Do smart people lean towards Libertarianism or National Socialism?

Sup Forums is split exactly 50/50 between libertarians and national socialists, but who is smarter?

strawpoll.me/12627653

Other urls found in this thread:

quora.com/Are-there-disproportionately-many-libertarians-with-Aspergers-Syndrome-If-so-why
reece.liberty.me/an-overview-of-autistic-libertarianism/
strawpoll.me/12581497/r
en.wikipedia.org/wiki/Bertrand's_box_paradox
twitter.com/SFWRedditImages

#include
#include
#include

int main(int argc, char* argv[]) {
int iter, i, g, s, box, ball;
char boxes[3][2] = {{'g','g'}, {'s', 'g'}, {'s', 's'}};
float res;
time_t t;
srand(time(&t));

iter = atoi(argv[1]);

printf("Doing %d iterations...\n", iter);
g=0; s=0;

for(i=0;i

it's 1/2, inb4 autists read it wrong and don't realize it's the same box and you don't put the first ball back

that's what I thought too, but
experimental results say otherwise

before this thread gets an heroed

>you dont put the first ball back

lolwut, of courses not. and you are wrong, its 2/3 because there were three possible gold balls you could have RANDOMLY chosen first, and two of those are in a box (yes the same box) that contains a second gold ball.

could the (((libertarians))) voting
1/3 - NatSoc
please fucking cease and desist.
you're not fooling anyone.

this
it's 50/50 why are people so fucking retarded

Every time you post that image, all Sup Forums will fixate on is the gold and silver balls. I can't tell if it's because they're all in precious metals or if they just have an obsession with grabbing their balls, or maybe stuffing their hands in boxes.

The Monty Hall problem is entirely fallacious.

Honestly baffles me how people can get so confused about a problem that is so trivial that you can actually just fucking draw a diagram.

I would choose not to pull the lever.

jesus dude. you went full retard

you know you picked a gold ball therefore there are only 2 boxes with gold balls and then you have a 50/50 on what box you went in

fucking explain then

>bin/balls 100000
>Doing 100000 iterations...
>16876 gold then silver, 33257 gold then gold.
>Probablity of gold then gold ~= 0.663375.

trends towards 0.6666... as you increase iterations. counterintuitive but there it is.

Because if you stick your hand in double gold then you are guaranteed to pick out a gold, but you only have a 50% chance if you take it from silver/gold.
Can you see it?
If you take out a gold ball then it's more likely that you took it from double gold, so your chances of pulling out a gold ball again are higher.
The answer is 2/3

This looks more like the monty hall scenario.

At first you had a 1/3 shot at the box with both gold balls. You made a selection. So you can now eliminate the 1/3 possibility that the box you picked was the one with both silver balls.
Which means there's a 1/3 possibility you have the box with the silver ball, and a 1/3 possibility that there's another gold ball in it.

So it's a 66% chance.

wait I did that wrong.

it's a 1/3 chance.

>counterintuitive

It's a monty hall paradox that's why.

so the nazis are right?

The 2/3 shills engage in the Monty Hall fallacy.
It's a cult for retards who don't realise that probability has to be recalculated after every round.

This logic is even more whacked out. The gold ball you took out confirms that it cannot possibly be the box with two silver balls, so that entire box is eliminated. Whilst you originally had a 1/3 chance of picking the 2 silver ball box at the beginning, removing a gold ball first means you have absolutely zero chance of pulling a ball out of the 2 silver ball box the following round.
It's 50/50 between having chosen the 2 gold, and 1 gold 1 silver, boxes.

It's 1/2

FUCK OFF MONTY HALL RETARDS REEEEEEEEEEEEEEEEEEEEEEE

you have a box and you pick a gold ball
the next ball in the same box is either silver or gold
it's 50/50

yeah, the scenario artificially eliminates all the times when you draw a silver ball first - so by the time you're evaluating the probability, there's no chance you've chosen box 3 and only 1/3 chance you've chosen the middle box.

The baits go off.

