Mathematically unsolvable problems

[Anonymous No.12246941]

I'll start.

[Anonymous No.12246949]

What? That's easy, 50%.

[Anonymous No.12246953]

>>12246941
Just because you don't understand a problem doesn't mean it's unsolvable, anon.

[Anonymous No.12246959]

>>12246953
Stop trying to sound smart and profound.

[Anonymous No.12246962]

>>12246949
>>12246953
You have no way of knowing if you picked a gold ball out of the 2-gold box or the 1-gold box, so you can't accurately calculate the correct probability.

[Anonymous No.12246965]

>>12246941
not science or math.
pure bait and cancer.
saged and hidden.

[Anonymous No.12246969]

this is an ill-posed problem

[Anonymous No.12246971]

NOT THIS FUCKING SHIT AGAIN
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
It's 2/3

[Anonymous No.12246978]

>>12246965
Can you give me some sage too anon, I only have parsley left.

[Anonymous No.12246979]

>>12246971
Wrong.

[Anonymous No.12246982]

>>12246959
The fact that you think a fact as basic as that is "smart" speaks volumes.

[Anonymous No.12246991]

>>12246982
Shut up silver tongued demon. Away with thee.

[Anonymous No.12246992]

>>12246971
mods should post this link automatically every time this picture is spammed. fucking imbeciles.

[Anonymous No.12247008]

>>12246979
Did you read the article nigger?

[Anonymous No.12247010]

>>12247008
Did you read the OP problem tranny?

[Anonymous No.12247027]

>>12246992
>t. imbecil

[Anonymous No.12247040]

2/5

[Anonymous No.12247048]

>>12246962
Either it's gold or silver, so 50%.

[Anonymous No.12247093]

>>12247048
That's... not...

[Anonymous No.12247248]

Is it 5/6?

[Anonymous No.12247415]

>mathematically unsolvable problems
my virginity

[Anonymous No.12247895]

>>12246941
It doesn't specify that u can't pick the same box twice

[Anonymous No.12247950]

The two gold balls in the leftmost box are distinguishable. That’s what confuses people.

[Anonymous No.12248007]

>>12246971
That used coins, this uses balls. Totally different!

[Anonymous No.12248390]

>>12246941
P is N/(1+N)
with N=2, P=2/3

[Anonymous No.12248435]

>>12246941
(1/2)*(1/2)/((1/2)*(1/2) + (1/2)*1) = 2/3

[Anonymous No.12248462]

>>12246941
Don’t think about separate boxes. You picked 1 ball out of 6. It’s gold. What’s the probability it was in the box with two golden balls? 2/3.

[Anonymous No.12248475]

>>12246941
My balls are in your mouth though.

[Anonymous No.12248481]

Are there no mods on this board? Or do they just no care about low effort bait posts?

[Anonymous No.12248483]

>>12248481
>too low IQ to solve the problem so MODS MODS
Embarrassing.

[Anonymous No.12248490]

Fuck you and fuck this picture, I TeXed a solution now and whenever I see this stupid picture I want to see it posted.

[Anonymous No.12248588]

>>12248462
seems great until you think about it.
the odds are 2/3 even if the gg box has a million gold balls - now explain that

[Anonymous No.12248603]

>>12248490
Nice reading comprehension bro.

[Anonymous No.12248618]

>this shitty png
I. The correct answer, if you read the question as written in this shitty png, is 1/2.
II. The incorrect answer, if you hallucinate Bertrand's box into this shitty png, is 2/3.
III. The "correct" answer, if you ever see this shitty jpg in a standardized test, is the incorrect answer, 2/3.

Your mission, should you choose to accept it, is to beat the life out of this shitty png until it disappears from every last corner of the internet.

N.b. The last time this shitty png was posted, it was kill-switched after two probability 101 brainlets blew themselves the fuck out by answering (no,no) and (yes,yes) to the same two questions.

[Anonymous No.12248652]

>>12248618
low iq post

[Anonymous No.12248655]

>>12248652
Meta.

[Anonymous No.12248692]

>>12248588
The idea of mixing all balls in just one box won’t work if the numbers of balls in GG and SG boxes are different because then probabilities to pick from boxes are not remain equal.

[Anonymous No.12248696]

>>12248692
speak english

[Anonymous No.12248729]

Shouldn't it be just 2/5 cause you took one golden ball away and you are left with 2 golden and 3 silver balls?

