# The Monty Hall Problem – Explained

So you’ve made it to the last round of a TV game show and have the chance to win a Brand-new car. It sits behind one of these three doors, but the other two have a sad little goat behind them. You make your choice, and the host decides to reveal where one of the goats is. He then offers you a chance to change your door. Do you do it ? Does changing your choice even make a difference? The short answer is yes, even though it seems counter-intuitive. Changing your door choice actually doubles your odds of winning the car, but how is that possible? This is the Monty Hall Problem. At the start, most people correctly assume that you have a one in three chance of choosing the correct door But it would be incorrect to assume that when one door is removed. Each door now holds a 50/50 chance of having the car. Let’s use a deck of cards to understand why. Pick a card from this deck without looking this card has a 1 in 52 chance of being the ace of spades, but now I’m going to flip over all the other cards except one none of which are the ace of spades of the two cards left. Which one seems more likely to be the ace of spades? The one you chose randomly out of a deck of 52 or the one I purposefully and suspiciously left turned down? It turns out your card remains at a chance of 1 in 52 where my card now has a 51 out of 52. Probability of being the Ace of spades. The same principle is true with the three doors you see when I removed the door. I did so with motive knowing there was a go behind it the only two scenarios that exists are A. You chose the correct door and I’m arbitrarily picking one of the wrong choices to show you in which case staying will make you win or B. You pick the wrong door, and I show you the other incorrect answer in which case switching will make you win. Scenario ‘A’ will always happen when you choose the winning door, and ‘B’ will always happen when you pick a losing door. Therefore, ‘A’ will happen one-third of the time and ‘B’ will happen two-thirds of the time. As such, switching your door wins two out of three times this Paradox has perplexed many people including scientists and mathematicians. To this day, because our gut tells us, that switching will have no consequence but when using formal calculations or computer simulators the results don’t lie. Switching your door increases the probability of winning. Let’s see it one more time using a chart. Here are all the possible scenarios the car is behind door 1, 2, or 3 , and you have the choice of each three doors. This means there are nine possible outcomes. Let’s tell you them up quickly if it’s behind door 1 and you chose door 1 you should stay but if you chose Door 2 or 3. You should switch if it’s behind door 2 and you chose door – you should stay but the other two you should switch. Add it all up and you should switch six out of nine times, so do you still trust your gut feeling? Got a burning question you want answered ask it in the comments or on Facebook and Twitter and subscribe for more weekly science videos.

but wouldnt they still say the so you want to switch even if you got the right door?

Skia SurkeIf you keep the door you picked, there’s a 1/3 chance of being correct. If the car is in EITHER of the other two doors, the goat that is also there is eliminated. Which gives you a 2/3 chance when you switch.——————————————————————-Which means if you choose the door you picked <Door 1>, then there's a 1/3 chance behind it. If the car is in Door 2, then the goat in Door 3 get's removed from the options. If the car is in Door 3, then the goat gets removed from Door 2. So if the car is behind either 2 OR 3 that will be what you switch to.——————————————————————-Which means if it's behind the door you picked there's a 1/3 chance. But there's a 2/3 chance the car will be in either of the two other doors. And it won't matter which one, because the one with the goat will be eliminated.——————————————————————I don't even want to know how confusing this sounds, but that's what helped me understand it when I first heard this puzzle.

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 = winswitch = lose

you could also pick the middle goat box initially:

[CAR]

[GOAT]~~[GOAT]~~stay = lose

switch = winyou could also pick the third box initially:

[CAR]

~~[GOAT]~~[GOAT]stay = lose

switch = winso it is clear from all the possibilities that switching lead to winning twice, while staying lead to win only once…

Switching > Stay

Maths and logic would suggest switching. But social intelligence will make you realize if the host is trying to effectively change your choice and therefore suggest staying.

Amy showed this video to Captain Holt, but all he had to do was BANGGGG

So Kevin was right…

Anyone else come to this video after watching the Brooklyn Nine-Nine episode about it? XD

I had 66% chance of LOSING in the beginning….until the host said I can switch…. which reversed my option into a 66% of WINNING.

this explanation finally makes sense.

That’s crazy! Nvr thought of it that way

But I didn't get why switching changes the odds. If you do the maths AFTER one of the 2 other doors is opened, both remaining doors have a 1/2 chance surely? It seems to me that one of these doors is insignificant so you really have a 1/2 chance at the start anyway because you're effectively choosing between two doors.