Nope. Monty hall says it's 1/3 if you don't change your selection
And 2/3 if you do.

Which makes sense because initially you made a 1/3 chance at it.

why is it counterintuitive?
when you pull out a golden one, then from that point forward there are 2 golds and 1 silver in the possible subset

>artificially eliminates
>what is conditional probability
Fucking british education.

reason.com/blog/2011/07/20/being-libertarian-may-cause-au
www.zerothposition.com/2016/07/22/an-overview-of-autistic-libertarianism/
quora.com/Are-there-disproportionately-many-libertarians-with-Aspergers-Syndrome-If-so-why
reece.liberty.me/an-overview-of-autistic-libertarianism/
www.autismpolicyblog.com/2012/10/libertarians-and-autism.html
Poll is fucking useless. We already know that Libertarians are confirmed autists.

You can only pick from the SAME BOX SO THERES ONLY TWO POSSIBILITIES REEEEEEEEEEEEEEEEEEEEEEEEEE

the question is phrased to be counterintuitive, that's the whole point of puzzles like this. it's more interesting linguistically than mathematically desu

You took a 1 in 3 chance. Nothing has physically changed since then. Therefore it's still a 1/3 chance.

i'm a nazi and it's 50%

make it look like libtards voted for 2/3 lmfao

read the question ya dumb leaf
it asks: >what is the probability that the NEXT BALL of the same fucking box will also be gold.
you have gold so the NEXT BALL IS EITHER SILVER OR GOLD
50/50

this.

it's 1/2 because elementary probability

>there's no chance you've chosen box 3 and only 1/3 chance you've chosen the middle box.

ANOTHER Monty Hall RETARD

If there is NO chance that you have chosen the 2 silver ball box, the chance you picked the 1 silver 1 gold box is 1/2!

Before you take a gold ball out, it's 1/3. Having taken the gold ball out, you now have additional information to consider, THE PROBABILITY HAS CLEARLY CHANGED.

You initially made a choice with a 1/3 chance, but having pulled a fucking gold ball out you know retroactively that there couldn't possibly be a chance that you picked the 2 silver box, so you must have picked one of the boxes with at least one gold ball, that is CERTAIN, therefore the chance following having picked a gold ball is 1/2 for either of the boxes containing gold balls

Bait? you replied to a picture that explains why it is 2/3, and you some how turned that into
>it's 50/50 why are people so fucking retarded

I wouldn't even pull the lever if I weren't on probation. I don't owe them shit.

>muh survivorship bias
KYS faggot.

yes my bad i read it wrong

compile the code and try it my dude

again - this is closer to a riddle than a math problem.

>You put your hand in and take a ball from that box at random
>It's a gold ball

So why even draw or mention the silver/silver box? They tell you:

>you picked 1 of 3 at random
>but not that one

misdirection.

Okay I'll change my answer. I'm with polan. 2/3
Monty hall ftw.

they only ask about the probability of the second pick. the first is irrevelant

Same thing here, it`s 2/3.

clc
clear all
x=[0,0;0,1;1,1];
a=2e3;
p=zeros(1,a);
for l=1:a
g=0;
s=0;
t=0;
for k=1:l
i=randi([1,3],1);
j=randi([1,2],1);
if x(i,j)==0
t=t+1;
j=mod(j,2)+1;
if x(i,j)==0
g=g+1;
end
end
end
p(l) = g/t;
end
plot(p);

also lol at Germany confusing "intuition" and "remembering the right answer from math class"

never change Hans

2/3 Fags think the initial draw is choosing one out six because they can't read.

WRONG. The question is about boxes, not balls themselves.

Another way to word the question:

>What is the chance of choosing the box with two gold balls? You've already picked one gold ball, which eliminates the third box.

IT'S LITERALLY A MATTER OF PICKING ONE OF TWO BOXES

>1/2
>50%
>1/2
>50%
>1/2
>50%
>1/2
>50%
>1/2
>50%

why are you so autistic about this stupid 2/3
read the fucking question or go back to first grade

Polan's diagram here
That says it all.
There are 3 unidentified balls left in play. 2 of them are gold.

this

very intuitive

>There are 3 unidentified balls left in play. 2 of them are gold.