[Anonymous No.12248838]

>>12246941
2/5
I don’t see the difficulty, what am I missing?

[Anonymous No.12248848]

>>12248490
but there are three of them, so it's 6/3

the is, by simplifying to a algebra, is now 2

[Anonymous No.12249583]

Isn't this just the same as the Monty Hall problem?
In excluding the box with the two silver balls, the probability that the selected box is the one with two golden balls is 2/3s.

[Anonymous No.12249704]

>>12246962
>You have no way of knowing if you picked a gold ball out of the 2-gold box or the 1-gold box, so you...
... have to account for both possibilities.
Answer is 2/3

[Anonymous No.12249737]

Solved for gold

[Anonymous No.12249832]

>>12249737
Wrong answer brainlet.

[Anonymous No.12249858]

>>12249737
"it's a gold ball"
3 ways to start: g1,g2 or g3
3 ways to continue
g1 --> g2
g2 --> g1
g3 --> s
two favorable cases
2/3

[Anonymous No.12250907]

>>12246941
>Mathematically unsolvable problems
How many cocks can OP suck if OP sucked even more cock?

Exact answers only.

[Anonymous No.12252554]

3 boxes
2 gold balls in box one so 2/6
1 gold ball in box two so 1/6
2/6+1/6 = 3/12
simplify 3/12 = 1.5/6
Divide by 2 because picking twice = 0.75/3
Answer is 0.75/3.

[Anonymous No.12252724]

>>12246941
>Mathematically unsolvable problems

[Anonymous No.12252761]

>>12246941
(1/3)*1 / ((1/3)*1 + (1/3)*(1/2))

[Anonymous No.12252785]

Easy peasy, it's 23/45

[Anonymous No.12252801]

>>12249737
~39,25%

[vz !!OhpVRMX9e1L No.12252817]

God I wish they taught how to actually use Bayes's theorem in high school... Look, it's easy, there's nothing to think about and no intuitions to debate as to what counts as a "distinct possibility" or some bullshit like that, just apply Bayes's theorem to the probability you've been given in the question and do the primary-school-level calculations.

Let [math]B_n[/math] stand for "having picked box [math]n[/math]", [math]G,S[/math] for "having picked a golden/silver ball" respectively, and [math]N_{G,S}[/math] for "picking a golden/silver ball next", respectively. Then, since the [math]B_n[/math]s are a mutually exclusive complete set of alternatives, [math]P(N_G | G) = P(N_G & B_1 | G) + P(N_G & B_2 | G) + P(N_G & B_3 | G)[/math].

Now, the probability with [math]B_3[/math] is 0, since there's no way that the box chosen was the third if we already took out a golden ball (you can kinda ignore this if you're fine saying [math]0 \cdot P(\dots | B_3 & G) = 0 [/math], even though that conditional probability isn't well defined). So:
[eqn]\begin{align}P(N_G|G)=&P(N_G|G)=P(N_G&B_1|G)+P(N_G&B_1|G)\\=&P(N_G|B_1&G)P(B_1|G)+P(N_G|B_2&G)P(B_2|G)\\=&1\cdot\frac{P(B_1)P(G|B_1)}{P(G)}+0\cdot\frac{P(B_2)P(G|B_2)}{P(G)}\\=&\frac{P(B_1)P(G|B_1)}{P(G&B_1)+P(G&B_2)+P(G&B_3)}\\=&\frac{P(B_1)P(G|B_1)}{P(G|B_1)P(B_1)+P(G|B_2)P(B_2)+P(G|B_3)P(B_3)}\\=&\frac{^1/_3\cdot1}{1\cdot^1/_3+^1/_2\cdot^1/_3+0\cdot^1/_3}\\=&frac{2}{3}\end{align}[/eqn]

Where I used the values:
[eqn]P(B_n)=^1/_3\foralln,P(G|B_1)=1,P(G|B_2)=^1/_2,P(G|B_3)=0[/eqn]
which is the "pick a box at random then pick a ball at random" part, and
[eqn]P(N_G|B_1&G)=1P(N_G|B_2&G)=0[/eqn]
which are simply stating that if you draw a golden ball out of box 1 you'll find another one, and if you draw it out of box 2 you're not gonna have another one.

It's just mindless calculation that follows straightforwardly from using Bayes's theorem, there's no controversial intuition to draw upon.

