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No, because you picked one of the 6 balls at random.
And since it's gold there is a 2/3 chance it's from the box with two of the three gold balls.

"you pick a box at random"

It was assumed that you have equal chance of picking from either box.

It's 1/3, do you idiots have poor reading comprehension? Only 1 box contains 2 gold balls

>It was assumed that you have equal chance of picking from either box.
lol brainlet

mfw more than half of this board is stupid. these are the people giving me advice on gadgets.

>being stupid enough to ask Jow Forums for advice

No. We know there are 2 boxes with gold balls, and the one you picked had a gold ball. So, the one without them is irrelevant. If you have picked one gold out of a box, there is a 50% chance you picked from the one with half gold, meaning there is none left in that box, and there is a 50% chance you picked from the one with two gold.

2/3

The question eliminates the left box. There are now 4 possibilities: 1st gold, second gold, 3rd gold, and 4th grey. But you can't pick the 4th grey, so you only have the 3 choices.

Now at this point, after picking a gold ball, you are asked the probability for the next pick. Since there were only 3 choices, and 2 of them will result in a gold ball, there is a 2/3 probability the next one will be gold.

If you are picking a box at random, it's implied that you dont have the bias for picking the first.

Brainlets who can't into highschool mathematics.
Nice digits but still brainlets.
See monty hall problem.

>left box

Meant right box

No, there are two potential cases here. Either you picked from the box with 2 gold or the one with 1. As you already have a gold, if you picked from the box with 1, there is none left and you wont get another from that same box. But if you picked from the one with two, you will get another gold. There is only one degree of freedom here: which box with gold did you take from? There are only two boxes with gold, and taking from either is equally likely, so the probability is 50%

>If you are picking a box at random, it's implied that you dont have the bias for picking the first.
dude, you just don't get it. you're gonna be a janitor

You didn't pick a box with at least one gold ball.
You could have picked the box with no gold balls.

Each box has a 1/3rd chance of being picked, But the fact the first gold ball is gold tells you the likelihood of having picked the box with two silver balls is now zero but the likelihood of having picked the box with one gold ball remains the same 1/3rd.

It's asking for the probability of you drawing a second gold from the same box you got your first. The third box is a brainlet red herring and the correct answer is 50%.

I think I do... also based on the wording of the question the order of the boxes is arbitrary anyway. Human bias for picking the first over another is pretty irrelevant for a logical puzzle.

No, it's not arbitrary. You withdraw a ball. The chance you DID, past tense, withdraw that ball is not the same for both boxes. In other words, the question explicitly states that you withdrew a gold ball and this implies that the chance of withdrawing from box A was higher! so the answer is 2/3

I'm not a native speaker but shouldn't be GOLDEN BALL? Wtf is wrong with proper English? Nobody care about grammar anymore or what?

If you are picking boxes at random, then the 2 left boxes will have equal chance of being picked, right?

However, when you pick a ball, the middle box has a 50% chance of being eliminated. So you end up with twice the probability of picking the left box, since half the picks of the middle will be thrown out.

both are correct and 'gold ball' is more common parlance

Not him, but it's actually more likely that you previously picked the box with 2 golds.

Initially, you had 6 possible choices, all equally likely. The problem stated that you picked gold first, so you picked either left from the gold+gold box, right from the gold+gold box, or left from the gold+silver box (those 3 are still equally likely).

2 out of those 3 choices led to you picking gold+gold box, so there's a 2/3 chance you picked that box initially. Which implies there's a 2/3 chance that the second ball you choose is gold.

The question is confusing and misleading because your initial choice is stated to be completely random, but by telling you what you ended up choosing, the question makes the randomness weighted towards the gold+gold box. It's not really intuitive because in a real demonstration of this, you would have to discard every trial where a silver ball was drawn first.

No, the third box is completely irrelevant. We already know you picked from one with a gold, so it's impossible that you have picked from the third box. You either picked from the first or second, so you either have a gold left or not. As you are taking another from the same box, and there is one left in this box, and there is either a gold or a silver left in this box, and the likelyhood of you being in either box is equal, the probablility is 50%.

Oh shit, the probability is 2/3 not 1/2...

Thats a pretty unintuitive question.

it's 50 %
you take it from _the SAME box_

>We already know you picked from one with a gold, so it's impossible that you have picked from the third box. You either picked from the first or second

What's exactly what I said:
2 silver = 0 chance
1 gold, 1 silver = 1/3 chance
2 gold = 2/3 chance

it's really not ambiguous at all but it is indeed unintuitive

Statistics are nothing more than a recreational hobby made so that people can go "huh that's interesting" and serve no practical value in the majority of cases that matter.
>but stocks and marketing
>look I know this isn't accurate but it should be good enough right
>did you know that if you throw a ball in the air there's a chance it won't come down!
wow

But it's more likely that you took a gold from the first than the second. To demonstrate this, I think hyperbole is useful. Imagine each box has 1 million balls, the first has 100% gold, the second has one silver, and the third has all silver. Then, it's waaaaay more likely you're in the first than the second if all you know is you have picked a gold ball.

