Is this really an impossible puzzle?

[Anonymous No.16022741]

[Anonymous No.16022751]

It's a 50% chance. This is because it is either the two gold balls box or the mixed box.

[Anonymous No.16022760]

>>16022751
But what if we take it from the center box first?

[Anonymous No.16022767]

50%
It's either gold or it isn't

[Anonymous No.16022769]

>>16022767
But what if you pick both gold first ?

[Anonymous No.16022797]

g1 --> g2
g2 --> g1
g3 -->s

2/3

[Anonymous No.16022831]

>>16022797
Wrong, despite putting most effort into answer

[Anonymous No.16022835]

>>16022831
It's the right answer, though the implied method doesn't scale well to different permutations of the problem.

[Anonymous No.16022852]

>>16022741
P(you picked the box with 2 gold | you drew a gold ball) = P(you drew a gold ball | you picked the box with 2 gold)*P(the probability you picked the box with 2 gold)/P(You drew a gold ball)
=1*(1/3)/(3/6) = 2/3

[Anonymous No.16022861]

>>16022835
Can't believe you are being challenged since you are the only one with a brain in this thread.

[Anonymous No.16022874]

Unironically, statistics is satanic.
Unironically.

[Anonymous No.16022910]

>>16022741
1/3
one box is a guaranteed gold pull

[Anonymous No.16023039]

>>16022741
I can't see any reasonable way it can be anything other than 1/2 chance the second ball pulled is gold.

[Anonymous No.16023067]

its a 0% chance
i'm not a fag so i aint touchin no balls

[Anonymous No.16023099]

>>16023039
Have you considered that you might be a brainlet?

[Anonymous No.16023131]

>>16022741
It's 20%. don't ask how I know. I don't have time to explain

[Anonymous No.16023169]

>>16023099
It's always something I consider. It's called imposter syndrome and you might know what it is if you had ever achieved anything

>>16023067
Kek.webp

[Anonymous No.16023178]

>>16022741
>A=g+g
>B=g+s
>C=s+s
>g=ball of gold
>s=ball of silver
The moment I have the golden ball in my hand, box C is automatically excluded. If I took it from box A the chances are 100%, if B 0%. Taking the average I get 50%. Convince me otherwise.

[Anonymous No.16023206]

>>16022741
The trick is it's given you take a golden ball. 2/3 of the time it's the box with 2 balls, so the next draw from that box will be gold. 1/3 of the time it's the mix box, so the next draw will be silver. 2/3 of the time you draw both golds.

2/3, you're a brainlet if you think otherwise.

[Anonymous No.16023215]

>>16022741
I'll put it simply, this question cannot be answered with any mathematical certainty, because if any individual human in the world, all the billions of them, you cannot calculate the chance of any ball being drawn to a guaranteed number. The question is moot because all will draw different at chance according to their unique decision, which cannot ever be calculated. It's a trick question, based on absolute chance of a humans will. You're right. It's impossible

[Anonymous No.16023224]

>>16023215
What I'm saying is you can factor in the chance of one human being tasked with picking another golden ball, at arbitrary chance. You cannot calculate all of the human race making the exact same calculable decision. It really is impossible to determine

[Anonymous No.16023226]

1/3 it's just the monty hall problem

[Anonymous No.16023229]

>>16023178
You're more likely to have taken it out of box A because it has more hold balls.
Imagine A had 100 gold balls, the rest being the same. Would it really be equally likely to pick one out of that as out of B (with 1 g)?

[Anonymous No.16023230]

>>16023226
wait I'm retarded

[Anonymous No.16023234]

No
the probability you pick a gold ball given you just picked a gold ball is 2/3
This is because you have 3 chances to do it initially then 2 chances left to do it again

[Anonymous No.16023238]

We don't even need math for this one dumbasses. The odds are twice as high that you are in box A than box B so 2/3 chance of doublegold.
If you even considered 50-50 you're hopelessly retarded and should just become a mechanic or something because you'll never amount to anything except fucking everything up and forcing others to fix your endless mistakes.

[Anonymous No.16023243]

>>16023229
The number of balls in A doesn't matter. It's still 2/3.

[Anonymous No.16023246]

>>16023238
Maybe a detailer or those mexicans that put the sticker paint film over cars, I wouldn't trust someone that doesn't know what double odds are to fix my car.
2 rolls out of 3 possibilities, some people are just unredeemable retards, what can ya do?
Hey to all the people that said 1/2 tell me your field of study so I can know which field contains the biggest retards.
For me it's accounting, but I like doing math problems for fun.

[Anonymous No.16023252]

>>16023246
Stop being a fool....
That's all the advice I'm giving you

[Anonymous No.16023258]

>>16023252
Getting strong Pajeet vibes from this one
t. other

[Anonymous No.16023272]

>>16023258
That's your rhetoric? OK, I wish the best for you. The only thing holding you back is you. I'm a white Christian Australian if it matters

[Anonymous No.16023316]

>>16023178
What about the balls of steel? Did you forget those in your analysis?

[Anonymous No.16023317]

>>16022741
Do you put the first gold ball back?

[Anonymous No.16023332]

>>16023169
>It's called imposter syndrome
Only if you're actually smart. Otherwise it's just self-knowledge.

[Anonymous No.16023334]

>>16023243
>>16023229
If B had a 100 silver balls it'd be 100/101 though

[Anonymous No.16023335]

>>16023178

The moment I take a sphere, I have these possibilities:

A gold1 gold2
B gold1 gold1
C silver1 silver2
D silver1 silver1
E gold3 silver3
F silver3 gold3

Assuming that the first sphere is the one I extract, if it is silver I eliminate C, D, F.
Therefore the percentage of having another gold ball is 2/3.