This is only if you're flipping the cards or opening the doors at random…… not if the host knows…

You explained it wrong. Very simple, he is offering you two doors for your one. Irrelevant that he shows you one. The only question is, is he is making the offer because you picked correctly? Same if there is 100 doors and he offers you the other 99 for your one.

i'm going to explain my point of view because i dont think the video is right, please explain why I'm wrong.

You have 2 door, goat and car dor. There is logically a 1/2 chance. Now just add another door that is already opened and has goat in it. What changes? Nothing, just a door that doesn't effect anything, how is this diffrent from the second part of the choosing part then?

this sounds like something i would get on my maths test

BONE

The maths explanation on this problem is always the same lameness…the key point is: The Host or Showman of the cards has already [manipulated] and [focused] your option so that you lose if you stick to your initial card (your random choice)…it's just a manipulation of the mind. Before you even play the game, the Host knows more information than you so he can then steer you away from the winning card (or door). This is not even a pure math problem. Even just expressing this troll-concept is me manipulating you…

IF THERE IS THREE DOORS, THEN THERE IS A ONE THIRD CHANCE FOR EACH DOOR BEING A CAR. IF ONE DOOR GETS REVEALED THAT ITS A GOAT, THEN THERES A FIFTY FIFTY CHANCE OF YOU GETTING IT RIGHT.Still here? Ok. Say that u picked door 1. Then Monty revealed that door 2 was a goat. There are 4 scenarios that could happen from here.

1. You stay and get a car.

2. Stay and get a goat.

3. Switch and get car.

4. Switch and get goat.

WHY IS EVERYONE SO STUPIDNINE NINE

So Kevin was right

Ah from 1:16 onwards I got it. Though it's definitely a mind stretcher!

I was asked this question in an interview and I failed it miserably. I felt so stupid haha.

“Where one of the goats is”

TO explain a reasonably difficult problem, you took cards, which not everyone understands 🙂

It appears to be ridiculous that switching increases your odds of winning, but then, that’s the paradox. In fact, it’s a perfect example of a paradox. Greater minds than mine say it is mathematically correct so who am l to argue? So, l can happily take this principle on board. Switching increases your chances of winning.

But what if you just walked behind the doors and cheated….Or you could just BONE!!

Switching is for cowards.

This does not explain why I should switch, just says 2 goats = 2/3

any brooklyn 99 fans?

I believe the odds with the cards are incorect. You draw first card out of 52 and yes, that card has a chance of 1/52 that it will be ace of spades. you will then show 50 cards with one left over, that potentialy might be the ace of spades, but that doesnt make it 51/52, it makes it 1/2. 50% Which means following the rule of this principle, the first of two cards has 1/52 chance to be the ace while the second one has 1/2 chance to be the ace, because you have only two unknown at that point. That is when two people are drawing. The first one has picked out of the whole deck and the second one is left with the second uknown card. But should only person be drawing, these numbers will not change.

When it comes to the door, I would like you to answer this. 3 people and 3 goats. Each person will pick a goat to eat. But two of those goats will cause a persons death. Number two will eat his goat and dies. Both number 1 and number 3 will start thinking whether they want to switch their goats or not and will follow your principle and start thinking, that while their initail chance of picking the healthy goat was 33.3%, now, number 3 has a chance of 66,6 percent of picking the healthy goat if he switches his goat with number 1 but number 1 has exactly the same chance of 66.6 percent for the goat of number 3, should he decide to change. But then number 1 realised, wait a second, If i have 66.6 chance on his goat, he must have the same chance on mine, and if mine has 66.6 chance of being healthy for number 3, shouldnt it has the same chance for me then? So sudenly both goats for both people has chance of 66.6 percent, which is imposible, but it shows that the odds are going to even and not shrink under the opposite choice in which case it must be 50/50.

To proove it mathematically the first guys goat is X, the third guys is Y.

X goat has smaller chance in favour of Y ergo x<Y but in the same time it works other way around for the 3rd guy X>Y.

so if two conditions are X>Y and in the same time X<Y then they must be equal.

That itself prooves that the odds will never shrink simply under the 3rd doors or goat in this example. They will always even up.

what is misleading is the fact thatn one person is picking out of 3, but put 3 people picking out of 3 and you will get this.