[Anonymous No.12252833]

[math]\mathfrak{c} = \omega_1 [/math] ?

[vz !!OhpVRMX9e1L No.12252835]

God I wish they taught how to actually use Bayes's theorem in high school
Look, it's easy, there's nothing to think about and no intuitions to debate as to what counts as a "distinct possibility" or bullshit like that, just apply Bayes's theorem to the probability you've been given in the question and do the primary-school-level calculations.

Let [math]B_n[/math] stand for "having picked box [math]n[/math]", [math]G,S[/math] for "having picked a golden/silver ball" respectively, and [math]N_{G,S}[/math] for "picking a golden/silver ball next", respectively. Then, since the [math]B_n[/math]s are a mutually exclusive complete set of alternatives, [math]P(N_G|G)=P(N_G\land B_1|G)+P(N_G\land B_2|G)+P(N_G\land B_3|G)[/math].

Now, the probability with [math]B_3[/math] is 0, since there's no way that the box chosen was the third if we already took out a golden ball (you can kinda ignore this if you're fine saying [math]0\cdot P(\dots|B_3\land G)=0[/math], even though that conditional probability isn't well defined). So:

[eqn]
\begin{align}
P(N_G|G)=&P(N_G|G)=P(N_G\land B_1|G)+P(N_G\land B_1|G)\\
=&P(N_G|B_1\land G)P(B_1|G)+P(N_G|B_2\land G)P(B_2|G)\\
=&1\cdot\frac{P(B_1)P(G|B_1)}{P(G)}+0\cdot\frac{P(B_2)P(G|B_2)}{P(G)}\\
=&\frac{P(B_1)P(G|B_1)}{P(G\land B_1)+P(G\land B_2)+P(G\land B_3)}\\
=&\frac{P(B_1)P(G|B_1)}{P(G|B_1)P(B_1)+P(G|B_2)P(B_2)+P(G|B_3)P(B_3)}\\
=&\frac{^1/_3\cdot 1}{1\cdot^1/_3+^1/_2\cdot^1/_3+0\cdot^1/_3}\\
=&frac{2}{3}
\end{align}
[/eqn]

Where I used the values:
[eqn] P(B_n)=^1/_3\forall n,P(G|B_1)=1,P(G|B_2)=^1/_2,P(G|B_3)=0 [/eqn]
which is the "pick a box at random then pick a ball at random" part, and
[eqn] P(N_G|B_1\land G)=1P(N_G|B_2\land G)=0 [/eqn]
which are simply stating that if you draw a golden ball out of box 1 you'll find another one, and if you draw it out of box 2 you're not gonna have another one.

It's just mindless calculation that follows straightforwardly from using Bayes's thm, there's no controversial intuition to draw upon.

[vz !!OhpVRMX9e1L No.12252842]

God I wish they taught how to actually use Bayes's theorem in high school
Look, it's easy, there's nothing to think about and no intuitions to debate as to what counts as a "distinct possibility" or bullshit like that, just apply Bayes's theorem to the probability you've been given in the question and do the primary-school-level calculations.

Let [math]B_n[/math] stand for "having picked box [math]n[/math]", [math]G,S[/math] for "having picked a golden/silver ball" respectively, and [math]N_{G,S}[/math] for "picking a golden/silver ball next", respectively. Then, since the [math]B_n[/math]s are a mutually exclusive complete set of alternatives, [math]P(N_G|G)=P(N_G\land B_1|G)+P(N_G\land B_2|G)+P(N_G\land B_3|G)[/math].