[Anonymous No.16023336]

>>16023335

A gold1 gold2
B gold1 gold1
-C silver1 silver2 Deleted
-D silver2 silver1 Deleted
E gold3 silver3
-F silver3 gold3 Deleted

[Anonymous No.16023337]

>>16023335
You can leave the thesaurus at home, here you impress with maths.

[Anonymous No.16023339]

>>16023317
paging dr schrodinger
dr schrodinger, you have a clean up in aisle five

[Anonymous No.16023343]

>>16023337
In my language (italian) the golden ball is the one in the pic

[Anonymous No.16023352]

>>16023337
>t. pederast shirtlifter

[Anonymous No.16023394]

>>16023352
Afraid I'll lift your shirt, pederast?

[Anonymous No.16023437]

>>16023335
To check, I prepared 3 small cards with G written on one end and S on the other, so that I couldn't see the part covered by the thumb. When G came out, if the other end was G I put 1, if S I put 0. The percentage of Gs is 68%, similar to 2/3

[Anonymous No.16023453]

>>16023437
To be precise GG GS SS

[Anonymous No.16023465]

>>16022741
It's a matter of perspective.

[Anonymous No.16023506]

>>16022741
If the gold balls are distinguishable - 2/3
If the gold balls are indistinguishable - 1/2

[Anonymous No.16023514]

>>16023506
Walk me through the latter

[Anonymous No.16023554]

>>16023506
nonsense, they are distinguished by their count

[Anonymous No.16023567]

>>16023506
Explain, friendo.

[Anonymous No.16023568]

>>16022741
Only way to know is to run a hundred simulations.
To prove it, post on youtube.

[Anonymous No.16023632]

>>16023568
Take 3 identical coins, on one write G on both sides, on another write S on both sides, finally write G on one side and S on the other. Put in a bag and take out a coin at random. You throw it in the air. If S comes out, put it back in the bag, if G comes out, look at what is written on the other side and take note. Repeat until you have 100 results.

[Anonymous No.16023678]

>>16023568
>Only way to know is to run a hundred simulations.
You can just use your brain? Imagine what's going to happen if you do this a hundred times.

[Anonymous No.16024059]

>>16023678
WOWOWOWOWOWOW that's incredibly rude, even for 4chan

[Anonymous No.16024061]

>>16023632
Can you explain further? You didn't really define what you want.

[Anonymous No.16024063]

>>16023554
Their count?

[Anonymous No.16024087]

>>16024063
every ball has a count, obviously when you take one of the gold balls, the respective count is going to see it and identify it as their ball and possibly get upset depending on the disposition of the particular count that owns the golden ball

[Anonymous No.16024090]

>>16024087
Fuck you, made me rea the whole thing.

[Anonymous No.16024259]

>>16024061
When you choose a coin it's as if you choose a box, when you throw it and see the result it's as if you catch a ball, when you look at the back side it's as if you catch the second ball. In fact, a coin is G G, another is S S, another is G S. You ignore the toss when S appears on the front face, if G comes up, take note of the back face, writing for example 1 if G, 0 if S. With 100 results you have a percentage, I got 70%.

[Anonymous No.16024542]

>>16023169
I've acheived the ability to solve a very simple conditional probability problem

[Anonymous No.16024545]

>>16024259
it can be simplified further by using a die. since the choice of boxes is random, we could consider a box the values of the opposite faces: 1-6, 2-5, 3-4. We could consider the first box 1-6 with both balls of gold, the second box 2-5 gold-silver, the third box 3-4 silver-silver. So when the value 1 or 6 comes out I consider finding the gold ball, when the value 2 comes out I know the opposite face is 5 and I consider it silver. when 5 comes out, 3-4 I ignore them because it is silver. at this point there is no need to do tests because the values I am interested in are 1-6 for gold and 2 for silver and the proportions are already done.

[Anonymous No.16024768]

>>16022741
>>16024545

I think it's confusing for people because all situations are relevant, even those althat are not part of the core scenario.
So we have three boxes left, middle as right (as drawn) and two balls each, left or right (also as drawn) giving us two possibilities in total:
right box is drawn, right ball is drawn, which is silver, and not part of the scenario
right box is drawn, left ball is drawn, same
middle box is drawn right ball is drawn, and it doesn't count
-------------------------------------------------
middle box is drawn, left ball is drawn, silver is obtained
left box is drawn, right ball is drawn, gold is obtained
left box is drawn, left ball is drawn, gold is obtained

Those below the line count, and give us 2/3 for gold

[Anonymous No.16024772]

>>16022741
>>16024545

I think it's confusing for people because all situations are relevant, even those that are not part of the core scenario.
So we have three boxes left, middle and right (as drawn) and two balls each, left or right (also as drawn) giving us two possibilities in total:
right box is drawn, right ball is drawn, which is silver, and not part of the scenario
right box is drawn, left ball is drawn, same
middle box is drawn right ball is drawn, and it doesn't count
-------------------------------------------------
middle box is drawn, left ball is drawn, silver is obtained
left box is drawn, right ball is drawn, gold is obtained
left box is drawn, left ball is drawn, gold is obtained

Those below the line count, and give us 2/3 for gold

[Anonymous No.16024774]

>>16022741
>>16024545

I think it's confusing for people because all situations are relevant, even those that are not part of the core scenario.
So we have three boxes; left, middle and right (as drawn) and two balls each, left or right (also as drawn) giving us six possibilities in total:
right box is drawn, right ball is drawn, which is silver, and not part of the scenario
right box is drawn, left ball is drawn, same
middle box is drawn right ball is drawn, and it doesn't count
-------------------------------------------------
middle box is drawn, left ball is drawn, silver is obtained
left box is drawn, right ball is drawn, gold is obtained
left box is drawn, left ball is drawn, gold is obtained

Those below the line count, and give us 2/3 for gold

[Anonymous No.16024800]

>>16024774
Well, you can safely ignore the all-silver box, though. There could be 100 all-silver boxes and it wouldn't affect the odds. As long as each box has an equal chance of being picked.

