Film has the monty hall problem

>film has the monty hall problem
>protagonist doesn't switch

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>Film explains some scientific or mathematic concept in the first half hour (Turing Test, Monty Hall Problem, Prisoner's Dialemma, whatever)
>Film acknowledges it in the last 10 minutes

Are you talking about 21? Is there another movie where this happens?

it literally doesn't matter

>protagonist doesn't switch
>protagonist is really good at math

It doesn't matter. His chances are the same.

>amerifats fell for the 'you should always switch meme'

why are ameridumbs so bad at math?

>film introduces chaos theory
>explains butterfly effect as an actual phenomenon

Mathematically, objectively, demonstrably, false

For the normies do the same principle with 100 doors and it makes more sense.

I'm talking about the real world

>being this much of a nerd that you know or care when a movie "explains the butterfly effect as an actual phenomenon"

give up user

>Science teacher shows "The Butterfly Effect 2" during class

I still, to this day, do not understand why.

The Last Crusade.

Indy chooses the plain cup as the Holy Grail; Donovan's choice reveals that one of the more ornate cups is not the Holy Grail; Indy does not switch and his choice is correct.

He was hung over. Literally the only reason you had surprise movie days in class.

Why doesn't the chances fo your choice of door also go up with the door you didn't choose? Wouldn't they both go to 50%?

Is that really the monty hall problem though? Indy chose the plain cup on an informed guess, not just at random.

If all the potential grails were identical you would have a point.

100% chance to check em

Because when you chose it you had a 1/3 chance. Once monty shows you whats behind a door you are weighing a 1/3 chance for a 1/2 chance at winning the prize.

Its supposed to be nonintuitive.

That makes as much sense as the people that try to explain 1+2=4 or whatever the stupid shit it. He chances for both doors go up to 50% regardless of what happened beforehand. Knowing 1 false door changes nothing. Fuck anyone that believes in switching, you chose that door for a reason now fucking stick with it faggot.

You're right. This well documented and proven statistical fact is not true after all because you can't understand it.

It's demonstrable in practice that you win more often by switching, look it up before posting like a retard.

You have more of a chance of picking the wrong door when there are 3, That's why you switch when there are two.

Simple enough, nigger?

It's about 18 minutes of spider man flinging web all over lady boy's in Thailand, very out of place.

But if the other door now has a 50% chance of being right, it means you also have a 50% chance being right in sticking with your door. The probability does go up, but for both doors

Okay which Film actually features the Monty Hall problem?

M8 all you need to understand is that your first choice was probably wrong. The first time you pick there's 2 goats and 1 car, so you probably picked a goat.

So you most likely picked a goat, then they reveal the other goat, car is probably behind the other door then and you should switch. Won't always work because sometimes you did pick the car the first time but usually you didn't.

Understanding that your first choice was probably a goat is all you really need to get to see how its right to switch.

Wrong. If you started at two doors, then yes it would be 50/50. However the three doors are valuable information that change the odds.

Where does the rest of the percentage go then?

You're making this more complicated than it is.

Your first choice was probably a goat, that is why you switch. That's all there is to it.

Let's say you have the same problem, but with 100 doors. The host opens 98 of the doors, and the host will never open the door with the prize behind it. Do you switch to the one door you didn't pick or stay with your first choice?

Same idea but it only improves your odds of winning by 33%

This

If you pick a goat at the beginning and switch you are guaranteed to get a car.
If you pick a car at the beginning and you switch you get the goat.
So what are the chances you get a goat?

>Implying you need to change
If you were right, you were right.

does this account for if they switch shit behind the doors depending on your choice?

Scenario one: You pick the car. Host reveals a goat. You change doors and lose.
Scenario two. You pick goat 1. Host reveals goat 2. You change doors to win the car.
Scenario three: You pick goat 2. Host reveals goat 1. You change doors to win the car.

2/3 times you win when you change doors.
You're only going to win 1/3 times if you don't.

Understand now?

But you can't know if you were right and if you're not a moron you can see your first choice is probably wrong.

There's 2 goats and 1 car so your first guess you probably pick a goat, then he reveals the other goat, gee I wonder where the car could be?

YO THIS IS BULLSHIT. WHEN LEFT WITH TWO DOORS. YOU HAVE TWO CHOICES.

>CHOOSE THE DOOR YOU PICKED : 50%
>CHOOSE THE SECOND DOOR : 50 %

Ergo, it changes nothing.

>Implying I'm unlucky as you

This is the best explanation I've seen. I'm not entirely sold, but this makes enough sense that if I were in the situation, I'd switch

>when the door 1 was chosen by the player: the host's deliberate action adds value to the door he did not choose to eliminate, but not to the one chosen by the contestant originally. Another insight is that switching doors is a different action than choosing between the two remaining doors at random, as the first action uses the previous information and the latter does not. Other possible behaviors than the one described can reveal different additional information, or none at all, and yield different probabilities.

sure is lots of assumptions on why the host opened the door

Scenario one : After picking one out of three choice, only a car and a goat remain. You choose to switch : it's a car !!
Scenario two : After picking one out of three choice, only a car and a goat remain. You choose to switch : it's a goat !!
Scenario three : After picking one out of three choice, only a car and a goat remain. You choose to not switch : it's a goat !!
Scenario four : After picking one out of three choice, only a car and a goat remain. You choose to not switch : it's a car !!

