Are there any films that deals with this concept?

Are there any films that deals with this concept?

The Witch

3rd entry is goat?

21

The Witch

Winner winner chicken dinner

There's lots of movies about savantism and autism.

Adaptation

Annie Hall

Keit-ai.

tell me about it?

That's a pretty long goat.

I love that statistics problem.

I still don't fucking GET IT WHY IS IT NOT 50%

Primer

lol

because monty hall is a dick.

Because if you have a 33% chance of the big prize and you know that you didn't hit the zonk the 66% is the most likely pick.

Since when you first made a choice the chance of you being right was only 33%. It doesn't magically become 50% because when you first chose you didn't only have 2 options.

it's pretty simple 3 doors all together, two doors are goats, only one is a winner so you are more likely to start off on a goat so always swap

... you're right, the chances don't "magically become 50%," they become "50%" because one option was removed. Without magic.

This concept is bullshit, math doesn't accurately reflect reality.

The variable that's being ignored here is if monty hall had shown a goat every time someone was on a zonk, lets make a deal would've been too predictable.

Think about it with 100 doors

The guy slowly removes 98 incorrect options and leaves you with 2 options. The chance that you are initially correct is the door you picked, that is there is a 1% chance the door you picked initially was right. There was a 99% chance that you picked the wrong door, but since all the other incorrect options are removed then that 99% chance that you are incorrect is condensed into that door remaining.

Think about it like "what is the chance I was initially correct" vs. "what is the chance i was INitially wrong"

>implying you don't start with a 66% chance to win

>pick a door
>it can be right or wrong
>removed doors can't be right

odds are still 50%

expand the problem to 1 000 000 doors. You choose one. You had a one in a million chance. It doesn't change only because they show you which 999 998 of the other doors are also empty.

Or just draw a fucking probablility tree for the base example of 3 doors and try to comprehend it.

> removed doors can't be right

The show host knows which doors are wrong retard, it's why this thing works

and that's the problem with trying to make a statistical model of human behavior.

The host knows which doors are wrong. Therefore the host is the deciding factor of the end-odds. Not the fact doors were removed.

The important part is that the door you've chosen is not kept because it could be right, it was kept because you've chosen it.

What's behind reminder door is not random, but the opposite of your original choice.

Why don't you just look up the mathematical proof online jesus.

Your problem is you're acting as if the removal of doors is some new factor to the problem. it isn't. If all but two doors are going to be removed every time, then the odds are always 50%.

haha

did you read The Curious Incident of the Dog in the Night-Time too?

I have. Its crap. its borderline numerology.

it's 66% because of the information the host gives you. Think of it as you purposely wanting to pick a goat first round, since that would assure you winn if you plan on switching. There's 2 goats so it's 2/3. However, if the host has no knowledge of where the goat is and merely eliminates a goat door by chance, the chance is 50/50 whether you switch or not.

2 doors are always wrong 1 is right

if you switch you have two chances of winning if you stay only one chance

it's not fucking difficult

Literally retarded

He always chooses the wrong one though, that's how the game works

The Prestige

that operates on the assumption monty will never show you a goat if you picked right the first time.

But you don't. You have no way of knowing. You can't purposefully pick a goat then switch when he shows the other goat. If hes always going to show a goat, then you have a 50/50 chance of being right.

how does it? if you pick a car the first time you lose if you switch. He'll still show you a goat in one of the two doors. But the chance of picking the car the first time is 1/3 QED

Goats and doors?

>You can't purposefully pick a goat
but you have a 2/3 chance of picking one...how are you this dense?

I've had this ride with my boss before and he too couldn't comprehend it.

Math is right and that's why Monty Hall tv shows aren't airing anymore.

If the extra doors (that had the very same chance of having the prize) stayed unopened and you were asked if you want to switch to any other of 999 999 doors you would argue that you don't switch cause your chances are 50%? No, you would be right in thinking that you have a 1 in a million chance.

Host removing doors and asking if you wanna switch is equal to host telling you "Hey, you want to stay with your 1 door or take 999 999 of my doors?".

Really, just read the wikipedia page over and over again until you comprehend why it isn't 50-50

If hes always going to eliminate one of the options, and that option is never going to be the car, whether you picked a car or not, then one option is always irrelevant and is meaningless. Meaning the odds of picking the car are always 50/50.

This applies if its 3 doors, 5, 50, 50000000, etc. If the removed options are never the winning option, then they are all irrelevant.

I think it's more clear this way. I've got 3 playing cards - one red and two black. I ask you to take one card from me without looking, then look at the reminder and reveal a black card. Even though there's now only two cards in game, chances you are holding to red card remain the same. Because this is the probability you grabbed it on your first turn.

Now if I revealed card without looking and it happened to be black, your chances got bumped up.

the second 1/3rd doesn't matter if its going to be removed from the pool. So with two options, the odds are 50%.

Another take:

2/3 chances of picking a goat initially. host shows the other goat, thus switch gives you a car.
1/3 chance of initial pick of a car. Switch looses.

Switch wins (n-1)/n % of the time where n is the number of doors.

>then one option is always irrelevant and is meaningless.
no it's not. Because if you pick the car first he may pick either door. say you pick door C, half the time he picks A and half the time he picks B. Plus there's also the option that you could have picked that "irrelevant and...meaningless" door first.

>if its going to be removed from the pool.
AFTER you pick one of the doors.

It is 50%, read de Finetti.

If I ever run a business I'll ask for Monty Hall problem on job interviews and reject the dumb fucks who don't get it that you should switch.