[Anonymous No.16024872]

>>16022741
>>16024774
it's confusing because the wording makes it counterintuitive, so people assume it's an independent variable instead of a conditional one
I thought it was 50% until it was presented in probability terms
a related paradox is this
https://en.m.wikipedia.org/wiki/Monty_Hall_problem
if Paul fucking Erdos couldn't figure out the correct answer then noone should beat themselves up if they don't get it immediately

[Anonymous No.16024890]

>>16022741
Since you know you're in either of the first two boxes, you can think that you've removed one gold ball from the superposition of the boxes and have 2:1 chance for another gold, so 2/3

[Anonymous No.16024925]

>>16024872
Thanks for the link, very interesting. We are on /sci/ to learn, and when we see connections with other situations we can say we have assimilated the question. Then we can find it in some strange contract that at first glance seems favorable to us.

[Anonymous No.16025065]

>>16023099
This desu Desu

[Anonymous No.16025067]

>>16024925
Gayest post of this thread

[Anonymous No.16025175]

>>16025067
I will try to answer, not for your benefit, but for others who will have the patience to read me.
First, homophobia can be an indicator of hidden homosexuality. I would love to see your internet history, it would explain so much.
Second, in my country it is an insult and can have unpredictable consequences, think twice before using it outside an anonymous chat room.
Third, what is your financial education? Do you have any idea how a financial product you are offered works? Can you recognize patterns?
I've already contributed enough here, I don't want to waste any more time, try to grow, goodbye.

[Anonymous No.16025203]

>>16025175
if an autist anon's random one-liner in an anon board triggers you so much, you need to work on your angst and self esteem for your own good

[Anonymous No.16025307]

>>16022741
[math]2/3[/math]

https://en.wikipedia.org/wiki/Bertrand's_box_paradox

Thank you OP for making sure the problem is always present on /sci/

[Anonymous No.16025500]

>>16025175
>When your autism has you misinterpret context
I may be gay, but you're a fag

[Anonymous No.16025856]

G

[Anonymous No.16025873]

It has to be a higher percent chance that its gold because if you picked a gold ball then there is a higher percent chance that it was from the box with two gold balls

[Anonymous No.16025877]

>the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?
at last I see, its 2/3. humanities major by the way

[Anonymous No.16026194]

>>16025873
Too complicated, can you gestalt me?

[Anonymous No.16026195]

>>16025877
>When humanities major mogs sci tards

[Anonymous No.16026230]

>>16022852
This is the first time ive seen the correct answer with the correct reasoning posted.

[Anonymous No.16026263]

>>16026230
Despite our thinking, so much is still available to be known and achieved!

[Anonymous No.16026847]

>>16023206
>2/3 of the time it's the box with 2 balls
What does that mean? All three boxes contain two balls, not just two.

[Anonymous No.16026946]

>>16026847
Exactly, answer isn't 2/3, it's 4/6.
Brainlets think they can reduce the fraction, thereby eliminating 2 possibilities, baka

[Anonymous No.16026958]

>>16022751
correct answer, this puzzle filters ESLs.

[Anonymous No.16027490]

>>16026946
But... it really is 2/3. There are three scenarios where you draw a gold ball first and two of those scenarios lead to another gold ball.
>>16026958
It's not even a language problem, you are just stupid.

[Anonymous No.16027540]

According to the question, the state at which the probability is being calculated is where the third box is discarded altogether due to the first outcome. The probability you now calculate is from the set including only the first two boxes.

[Anonymous No.16027591]

>>16027490
>But... it really is 2/3. There are three scenarios where you draw a gold ball first and two of those scenarios lead to another gold ball.

It's 4/6ths, brainlet, fractions can't always be reduced

[Anonymous No.16027763]

>>16022741
Looking at the majority of posts in this thread, /sci/ is vastly overrated. Absolutely avoid.

[Anonymous No.16027799]

>>16027763
Maybe, but only a bit.
B is not that dumb
Pol is mostly shills, doubt they even belong on the graph

[Anonymous No.16027916]

>>16027540
Of which one is twice as likely as the other
>>16027591
Literally three possibilities tho

prove me wrong

[Anonymous No.16027937]

>>16027799
The level of /pol/ when they discussed the pic question was not at all lower.

[Anonymous No.16027964]

>>16027591
>fractions can't always be reduced
Yes they can.

[Anonymous No.16027966]

>>16022741
Why are you assuming I would put another ball from box, if I already picked the gold one?

[Anonymous No.16027968]

>>16027966
Hitler had one ball, you don't want to be like Hitler, do you?

[Anonymous No.16028099]

>>16027964
>>fractions can't always be reduced
>Yes they can.
Imagine being from there (plebbit

[Anonymous No.16028102]

>>16027966
>>16027968
Quality discourse

[Anonymous No.16028211]

>>16028099
At this point you have to reveal at least a little bit of your troll logic to keep the engagement going, just a tip

[Anonymous No.16028335]

>>16028211
>>16027964
Not when they disregard 2 of the scenarios, picking from the box sans gold. Reducing fractions is situational, friendo

[Anonymous No.16028558]

>>16028335
According to this logic, can 0.1 (1/10) be different from 0.10 (10/100) or 0.100 (100/1000)?