Now let's add up those %. 2 : 2. 50 %. WHatever you choose to do once a door has been opened has no bearing on your actual chances to get something.

50/50

Name 5 kinos, 3 flicks, 16 cinemas, and 45 joints where this happens.

Two doors have goats one has a car.
What are the chances you pick a goat?

Your first choice is a goat 2/3 times that's why you should switch user.

It has no bearings on anything since you make a new choice.

But it does have bearing because you know your first choice was probably a goat.

It's a 50% chance, idiots. Either you pick the car or you don't.

Irrelevant since no matter what you actually end up doing, you make a new choice.

You don't make a new choice.
You don't randomly pick a new door.

The host opens a door with a goat.
He never opens a door with a car.

You don't get another choice between two doors.
You get a changed choice of switching or staying after the host removed a failure.

But that's irrelevant. You shouldn't see it as "keeping your choice". You should see it as choosing one out of two door.

pretty much

this is just probability meming WE IZ REAL SCIENCE

"switching". Aka, choose door 2. Or choose not to choose door 2 and go with door 1.

MEESA SO CONFUSED

more important question

what is the probability of the host showing one wrong door to fuck with the player?

>Monty hall problem
>implying the theoretical problem wasn't an autistic solution to a theoretical problem that didn't actually exist
The show withe the problem would have had a different answer than the perfect Monty hall problem dictates

Furthermore 99% of the "Monty hall problems" are not the real problem.

Read the Wikipedia article on it and it becomes evident the mathematician who made it into a big deal was just being autistic

Yes but switching depends on the first door choice.
Switching guarantees you get the opposite of your first choice.
And the chance to get a goat as your first choice is 2/3.

The host knows what's behind all of the doors and what door he offers to let you switch to depends on what's behind the door you picked originally. Namely, Monty's door has to be the prize door unless you picked the prize door on the first attempt, in which case the door would have to be a goat.

It's not a "new" choice because the door you get offered to switch to is dependent on your first choice.

You should look up at what "switching" actually means in term of concrete action. It's actually choosing a new door, or not choosing the new door. It's no different than the choice the person had when there were three doors.

>If we assume the host opens a door at random, when given a choice, then which door the host opens gives us no information at all as to whether or not the car is behind door 1.

Only when the decision is completely randomized is the chance 2/3.

based wikipedia for the truth

It's not though, since, whatever you pick, one door will be opened (which won't have the "car").

ITT people who can't into bayesian probability
This is really the best way to make sense of it intuitively. I refused to believe the Monty Hall problem until I actually ran a simulation of it.

The door that is left is dependent on the door you initially pick. This means that your initial pick has a bearing on the second "choice".

Here's a slightly different scenario: imagine that you are given the same initial choice of picking a door. The host then still reveals one of the goat doors. However, this time you're never given a choice to switch or not, instead the host forces you to take the leftover door, the one that wasn't picked by your nor revealed. What are the odds this one contains the car? The key is that because a goat is always revealed during the reveal step, the final dilemma is guaranteed between a car and a goat. If your initial choice was a goat, then a switch guarantees a car, and the converse is true (picking the car initially guarantees a goat). You had a 2/3 chance of picking a goat initially, so in this scenario where the host forces you to switch and there is no second "choice" you will get a car 2/3 of the time. The principle can be applied to the real problem: if you set your default mindset to always switch regardless of what is revealed (i.e., you're forcing yourself to switch and eliminating the notion of a second choice), then you will land on a car 2/3 of the time.

That's fucking stupid.

Computer simulations have demonstrated that switching is objectively the best choice.

But in this situation switching means you always get the opposite of your first choice.

You understand it so far do you?

It is because the host does not open a random door. He always opens a goat door.

Imagine watching the game from the other side and knowing what is behind all doors.

I don't know how I could further help you understand it.

>Schrödinger's cat

maybe he was feelin' real lucky

I'm so sorry. Butterfly Effect 2 is dramatically worse than both the first one and BE 3, and neither of those are really "good".

It doesn't fucking matter because 1/2 goat remains.

Always switch.

Here's an example that makes this problem and solution MUCH clearer.

Imagine FIVE HUNDRED doors instead of three. There is a prize behind just one of these doors.
You pick one of the 500 doors. The remaining 498 doors are removed, the door with the prize is still left. Switching now is almost guaranteed to get you the prize. The first time you picked, the chance to pick the prize was very slim, 1/500.

Now imagine this with 3 doors and the logic is still the same as with 500 doors.

>The door that is left is dependent on the door you initially pick
It's not tough, since whatever you pick, one door will be opened and one door will be left open.

Let's run it out.