>There are two BOXES left in play. One of them has 2 gold, one of them does not.

Fixed

durr

Did you read the code? That's exactly what it does, picks one of the boxes, then it picks one of the balls in it, if it's a gold ball, it check what's the other ball in the box.
Then the probability is the number of times it picked a second gold ball divided by the amount of times it picked at gold ball first.

SAME BOX YOU FUCKING RETARDS!!!!!

Monty Hall doesn't apply, because in this scenario you can't pick the goat (the box with 2 silver balls).

>pick a box at random
>then pick a ball at random from that box

its one of six on the initial draw, burgertard

you have a box with a gold ball
next ball is either gold or silver
50/50
>tfw too smart for a code

...

Nope. Imagine you have 6 people run this test. Who have somehow beat the law of averages boss and they each picked a different ball when each turn came up.
So Draw 1:
>S1
>S2
>S3
>G1
>G2
>G3.

The first 3 people don't get a second draw, they lost round one with their faggy silver ball.

So 3 people go on to check the second ball.

The guy who picked G3, will Draw S3, because that is the box with one of each.
So he loses.

Which leaves the 2 who picked G1 and G2. They both draw G2 and G1 respectively on the second draw.

2/3

Fuck you. It's fucking monty hall
Your nigger-like inability to understand that is what's called survivorship bias.

You forget about the 3 people who didn't make it to round 2 and lose sight of the actual odds because of it.

After thinking about this for a while, it's clear to me that drawing a gold ball initially means it's more likely you've picked the box with two gold balls, but I still don't see how you calculate an exact probability...

P(box1 | Gold) = 2/3
This is because if you randomly selected the middle box there is only a 50% chance that you grabbed the gold ball from it.

it doesnt matter
what's the next ball? gold or silver
50/50

Sup Forums is split exactly 50/50 between libertarians and national socialists, but who is smarter?
>proof not found

the proof is having a brain

Fools.

If you extracted a fucking GOLD ball it can't be the box with two silver balls.

So that box is out of the game.

You now have to extract the second ball from one of the two remaining boxes. So the next ball is either a gold ball or a silver one. 1/2 is the right answer.

Ultraconservative here, by the way.

>What is the chance of choosing the box with two gold balls? You've already picked one gold ball, which eliminates the third box.

it also eliminates half of the possibility of having chosen box B.

You have to be trolling at this point

>Imagine you have 6 people run this test.

Literally why? That is totally irrelevant to the problem in question. Read again.

>You've already picked one gold ball.

THAT IS THE START POINT OF THE QUESTION. It means you've either picked gold/gray or gold/gold. 50/50

proof: strawpoll.me/12581497/r

This would be true if the events were independent, but they're clearly not, because the ball you picked first says something about the ball you will pick next, unlike a coin toss or a dice roll.

And that is exactly what is written on the code:

>if x(i,j)==0
ie if the ball is a golden ball
> t=t+1;
increase the count of golden balls picked

>j=mod(j,2)+1;
>if x(i,j)==0
checks the other ball in the box
>g=g+1;
if it's gold too, increment the count of 2 golden balls in a row picked


So the code does discard the 3rd box, and only functions on said situation of "what's the next ball? gold or silver"

Yet that gives us 2/3 experimentally, with no other way to dice it.

This code is garbage for a probability test. The call to rand() does not produce a random value. It uses the C standard library call to a pseudo-random number generator. Totally useless.

Exactly right

Like wow. Seriously wow.

cause you can't read the question?
you have a box with a gold ball
the chance that the NEXT ONE is also gold is 50/50

Why would it? B contained a gold ball before you picked it up. It does not have any gold left now, which makes the probability of getting another gold ball lower from 100% (A & B) to 50% (only A)

Prove me wrong or STFU

>Half libertarian half natsoc
You're forgetting our Trumpers, conservatives and monarchists

>It means you've either picked gold/gray or gold/gold. 50/50
and that means 2/3

see

Real answer is we're talking about odds. I just explained that.