[Anonymous No.16028653]

>>16028558
Yes
Ever heard of significant figures?
Smh baka Desu

[Anonymous No.16028839]

>>16028335
So what does the 4 represent in your mathematics? How do you get four gold balls?

Insisting on the denominator being six because there are six balls is asinine and shows an inability to abstract. If something happens 50 times out of 100, it happens once every two times by definition.

[Anonymous No.16028847]

>>16022741
you can just simulate this

from random import shuffle

boxes = [
['G', 'G'],
['G', 'S'],
['S', 'S']
]

req_samples = 100_000
cur_samples = 0
successes = 0
while cur_samples < req_samples:
shuffle(boxes)
rand_box = boxes[0]
shuffle(rand_box)
rand_ball = rand_box[0]
if rand_ball == 'S': continue
successes += rand_box[1] == 'G'
cur_samples += 1
print(f'{successes/req_samples:.2%}')

[Anonymous No.16028875]

>>16027964
are there bots arguing about very basic shit to make people more despondent? that's some pretty good hell tech

[Anonymous No.16028919]

>>16028875
Hasn't 4chan always been people arguing about very basic shit? Why should it have to be bots?

[Anonymous No.16028962]

>>16028919
that's always gonna happen, but in the past the arguments would have resolution every now and then, which seems much less common now
it would be a pretty good, low-cost way to distract people & spread nihilism. "if people are this stupid/evil, why should i care about anything?".
it's a self-fulfilling prophecy, just like everything else. if you can just jump-start the reaction by making a small portion of real people into antisocial nihilists who think they want to make things worse, other people will quickly stop giving a fuck as well in the face of that brutality

[Anonymous No.16028975]

>>16028962
>in the past the arguments would have resolution every now and then
Nah. You may have thought so, but then someone would just remake the same thread a month later and repeat all the same arguments that had been debunked in the previous thread and act like it was all new. You should see the portal A vs. B threads that have been going for almost a decade and a half.
Also, if you really think about what you're implying, what's the point? I can get individual trolls being cunts, but a concerted effort to spread nihilism backed by advanced technology? When it very likely isn't even necessary to intervene in what 4chan has already been doing on its own for decades? What are you suggesting here?

[Anonymous No.16029006]

>>16028962
>>16028975
And you know what? It can be pretty disheartening, yeah. But I've come to realise that part of the problem is 4chan itself. Ephemeral and anonymous, nowhere is it as easy to erase facts and history and substitute your own reality, to put lies and truth on exactly equal footing and render them both equally meaningful. When people wonder why this place appeals so much to fascism, it's not the commitment to "free speech" (a flattering lie) but this.

4chan is the postmodern condition distilled.

[Anonymous No.16029019]

>>16028975
the technology isn't advanced, it's beautifully simple, but it relies on an advanced understanding of self-regulating systems (read about permaculture design principles)
the internet made people too smart fo
it works in the interest of those in power for the world to be filled with chaos and noise, it makes people dependent & reduces competition.
they have to fight an arms race with human nature, the tendency of the global population, when not interfered with, to increase quality of life.
4chan never used to be an efficient factory of suicidal, homicidal, genocidal walking dead goombas.
>>16029006
sure, the anonymity lends itself to power games, but no one PLAYED those games in the past because it was all just regular humans using the site.
the anonymity only leads to this hell when something is wrong with the people using it.

[Anonymous No.16029020]

>>16029019
>wrong with the people using it.
wrong with the interests of the people using it, specifically

[Anonymous No.16029031]

>>16029019
>the anonymity only leads to this hell when something is wrong with the people using it.
Chicken and the egg innit
I will admit that there is something profoundly wrong with the vast majority of 4chan users but I'm not sure what to conclude from that as to the causes.

[Anonymous No.16029475]

>>16028839
>So what does the 4 represent in your mathematics?
Nothing in this particular puzzle desu.
>How do you get four gold balls?
RTF OP
>Insisting on the denominator being six because there are six balls is asinine and shows an inability to abstract. If something happens 50 times out of 100, it happens once every two times by definition.
Ask me how I know you're from plebbit

[Anonymous No.16029476]

>>16028875
>are there bots arguing about very basic shit to make people more despondent? that's some pretty good hell tech
No sadly, it takes effort that I could use to help the homeless or smth

[Anonymous No.16029478]

>>16028962
Maybe it's a form of accelerationism.

[Anonymous No.16029490]

>>16028847
Sorry, I don't speak virgin

[Anonymous No.16029524]

>>16022852
this assumes all 3 boxes were equally likely to be picked. you cant pick a gold ball first (given) out of a box with two silver balls.
>>16022751
is correct
>>16022760
thats the 50% that fails
>>16022767
is, as usual, the correct answer.

[Anonymous No.16029533]

>>16029524
>this assumes all 3 boxes were equally likely to be picked. you cant pick a gold ball first (given) out of a box with two silver balls.

Is the second sentence supposed to be related to the first?

The boxes were all equally likely to be picked.
Then, we draw a gold ball. This updates our knowledge of what boxes we drew: NOW they're no longer equally likely. Updating the probability with the information we have (we drew a gold ball), we have a 2/3 chance of having the first box, 1/3 chance of having the second, and 0 chance of having the third.

[Anonymous No.16029557]

>>16029475
>Ask me how I know you're from plebbit
Probably because you're making shit up so you might as well go the whole hog
There are three gold balls you could have selected initially. Two of the three yield the desired outcome. 4/6 doesn't mean anything.

[Anonymous No.16029598]

>>16028847
It's a good idea, if there was an online simulator for this language, everyone could check it out.