I fell for this initially. I was the guy arguing above that it's 50% until i realized that makes no sense. If we know you have picked a gold, there's a 66% probability that you are in the first, as tehre are three golds and two of them are in the first.

Real world statistics is very often unintuitive.
That's why you have to calculate rather than guess.

For example:
You have a test for a rare disease that is 99.9% accurate.
1 in 1.000.000 people have the rare disease.
The test is positive.
Would you say the chance of having the rare disease is more or less than 1%?

language is naturally evolving man xD
its how people write it NOW xddd

Statistics is extremely important in science and in medicine.
Maybe not in brainlet fields, no.

While both are correct, I was under the impression that "golden" implies not exclusively gold, like a gold plated ball would be a "golden ball", but a "gold ball" is simply a ball of gold.

The problem too is that people use "gold" as a color so a gold ball is golden as well.

English a shit.

>there's a 1 in 1gorillion chance we can cure this disease, the statics really help us here
>spin up a worldwide distributed calculation platform so we can bruteforce a solution for decades
But at least we knew it was possible, as if anything is strictly impossible to begin with.

Do you want the box with the 2 gold balls? 33% chance
Have you already picked a gold ball out of your box and do not know the color of the 2nd ball within? 50% chance
Arguing about the problem just because it was phrased by a retard? 100% this thread

>there are two potential cases

Incorrect. There are 3. You could have picked Box1-Ball1, Box1-Ball2, or Box2-Ball1. In 2 out of those 3 possible scenarios, the other ball is gold.

There's a .1% chance I get a false positive and ((1/1000000) * 99.9) = basically nothing probability that I get true positive, so the test is really quite inaccurate.

Option 4) leave, they are not paying you for this observation

It's very important when diagnosing.
see And doctors often do fuck up because of a poor grasp of statistics, causing many deaths.

>Arguing about the problem just because it was phrased by a retard? 100% this thread
Let me tell you about factorials and /v/

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>It's very important when diagnosing.
Fair enough.

It's 50% you fucking retards.
t. BSc. Mathematics

Heres the thing, its not asking for the probability of picking 2 gold balls in a row, its asking for the probability of getting another gold ball after already selecting a gold ball. The chance of which box you chose it from doesn't matter when you already have selected the ball.
50%

Kek!

Neither is exactly right because of the way the question is phrased. This is like the gold/black vs. brown/blue dress, arguing over a pointless problem that was intentionally made confusing.

>You pick a box at random
>It's a gold ball
These are contradictory. One implies randomness, the other implies a fixed outcome (not random)

Now, depending on how you interpret it the answer is different.

2/3 is correct if you ignore the "pick a box at random" part and just assume you got a golden ball and start calculating from there.
>example: You got a gold ball. Since there is 2 balls in the first box, and one in the second, it's a 2/3 chance that it was box one.

1/2 is "correct" if you start calculating at the "pick a box" stage.
>example: You pick a box at random. 1/2 chance between box one and two. It just so happens that you get a gold coin because the text says so.

The silver-only box can be ignored in both cases.

We should focus more on why the asker doesn't want us to grasp the silver balls. What secrets are they hiding?

Attached: vO7lRZ7[1].png (621x702, 56K)

You're picking from the same box you initially picked from. Effectively you've been given that there is a gold ball in the box so the probability is P(Two Golds|G) = P(G&G)/P(G) = 1/3 / 2/3 = 1/2

holy shit the state of higher education

Read the rest of the thread.
We already established the correct answer is 2/3

>You're picking from the same box you initially picked from
Yeah, and the box you initially picked from has a higher probability of having been the first box, you complete brainlet

It's not contradictory at all.

>You pick a box at random
>It's a gold ball

This tells you that it could be either of the boxes containing a gold ball. Without being told it's random you have no way of knowing the probability of which of the boxes with a gold ball it could be.

>This tells you that it could be either of the boxes containing a gold ball. Without being told it's random you have no way of knowing the probability of which of the boxes with a gold ball it could be.
lmao!
No, there are 2 boxes with gold balls, you are told you certainly picked from one
one contains 2 gold balls and one contains 1 gold ball
use your fucking brain

He's obviously just some highschool kid lying on the internet.

t. billionaire with 11 PhD's who is also a formula 1 driver and a movie star.