Its not possible to pick a door that will be removed.
Its not possible to remove the winning door.

Meaning the odds of picking a goat are 50% and car are 50%.

Instead of stubbornly accepting an incorrect solution that flies in the face of mathematics, try keeping an open mind until it clicks.

It's very, very simple

>one of three doors has a prize
>you have a 33% chance of having selected it
>one of the other doors, guaranteed to be a loser since they obviously can't get rid of the prize door, is removed
>now there is one prize door and one loser
>you can now either assume that the door you picked when the odds were 33% is the prize or take the 66% chance that you were wrong

If you still don't understand it, imagine there are a hundred doors, giving you a 1% chance of having picked it originally, and then they get rid of all of 98 of the doors, leaving your pick and another one. The only way the door they left for you could be a loser is if your 1% chance of picking the prize on the first selection panned out. In the 99% chance that you were wrong, the other door would have to be the prize.

In essence, you're betting against the odds that you were right on your first attempt at picking the doors.

Since door you've picked will stay no matter what, it not being removed does not actually bump its probability of being correct.

I hope I get to be in that interview so I'll roast you as I've done with many Math illiterate fucks before.

You're confusing yourself with the sequence of events. This is basically a pemdas error that's run wild over the internet until someone with sufficient education and insufficient sense tied a circular logic proof to prove his proof.

If a guaranteed useless variable is present and going to be discarded, at any point, then its presence or disappearance is irrelevant to the end result.

underrated comment

Simplest explanation:

You show me a goat.

What is more likely:

1. That I picked the car and the other door is a goat (There was only a 1/3 chance of picking the car)

2. That I picked a goat and you just showed me the only other goat (There was a 2/3 chance of this happening).

Since situation #2 happens 2/3 of the time, and switching in situation #2 makes you win, you have a 2/3 chance of winning if you switch, simply because that's the chance you're in situation #2.

but you can pick one of the choices has the potential of being removed, and if you do you win.
I realize you're being purposely obtuse, so let's clear up some things
do you accept that you have a 2/3 chance of picking a goat initially?
do you accept that if you plan to switch, picking a goat the first time guarantees you win?

Nice.

If there are two goat doors, and one is always going to be opened, the odds were never 1 in 3.

Nigga what?

If monty is always going to open a door with a goat, if you choose right or not, then the odds were 50/50, because one door never mattered.

...

Three doors, A B and C. For simplicity, C will always have the prize, and A and B will always be losers.

SCENARIO 1:
You pick door A
Door B is revealed as a loser
You switch to door C
You win!

SCENARIO 2:
You pick door B
Door A is revealed as a loser
You switch to door C
You win!

SCENARIO 3:
You pick door C
Either of doors A or B is revealed as a loser
You switch to the other door
You lose!

In two of the three possible scenarios where you switch, you win, and therefore the odds are 66%

CIA Rises

pretending to be retarded is not trolling

a has a goat, b has a goat, c has a car.
which goat do you think never matters?

>because I can imagine two winning scenerios and one losing, the odds of winning are 2/3.

creationist math huh

One of them.

Except it's completely correct, and was actually verified with practical testing?

No how about just thinking about it logically.
3 Doors A B C

let's say
Door A is correct

You pick A
Monty opens B
If you switch to C you lose
If you stay on A you win

Let's say
Door C is Correct

You Pick A
Monty opens B
If you switch to C you win
If you stay on A you lose


Let's say
Door B is correct

You pick A
Monty opens C
If you switch to B you win
If you stay on A you lose


In two out of 3 scenarios switching allows you to win.

>I hope I get that interview so I can exhaust you with stubborn idiocy and walk out thinking I've won the argument once you realize that my capacity for comprehending simple logic games is pathetically close to null
ftfy

Then what's your counter scenario?

The only way Monty's door can be a goat is if you picked the prize door on your original selection. If you picked one of the goat doors, his door HAS to be the prize door.

Therefore, the odds that you'll win by switching to Monty's door are the inverse of the odds that you picked the prize door originally (66% versus your 33% chance of picking the prize door initially), because that's the ONLY way Monty's door can have a goat behind it.

You can't pick either one of them to be useless because if you do, you're removing the case where the player picks that door.

I mean if you're gonna troll at least pick something that's not provably wrong from logic alone. Go say anthropogenic global warming is a myth or something.

>still wont answer
kek well i'm out
everyone else have fun if you want, it's obvious he's not going to accept anything you say cause he just wants to make you mad. I'd suggest not feeding him and pushing to front page but meh.

Well, the scenarios where monty opens A and B and you switch get slammed into one and the scenarios where you don't switch are discarded completely.

So replace your set three with

Set 3
Choose C
Monty opens A
you switch to B
you lose

then go on to

Set 4
Choose C
Monty opens B
You switch to A
You lose

Set 5
Choose C
Monty Opens A
You don't switch
You win

Set 6
Choose C
Monty opens B
You don't switch
You win

and add the non-switch losers

Set 7
Choose A
Monty opens B
Don't switch
You lose

Set 8
Choose B
Monty opens A
don't switch
You lose

So that's eight sets and four winners. 50/50.

dude. there are two goats. It doesn't matter what door you pick, monty can (and will, cause hes an evil prick) always show a goat.

I'd watch a game show with this concept

its easy if you open the other two doors first tho

They would be, if after doors were removed you chose again on random as before.

I'm sorry I took 17 minutes to answer another more articulate post oh kek'in one.

You know whats behind the other two doors?
Dogs

see You're letting some aspect of the scenario confuse you. Not sure what.