[Anonymous No.16029605]

>>16022852
You define P(You drew a gold ball) as 3/6, presumably because that's the ratio of gold to total balls. Does that mean that if we add one gold ball at random, the odds of getting gold are now 4/7?

[Anonymous No.16029704]

>>16029557
The confidence of plebbitors is unworldly.

[Anonymous No.16029706]

>>16029605
I feel you're onto something

[Anonymous No.16029737]

>>16029704
You would know, I suppose, considering you display it. The unwarranted confidence that lets you proudly state a blatantly wrong answer and cry "plebbit!" when even slightly pressed to justify yourself. All you're convincing anyone of is what a complete dolt you are.

The answer is 2/3 as given by the person who first formulated the problem. That is what 4/6 reduces to, always. But it doesn't matter because the answer is never 4/6 to begin with. You yourself said the 4 means nothing. Where are you getting 4/6 from? Probably just working backwards from 2/3 to six balls. It's bullshit delivered with enough confidence to masquerade as genuine insight. But it seems you've exhausted your well somewhat and resort to repeating yourself.

[Anonymous No.16029802]

>>16029533
if i added 100 more boxes with only silver balls would it change things? its a given you have a golden ball on the first try. 50-50

[Anonymous No.16029838]

>>16029802
>if i added 100 more boxes with only silver balls would it change things?
No, it would still be 2/3

[Anonymous No.16029848]

>>16022797
>>16022835
>>16022861
>>16022852

>1/3
>you already picked a gold ball
>you are picking a ball from the same box
>how many boxes have at least one gold ball in them?

[Anonymous No.16029850]

>>16029802
>its a given you have a golden ball on the first try
Ah, the classic mistake. That's poor reading comprehension on your part. No, your first pick is explicitly random. It HAPPENS TO BE a gold ball.

[Anonymous No.16029854]

>>16029850
And now, only two gold balls remain.

[Anonymous No.16029899]

>>16029605
actually thinking about it more, I think 1/2 is still the correct ratio but I derived it wrong. It's not that there's 3/6 balls, it's that there's 1/3 chance of choosing the 100% gold box, 1/3 chance of choosing 50% gold, 1/3 of choosing 0% gold. So 1/3+1/6=1/2. If you added a gold ball at random it turns into:
1/3 chance of box 1 (always gold even if the new ball was put there)
1/3 chance of box 2 (2/3 chance of being 1/2 gold, 1/3 chance of being 2/3 gold if the new ball is there)
1/3 chance of box 3 (2/3 chance of 0 gold, 1/3 chance of 1/3 gold)
So the chance would be 1/3+1/3(2/3*1/2+1/3*2/3) + 1/3(1/3*1/3) = 5/9. Thanks for pointing that out.
>>16029802
If you repeat my math with 100 more silver boxes, you see I still get the same answer of 2/3. The problem you and >>16029848 are having is that you're ignoring information. It's true that you know you picked one of the boxes with gold balls in it, so you're choosing between box 1 and 2. But if you stop here and just say "only two boxes are possible, it's 50/50", you're ignoring the information you have that box 1 has more gold balls in it than box 2. Imagine that box 1 had 1,000,000 gold balls, box 2 had 1 gold ball and 999,999 silver balls, and box 3 had 1,000,000 silver balls. If you drew a gold ball, you'd still narrow it down to two boxes, but it's much more likely that you picked some random ball from the all-gold box than getting the one-in-a-million chance to get a gold ball from the mostly-silver box. The same idea applies to the original problem, just with less extreme numbers.

[Anonymous No.16029904]

>>16022741
>"how likely are you to choose the middle box"
1/3

[Anonymous No.16030177]

>>16022741
Well you have either 1/3 or a 2/3 chance of picking a gold ball depending on if you pulled the gold ball from the box with the one silver ball and one gold ball first.

You can't really give a definitive answer without knowing which box you pulled the gold ball from first.

It is either 1/3 (33.3%) or 2/3 (66.6%)

[Anonymous No.16030185]

>>16029598
https://www.onlinegdb.com/0DwS2aWiE

[Anonymous No.16030196]

>>16030185
Thank you very much, 66.64%

[Anonymous No.16030203]

>>16029854
Yes, and? Picture where they are, friend.

[Anonymous No.16030204]

>>16029899
>Thanks for pointing that out.
Well done, you avoided my trap expertly and showed a true understanding of the problem.
>Imagine that box 1 had 1,000,000 gold balls, box 2 had 1 gold ball and 999,999 silver balls, and box 3 had 1,000,000 silver balls.
Incidentally, it works out the same whether Box 1 has two balls or two million. It's already at 100% chance of gold balls. It's really the ratio of the mixed box that affects the odds.

[Anonymous No.16030219]

>>16030196
di nulla

[Anonymous No.16031444]

to the 50/50 crowd: you do increase the chances from 1/3 to 1/2 by choosing a gold first. that ball was one of 4 balls in 2 boxes, you have removed one ball, and 2/3 of the remaining that could be switched to are gold.

[Anonymous No.16031625]

>>16029737
Tldr friendo

[Anonymous No.16031733]

>>16031625
tl;dr you're full of shit and if you weren't you'd have demonstrated mathematically how you aren't, but you can't

[Anonymous No.16032043]

I'm willing to entertain the notion that 4/6 can't be reduced if someone can demonstrate how they would arrive at 4/6 in the first place.

[Anonymous No.16032052]

>>16022741
You can simply count the possibilities: you know you're in box 1 or 2, and you have taken a gold ball out, for which there were 2 possibilities in box1 and 1 in box2; so being in box1 is 2 out of 3.

