it isn’t a sprain”. Given that your foot still hurts, is it now more or less likely that it is broken?

Monty Hall is NOT acting randomly. This is the thing that needs to be understood. Monty Hall cannot open the door the contestant chooses and he cannot open the door with the car behind it. The information he reveals by opening one door isn’t giving you EQUAL information about what lies behind the remaining closed doors.

The explanation that clicked for me was to think of them as two groups. Group A is the door you choose. Group B is the other two.

After your first choice, there’s a 33% chance it’s in A and a 66% chance it’s in B.

Now what if they said, “Do you want to stick with A, or do you want to switch to B?” Of course you’d switch, because group B has two chances.

When they remove one of the bad ones, there’s still a 66% chance it’s in group B. By letting you switch after getting rid of one of the losers, they’re still giving you *all* of Group B.

It’s kind of like they let you pick two doors instead of one.

Posting this as a top level comment in case this explanation helps anyone;


The break down in logic most people have is not realizing the way the initial condition is set up. Look at it from the first pick.

You have a 2/3 chance of picking the wrong door initially, so that is most likely what you do. Now after you pick he eliminates one of the wrong options, but you already picked based on 2/3 probability, so you likely already picked the wrong door, thus meaning that if you were to switch doors 100% of the time after given the option you will switch from a wrong door to a right door 2/3 of the time.

The key thing is that the host knows which door has the car.

Imagine a more extreme case. There are 100 doors with a car behind 1 of them. You pick door 7. The host reveals the goat behind every door except door 7 and door 38. Suddenly, door 38 looks suspiciously appealing…

You had a 1% chance of getting the right door. This means there’s a 99% chance that door 38 is the right one.

Again, the trick is that the host knows the right door. If the host didn’t, then most of the time the host would accidentally reveal where the car was. It’s the *host’s* knowledge that changes the odds.

It’s easier to explain backwards. When you pick one option out of three there is a 2/3 chance that the correct answer was behind the other doors. You picked one there were two left. The odds say that one of the ones you didn’t pick has the prize. If you could pick 2 doors instead of one you would have a 2/3 shot of winning. However if you pick 2 doors you know one has to be empty. In this scenario it is like picking 2 doors and the host tells you which one is empty and then let’s you have the last one.

Let’s do the experiment:

Take 3 cups and 1 ping pong ball. Line the cups up in a row. Place the ball under cup 2. You switch cups every time offered by the host.

First experiment you pick cup 1. The ball is under cup 2. The host shows that cup 3 is empty and asks if you want to switch to cup 2. You do and win.

Second experiment you pick cup 2. The ball was under this cup. Host shows you cup 3 is empty and asks if you want to switch to cup 1
You do and lose.

Third experiment you pick cup 3. The ball is under cup 2. Host shows you cup 1 is empty and asks if you want to switch to cup 2. You do and win.

You win 2 out of 3.

On your first choice you have a 33.3% of being right. But you also have a 66.6% of being wrong.

The opening of another door does not change the fact you only had a 33.3% of choosing the right door in the first place.

But with one door opened the full 66% chance of you being wrong “collapses” to the other closed door you didn’t choose.

Imagine there was 1,000,000,000 doors. You pick one. The host then opens up 999,999,998 doors. And asks if you want to switch. You clearly did not have a 50:50 chance of picking the right door even though there’s only two closed doors out of a billion. You had only a 1 in a billion chance of picking the correct door. And a 999,999,999:1billion odds you chose wrong.

Which is more likely?

You picked the one door out of three that has the prize.


You picked one of the two doors that does not have the prize.

Chances are that you picked one of the two doors that does not have the prize, it is a 2/3 chance that you chose wrong. It is a 1/3 chance that you chose the one with the prize.

Since it is more likely that your choice was wrong, it would be wise to change your choice.

I still get confused by it, but the simplest way I can explain is that, when you pick a door, the chances you picked an empty door (no prize) is 2/3. The fact that the host reveals a door doesn’t change much. You still know there is a 2/3 chance that you picked the wrong door on your first choice and a 1/3 chance that you picked the right door. So when he asks if you want to switch, he’s really showing you that there were really always only 2 options: opening the 1 of 2 incorrect doors you most likely picked or picking the 1 door left that the host knows is correct.


|___ **your door** (which is most likely empty) ___|


|___**the last unopened door** (that is most likely correct)___|


hope this helped

If you don’t switch, your odds are always one in three of being correct.

Now break down the odds when you switch:

There is a one in three chance you picked the right door originally and are losing it by switching.

There is a two in three chance you picked the wrong door originally, and since Monty eliminated the other incorrect door you are switching to the correct one.


One way to solve it is to list all the scenarios:

Assume car is behind door #1.

Car | You choose | Monty opens | You switch | Don’t switch
1 | 1 | either 2 or 3 | lose | win
1 | 2 | 3 | win | lose
1 | 3 | 2 | win | lose

Doors 2 and 3 are the same.

The reason is quite simple.

If your initial choice was incorrect, and you switch, you have a 100% chance of getting the car.

Since the odds of your initial choice being incorrect were 2/3rds, therefore the odds of winning if you switch is *also* 2/3rds.

Caveat to what everyone else is saying: if the problem is not explained very precisely, the true answer is undetermined as there is not enough information to compute.

In particular, it is important that

1. the host knows where the car is, and

2. the host deliberately chooses never to reveal the car

Saying that the host knows where the car is and happened to open nothing in this particular instance is not sufficient information to create the 2/3, 1/3 solution. It is important that the host deliberately avoids the prize, as that is what causes the door he didn’t choose to become special. He has special knowledge, and you can gain partial knowledge from observing his behavior if his behavior depends on his knowledge.

If, instead, the host opens doors randomly and just happened to open nothing this time, then we are not in the true Monty Hall problem, but the “Monty Fall” variant, in which the answer is 1/2, 1/2 as people intuitively suspect.

In a lot of cases, the problem is difficult not because people being given the problem are bad at math, but because the presenters are, and provide an ambiguous scenario which could either be the Monty Hall problem or the Monty Fall problem, and people assume it’s the Monty Fall problem since that’s a reasonable assumption someone would make if not given information that rules it out.

Monty always reveals the dud prize. If you did not originally choose the car, you will always get it by switching. The probability of getting the car initially is 1/3 so switching is 2/3.

It becomes easier to imagine if you think about 100 doors. You pick one. I open 98 donkey doors. Do you want to switch?

Here is a better way to look at it.

First, understand that Monty only reveals doors that do NOT have the car. This is very important.

Let’s say you are playing with 1 Million doors. You choose a number let’s say door 2. Monty opens all the doors except door 789543. Do you keep your original door or do you choose door 789543? The car is either in door 2 or door 789543. The probability of it being door 2 was 1/Million. The probability of it being door 789543 is 999999/million. This is because Montey has only opened doors that do NOT contain the car.

Now let’s say you’re playing with 10000 doors. You again choose door 2. Monty opens all the doors except door 3897. Again, the probability that door 2 was correct was 1/10000. The probability that door 3897 is correct is 9999/10000 because Monty only opens doors without the car.

Now let’s say you’re playing with 10 doors. You again choose door 2. Monty opens all the doors except door 10. Again, the probability that door 2 was correct was 1/10. The probability that door 10 is correct is 9/10 because Monty only opens doors without the car.

Now finally let’s say you’re playing with 3 doors. You again choose door 2. Monty opens all the doors except door 1. Again, the probability that door 2 was correct was 1/3. The probability that door 1 is correct is 2/3 because money only opens doors without the car.