Now, the probability with [math]B_3[/math] is 0, since there's no way that the box chosen was the third if we already took out a golden ball (you can kinda ignore this if you're fine saying [math]0\cdot P(\dots|B_3\land G)=0[/math], even though that conditional probability isn't well defined). So:
[eqn]
\begin{align}
P(N_G|G)=&P(N_G|G)=P(N_G\land B_1|G)+P(N_G\land B_1|G)\\
=&P(N_G|B_1\land G)P(B_1|G)+P(N_G|B_2\land G)P(B_2|G)\\
=&1\cdot\frac{P(B_1)P(G|B_1)}{P(G)}+0\cdot\frac{P(B_2)P(G|B_2)}{P(G)}\\
=&\frac{P(B_1)P(G|B_1)}{P(G\land B_1)+P(G\land B_2)+P(G\land B_3)}\\
=&\frac{P(B_1)P(G|B_1)}{P(G|B_1)P(B_1)+P(G|B_2)P(B_2)+P(G|B_3)P(B_3)}\\
=&\frac{^1/_3\cdot 1}{1\cdot ^1/_3+ ^1/_2\cdot ^1/_3+0\cdot ^1/_3}\\
=&frac{2}{3}
\end{align}
[/eqn]
Where I used the values:
[eqn] P(B_n)= ^1/_3\forall n,P(G|B_1)=1,P(G|B_2)= ^1/_2,P(G|B_3)=0 [/eqn]
which is the "pick a box at random then pick a ball at random" part, and
[eqn] P(N_G|B_1\land G)=1P(N_G|B_2\land G)=0 [/eqn]
which are simply stating that if you draw a golden ball out of box 1 you'll find another one, and if you draw it out of box 2 you're not gonna have another one.

It's just mindless calculation that follows straightforwardly from using Bayes's thm, there's no controversial intuition to draw upon.

[vz !!OhpVRMX9e1L No.12252846]

God I wish they taught how to actually use Bayes's theorem in high school
Look, it's easy, there's nothing to think about and no intuitions to debate as to what counts as a "distinct possibility" or bullshit like that, just apply Bayes's theorem to the probability you've been given in the question and do the primary-school-level calculations.

Let [math]B_n[/math] stand for "having picked box [math]n[/math]", [math]G[/math] for "having picked a golden ball, and [math]N_G[/math] for "picking a golden ball next". Then, since the [math]B_n[/math]s are a mutually exclusive complete set of alternatives, [math]P(N_G|G)=P(N_G\land B_1|G)+P(N_G\land B_2|G)+P(N_G\land B_3|G)[/math].

Now, the probability with [math]B_3[/math] is 0, since there's no way that the box chosen was the third if we already took out a golden ball (you can kinda ignore this if you're fine saying [math]0\cdot P(\dots|B_3\land G)=0[/math], even though that conditional probability isn't well defined). So:
[eqn]
\begin{align}
P(N_G|G)=&P(N_G|G)=P(N_G\land B_1|G)+P(N_G\land B_1|G)\\
=&P(N_G|B_1\land G)P(B_1|G)+P(N_G|B_2\land G)P(B_2|G)\\
=&1\cdot\frac{P(B_1)P(G|B_1)}{P(G)}+0\cdot\frac{P(B_2)P(G|B_2)}{P(G)}\\
=&\frac{P(B_1)P(G|B_1)}{P(G\land B_1)+P(G\land B_2)+P(G\land B_3)}\\
=&\frac{P(B_1)P(G|B_1)}{P(G|B_1)P(B_1)+P(G|B_2)P(B_2)+P(G|B_3)P(B_3)}\\
=&\frac{^1/_3\cdot 1}{1\cdot ^1/_3+ ^1/_2\cdot ^1/_3+0\cdot ^1/_3}\\
=&\frac{2}{3}
\end{align}
[/eqn]
Where I used the values:
[eqn] P(B_n)= ^1/_3\forall n,P(G|B_1)=1,P(G|B_2)= ^1/_2,P(G|B_3)=0 [/eqn]
which is the "pick a box at random then pick a ball at random" part, and
[eqn] P(N_G|B_1\land G)=1,P(N_G|B_2\land G)=0 [/eqn]
which are simply stating that if you draw a golden ball out of box 1 you'll find another one, and if you draw it out of box 2 you're not gonna have another one.

It's just mindless calculation that follows straightforwardly from using Bayes's thm, there's no controversial intuition to draw upon.

[Anonymous No.12252859]

>>12248618
>the answer is 1/2
Okay retard

[Anonymous No.12252886]

>>12252859
But he is right? The question at the end says "a box", meaning any of the three boxes. Bertrand's paradox is for the the specific box from which you already took one golden ball.

[Anonymous No.12252977]

It's 4/9
This board can't into math

[Anonymous No.12253038]

>>12252886
Oh, I see, OP is being clever. He posted an edited version of the original thread-derailer. In it's original (and most common) incarnation, it says "from the same box", not "from a box"

Yes, in this particular (ambiguous, since it doesn't explicitly state whether its the same box or a randomly selected box) case, the answer would be 1/2.