[Anonymous No.16032074]

>>16022741
We already solved it, the first ball is always gold, so that only leaves 1 and 2 or a gold or silver ball, so 50/50, now stop making this fucking thread

[Anonymous No.16032081]

>>16032074
No, we only count the times when the first ball is gold. Important distinction.

[Anonymous No.16032083]

>>16032081
Yes, that is the framework in which the problem is presented, you crayon eating tard

[Anonymous No.16032093]

>>16032083
Ah, to be confidently called a "crayon eating tard" by someone who fails to grasp a simple statistics problem. That's what I come to 4chan for.

Hint: just because there are two possibilities that does not mean they are both equally likely.

[Anonymous No.16032102]

>>16032093
There is no problem, you have two boxes to choose from, you're guaranteed to pull out a golden ball from said box, then depending on which box you chose you will either have a gold or a silver ball, so the choice simply boils down to which of the two boxes you pick, ergo 50/50
Hint: try reading the image in the op

[Anonymous No.16032113]

>>16032102
>you're guaranteed to pull out a golden ball from said box
No, you're not. You're picking at random. You're guaranteed to HAVE pulled out a golden ball at the point where the problem is posed, because the random pick is now in the past. This is what we're looking at. The outcome of a random event, and what it tells you about the situation you are currently in.
Hint: how wilt thou say to thy brother, Let me pull out the mote out of thine eye; and, behold, a beam is in thine own eye?

[Anonymous No.16032116]

>>16032113
>You're guaranteed to HAVE pulled out a golden ball at the point where the problem is posed
That is, if you hadn't randomly pulled out a golden ball, we wouldn't be talking and you try again until you do.

[Anonymous No.16032126]

>>16032113
Here's your (You), bodhi, you can start namefagging again now

[Anonymous No.16032133]

>>16032126
I wasn't a schizophrenic nazi when I checked this morning.

[Anonymous No.16032154]

>>16032133
But you sure as hell are retarded, so basically the same thing

[Anonymous No.16032180]

>>16032154
Well, you're too retarded to read the OP or google the problem or work it out yourself and I... I'm still trying to get you to see reason, so perhaps I am retarded.

[Anonymous No.16032183]

>>16024872
The gold ball puzzle is completely unambiguous, unlike Monty Hall, the actions specified will result in a 2/3 chance, and anything otherwise is just wrong.

[Anonymous No.16032203]

>>16032183
Monty Hall is also unambiguous. The boy/girl paradox is the one that has different answers depending on how the information was obtained.

[Anonymous No.16032238]

>>16032180
I already did, you pick a box, pull a gold ball out of it and the odds that the other ball in your box will be gold is 50/50, because you only have two boxes to choose from and removing a gold ball from either box will leave you with a choice between a gold ball or a silver ball, ergo 50/50

[Anonymous No.16032260]

>>16032238
No, Anon, here is how it works:
One box will give you a gold ball 100% of the time
One box will give you a gold ball 50% of the time
Therefore, when you get a gold ball, it is twice as likely that it came from the first box
And if the ball came from the first box, your second ball will also be gold

Like I said, just because there are two boxes it could have come from that doesn't make them both equally likely.

[Anonymous No.16032341]

>>16032260
That's not what the question asked you illiterate troglodyte

[Anonymous No.16032349]

>>16032341
It is precisely what the question asked and your repeated insistence that it is I who is illiterate between is is becoming rather too ironic. One box will give you another gold ball. One box will not. Therefore, if you the likelihood of which box you picked, you know the likelihood of getting another gold ball. That is the answer to the question.

[Anonymous No.16032355]

>>16032349
The probability is still 50%, it will either be a gold ball or it won't, you have a 50% chance of picking either box

[Anonymous No.16032361]

>>16032355
Yes, and a coin will land either heads or tails, even if it's weighted. So a weighted coin and a fair coin therefore have the exact same odds. Right?

You actually never had a 50% chance of anything, though. You had a 1/3 chance of picking any of the three boxes, and after you take out gold, there is a 2/3 chance that you selected the all-gold box.

[Anonymous No.16032382]

>>16032361
>So a weighted coin and a fair coin therefore have the exact same odds. Right?
No, because weight, and this applies to a regular coin as well, is a physical influence, but in the case of a coin flip you would also have to take into consideration the strength at which you flip the coin, the air resistance, how high do you flip it, how fast of a spin do you give it, none of this applies to you picking a box, there is no physical law or mechanism which forces you to choose one box over the other
>You had a 1/3 chance of picking any of the three boxes
But there are only two boxes, the question states you pull out a gold ball, that only leaves you with two choices, the only outcomes are that the second ball is gold or it is not, ergo a 50/50 probability

[Anonymous No.16032390]

>>16032382
>But there are only two boxes, the question states you pull out a gold ball, that only leaves you with two choices
Your choice has been made at this point. You are not making another choice. You are trying to determine the outcome of your first random pick between three outwardly identical boxes, based on the additional information you now have regarding their contents. You know one box never produces this outcome. So you dismiss that possibility. You know one box sometimes produces this outcome. So that is a possibility. And you know another box produces this outcome all the time. So that is the likelier possibility.

[Anonymous No.16032406]

>>16032390
But you do not know the contents of the box, you can not make an accurate prediction

[Anonymous No.16032413]

>>16032406
You can assign it a probability. That's what we're doing here. Do you think everything you don't know for sure has a 50% chance of happening because either it does or it doesn't?

[Anonymous No.16032419]

>>16032406
>>16032413
Also, since you don't know anything about the contents of the box: do you think there is a chance it could be the all-silver one?

