2/3 of conservatives get this correct

2/3 of conservatives get this correct
1/2 of liberals get this correct

What is your answer Sup Forums

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Other urls found in this thread:

en.wikipedia.org/wiki/Bayes'_theorem
en.wikipedia.org/wiki/Bertrand's_box_paradox
twitter.com/NSFWRedditGif

The ball is both dead and alive?

Yes

5/8

1/2
Double silver is removed as you already pulled a gold. Your only possible results are as silver from the gold and silver box or a gold from the double gold box.
Also put me down for conservative

50/50

10/16

WHAT

We win.

dis, iss fiddy fiddy bruddah

1/3

How could you come up with 16, rats
Do you not math?

/sci/ conservative here

case 1: you pick the 1 gold 1 silver box, and select the gold
case 2: you pick the 2 gold box and pick the "left" one
case 3: you pick the 2 gold box and pick the "right" one

2/3

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40%?

The absolute state of canada

0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001%

2/5

100% as we live in a world governed by predeterminism and I am white, therefore more likely to succeed

It's confirmed 50% you bunch of dumb faggots

50/50

1/3

2/3? Considering that you can blindly take the same ball.

>/sci/ fooled by a simple word problem
Checks out.

Best answer so far

did you come here just to post that?

2/5 you guys are retarded

never mind its 1/2

50/50

You've already drawn 1 gold ball. There are only 2 boxes with gold balls, 1 with gold-gold, 1 with gold-silver.

So if you reach into the same box the only remaining ball is either gold or silver. There only two possible outcomes.

It would be more clever to say you pick a ball at random from a box at random and ask what are the chances of drawing a gold ball.

14/88

The answer is 100% since there's no such thing as random and I know I will pick the right one. That is what it means to be me. My hand is only capable of victory.

Prove that random exists or else i win.

/thread

2/3 because if you get a gold ball to start you are more likely to get it from the box with two gold balls in it.

i'm not good at math, all it did was look at the image and it looks symmetrical of half and half so I thought it's just half and half

this

en.wikipedia.org/wiki/Bayes'_theorem

2/3

By Allah I should hit you with my shoe for such comment. You are lower than a Swede cuck you filthy leaf.

It was 50% before you grabbed the gold ball. 40% chance of grabbing another one on the next pull. Though depending on the statistical model you use, could be as low as 30%

Quit getting us to fondle your damn balls!

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if you picked gold, you picked one of:
box 1 ball 2 (next will be gold)
box 1 ball 2 (next will be gold)
box 2 ball 1 (next will be silver)

the answer is 2/3

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Bertrand paradox

so I don't know if the wording threw anyone off but this is what I've got:

en.wikipedia.org/wiki/Bertrand's_box_paradox

2/3

Monty hall niggers

Does not apply because a gold ball was already taken. You have to multiply the probabilities together.

>macedonian education

>>>/highschoolstatsclass/

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My grand father's hair is on fire. The turnips in the street grow from salt mined in Texas coasts collected from the feet of prostitutes on the beach. There is no such thing as things. The to for went had go round in had for that she having with am to the moon.

Also, salsa.

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Are you a cool fucking dude OP giving both of the most common answers in your first two lines?

this
it is the resolution of the Bertrand paradox. 2/3

trick question the ball identifies as copper

Chances don't matter in a non-binary world.

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I was having a laugh, curious if anyone would doublethink their answer based on that

JUST AS I SUSPECTED. YOU CANNOT PROVE THAT RANDOM EXISTS. HENCE THE TRUE CHOOSER IS ME, NOT "RANDOM". I HAVE CHOSEN TO GET THE CORRECT GOLDEN BALL.

2/3rd

Fuck Trump and Fuck White People!

i think it's 100%

it depends

what would've happened if i pulled a silver ball?

I am a golden ball user.

If it was a black ball I calculate 3/5

because of how it is worded the expiriment only starts when you pick a gold ball therefor the box with 2 silver balls is not part of the calculation from the begining and just there to confuse you. its 50/ 50.

Then you'd get the rake

14/88
jk it's 0.5

HOWEVER, KEK HONORS MY WISE CHOICE WITH THE NUMBER 5 (5 GOLDEN RINGS)


THIS IS WHAT IT MEANS TO BE ME.

Look at this dumb nigger and laugh!

from the same box. ..

