Prove yourself, surely the geniuses of Sup Forums could solve this simple puzzle?

Prove yourself, surely the geniuses of Sup Forums could solve this simple puzzle?

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en.wikipedia.org/wiki/Bertrand's_box_paradox#Solution
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2/3

50%

33% ya?
only 1/3 of the boxes have 2 gold balls so thats the only box that could lead to the double gold pull

1/3

50%

1/3 fags ez

50%

I'm confused. Do the balls weigh the same?

yes

odds are 3/6 (50%) that your 1st ball is gold.
If it is, the odds of the next one being gold is
100% left box
50% middle box
you know you didn't pick the right box.

So, odds of the next being gold are
(100%)(50%)=50%

...

Ahhh, I am the fool. Thought it was 50%, did the calculation wrong. It is, in fact, 1/3.

in that case my answer is 2

50% I graduated #1 in my advanced particle topological physics theory class at princeton and I can see that is clearly the answer.

50%

One ball remaining, two possible results: Gold or silver.

The silver/silver box is a red herring since the one gold ball is already given.

2 is indeed the correct answer.

Congratulations, you are Sup Forums's very own intellectual.

you didn't have to figure out the first part.

label the balls left to right 1-6.

Scenario 1: Pick up gold ball 1. You will then pick up gold ball 2.
Scenario 2: Pick up gold ball 2. You will then pick up gold ball 1.
Scenario 3: Pick up gold ball 3. You will then pick up silver ball 4.

Out of the 3 scenarios in which you can pick up a gold ball first, 2 give you another gold ball.

Answer: 2/3

No it says you pick a gold ball so the last silver silver box is eliminated. This leaves 2 boxes gold gold and gold silver. So the probability of the next ball being gold is 50%.

this is a joke right?

It would be 50% right? By grabbing a gold ball you eliminate the possibility of the box having two silver balls. Meaning that you would have a 50/50 chance of either picking up a gold or silver ball.

50%

-_- that is unnecessary lies. also two of those words have nothing to do with each other?

That was a bonus.

If I return the gold ball to the box the chance is 75%
Otherwise it's 66%

nope. scenario 1 and 2 are the same as far as the odds are concerned what matters is the box. not the specific ball in the box

yes

Pretty sure you have zero percent chance of finding a second gold ball in the middle box.

Came to the same final percentage of 50%, however. Because...obviously the box on the right isn't in play since you couldn't have found a gold in there. So, there are only 2 boxes that matter at that point. 50/50 shot at getting a gold ball between the left and middle boxes.

which two? topological and particle physics? It's all based on group theory and topology nigga.

THIS is a classic b8 thread. The question is worded so vaguely you have to guess one way or another on the point of the question. No way to tell if the question is asking the probability of getting the gold ball BEFORE or AFTER already picking the gold ball.

If the probability being asked starts before any balls have been chosen, it's 1:3.

If the probability being asked starts with the assumption the first ball is going to be gold, it's 1:2.

But this is Sup Forums so you guys will argue over interpretations.

what you just said makes no sense

Plus the topology implicit inside the theory of Lie groups, nigga

Nah. You have two boxes indeed, but you have three gold balls, let's call them 1, 2 and 3. 1 and 2 are in box A and 3 is in box B with a silver ball. If you extract 1 or 2, then the other ball from the same box is 2 or 1 respectively, which are gold. Only if you pick ball 3 you can then pick the sliver ball. So, your chances of picking a gold ball which has another gold ball as a roommate is 2/3. Faggot you are.

Not sure if next level baiting or retard, but neither of your answers is actually correct. And it's not really ambiguous (although it does sound like a similar problem).

No

No, they're not. You're twice as likely to pick up a gold ball which gives you another gold ball as you are likely to pick up a gold ball that doesn't give you another gold ball. 2/3 is correct.

lrn2math.

>reading comprehension
It literally says what's the probability of the "next" ball.

The probability isn't 1/3 or 1/2. It is 2/3 and you're an astonishing moron.

But the first two scenarios play out the same way. Since scenario one and two give an equal result, you would have a 50% chance on choosing a golden ball

This

right bro

>Same box
It's 50% faggot

are there people who actually think that the first interpretation is what's being asked? it's clearly not. I think it's just that people are complete shit at probabilities since it's not taught in school to any degree

You're correct. It assumes you've already selected the first gold. So you have a box with 1 gold 1 silver. Your probability of it being box A or box B is 50/50. 50% chance.

>only 1/3 of the boxes have 2 gold balls so thats the only box that could lead to the double gold pull
By that logic, there's a 33% chance of picking a gold ball from a box with 2 silver balls.