[Anonymous No.16032469]

>>16032413
>Do you think everything you don't know for sure has a 50% chance of happening because either it does or it doesn't?
There are infinite parameters, so yes, if you could repeat the same thing over and over again you can just skew the parameters, you can't just compare the results without the actions, it's an inherently biased system, you can toss a coin in an infinite amount of ways an infinite amount of times, both are equally likely to occur, but if you toss a coin the exact same time 49 times and differently 1 time you would say that one occurs more than the other, same is with picking the box, you can not trace the causal influence which resulted in you choosing that specific box, since you don't know all the parameters, which means you have to guess, which means it's completely random, outside of your influence
>>16032419
If there isn't a question which forces you into two boxes, sure, but that's not what the question asked.

[Anonymous No.16032479]

I propose a pedantic solution, showing all the possibilities

[Anonymous No.16032480]

>>16032469
>you don't know all the parameters, which means you have to guess, which means it's completely random, outside of your influence
Right, I don't think you're cut out for probability. It's not fucking schrödinger's cat in the box. It's a simple question.
>If there isn't a question which forces you into two boxes, sure, but that's not what the question asked.
Well, you picked randomly from three boxes, and one of them was all silver. So, yes or no, could it be that box, do you think?

[Anonymous No.16032485]

>>16032480
Just saying one number is bigger than the other does not mean one is more likely to occur than the other without taking into account the influence you fucking retard, please tell me, what physical influence causes you to pick the box with two gold balls more times than any of the other ones given an infinite amount of time, I'll wait

[Anonymous No.16032495]

>>16032485
None, that's precisely the point. If you do this 6,000,000 times, then:
You pick box A ~2,000,000 times and get a gold ball
You pick box B ~2,000,000 times and get a gold ball 1,000,000 times (that is half the amount of times as box A, for those keeping track at home!)
You pick box C ~2,000,000 times and get no gold

It is precisely because there is no special influence that is causing you get gold every time that you can conclude that gold is more likely to come from box A. If, on the other hand, your hand was magically guided to a gold ball by forces beyond your understanding, then you might have a case for 50-50. Get it? You don't pick Box A more often. But you WILL HAVE PICKED box A more often if we're only looking at the times where you got gold on your random first pick.

[Anonymous No.16032534]

>>16032479
Sorry, wrong pic.
2/3 Gold

[Anonymous No.16032539]

>>16032495
>It is precisely because there is no special influence
There is, it's the question in OP
>If, on the other hand, your hand was magically guided to a gold ball by forces beyond your understanding
Yes, the fucking question!!!
>then you might have a case for 50-50. Get it?
That's what I've been telling you, it's a 50/50 that it's either a gold ball or a silver ball, since there are only two boxes you can choose and the gold ball is removed from either one automatically, that only leaves one other choice in your box, either a gold one or a silver one, there is no scenario in the question where you pick the box with the silver balls, the question doesn't ask anything about the silver balls besides the only one in the box next to the golden one

[Anonymous No.16032549]

>>16032495
see, this >>16032534
is biased, it doesn't matter which ball you pick from the all gold box, it's gonna be gold either way, it's the exact same event, you're throwing the coin the same way twice

[Anonymous No.16032554]

>>16032539
Right, and we're back to the reason it's so incredibly ironic for you to be calling anyone illiterate: you've blatantly misunderstood what is being asked here. The OP actually explicitly specifies that your first pick is random from among three boxes, and the result thus randomly produced happens to be, by mere chance, a gold ball. And that exact outcome is the result of picking the all-gold box two times out of three.

In case you're not sure how the numbers work out, that's not because there's three boxes. It's because you get ~3,000,000 gold balls and 2,000,000 are from Box A.
Also, regarding Box C: yes, we know we can disregard it because we picked gold. So you see how picking a gold ball actually does give you information about the contents of the box you picked? Even more than you realise.

>>16032549
>it doesn't matter which ball you pick from the all gold box, it's gonna be gold either way
Yes, compared to the other box, which is not gold either way. So there was another way it could have gone, which would not have given you gold. Which had an equal chance of happening. So you see, "gold either way" is actually twice as much as "gold half the time".

[Anonymous No.16032563]

>>16032554
>So you see, "gold either way" is actually twice as much as "gold half the time".
But you don't know the contents of the box, so you can't intentionally choose "gold either way" more than "gold half the time" unless you already knew which box it was, you're equally as likely to choose any box in any conceivable scenario should you do it randomly, thus every box would have an equal amount of chance to be picked, but since the question forces us into a scenario where the choice is already made for us, that leaves us with only two options, either we have the box with the gold ball, or we don't.

[Anonymous No.16032574]

>>16032563
>But you don't know the contents of the box, so you can't intentionally choose "gold either way" more than "gold half the time"
Exactly. That's just how it happens randomly. If you knew the contents of the box, THEN you could choose gold every time for your first pick, see? But the OP explictly says that's not the case. So you misinterpreted it. Clearly. It tells you exactly how you came to be in this situation, and that was by randomly picking a ball from a random box without being able to see inside.

Again, we have two possibilities, it's either one or the other. But one is twice as likely. Do you know how Monty Hall works, incidentally?

[Anonymous No.16032581]

>>16032574
>That's just how it happens randomly
And that's how you choose between the boxes, randomly, there is no higher chance for you to pick any box over the other
>But one is twice as likely
But it isn't, you're equally likely to pick any box out of the three

[Anonymous No.16032586]

>>16032563
I'm not the guy you're "arguing" with, if that's what to call it, I just want to point out that it makes me melancholic about the human race that people who cover their utter incapacity for abstract thought by word salad (like you) sometimes get to responsible positions in society.