Scenario 1 : Pick Door C (for CAR) : Host choose door A, door B has a Goat.
Scenario 2 : Same one but he chooses door B. Also a goat.
Scenario 3 : U pick Door A. Host opens Door B : GOAT
Scenario 4 : U pick Door B, Host opens door A : GOAT.

ADD IT UP.. In TWO scenarios you are left with Door A. And in the two other scenarios, you are left with Door B. 2 : 2. Whatever you pick, you have the same chance of having Door A or B left.

No. See. In your hypothesis you don't have a choice.

top b8 m8

Hey. At least you got some hot Erica Durance moments.

No that's you

You picked a goat, they showed the other goat, remaining door is the car unless you picked the car the first time but you probably didn't because there were 2 goats then and only 1 car.

Switching does not guarantee a win but it does give you the best chance.

Its simple:
You start with 3 doors and make a random choice. You did not "choose that door for a reason". The doors are identical and you have no information to work with.

The host opens one door, revealing a goat. There are now only two options. What is important, however, is to consider why he opened the door he did.

The host knows where the prize is. He would only open one of the two goat doors. However, he cannot choose the door you have chosen because then the question "Would you like to switch doors?" would not make sense.
The host made a conscious choice not to eliminate one of the doors, but was obligated not to choose your door. The fact that he knows where the prize is and chose not to open the door you did not choose means it has a greater probability of concealing the prize.

This is not why switching is a better choice. If that were so, switching and not switching would be functionally identical actions.

>2016
>there are still people who don't understand the Monty Hall problem

no wonder the world is getting conquered by chinks and goat fuckers

>You picked a goat
No, I picked a car.

You're conflating that each of those scenarios has the same probability of happening, but they're not. You have a 1/3 chance of picking each door, so scenarios 3 and 4 each have a 1/3 chance of happening. Scenarios 1 and 2, however, need to combine to have a 1/3 chance of happening, because your choice (picking door C) is the same as picking the other doors. The way you have it set-up is that scenarios 1 and 2 both have a 1/6 chance of happening.

Show me where there's a mathematical difference between the host forcing you to switch and yourself forcing you to switch.

I think I understand it but it's wrong because it's assuming the host choice to open a wrong door is truly random

en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solutions

The table here explains it the best, if you can't figure it out after reading that then you need to go back to school.

You have yet to prove why, when you make the second choice, it's as if your first choice had any relevancy. You sound like one of those dumbfuck playing roulettes saying: IT HAS BEEN RED 8 TIMES ALREADY, NEXT TIME WILL BE BLACK. IT HAS TO".

New Monty Hall problem:

You're the host of a game show. There are 100 doors, one of which has a prize behind it. A contestant picks door 76 at random. You know the prize is behind Door 1. You must now pick a door to offer to the contestant for him to possibly switch to. The catch? It has to be the door with the prize behind it.


Which door do you pick?

The real issue here is if the host is going to open a door anyways why don't they just do it before you make your first guess? Its the only fair way of doing it.

Nope, you are wrong here, since the host has to open one of the wrong door, he can pick either door in scenario 1 (and 2). Which, have the same probability of happening. BUT, scenario 3 and 4 have the same probability of happening. If you add up scenario 1 and 4, the probability of it happening is EQUAL to the sum of scenario 2 and 3.

there are like 8 people in here who have already demonstrated why switching is preferable, and you can even simulate yourself if you don't understand them

You have yet to refute anything. People spouting bs they read somewhere but are unable to actually explain it is meaningless.

Whatever the host does is dependent on what you pick first. It's only your initial pick that matters, what the host does is irrelevant because you'll switch from a goat to a car or from a car to a goat.

The whole point of the problem is that it ISN'T assuming that. The value of the switch is based on understanding that the host knows where the car is and makes an informed decision regarding which door to open. The fact that he didn't choose the door you didn't choose makes it more likely to be the car door.

As I've shown you, what the host does is actually NOT dependent on what you do. Whatever you choose, he will always open either door A or B.

Because whether or not Monty offers you the prize door is directly fucking related to the chance that you already picked said prize door. In the one in three chance that you did, Monty's door will have the goat. In the two in three chance that you were wrong, Monty's door will have the prize.

Monty does not actually get to make a decision. The first choice completely dictates his actions.

but the host doesn't have to open a door

It absolutely is dependent because he can't open your door. If he did, the question of whether or not you wanted to switch would be moot and the whole scenario couldn't exist.

It doesn't matter which door he opens, he always reveals a goat. You're not trying to choose a particular door, you're trying to pick a reward behind the door.

It's far more intuitive if you imagine there are 100 doors. You pick one, and the host eliminates 98 of the doors that don't have a car. Which is more likely; that you picked the correct door out of 100, or that you now pick the correct door out of two? Obviously you would swap.

yall niggers adding more numbers aren't helping us stupid people

Or 97*, whatever. It doesn't really matter how many

Yes, that's why he will open Door A or B. Ine one scenario, that has a 1/3 chance of happening, he can choose either one, in the two other he has to pick A or B.