Prove you are white

Surely the high-iq white master race ubermensch alpha geniuses of Sup Forums can answer this simple brain teaser right?

if not, you have to go back

Other urls found in this thread:

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

50 percent fuck off

fuck off with your non-threads

The answer is: fried chicken

right box has no gold bals, so it can be ignored.

left box has 2 gold balls, mid one has 1, so 2/3 chance left box is selected. so, p =2/3

2/3?

Answer: I run off because I just got a ball made of gold fuck you I'm rich.

jew detected

>running away when there are at least a silver ball for the taking

50%
Chance of box 3 is 0% since it has no gold ball, so its 1/3 chance gets evenly distributed to the other two
1/3+1/6=3/6=1/2=50%

50%

All we have to do is determine the probability that you have the box with two gold balls in it.

Because your box has (at least) 1 gold ball in it, it's definitely not the box with two silver balls. There are now only two possible choices, so the probability is 1/2.

>Canadian intellectuals

>non-country that wishes it was a continent spotted.

probability is 2/3. This is a classic logic question posed by Bertrand Russell. Look it up.

TLDR:^)

>Posting the low hanging fruit of bait that is math threads

You'll get no satifaction from this op.

Fuck off explain the logical flaw then you stupid kiwi
I'm willing to admit I'm wrong if you present an argument

fuck you nigger

Sup Forums tier thread
Reported faggot

There are 2 gold balls in the left box and 1 in the middle box. If you get a gold ball there's a 2/3 chance it's from the left box.

Well I don't think they would let me keep the golden ball, so instead I'd just put it back and pick it up again for a 100% chance.

Yeah, after reading online, I realize my mistake
Say your argument the first time though

There are three gold balls to pick. For ball 1 in the first box, the other ball, ball 2, is gold. If you pick ball 2, then the other ball, ball 1 is gold. If you pick ball 3, then you get a silver ball. So, 2 gold balls lead to another gold ball while one leads to a silver ball, hence 2/3.

50% it either will be gold or it won't be. There are only two different balls it could be.

3/6 = gold
3/6 = silver

you take one gold out

2/5 gold
3/5 silver

40% chance it will be gold assuming you dont keep the other gold ball in

Not even close.
Its 50%

/thread

There are three gold balls at first, but you're not randomly picking a gold ball, you're randomly picking a box and then randomly picking a gold ball.

I read Codreanu's book and liked it.

you already picked a box and took a gold ball from it and it had a gol ball, therefore there are only 2 boxes you could take the next ball from, in one box the other ball is gold and the other it's silver, aka it's 1/2 of gold 1/2 of silver for the next ball.

basically, its a paradox of a question because they tell you the rules, tell you whats in the boxes, and then they say you have to forget whats in the boxes "Note: you cant see into any of the boxes" and just guess a random percentage so that anything you say is probably going to be wrong. the question would make sense without the red text at the bottom. that was added on by some autist, probably OP

2/3

You got a gold ball. That makes it more likely that you selected the box with two gold balls. The probability is higher the box has two gold balls.

Write out the ways you can pick a gold ball from a box:
Box 1: 2 ways
Box 2: 1 way
Box 3: 0 ways

Hence 2/3 you got from box 1.

1/3

No nigger it's 2/3 because there's 1 silver ball and 3 gold ones and you just took out the golden one
>but muh 50%
This is not how it works, the question is "what's the probability that the next ball will be gold?" not "what's the probability that you picked a box with 2 gold balls initially?"
The situation changes once you take the ball out you fucking favela nigger

OP here,
Good job to all the 50%'ers, it was indeed 50%.

I'd argue those two questions are the same. They both come out as 2/3.

50% You are picking from the same box. If you pick up a gold ball then it has to be box 1 or box 2. No chance of box 3.

remove yourself from the gene pool

As others have already stated why 50% is the correct answer, I want to agree with them.

On the other hand, I'm also thinking that you have to account for the fact that it is more likely that if you have drawn a yellow ball, you have a higher probability to have drawn it from the first box (i.e. 2 yellows; 100% chance) than the second (1 yellow; 50% chance).

This slightly reminds me of the Monty Hall math problem: en.wikipedia.org/wiki/Monty_Hall_problem

So if I have a box with a million gold balls and a box with a million silver and 1 gold it's 50%?