[Anonymous No.16032593]

>>16032581
Yes, you're listing all the reasons it's 2/3. Each box is equally likely to be picked. Each ball is equally likely to be picked. And that is why, when you see that you've picked a gold ball, it is more likely to have come from the box that is more likely to give you gold when you pick it. That's pretty much a tautology isn't it?

[Anonymous No.16032596]

>>16032593
>Each box is equally likely to be picked
>it is more likely to have come from the box that is more likely to give you gold when you pick it

[Anonymous No.16032606]

>>16032596
That you think there is a contradiction there is probably why you aren't getting it.

Okay, I have three dice: a d6, a d20, and a d100. I pick one at random and roll it ten times. I get 6, 11, 13, 8, 9, 17, 20, 20, 14, 3. Which one do you think it probably was that I took?

[Anonymous No.16032609]

>>16032606
Whatever you roll does not matter, you are equally likely to pick all of them

[Anonymous No.16032612]

>>16032609
You think, with the information you've been given, that I am as likely to have picked the d6 as any other die?

[Anonymous No.16032634]

>>16032612
In vein of the question you do not know which die you pick, you are unable to see them, you pick at random, by what influence do you pick d6 more likely than d100

[Anonymous No.16032645]

>>16032634
You don't. Each die is initially equally likely to be picked. But then I roll it, see? I get a result. What does that tell you?

You're not being consistent here by the way because you already admitted that we can exclude the all-silver box after getting gold but now you won't even allow us to exclude the d6 after rolling 11?

[Anonymous No.16032655]

>>16032645
>But then I roll it, see?
You don't have to, you can tell by the shape of the die
>>16032645
>But then I roll it, see? I get a result. What does that tell you?
But you don't have to roll it, the question isn't what die you picked, the question is about how likely you are to pick it, and you pick at random, how can you pick any more likely than the other

[Anonymous No.16032679]

>>16032655
>You don't have to, you can tell by the shape of the die
Pretend it's a random number generator with three random settings if you're going to be pedantic.
>the question isn't what die you picked
It is. I asked it. I know what the question is.
>the question is about how likely you are to pick it
Each die is as likely to be picked as each box. So how likely are you to get each box, then?

[Anonymous No.16032716]

>>16032679
>Pretend it's a random number generator with three random settings
But reality doesn't have three random settings
>So how likely are you to get each box, then?
As likely as any other

[Anonymous No.16032724]

>>16032716
>As likely as any other
So you think the odds that you got the all-gold box is equal to the odds that you got the all-silver, i.e. 1/3?

[Anonymous No.16032734]

>>16032724
Why wouldn't it be, what would influence you picking the gold box over any other

[Anonymous No.16032744]

>>16032734
Okay, so, you no longer think the odds of getting another gold are 50-50? Or you don't see how these things are connected?

[Anonymous No.16032811]

>>16032734
I'm gonna call it a night but, just to be perfectly clear, you are now saying that, after you drew a gold ball from a random box that you had 1/3 chance of selecting among a total of three boxes, the odds that you selected the all-silver box have remained entirely unchanged?
Because if you are, that's clearly fucking stupid. But if you aren't, then you have to admit that you have to update your probabilities after you've obtained a result and learnt something about the contents of the box, and you can answer my question about the dice as well >>16032606.

[Anonymous No.16032828]

>>16027763
You should have seen the /v/ thread before it got deleted, Jesus christ

[Anonymous No.16032959]

we have box 1: [G_1, G_2], and box 2 [G_3, S_1]

If we know that the ball is gold, it could be G_1, G_2, or G_3.

If you pick G_1 or G_2 initially, the other ball is gold. If you pick G_3 initially, the other ball is silver.

So out of G_1, G_2, and G_3 as your initial ball pick (which is the condition that the problem states), 2/3 of them will lead to you picking up another gold ball.

this is not fucking hard you retards

[Anonymous No.16032979]

>>16022741
Increase the silver balls number in the second box, how likely is it you put your hand in it and pull the lucky golden ball?
It is not 50%

What is even funnier is that the explanation for the 2/3 case I sometimes see, like this one >>16022797,
is wrong as well. Increase the number of gold balls in the first box and the chance remains the same, picking either of them doesn't matter, what matters is the relative proportion of gold inside each box. Make the third box be 50% gold, now what are the chances c the next pick is gold? c = 1/3*100% / (1/3*100% + 1/3*50% + 1/3*50%) = 1/3 / 2/3 = 50%

Both midwits and brainlets owned

[Anonymous No.16033587]

I just wanna hear back from the guy who got 4/6

[Anonymous No.16033609]

>>16022751
now solve for this:
box1 contains 1000000 gold balls
box2 contains 999999 gold and 1 silver ball
box3 contains 1000000 silver balls

[Anonymous No.16033632]

>>16033609
0.5*1+0.5*999998/999999

[Anonymous No.16033661]

>>16033609

Gold 1'999'997/1'999'998
Silver 1/1'999'998

[Anonymous No.16033828]

>>16033609
>>16033632
>>16033661
It's (1000000+(999999*(999998/999999)))/1999999 = 1999998/1999999
Three million trials, 1999999 gold balls, of those, 1 million guaranteed second gold, 999999 shots at a 999998/999999 chance of second gold. Silver is the inverse, 1/1999999.

The denominator is the amount of gold balls you get after running enough trials to exhaust all the possibilities. The numerator is the number of balls you get from box 1 + the odds of getting gold from box 2 after picking gold, times the amount of gold balls you expect to pick gold first from box 2.

[Anonymous No.16033930]

>>16033587
I'm he
Any other queries I can help you with?