Sup Forums is way too stupid to answer this question correctly

Sup Forums is way too stupid to answer this question correctly.

Other urls found in this thread:

en.wikipedia.org/wiki/Monty_Hall_problem
en.wikipedia.org/wiki/Bertrand's_box_paradox
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1/2

Point proven. Sup Forums is filled with mongoloids.

probability questions are a great way to make no-life autisitic nu-males feel superior cus they were good little boys during math class.

good luck applying this to an actual mans job faggots. you have been replaced with computers for a reason, nobody likes smarter-than-thou aspie brainlets

-1/12

the odds are two outta five or a forty percent chance of pulling a gold ball next

What algorithms do you think artificial intelligence use to learn?

They use probabilistic models, more often than not Bayesian models. This question is a basic Bayesian probability question. If you don't understand this you cannot build any probabilistic machine (control system, AI, recommendation system like Netflix, ranking system like Google.)

Enjoy your "man's" job in the woods like a savage while I continue running your shit.

Both wrong.

Common Sup Forums, this is an easy question.

OP did some maths today in class and thinks he's superior

Even if Box 1 has a million gold balls, if you choose a box with one gold ball, the only two options are either the next ball is or is not gold, since there are only two boxes, one with another gold ball to pick, and one without.

/thread

2/3 last time this thread was posted, 2/3 today.

it's 50%. if you say anything different you are literally retarded.

Well justifying a wrong answer with an incorrect intuition doesn't make it anymore correct.

If you pick a random box. If after the first pick you have a gold ball, then it is most probable you have the box with 2 gold balls. Thus, it has to be greater than 1/2. What is the answer then?

enjoy being a weak little faggot with no pussy because women dont give a fuck about autistic number running bullshit, they want a MAN who can PROTECT and PROVIDE for her.

replace yourself with a machine and she'll replace you with a machine too. just watch.

Correct.

5 wrong answers
2 people thinking they are better than math because they are savages
1 correct answer

Sup Forums is looking rough or maybe just underage

Given the 2 silver ball box can be eliminated from the initial 1/3 chance it is a 66.666' chance to pick up another golden ball.

Lol, protect and provide. As though we live in a feudal society. Protect and provide means having money. Tech makes more money than your ancestral occupation. Enjoy your alcoholism and white trash.

/rcon !restart /r alternatereality.cfg

two out of five possibilities also expressed as a forty percent chance are both correct, are you high?

P(left box | gold) = P (left box ^ gold) / P (gold) = P (left box) / P (gold) = (1/3) / (1/2) = 2/3
P(second gold | gold) = P (left box | gold) = 2/3

But that's not the question

thats not how this works you fucking retard

You're retarded.

Correct

It is. Learning how to read is the first step to answering a problem.

there is no answer for your level of stupid.

It is. Refer to this user who proved it correctly: This is a conditional probability question.

lol you are so stupid and autistic. you have no idea what real life is about because you spend all your time posting on image boards and playing video games.

like i said faggot, in the end, what women REALLY want is good dick, and im 100% sure you cannot possible provide that. its ok, your bull will be very nice to you while you prep him.

It's actually 75%. It's impossible for you to be on the 3rd box. If you're on the first box (which you have a 50% chance of being in), then you have a 100% chance of pulling a gold ball. If you're in the 2nd box (again, 50% chance), you have a 50% chance of pulling a gold ball. So it's 75%.

50% + 50% of 50% = 50% + 25% = 75%

It's impossible to answer because it was not specified if the ball was replaced or not.

For any half educated mathematician, this is the equivalent of setting a question where you forgot to put a pen in your hand before you wrote it, and also somehow accidentally getting your dick caught in the desk fan.

no, it is basic probability. you have two boxes, one box contains a gold ball, the other contains a silver ball. 50% chance everyday.

Simpleton white trash women enjoy only good dick. Even those enjoy financial security more than "good dick". Despite me doubting you are above average in bed given this puzzle is too demanding for you I am sure you will never achieve more than working a dead end blue collar job for the rest of your life.

>Learning how to read
But you have to draw a gold ball for it to work, so you must eliminate the silver ball box, making it only 2 possible boxes you could have, one with silver, and one with gold

Denial is a beautiful thing isn't it. The answer is 2/3. If you do not believe it refer to this user: . If you still dont believe it you're a mongoloid too.

Stay mad you subhuman. As I said enjoy your white trash. It's emanating from your posts.

We all know that tech jobs are alpha, while you are a beta who works for a poverty wage and kills yourself overworking while being proud to be wearing blue collars.

>calls -1/12 wrong
Are you aware that -1/12 is a meme good sir?

2/3 is correct btw

No, there are 3 gold balls, not 2. So it's X out of 3.

And since 2 of the those 3 gold balls are in the same box (i.e. in a box with another gold ball), it's 2 out of 3.

And to add to this, the people counting the 3rd box in their equations are the actual morons, even though you may know math, you don't have any critical thinking. You already know you must be in box 1 or 2, even though there is a 3rd box, you can safely ignore it, because there are no gold balls in it. There could be 500 boxes, but if there is 1 box with 1 gold and 1 silver, another with 2 golds, and 498 with 2 silvers each, you can safely ignore them all because you will NEVER be in any of them (you already pulled a golden ball, for god's sakes).

