The Monty Hall Problem - Explained

You've made it to the final round of a game show, and get to pick between 3 doors, one of which has a car behind it! You make your choice, and then the host decides to show you one of the wrong answers. He then offers you the chance to switch doors. Should you do it? Find out just how wrong human intuition can be in this classic Monty Hall Problem.

Written and created by Mitchell Moffit (twitter @mitchellmoffit) and Gregory Brown (twitter @whalewatchmeplz).

Music by Mitchell Moffit

Art by Gregory and Mitchell

Further Reading----

Wikipedia also has quite a few more explanations:

[photoniccannon]

This FINALLY makes sense. I couldn’t wrap my head around why the probability wasn’t 50/50 on new information. But it turns out that it is more likely you randomly picked a goat, and therefore the host has to show the door with the other goat. And then the unselected one would have the car 2/3 of the time. In other words, it’s not an independent probability.

Clearest explanation I’ve ever seen. And the only one that actually makes sense.

[Mizohal]

How dare you Detective Diaz, I am yoUR SUPERIOR OFFICER!

[seignee]

Guys, switching always gets the opposite result of what you originally chose.

Choose bad door originally = End up with good door

Choose good door originally = End up with bad door

Since it's more likely that you chose the wrong door in the beginning (there are 2 wrong doors and only 1 good door)

then switching is more likely to get you to the right door.

[lafrashenning]

The confusing part of the Monty hall problem is that the "host" reveals a door he knows is not the prize. This is often mentioned only briefly in the setup, but it is key to understanding why this happens. If the "host" chose the revealed door at random, the whole game falls apart.

[lisettedejesus]

Here’s how I finally understood it: It is more likely that you chose the wrong door than not. Switching then is more likely to have desired result.

[salmanhussain]

here is another way to understand it:
Since there is only one of the three doors you can pick at the start,

say you pick the actual car door as the initial pick...
*[CAR]* [GOAT] -[GOAT]-
*stay = win*
switch = lose

you could also pick the middle goat box initially:
[CAR] *[GOAT]* -[GOAT]-
stay = lose
*switch = win*

you could also pick the third box initially:
[CAR] -[GOAT]- *[GOAT]*
stay = lose
*switch = win*

so it is clear from all the possibilities that switching lead to winning twice, while staying lead to win only once...
Switching > Stay

[Annadog]

If you open the door and get the goat, Do you get to keep the goat?

[grain]

Hopefully this explanation helps:

If you pick one door, you have a 1/3 chance of getting it right. Logically, the other two options would be 2/3 since its 2 options out of 3. Now, assume there are two sides. Your 1/3 side for your pick and the 2/3 side for the two other options. The host reveals a goat from the 2/3 side, this reduces the options from the 2/3 side to one option. So now you are left with your 1/3 side one option vs 2/3 side with one option. Its better to switch.

[jaybond]

Who's here because of Brooklyn 99






If yes: BOOONNNNEEEE?!!

[mayakosaraju]

i have no clue of what just happened...

[Max-xsdv]

That card explanation and visual was so much more useful than any verbal explanation I've heard

[r-pupz]

I FINALLY GET THIS! I honestly have broken my brain with this problem so many times. It still feels super weird of course, and I always believed the calculations but assumed the answer of why that is true would allude me. Using the cards really helped me to grasp it, thank you!

[thamastha]

EASY EXPLANATION
Option 1: you pick goat a, Monty picks goat b, you swap and win the car.
Option 2: you pick goat b, Monty picks goat a, you swap and win the car.
Option 3: you pick the car, Monty picks goat a/b, you swap and pick the remaining goat.

SO! 2/3 you win the car when you swap. See? Easy.

[jonathanz.]

About 4 years later, I know how the math works but my brain still doesn’t see why it makes sense but at the same time it does

[tetleyT]

The way I personally understand this problem is that there is always 3 doors involved, so it doesn't change to 50/50 at any point - the probability is "locked in" from the start. When you first choose a door it's "unlikely" you have selected the correct door (because you only have a 1 in 3 chance of selecting the correct door). It's more "likely" that you have selected an incorrect door - because there's a 2 in 3 chance of that happening. So it's actually "twice as likely" that you've selected a wrong door than the right door (2 out of 3 chance versus 1 out of 3 chance). Therefore you should already be thinking of switching doors, right? But which other door to choose? At this point the host does you a huge favour by revealing a door he already knows has a goat behind it. He's essentially saying: "Here, let me show which of the doors you should switch to." The door you're currently sitting on (the first door you picked) STILL has a 1 in 3 chance of being the correct door. Probability-wise nothing has changed in that regard. But since it's already been established that you've likely picked the wrong door - and you now know which door to switch to - you should switch. Here's the kicker though. Making the switch won't always be the right decision. 2 out of 3 times it will be the right decision, and 1 out of 3 times it will be the wrong decision. But which odds do you prefer?

[richturtle]

I think I get it,

You have a 2/3 chance of picking the goat before the host reveals a goat. The car has a 1/3 chance. Chances are likely that you picked the goat before the host revealed a door. So if you swap, it would bring you to the car likely since your original choice was more likely to be the goat.

[parallel]

This is a much better explanation than the others that I've watched. A great source of science as always, AsapSCIENCE.

[zhezhang]

SO SIMPLE. The big question is, do you want a brand new car or a goat?

[cstill]

Short answer: You have a higher chance of picking the wrong door. Theres a 1/3 chance that you'll pick the car and a 2/3 change that you'll pick a goat. If you pick a goat (which you most likely will) and another goat is revealed behind another door, if you swap, you get the car.

[richpaul]

The scenario with the cards only works if the dealer knows where the Ace is. If I picked a card off the top and you flipped over 50 other cards randomly and none was the ace, both cards left would have the same probability of being the ace. The knowledge of the dealer or game show host is the only reason this works.