...

If you understand the Monty Hall problem you'd know why it's 2/3
There's even an easy explanation with 1000000 doors

this IS the monty hall problem
instead of monty "fixing" the odds, here's the first ball doing just that

Thats a moody explanation

Ummm, sorry sweeties.
If you do the experiment at home, you'll get 1/2.

You've eliminated the probability that it is the box with two silver balls, hence it can only be one out of two boxes, leaving it to be 50/50.

congratulations OP, you have proved that half of Sup Forums does not understand math

no, you see if you've already removed one of the gold balls from the equation, then the amount of gold balls drops to 2, there can only be a 50% chance that the other ball in the box you just took a gold ball out of is gold

Without reading the responses I came up with 2/3 at first. I thought the 50-percenters in the replies made a convincing argument though. I am a weakling.

It's deeper than that.

There are two ways to select the first box. One way to select the second.

Ergo, 2/3 that you picked the first box.

Suppose I had 100 gold balls in the first box. And I had 99 silver balls and 1 gold ball in the second. It is surely not 50%, my friend. If I do the experiment again and again, throwing out times when I get a silver ball on the first try, I am far more likely to have selected a gold ball from the box with 100 gold balls on the first try.

SAME BOX you fucking retarded Aussie.

Remove a gold ball from the left and middle boxes and you're left with 1 silver and 1 gold.

50 fucking percent

this post smells of bait, either that or the failure of the american educational system

It's 50%, people saying 2/3 are being retarded and not realizing that you can only be picking from one of the two boxes on the left and the only 2 scenarios after taking out one gold is there being either one gold ball left over or one silver ball left over, ie a 50/50 chance of either one being chosen.

I'm not good with numbers or probability, but I know that there is a 100% chance that OP is a homosexual.

that's not how probability works kiwi. each scenario plays out separately, as shown here. there's still three scenarios in which a gold ball is chosen and three different results, two of which result in the second ball being gold.

Good thing Weebmoot blocks social security numbers - x'es them out
Ryan Hill
xxx-xx-xxxx

I miss these math slide threads

You missed his point.

You will have a second gold ball if you selected the box with two gold balls. That is obviously true.

There are two ways to select the box with two gold balls and merely one way for the box in the middle. It's 2/3, my friend.

There's only two scenarios in which a gold ball is chosen, which gold ball you take from the box containing 2 of them is totally irrelevant.

must suck being this stupid

Not sure what you mean. Do you know about the Monty Hall problem and why it's a somewhat interesting scenario if you aren't into or majoring in math?

I don't think there's been a single non-math person I've told the Monty Hall problem to who has gotten it correct initially. I think this problem is pretty similar (i.e. it seems like it should be 50%, but I think it's actually 66%).

Please google the scenario if you are not familiar with it.

There are two ways to select the left box such that the first ball is gold. There is one such way for the middle box.

2/3

50%

20%

it's 1/3

the question is essentially "what are the odds that you picked the box with two gold balls in it at the beginning"

There are 6 balls, therefore 6 possible scenarios. Play out each scenario and you'll get it.
The problem states that the first ball is gold, making the 3 scenarios where you take silver first not part of the problem. Of the remaining 3 scenarios, 2 result in getting a gold ball second.

Question assumes you've picked a box with a gold ball in it.

2 boxes are viable in order for you to initially pull a gold ball from it. Only 1 has what is necessary to pull another gold ball. So 1/2.

the box is randomly selected, the fact that you pull out a gold ball is inevitable, the chance of you selecting one of the two boxes with gold balls in them in 50% and the chance of there being another gold ball in the box you selected is also 50%

This is correct.

This is also correct.

Its 50% based on real world logic but 66% when you apply statistical voodoo to combine boxes 1&2.

2/3?

Most retarded question of the century

wew.

if you take the 2xGold Box = 100%
If you take from the Gold/Silver Box = 50%

It's retarded because it asks for chance from "the same box" and ittnroduces weird conditions such as "You pick box at random but get a result that is not truly random" by saying you draw a gold ball

(there's a chance you WON'T get a gold ball if you pick a box at random ffs, so that makes the third box completely irrelevant to the question at hand)

Whoever made this image is fucking retarded.

My friend, please realize that the problem states you have gold on your first grab.