The scenario starts in round 1, but eliminates anybody who picked a silver ball in that round.

It starts with you. In my perfectly averaged example of the experiment being run multiple times, (hence why 6 people) if you made it to round 2, where this supposedly begins you picked one of three balls in round 1.
G1, G2, G3.
If you picked G1 or G2 in round 1, you win. If you picked G3 you lose.
2/3.

>It means you've either picked gold/gray or gold/gold. 50/50
That is not the probability though. It's 2/3 and 1/3. This is conditional probability 101

en.wikipedia.org/wiki/Bertrand's_box_paradox

It says to pick the other ball that is in the same box nigger.

good catch, stdio RNG is garbage. would never use it for crypto but couldn't be arsed to use mt19937 for what is basically a shitpost

that is exactly it, user.

what is the chance of picking by random the box with one gold ball? one in three.

the box with two gold balls is twice as likely once you knoww you have one gold ball at first, so the total odds are two in three.

Furthermore, if you run the code but call arc4random() instead of rand(), the result is 0.5. Never be a dumbass and use a pseudo-random generator folks.

i mean stdlib ofc, fml

>said situation of "what's the next ball? gold or silver"
how many choices do you have? 2 right?
either gold or silver. that's what the pic is asking
50/50

>en.wikipedia.org/wiki/Bertrand's_box_paradox
AHAHA
50/50 NIGGERS BTFO

what does it matter, I own all three boxes

itt retarded Swede and Swiss education being btfo by Brit and Brazilian Education.

No, it's still useful. Pseudo-random shouldn't make a difference here.

You're right that once you pick up a gold ball, you're only deciding between one of two boxes, but you're forgetting that it's more likely that you picked a gold ball from the box with two gold balls in the first place. It's twice as likely that the first gold ball you chose came from the box with two gold balls than the box with one gold ball, so therefore it's twice as likely that you will get a gold ball on your second pick than a silver ball.

2/3 chance to pick the gold ball second
1/3 chance to pick the silver ball second

I'm neither a libertarian nor a national socialist. I'm a liberal.

>random
>you pick a gold ball
That's not random though.
It's already in the premise that you didn't choose the box with two silver balls. The third box serves only as a misdirection and is completely irrelevant.

I fail to see how your averages and multiple players are in any way relevant.

The question starts at round 2. Round 1 is irrelevant. You picked 1 gold, leaving the C out. Which means you either picked A or B. Of A and B, only one has two gold balls. If you picked A, you won, if you picked B, you lost. Therefore the chance of having picked the right one is 50/50

The initnial process of choosing is a combination of 2 choices.

>Picking 1 out of 3 boxes
>Picking 1 out of 2 balls

3*2 = 6 - there are six possibilities you can end up with

They are: Box1-GBall1; Box1-GBall2, Box2-GBall3; Box2-SBall1; Box3-SBall2; Box3-SBall3

In 3 cases you'll end up with a golden ball. In 2 out of 3 the other ball in that box is also gold.

You already picked a golden ball. That means in 2 out of 3 cases, the other ball will also be gold.

It's 2/3.

You didn't have an "Other" option, though.

I would have checked the natsoc box if I had to choose one, and I am not natsoc, it's just closer to my ideology.

6 options, but 3 of them are irrelevant.

If you grab the first or second ball, you will pick another gold ball, but if you grab the third gold ball you will get a silver. So the odds are 2/3

1-(1/2)-(1/3) = 1/3 of the results return a silver ball, 2/3 don't.

pseudo random generators are statistically random out to the fucking 26th decimal or more.

what that means, for people reading your troll bullshit, is that you would have to run this problem trillions of trillions of trillions of times before you could detect a significant deviation from random.

>two possibilities, therefore 50/50
nigger confirmed. By the same logic you have a 50/50 chance of dying within the week, since either you'll die or you won't.

The probapilities of initial process of choosing is irrelevant. You have gold. That's it. It's either A or B -> 50%