>the calculation
lol

see? like I said it's not about differing interpretations it's just a lack of understanding of how probability works

The real brain teaser is how many people actually think the answer is something other than 2/3. I know I troll the 50% every time I see this thread.

actually one of his answers is correct...

With Bayes theorem:
P(second gold | first gold) = P(second gold ^ first gold) / P(first gold)
P (second gold ^ first gold) = P(pick left box) = 1/3
P(first gold) = P(pick left box) * 1 + P(pick middle box) * 1/2 = 1/3+1/3 * 1/2 = 1/2
P(second gold | first gold) = (1/3)/(1/2) = 2/3

For Sup Forumstards who don't know math:
You picked the first ball, which is gold. There are 3 possible balls you picked, all equally likely. Of those, only 2 (the ones in the left box) will result in a second gold getting drawn.

Answer is 2/3

If you have special needs, feel free to google this problem and get it explained to you step by step, including stats background and physically relevant examples that validate the correct answer.

All bait. Exactly the shit I was talking about. Notice none of them give any explanation. Worthless posts

No retard. It says you already picked a gold ball. So the next ball can only be a gold or silver = 50%. How don't you understand it's simple.

the question isn't actually about the gold ball. it's about the box. what are the chances that you've picked the box that has two gold balls based on the knowledge that you've already got one gold ball. answer? 50%

1/2

so you do admit to trolling

this is next level retarded

I have explained this here but you are a faggot.

There's no point in talking to you f you can't even read my explanation.

I wonder if you have googled it to make sure that you are correct. although I lack faith in your ability to google this problem correctly

some people said this already

More worthless gibber gabber. Please delete ur account

good theorem applied incorrectly

sure. because you just say so right?

Of course I've googled it to make sure I am correct. It's easy to miss things with these silly problems. I've read published papers where authors miss little combinatorics factors. So yes, I always check.

explain your reasoning or gtfo

Don't spoil the fun :P

I can see why you think that would be right but it's not even though I'm not confident I could explain it so everyone could understand. I'll just say that only works if you ignore the boxes existence and pay attention only to the order they are laid out in.

lol @ the amount of people that think it's 50%

50%
>pull out gold ball
>same box
>either left box or middle box
>100%not right box
>so in same box next ball is either silver or gold
>50/50

cool

this isn't 'classic bait'

this is bertrand's box paradox. it's a classic riddle

oh. that's interesting. I blame you guys for not explaining it better. lol. sorry

Nope.

I wasn't thinking about the fact that the odds of it being the gold gold box was greater than the odds of it being the gold silver box because of the fact that we already have a gold ball. my bad.

This post explains it the best.

The odds of first picking a gold ball and then picking a second gold ball from the two boxes which contain a gold ball are 2/3. However, the question asks only the odds of picking a second gold ball having already selected the first. The odds of this are 1/2, since we now have 2 possible outcomes which are equally likely: either the second selection is the one possible silver or it is the one pissible gold.

...

you were right and then you decided to over simplify. getting rid of relevant information.

mfw i see answers that aren't 2/3

50:50

66%

again. probability isn't taught in school. not to mention that fact that it is labelled as a "paradox" you over melodramatic fuckwit

not to mention the fact that we're on Sup Forums

I already know I'm in box 1 or 3. So it's a 50% chance. Box 2 is irrelevant once I have the information from the first box.

Is the point that I also don't know the setup? If that's the case, the problem changes.

that too. I somehow keep forgetting that. I should leave and go somewhere else in this unlimited ocean of knowledge

>Is the point that I also don't know the setup?
The setup is a given, so you know.

The answer is 50% and everyone else is trolling

Well that's annoying. Why would you troll logic?

why am i still in this thread debating elementary probability with a bunch of teenagers?

Why does anyone do anything? The lulz of course!

Because you don't understand elementary probability?

Oh, that sounds pointless.

how many YEARS are we going to troll eachother with this gay ass question

i've got a degree in mathematics, and the answer is 2/3
i even simulated it here this is a well known and documented problem
see here for solution
en.wikipedia.org/wiki/Bertrand's_box_paradox#Solution

2/5, duh

>inb4 your program assumes the answer
>inb4 wikipedia can't be trusted
>inb4 muh gut feeling overrides conditional probability theory

This is the answer. Everyone saying 2/3 has had their brain tricked by this puzzle, which was the intention.

I'm sorry your degree in mathematics didn't teach you anything. You have to account for the information you have. Your simulation only accounts for the chance of pulling two gold balls, not the chance of pulling a gold ball after pulling a gold ball.

Holy shit guys here: there are 2 correct answers depending on instructions, and whether or not the first ball is replaced.

The bade assumption is not to replace the ball, so you don't. The only answer is 50%. The question is not vague.