2 ways to get gold from the first box.
1 way to get gold from the second.
0 ways to get gold from the third

Hence 2/3 that the box you picked had two gold balls.

I've been saging this whole time. I'm tired.

Ignore the number of balls total

You have a single box

You take 1 gold from it

The only remaining options are taking 1 gold or 1 silver, there is a completely equal possibility that you have the box containing either gold or silver and there is only 1 of each. 1/1 = 50/50. If you say 2/3 you're focusing way too much on le ebin math jew and not the actual scenario

No, the question is "what are the odds that the other ball in the box you chose is gold."

Those are two different questions. It's pretty much a variation of the Monty Hall paradox.

correct

All you guys who went with 2/3 are absolutely correct. Though I weep for the lack of critical thinking skills - not because people went with their intuitive response of 50%, but because not knowing they didn't even bother to see if they could find the answer somewhere else.

It's like one of the first rules of critical thinking: don't pretend to know something/admit when you don't know.

This is a particularly important thing to be able to admit to yourself.

For the answer: en.wikipedia.org/wiki/Bertrand's_box_paradox

>people are still replying when the solution has already been posted here

alternate solution is also here

...

>What is the probability that the next ball you take from **THE SAME BOX** will also be gold?

Answer: depends on the box.

Either 100% or 50%. It cannot be 66.6%, that wouldn't make any sense.

No, you're retarded, this is a really common problem used in intro statistics and probability to teach students about how probability works.

2/3

i see the error of my ways thank you for your correction

There is absolutely no fucking voodoo. There is no combination.

Just the count the ways you can get a gold ball. Write out the possibilities in if you absolutely need to.

All it means is that you discard the events in which you got a silver ball on the first try. This is high-school-probability-tier, my friend

Answer: we wuz kanggs

Let's hold a vote to see whether the majority of people on this board are autistic
www.strawpoll.me/12624051

Finally, a smart Swede.

However, I must admit that as we don't know which box is the double silvered one we can't factor that in, fuck. Ignore my shit.

>en.wikipedia.org/wiki/Bertrand's_box_paradox

That question and Bertrand's box paradox differ A LOT.

>The 'paradox' is in the probability, after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, of the next coin also being a gold coin.

vs.

>What is the probability that the next ball you take from **THE SAME BOX** will also be gold?

42

also mods

Going to rephrase If the question asked, "What is the probability that after picking a gold ball, the next will be gold?" then the answer would be 1/3. This is basically asking if you picked the box with the 2 golds.

However, the question states that we've already started in one of the boxes with a gold. It just asks what are the chances that the next is gold. Since there are two boxes to choose from, the answer is 50%.

There's a miscommunication here I think, I believe the picture in OP is written by a disabled retard and thus is confusing. As it is written in OP it is, 100%, a 50/50 chance, as you are only, solely accounting for the probability of you having one of two boxes with a completely equal chance of getting either. The actual scenario is probably written differently, and is where all these 2/3 answers are coming from, from people who didn't actually read OP itself.

Actually, that box doesn't matter, ignore this post rather. I'm too tired for this shit.

Yes, that is true once you have already pulled the first yellow ball, but what you're not factoring in is that it was more likely that you pulled from box 1 to begin with because there are more yellow balls there to choose from.

I think the mistake most people are making here is that since you pulled a yellow ball to begin with, box 1 and box 2 must have an 'equal' chance to have their 2nd ball pulled (i.e. yellow for b1 and grey for b2). That is not the case... Again, read the Monty Hall problem for more reasoning on this.

No, I'd say the error is the second to third line. There are two ways to get the box on the left. For example if you drew (gold nothing) (nothing gold) (nothing silver) that would more accurately represent the choices available.

>we've already started in one of the boxes with a gold.
great
so you have 3 gold 1 silver
1 gold gets taken out
what are the odds?

They say exactly the same thing. Judging by your flag I'm going to just assume English isn't your first language and give you a pass on not being able to comprehend the grammar.

I assure you, if you actually do this problem a bunch of times, you will get 2/3.
You cannot ignore the total number of balls. That's not how the world works.
You are picking a random box AND a random ball from that box. There are 6 possibilities and 3 of them (silver first) do not match the description of the problem.

Stupid nigger, go try it right now, do 100 iterations and record the results