Before you stuck your greedy jew mitt into the box, the chance was 50% (3/6) of getting a gold ball. If you get a gold ball on your first try, then you are down to two gold balls with five total remaining (2/5)--otherwise known as 40%

I remember my kike math teacher disagreed with me though. Can't remember the jew magic reason though...something about you don't know what's in the box and it is more likely that you'll get a gold ball since you've something something something.

that anons correct

1/1
Pee is stored in them so they are gold.

no because if you got a golden ball box, it couldn't have been from the one with two silver. That box is kill.

underrated

You think it's 50/50, equal odds that you picked box 1 or box 2. But here's the problem: If you pick a ball from box 2, what is the probablility you get a gold ball? It's 1/2. If you pick from box 1, what is the probability you get a gold ball? The probablility is 1/1.

So by the mere fact you picked a gold ball, doesn't that tip the odds in the favor of box 1? It's no longer a 50/50 chance between box 1 and box 2. Since you picked a gold ball, it is more likely you picked box 1.

And since it's more likely you picked box 1, it is a GREATER THAN 50% chance the next ball will be gold

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2 chances for gold 1 chance for silver.

If you get a gold ball from the box then you automatically disqualify the double silver box from the equation. Remember you're picking the remaining ball from the SAME BOX.

It's 50%.

50%

If it's a given that the first draw is a gold ball (i.e. it doesn't actually influence the probability), then it's obviously 1/2. Box 3 is right out from the start since there's no gold ball in it, and the remaining possibilities are either 1 silver or 1 gold ball in the box from which the first gold ball came.

In a complete probability tree, in which the first draw isn't predetermined, the overall probability of drawing first a gold, then a second gold from the same box is 1/4.

I understand conceptually why the answer is 2/3, but it still instinctively seems wrong. Has a simulation of this problem been run, with say 1,000 attempts, and proven that the probability is actually 2/3?

Like if I got these boxes and balls in real life and ran the experiment 1000 times, would I actually observe that 2/3 of the time after I pick 1 gold ball, another gold ball follows?

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i just got it lol

You already got the gold ball.
Start counting gold balls from the left.
1/3rd chance you got the 1st
1/3rd chance you got the 2nd
1/3rd chance you got the 3rd

If you got the 1st or 2nd then the next ball from the same box is also gold. That is 2/3rd of the chances.

you already picked the gold ball
meaning that your next choice is between a box with no golds, and a box with two golds
ie 50% chance

It has been run on computers. Look up Bertrand's paradox on wikipedia, they will give sources of the computer simulations

66.67%

>golden ball hoax
KIKE TEACHER TRICKED YOU user

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you forgot my point that the experiment only starts if you take a gold ball in the first place. picking the silver ball out of the second box is not even an option.

Yes but the fact that you picked a gold ball is evidence that you picked box 1.

Best

50%

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>After a box has been chosen, but before a box is opened to let you observe a ball, the probability is 2/3 that the box has two of the same kind of ball.

>If the probability of "observing a gold ball" in combination with "the box has two of the same kind of ball" is 1/2 then the probability of "observing a silver ball" in combination with "the box has two of the same kind of ball" must also be 1/2. And if the probability that the box has two like balls changes to 1/2 no matter what kind of ball is shown, the probability would have to be 1/2 even if you hadn't observed a ball this way.

>Since we know his probability is 2/3, not 1/2, we have an apparent paradox. It can be resolved only by recognizing how the combination of "observing a gold ball" with each possible box can only affect the probability that the box was GS or SS, but not GG.

1/2, because the all-silver box is eliminated.

what gold balls goyim?

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It's only 100 percent or 0... but if we use fake math statistics then there's a 66.66% chance that the next ball will be silver.

I'd say 50%

You obviously drew from 1 of 2 boxes, as only 2 contain gold.

So one box has 100% chance of drawing a gold again. The 2nd box has 0%. So depending on what box you drew from, determines the out come.

So 50%

In a Libertarian.

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2/3 is greater than 1/2 so if you're trying to insult conservatives you did it wrong

Well shit dawg I guess I better go do some relearnin

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Doesn't matter, as it stands, the fact that you picked a gold ball on the first try is evidence that you have picked box 1. Picking a gold ball on the first try is more likely from box 1 than it is from box 2

60% of the time it works every time

no its evidence that you picked from box 1 or box 2, since both have a gold
which means you've either picked from a box with a remaining gold, or a box with a remaining silver
its a 50/50 chance on which box you initially picked

false, the boxes are both alive and dead.

also 1/3

50% obviously, anyone who says otherwise is either a shill trying to make his/her country look dumb or a retard.

GIVE BACK MY BAWLS

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Consider the case that each box had 100 balls in it.

The first one has all gold balls,
the second 1 gold ball and 99 silver
the third has 100 silver

If you pick out a gold ball, you know almost certainly you have picked the first box.

the intention isnt to insult anyone
just fuck with them