Can I move in with you though?

I used Bayes, and ended up with a probability of 1/6. Ah well.

And? Without being told it's random then the choice could have been non-random. e.g. you could have known which box had two gold balls and chosen that one. In this scenario you'd have a 100% chance of picking a second gold ball. By being told it's random it's tell you that the laws of statistics are applicable.

hahahahaha

>implies a fixed outcome

In what universe was a fixed outcome ever implied?
You rally need to +1 your reading comprehension.

Clearly the first ball being gold was the result of chance: it could have been silver but it happened to be gold.

>hahahahaha
Yeah, that really dispelled my point.

>be OP
>questions shit about VMs
>adds an IQ test
>thread derails to said IQ test
>OP's question gets ignored
>entire Jow Forums board shits on each other debating whether the probability of the outcome is 50% or 2/3
You cannot get more Jow Forums than this

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yeah you're just dumb

>if you ignore the "pick a box at random"
WRONG because the probability of HAVING WITHDRAWN from one of the two boxes with gold balls is different from the second box you STUPID FUCK

So say the balls were numbered 1,2,...,6 in the order of the picture. If you pick at random and get a gold ball and then pick from the same box your picks look like this

1,2
2,1
3,4

Only two of these produce the desired outcome.

Faggots stop arguing

#!/usr/bin/env python3

from random import randint

boxes = [['gold','gold'],['gold','silver'],['silver', 'silver']]

alsoGold = 0
sample = 1000000
problemScenarios = 0

for _ in range(sample):
box = boxes[randint(0,2)]

initialChoice = randint(0,1)
if box[initialChoice] == 'gold':
problemScenarios += 1

secondChoice = (initialChoice + 1) % 2
if box[secondChoice] == 'gold':
alsoGold += 1

print(alsoGold / sample)
#0.334719
print(alsoGold / problemScenarios)
#0.6679218174743332

Attached: Screenshot_20180731_211520.png (882x661, 44K)

thank fuck

if you keep picking it'll be 2/3 but if you pick another box it'll be 1/2

Also, allow me to give a brief explanation:

If you take a golden ball, there's 3 possible events:
You're fucked and picked the box with another silver.
You landed on golden box and picked left ball.
You landed on golden box and picked right ball.

2/3 end up in gold ball here.

I aced probabilistic exams just by mentally imagining those scenarios. No math needed.

No, it implies that you will always get a gold ball as if by magic.

>In what universe was a fixed outcome ever implied?
The fixed outcome is that the first ball is gold.

Consider these two interpretations:
>You pick a box at random. You take a ball from it which is guaranteed to be gold. what's the chance of the next ball being gold?

>You pickED a box at random. IF you got a gold ball, what's the chance of the next ball being gold?.

you're wrong, proof

This doesn't address anything I said.

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You are equally likely to pick each individual ball. So, you are equallly likely to pick each individual gold ball. As the first box has twice the gold balls as the second, you are more likely to have picked a gold ball from the first one than the second. So, you are more likely to have picked a box with 2 golds.

you don't think so? what's your point then?

1/3 no?

>2/3 to get a box containing a gold ball
>1/2 to get a box with two gold balls
>2/6 = 1/3

Probability theory BTFO

66.(6)%

woah.. is this machine learning AI?

That the reason people disagree is because of the text interpretation.

there's only one correct interpretation and that is very clear, everything else is simply wrong, no amount of rationalization would change this

it is not ambiguous

No It's a python script that finds probability by collecting a ton of samples..

It's 1/3 if you don't have the constraint that the first ball was gold.

Care to elaborate?

it's satire about how similar they are as both throw samples at the computer until it works out, yet one is a hyped up buzzword, faggot

Yes, but it's considered a paradox for a reason.

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>You pick a box at random.
>You take a ball from it which is guaranteed to be gold.

Those two statements are incompatible.

Refer to this post all the possible picks would be

1,2
2,1
3,4
4,3
5,6
6,5

Only 1,2 and 2,1 produce two gold balls.

See proof.
First print is without constraint.
Second one is with constraint.

50%

please be bait.. anyway:

>two silver 0
>one gold 1/3
>two gold 1/3
thats not 3/3 or 100%, so it is wrong.

but here you (?) make another mistake:
so fuck it

>thats not 3/3 or 100%, so it is wrong.
Either you picked the middle box and have a 0% chance answer
Or you picked the first box and have a 100% chance anwser
So the final answer is 50%. Learn English and try again.

Exactly.
>You pick a box at random
>It's a gold ball

It should be "IF it's a gold ball", because otherwise it suggests that cases where you get a silver ball on the first try never happens.