That's incorrect.

If it doesn't say with replacement then it's not with replacement retard. Simple enough. Read the problem, solve it.

Enjoy your idiocy.

You don't eliminate it, you condition your set upon the gold ball having been pick. Seriously read this anons response:

this is the monty hall problem.

en.wikipedia.org/wiki/Monty_Hall_problem

i hope that at least a single Sup Forums tard will realise today that he is in fact fallible like all humans. to everyone saying it's 50/50 even after reading this, kys kthxbye

that formula doesn't apply to this problem. but that's ok, basic maths isn't for everybody, don't feel too bad for yourself.

You already pulled one out, genius... Think about it... Really think about it, strain your brain there for a minute.

I was unaware. 2/3 is correct.

That's the wrong idea. You got the correct answer by fluke. But, the way you justified it is incorrect.

Still wrong. Again justifying a wrong answer with an incorrect intuition doesn't make it anymore correct.

But you didn't choose the silver silver, it's eliminated you retard, you have to draw a golden ball

You can't pull a second gold ball out of the second box, though. If you drew one initially, that would be the only one there.

the boxes are actually irrelevant in this question, you can solve this just considering the balls alone.

>That's incorrect
>Can't explain why

What am I being doing trying to argue with either a retard or a baiter.

No it's not. The conditional probability affects the set differently in this problem. It is en.wikipedia.org/wiki/Bertrand's_box_paradox

Lol read the wiki please. Stop being retarded.

If you manage to achieve some moderate success in this life despite you're shit attitude, you can look forward to sucking actuary dick for the rest of your life because you don't actually know how anything fundamentally works.

Both incorrect assumptions.

The boxes are very important. If you disregard the boxes you are not solving the problem correctly.

An user already solved it in the thread with the correct proof. Refer to:

>isn't capable of critical thinking
>doesn't realise he's actually retarded

You are right, then it's 50%. Either you're in the 0% probability box (the 2nd) or you're in the 100% probability box (the 1st). I was actually the moron, thanks for pointing it out (but so was everyone else).

lol this is like looking at something with a 99% chance versus a 1% chance and saying its 50/50 because "it can either happen or it won't"

It's incorrect. Since we know you pulled a gold ball, there's a higher chance you picked box 1 as opposed to box 2. Keep restarting the experiment where you pick a random box and then a random ball. In the cases where you picked a gold ball, 2/3 of the time you will have chosen box 1.

these autistic faggots are so mad lol

> Can't solve a simple logic question
> Calls other people unable of critical thought
Interesting.

Lol you let an idiot convince you to go from having the correct answer to the incorrect answer. That's probably worse than originally having the wrong answer.

It's not 50%

So actually the very first post was the correct answer. Who would have thought.

>calls people idiots for getting the question wrong
>gets the question wrong

you are right of course. it's very similar to the monty hall problem, though

Box with gold ball: 2/3

Box with 2 gold ball: 1/3

Your odds: 50% of second gold ball.

Correct.

Solve the problem, it's easy. if you can't then admit it and learn something new. Don't be a useful idiot (blue collar worker).

put any number of gold/silver balls in those two boxes and you will get the right answer if you consider only the 2 balls.

If you drew the gold ball out of the box containing silver and gold, how would you then draw another gold without being in box 1? Thus, you cannot pull a second gold ball out of the second box. If you drew one initially, that would be the only one there (the remaining ball would be silver).

except it's not and you are wrong.

So I've taken a box, in that box is a gold ball. Knowing this we can discard the silver box entirely. We now have 2 boxes left that can be the one I have chosen. In this box that I have picked, there remains either 1 gold ball or 1 silver ball as we have already taken 1 gold out.
The odds are 1/2 (50%).

i think this guy is the op. he doesn't realise how stupid he is

only the balls* didn't mean to include the 2

The answer is 2/3

protip: you can ignore the box with two silver balls completely

50% is not the correct answer. Read the thread.
en.wikipedia.org/wiki/Bertrand's_box_paradox

Yeah very similar. Both are very confusing without some knowledge of probability theory.

lol wut. This is extremely wrong. How did you get any of those values.

>That's the wrong idea. You got the correct answer by fluke. But, the way you justified it is incorrect.
No, you're incorrect.

It's really quite simple. You draw a gold ball first. There are 3 gold balls, and there's an equal chance you drew any one of them. So the easiest way to visualize the solution is to look at the 3 gold balls independently. And for each of them ask, what color is the other ball in the box?

>Both incorrect assumptions.
But the problem isn't being done repetitively it is done only once in the problem, hence it is 1/2

lol dont use phrases you dont understand moron, you are the useful idiot here. i bet you fell for the overpopulation meme lol

Wrong. Initial condition is you are in 1 of 2 boxes which contain gold balls.

3rd box is irrelevant to initial condition.

50% chance.

