The Monty Hall math problem

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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

27 Answers

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.

Anonymous 0 Comments

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.