[Anonymous No.12253059]

>>12253038
Wouldn't it be 2/5 in that case? Because if the next ball can be selected from any box then the grouping of balls into boxes doesn't matter at all

[Anonymous No.12253091]

>>12253059
No; I'll explain this through enumerated outcomes, instead of bayes' theorem, because I feel its more obvious.

In the original version, you choose a random box out of the three. One box has two gold, one box has one gold and one silver, and one box has two silver (box 1 through 3 for simplicity) You do not know which box is which.

You pull one ball out of your randomly selected box. You then pull the second ball out of the same box. Since there are only two balls in each box, the second ball pulled is completely determined by the first ball pulled.

There are exactly six ways you can pull balls.

Box 1, gold 11, gold 12
Box 1, gold 12, gold 11
Box 2, gold 21, silver 22
Box 2, silver 22, gold 21
Box 3, silver 31, silver 32
Box 3, silver 32, silver 31

We are then given the condition that the first ball pulled is a gold ball. From this, we reduce the possible sample space to the following, as all others besides these have you pulling a silver ball first.

Box 1, gold 11, gold 12
Box 1, gold 12, gold 11
Box 2, gold 21, silver 22

Of these, only the first two outcomes (where you happened to randomly select box 1) allow for you to pull a second gold ball. Thus, the probability is 2/3.

[Anonymous No.12253098]

>>12253091
Oh I was talking about OP's modified version, not the original version. You said that the answer to the modified version would be 1/2 but I think it's 2/5 and I may be wrong.

[Anonymous No.12253123]

>>12253098
Hmm, let's see

There's six possible "first pulls"
Box 1, gold 11
Box 1, gold 12
Box 2, gold 21
Box 2, silver 22
Box 3, silver 31
Box 3, silver 32

Since the first ball pulled is gold, the first pull can only be

Box 1, gold 11
Box 1, gold 12
Box 2, gold 21

From this, there is three distinct cases for the second pull, in terms of remaining balls

>box 1, gold 11
Box 1, gold 12
Box 2, gold 21
Box 2, silver 22
Box 3, silver 31
Box 3, silver 32

>box 1, gold 12
Box 1, gold 11
Box 2, gold 21
Box 2, silver 22
Box 3, silver 31
Box 3, silver 32

>box 2, gold 21
Box 1, gold 11
Box 1, gold 12
Box 2, silver 22
Box 3, silver 31
Box 3, silver 32

The important thing to remember here is that, unlike the first part, each outcome no longer has an equal probability for the second pull, since you're still picking the box first. For each outcome that's the last remaining ball of that box, it has a probability of 1/3, while the other four are 1/6.

For the first two cases (first gold is pulled from box 1), the probability of a second gold is 1/2. For the third case (first gold is pulled from box 2), the probability of a second gold is only 1/3.

As each of the first gold cases are equally likely, the final probability of a second gold (given the first gold) is:

(1/3)*(1/2) + (1/3)*(1/2) + (1/3)*(1/3) = 1/6 + 1/6 + 1/9 = 3/18 + 3/18 + 2/18 = 8/18 = 4/9

[Anonymous No.12253196]

>>12247048
No, it's gold, another gold or silver so ~67%.

[Anonymous No.12253204]

>>12253123
wtf is wrong with you

[Anonymous No.12253227]

>>12253123
Great explanation. Thanks anon

[Anonymous No.12253275]

>>12246941
Solved

[Anonymous No.12253287]

>>12253204
I don't know anon, what is wrong with me?

[Anonymous No.12253317]

>>12252977
based nonretard

[Anonymous No.12253336]

>>12253275
This is obviously wrong lmao. The question tells you that a gold ball has already been picked. So you should use a counter variable c that increases by 1 whenever ball==1. And then you should divide finalvalue by c, not by iterations.

[Anonymous No.12253345]

>>12253336
I assume I already know if the next ball is gold or not. Since if I picked from box 1 I'm 100% sure it's gonna be gold, else if I picked it from box 2 then I know it's gonna be grey.
Simulations are here for a reason, brute calculations by testing out the outcomes.

[Anonymous No.12253363]

>>12253345
That part isn't wrong. What I'm saying is that you're dividing by the variable iterations which is wrini. Because not all iterations will lead to a gold ball being picked at first.

[Anonymous No.12253388]

>>12253363
Oh yeah, you're right!