IF the initial condition were not given as gold ball, but as silver ball, odds would be more complex.

zero because of the way you framed the question

i dont understand how people are so idiotic,

if you ALREADY have a gold ball how much chance is it to get a second one, obviously you have the first or the second box so 50%

Who mentioned a third box? You did. Nobody's considering the third box except you.

>So I've taken a box, in that box is a gold ball. Knowing this we can discard the silver box entirely.

You can also "half-discard" the mixed box for the same reason

thats why the probability is biased towards it being the gold+gold box and the answer is 2/3

This is intelligible.

So there is a possibility that the second ball drawn will be silver. However, it is more likely that it will be a gold ball, since it is more likely that the first gold ball came from the box with two gold balls inside.

Wrong.

I am the OP and I am right. Read the wiki en.wikipedia.org/wiki/Bertrand's_box_paradox

en.wikipedia.org/wiki/Bertrand's_box_paradox

en.wikipedia.org/wiki/Bertrand's_box_paradox

en.wikipedia.org/wiki/Bertrand's_box_paradox

en.wikipedia.org/wiki/Bertrand's_box_paradox

Nope. It was TRYING to be, but was set up wrong in the problem statement. Monty hall problem relies on single item and all 3 doors in play. This is simple 1 in 2.

If you RESTATE problem as PULL OUT SILVER BALL, then what is probability of 2nd ball being GOLD, this would not be 50%.

Okay, you have 2 boxes, GG and GS, you draw a gold from one, whats the probability you draw another gold from the same box?

Except there's a 2 in 3 chance the gold ball you just drew came from the first box. So the first box counts twice, the second box counts once. (If you're counting boxes, which is a very misleading approaching.)

Correct.

I think people didn't read the problem correctly. The solution is to ignore the box with two silvers and focus on golds. There's a 1/4 Chance of a silver in the beginning.

Removing one gold brings that chance to 1/3 for silver and 2/3 for gold.

You are all fucking retarded if you can't read.

correct
retarded
incorrect
incorrect
incorrect
correct
retarded
incorrect
retarded
correct
incorrect
incorrect
retarded

Yes, but my statement was exclusively referring to the 2nd box, and the fact that you can't draw two gold balls out of a box containing silver and gold. I never referenced the probability that the second one will be a gold ball. Incorrect assumption.

>autistic screeching

en.wikipedia.org/wiki/Bertrand's_box_paradox

33 percent

There is a higher chance you got the gold ball from the first box

Still 2 out of 3. Here:

G1 G2 | G3 S1

If you just drew a gold, you have an equal chance of drawing G1, G2, or G3.

If you drew G1, the next ball is G2. (Win.)

If you drew G2, the next ball is G1. (Win.)

If you drew G3, the next ball is S. (Lose.)

2 out of 3 chance your second ball is gold.

you don't understand basic probability. it's 1/2. you are fucking retarded.

Why is the box with 2 gold balls 1/3. That's not correct. That's not a correct way to solve the problem.

Wrong. It can be done repeatedly. If you repeat the experiment 100 times, how many times will you get 2 gold balls? That's a probability.

Please solve the problem or go read a book. It's the least you can do.

Wrong.

Lol, are you retarded?

Well then be puzzled upon you're own stupidity.

You can't discard your problem space.

it must feel good to be as retarded and arrogant as you

ITS REALLY EASY ACTUALLY:

IF YOU HAVE THE GOLD, YOU'VE BEEN 1.
THEN YOU NEED ANOTHER GOLD!!!!!
HELLO?! ARE YOU UNDERSTANDING ME?!

YOU NEED 2 GOLDS AND YOU HAVE 3 SILVER IN ALL!!!

SO THE ANSWER IS 2/3

>Why is the box with 2 gold balls 1/3. That's not correct. That's not a correct way to solve the problem.
What the hell are you talking about? Nobody said anything like that.

just give up, you're retarded and don't understand basic maths. maybe call your local preschool and ask to talk with the class retard, he might be able to teach you something.

well set out. I don't see how anyone can disagree with this logic, but I'd like to see the stubborn people in this thread try

The following reasoning appears to give a probability of 1/2:

Originally, all three boxes were equally likely to be chosen.
The chosen box cannot be box SS.
So it must be box GG or GS.
The two remaining possibilities are equally likely. So the probability that the box is GG, and the other coin is also gold, is 1/2.

The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the GG box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin. Specifically:

The probability that GG would produce a gold coin is 1.
The probability that SS would produce a gold coin is 0.
The probability that GS would produce a gold coin is 1/2.

From wikipedia.

you have two boxes left. one has a gold ball, the other has a silver ball. thats 50%. you are a fucking mongoloid if you say otherwise

2/3 you fucking idiots

see the monty hall problem

These anons can't read.

You can't be this retarded. It's not the same as the Monty Hall problem. But, it is still a valid problem.

No, that is not correct. You have the right answer, but your intuition about the problem is wrong.

en.wikipedia.org/wiki/Bertrand's_box_paradox

You're retarded

Sure.

Wrong.

Right.

Nope.

it is in fact you who does not understand basic probability

This

it's 1/2 you fucking idiot. see basic fucking logic