To be honest I am no matematician. I never really managed to get through quadratic equations, so that is where maths ended for me, but this make sense to me this way.

and one more

A. STAY. STAY. SWITCH. SWITCH. GOAT GOAT

B. GOAT. SWITCH STAY GOAT SWITCH STAY

C. SWITCH GOAT. GOAT STAY STAY SWITCH

AFTER SHOWING WHERE THE GOAT IS YOU HAVE 6X SWITCH AND 6X STAY. THAT IS 50/50

Not sure why scientists or mathematicians ever had to study this, it's incredibly simple. Pretend there's two contestants like the guys from the Labyrinth. One always switches, and one always stays. That means Switchy wins if he originally chooses a goat, and Sir Stays-always wins if he originally chooses the car. The odds of picking a goat is twice as likely so if you aim for a goat you're twice as likely to win.

so kevin

wascorrecti think this is pretty intuitive?

Thank you for the explanation.

How about looking at it like this – Monty's choice factors in a greater quantity/quality of information (than that of the contestant), so you should always go with Monty's choice. Is that logic flawed?

do you actually get the goat if you pick it

The goat is the greatest of all time so realistically the goat is what you want

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

Yeah I still don't get it.

BONE?!This was no help at all

1 in 3 comments still dont understand it

2 in 3 comments are trying to explain it

3 in 3 comments want the goat

So Kevin was right

The card explanation got cheers mate just watched 21 the movie

There’s 1/3 chance you’ll pick the right door and 2/3 you’ll pick the wrong door.

So if you ALWAYS switch then there’s a 2/3 chance of switching to the right door and 1/3 chance of switching to the wrong door.

So what if there were 4 doors and 3 goats huh ??

Hoooow dare you you, Detective Diaz. I aM YoUR SUPERIOR OFFICER!

……

BONE!

But consider this:

Let A be the winning option and B and C the losing ones.

Suppose, you choose A, the host reveals B. You swap. You lose.

ORYou choose A, but the host reveals C. You swap. You lose again

ORYou choose B, host reveals C. You swap and win.

ORYou choose C, host reveals B. You swap and win.

This means there are 4 situation events, and 2 you win, 2 you lose.

I am not saying that the solution as of now is not right, but what is the flaw I'm mine that says swapping still gives 50/50 chance? Can someone explain?

Are you stupid? 1 of the doors are eleminated leaving 2 doors left with one car and one Goat behind. 50%.

I understand the Monty Hall problem fully, but this video does a piss poor job of explaining everything.

BONE?

What if you got the car on your first try?

Awesome

This assumes the variables didn't change, which is incorrect. The fact that they showed yoruba goat eliminates one of the choices and you rest the problem to a 50-50 chance. Shown me a mathematical proof otherwise. Another way to think of this would be to have a second contestant take over the game after the first goat is shown and they don't know which door was picked in the first places. Are you telling me their odds are not 50/50?

You can get a better feel for the Monty Hall problem if, instead of 3 doors, assume there were a billion. You choose one, therefore having 10^(-9) chance to win. In that case, you could even say the prize "definitely" isn't in the door you picked. The 999,999,998 remaining false doors are opened, therefore leaving only two doors closed. The one you initially chose, that "definitely" isn't the prize, and the one remaining unopened that you can switch to, which as a result "definitely" is the price.

If you were to perform the problem a billion times and always sticked with your door, statistically you would only win once. If you instead switched every time, statistically again, you would only lose once. This applies in the same way to the problem with the 3 doors, where, after one of the two false doors is revealed, the remaining door "stacks up" that door's probability, rising its own winning chance to 66.6% while the one you have initially chosen still only has 33.3%.

This is incredibly convoluted video on setup (though other comments clarify the decision tree which had to be for Monty to play/ be nicely sustainably).. HE knew so odds are not that odd, just mentalism pseudo jedi mind tricks like horoscopes

A 1

B 1

C 0

Project forward where any 1 is removed does not change probability of the others that already are what they are

I saw this video a couple years ago and did not understand, but now I get it lol

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.

but goats are usefulll… Stop shaming the goat. The goats better than you— dont lie.

Who's here because of Brooklyn 99

If yes: BOOONNNNEEEE?!!

This is pissing me off and no matter how many times it's explained to me I still think it's 50/50.

I seen many videos about this problem,i can't understand. But this video make me understand

I hate the deck of cards/100 door analogy. You can't compare the tv host revealing 33% of the 3 doors to you revealing 96% of the unknown cards in a deck. You're both leaving only 2 answers but since the number of total guesses differ, the 2 answers left for each situation don't hold the same weight. Revealing 50 out of 52 cards would be much more comparable to revealing 2 out of 3 if you're proportionately receiving the same amount of information for each equation.

Correct me if I'm wrong.