[Anonymous No.12253391]

>>12253363
>which is wrong
which is incorrect*

[Anonymous No.12253396]

>>12253388
Great. Glad that code worked

[Anonymous No.12253420]

>>12253388
I do not understand. Why did ya call the function pickball() again when the ball was equal to 0?

[Anonymous No.12253562]

>>12253038
>>12253059
Hadn't even noticed the slight change in wording in OP's image. It can't be answered. If you had one box with three of each color, and you drew a gold, it would be 2/5 of the next draw being gold. However, we have three boxes, and if the person understands odds, then they'll know their best bet is to stick with the box they already pulled from if they want a second gold ball. We can't know their intent of which box they'll draw from the second time. So OP's title is actually true.

[Anonymous No.12253594]

>>12253420
Read
>>12253363

When the ball is gray I just re-run the function to get a golden ball since I only want the gold one.

[Anonymous No.12253926]

>>12246941
Literally a 2 in 5 chance

[Anonymous No.12254009]

>>12253926
Figuratively 1 in 2 chance.

[Anonymous No.12255233]

>Original prompt never says you take from the same box again
Fs for everyone, try learning to read before you learn math nerds

[Anonymous No.12255249]

>>12255233
To be fair, the same image has been posted here dozens of times without the slight tweak in wording, so I doubt anyone is bothering to read it thoroughly at this point.

[Anonymous No.12257051]

>>12246941
Try this one brainiacs!
>inb4 i fucking love science!

How many genders are there?


>>12246949
>>12246953
>>12246959
>>12246962

[Anonymous No.12257068]

>>12257051
>How many genders are there?
None, we're all just a clump of cells. My body my right.

[Anonymous No.12257426]

>>12246941
So we want P(2 gold balls). The chance of picking any box is 1/3 and you picked a gold ball. The chance of getting a gold ball from the two gold ball box is 1. The chance of getting a gold ball from the one ball box is 1/2. So, either you picked a gold ball from the box with two gold balls (1/3) or the box with one gold ball (1/6= 1/3*1/2). You cannot pick from the same box, so to get two gold balls, either it's from 2b box then 1b box which is (1/3) (1/2) (1/2)=1/12, or it's the 1b box and then the 2b box which is (1/6)(1/2)=1/12. So the chance of getting two gold balls should be 1/12 + 1/12 = 1/6.

[Anonymous No.12257478]

>>12246962
>I've never taken an introductory probability class
>IT'S UNSOLVABLE!!!11!!!

[Anonymous No.12258182]

>>12246941
1/2 stupid nerds

[Anonymous No.12258304]

>>12258182
One of my female coworkers does this pose to me sometimes, does it mean she wants to have sex with me?

[Anonymous No.12259738]

>>12249583
Its more complicated and confusing, less useful in exposing total brainlets

[Anonymous No.12259786]

>>12259738
So far it has exposed 99% of /sci/ as illiterate.

[Anonymous No.12260066]

>>12246941
4/9

[Anonymous No.12260091]

>>12253123
Lmao at all the 4/9 retards in this thread. I can't even imagine how many milligrams of LSD I'd have to drop to come up with that answer.

[Anonymous No.12260241]

>>12260091
I can't imagine what it's like to be illiterate in XXI century.

[Anonymous No.12260254]

>>12246971
Nigger.

The problem tells you from the start you've picked a box with gold. You may as well ditch the other box out of the situation.

You have 2 boxes with 2 balls each. You find out either has gold. Which is the remaining ball?

Case 1: Picked 2 gold boxes. Remaining=Gold
Case 2: Picked 1 Gold box. Remaining = Silver

Tell me where the fuck is the third option supposed to come in.

[Anonymous No.12260264]

>>12260254
>third option
the prompt has 3 gold balls

[Anonymous No.12260270]

>>12260264
You have two boxes. The question is about the next ball. This is not the Monty python problem. You can't switch them.

[Anonymous No.12260284]

>>12260264
>the prompt
Fucking NPC.

[Anonymous No.12260462]

>>12260270
so what

[Anonymous No.12260466]

>>12260284
fuck off, tourist

[Anonymous No.12260512]

>>12260241
If you think the answer is 4/9, you don't have to imagine.

[Anonymous No.12260546]

>>12260512
You give me no reason to imagine. Read the OP question, cretin.

[Anonymous No.12260563]

>>12260546
I did, cretin. Assuming you don't return the first gold ball to any of the boxes, the answer is 2/5.