But if there are two doors to pick from there is a .5 probability of either door so just because some fat cunt introduced 3 doors he broke math

You should show this to captain holt and kevin.

https://ima.org.uk/4552/dont-switch-mathematicians-answer-monty-hall-problem-wrong/

I understand the premise but to say it's not a 50/50 chance once the first door is revealed is ridiculous. The fact is two doors remain of which one has the car. That is literally 50/50. I think the monty hall equation factors in the original door (hence the 2/3 probability once you switch).

Look at it like this: You have 1/3 chance you will get the car right. You have 2/3 chances you will choose wrong and get new pet goat. After revealing one goat by host, you can either stay with your 1/3 chance, or you can turn tables and say you wanna switch and then that 2/3 chance to miss turns into 2/3 chances of winning. It is 2/3 chances you will win the car then. But that is still probability only, and that is what confuses people. It doesnt mean you will 100% win after switching in laws of probability, but it is rather 66.666666666%.

That's retarded

That deck of cards analogy really helped me understand

So it's not to do with math it's to do with motive? Really confusing way of explaining this here..

My gut feeling still tells me to not switch..

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!

Shit, I'd take a goat. That would be rad

What if you switch doors 2/3 times?

Lol reading all these comments about how confusing the MHP is. There are several keys to why this problem is intuitively difficult to understanding, but I just came here to say this: in 2010, scientists presented the MHP to pigeons. The pigeons learned to always switch as an optimal strategy. Peeps, we're dumber than pigeons.

Most explanations in the comments are wrong. Just an FYI so you don't get more confused.

How dare you Detective Diaz, I am your superior OFFICER!!!

BONEEEEE!!!

This is just depressing to know…

retards jesus mans are mod

I do get it

For the most obsocated with 50% (like me at the beginning):

You never play alone with two doors, nor can a door ever

have 50%, always play with three doors, which opens not

the withdrawal of the game, the reuse to put anything.

Therefore, it is always 1/3 present at each door.

If they offer me two doors to change, the one not chosen and the one that they open,

2/3 always of probability of success. In which I choose, always only 1/3,

As much as one of the other two is open.

In all the two pairs of doors that remain, one to choose from, and another

to open, there will be a total at the end of the simulations, 2/3.

If I always change, I always win, use a group of two doors, of the three.

People get 50% obese and are never playing alone with two doors.

The possibilities are doubled not by changing, but by switching to two doors.

plot twist:

the goat is in a car

I have a problam.

it follows:

I have three doors. The first has a chance of 4/7 to have behind it gold. The second doorhas a 2/7 chance and the last has a chance of 1/7 to have gold behimd it. . Now someone opened the first door (4/7 chance)and behind it was nothing. Now they're asking what are the ods that the gold would be in the second door

. thank you for helping

You are wrong you prick learn basic maths then get back to me xoxoxoxoxo 😡😡😡😡😡😡😡😡

Imagine that the presenter, in another game with only two doors, always

puts the photo of the car in the door on the right. Yes, without being random.

Do you still think that two doors, being two doors, always have 50% chance?

Well, in the already random example of Monty Hall, the same thing happens,

and the presenter, can not avoid it if it is random, puts car 2/3 of the time

in one of the two doors it offers you.

Do you still think that two doors have 50%, when you know that

you choose they have only 1/3 and between the other two appears car 2/3 of the time, and

also shows you the one that doesn't?

Yes it is thought that they are at 50% each we are facing a new religion.

The 50% are continually ignoring how the car is placed in the doors, only see the doors.

Regards.

What if u got lucky

Basically it would be 1/3 chance if the door that was revealed was randomly selected. This was the only reason why it didn’t make sense to me. Since the door revealed WAS purposefully selected, then ig you could say it’s a 1/3 and 2/3 difference.

i still don’t understand

>:D what if my aim was to get a goat all along xD

2:15 Actually no, that’s wrong.

There’s not a 6/9 chance of switching because one of the 3 doors is opened.

This means there is a 1/2 chance that it will be in One of the Two doors left.

Bah…?

BONE!

I get it (after watching the video like 4 times) but on the contrary, I'm too young to drive and id love a cute little goat!

The way I understand it:

The chance increases because the moderator can't open the door with the car when he asks you to switch. That should be the insight that changes the odds.

BOOOOOONE

For short

s = sqrt{frac{1}{N-1} sum_{i=1}^N (x_i – overline{x})^2}=∀(x, y ∈ A ∪ B; x ≠ y) x² − y² ≥ 0= (x, y :- A u B; x != y) x^2 – y^2 >= 2/3

Thats why you have to switch

That's stupid it doesn't make sense

Somehow managed to make it even more confusing and complicated, congrats