[Anonymous No.12260568]

>>12260563
Ah sorry you're just retarded.

[Anonymous No.12260584]

>>12260568
Ok schizo

[Anonymous No.12260625]

>>12260563
>>12249858

[Anonymous No.12260645]

>>12260625
Wrong problem. In this version, the boxes are a red herring. The second box you pick has nothing to do with the first box you pick. All that changes is that instead of picking from 6 balls, 3 of which are gold, you're picking from 5 balls, 2 of which are gold.

[Anonymous No.12260659]

>>12260645
>you're picking from 5 balls, 2 of which are gold.
True, but you can manipulate the odds if you choose to draw from the same box you picked from the first time, so the question can't adequately be answered.

[Anonymous No.12260668]

>>12260462
So, If the thing says specifically you pulled a yellow ball, the next ball you'll pull from that box is either yellow or gray.

[Anonymous No.12260690]

>>12260659
I agree with this caveat. It also isn't clear whether the first ball is returned to a box or set aside. The question is flawed and the simplest answer is 2/5.

[Anonymous No.12260713]

>>12260645
>second box you pick
idiot

[Anonymous No.12260716]

>>12260690
>isn't clear whether
idiot

[Anonymous No.12260722]

>>12260713
>>12260716
wow another illiterate ass chimes in

[Anonymous No.12260731]

>>12260659
>>12260690
The simplest assumption is that you pick from a random box for your second draw, such that you don't remember if you drew from the box you already pulled from.

[Anonymous No.12260737]

>>12260091
Get dabbed on

[Anonymous No.12260738]

>>12260731
Correct, that's why the simplest answer is 2/5.

[Anonymous No.12260743]

>>12260738
No, if you're choosing from a random ball from a randomly selected second box, the answer is 4/9.

[Anonymous No.12260815]

>>12260743
Nope.
>>12260737
That's fascinating, but we already know it's a gold ball. Delete the bottom half of your rows and try again.

[Anonymous No.12260844]

>>12260815
Yep.

The bottom half of the rows are unimportant, but they're there explicitly for context.

That's what the "1st gold" probability of 1/2 means; it's when the first gold ball you pick is gold.

The "2nd gold ball (and 1st gold ball)" are those cells which are highlighted in the 4th choice column, and out of all possible outcomes, they have a total probability of 2/9.

P(1st ball is gold) = 1/2
P(2nd ball is gold AND 1st ball is gold) = 2/9

Thus, P(2nd ball is gold given 1st ball is gold) =
P(2nd ball is gold AND 1st ball is gold) / P(1st ball is gold)=
(2/9) / (1/2) = 4/9

QED, get dabbed on

[Anonymous No.12260860]

>>12260815
And just to appease you, pic related

>inb4 "hey you changed the probabilities for the first ball picked to 1/3"
Yes, because given we already know we will pick one of those three possibilities, and that they are all equally likely, they must have a probability of 1/3 each.

[Anonymous No.12260876]

>>12260815
>>12260860
and, just in case you think there's some trickery by indexing the balls, here:

[Anonymous No.12260885]

>>12260815
>>12260860
and, just in case you think there's some trickery by indexing the balls, here:

(posting fixed version)

[Anonymous No.12260894]

>>12260844
>>12260860
Lmao okay I'm going to assume cells F3, F8, and F15 are a joke, and that I'm laughing with you, not at you.

[Anonymous No.12260907]

>>12246959
He is right though...

[Anonymous No.12260911]

>>12260894
No, those are entirely serious, and it's explicitly due to the wording of the question and the fact that we don't replace the first ball chosen.

If our first ball chose is from box 1, there is naturally only 1 ball remaining in box 1, and it is guaranteed to be gold. Box 2 still has a gold and a silver.

If our first gold ball chosen happens to be from box 2, then there are two gold balls remaining in box 1 (meaning its guaranteed to pull a gold from box 1 if we pick it), but ALSO that there is only a singular silver ball remaining in box 2, making it a guaranteed pull if we choose box 2.

[Anonymous No.12260912]

>>12246949
Ain't it 1/3? If you pick a random box and you get a gold ball there is a higher chance you picked the one with 2 golden balls in it than the one with 1?

[Anonymous No.12260932]

>>12260254
never says you have to pick from the same box twice

[Anonymous No.12260950]

>>12260911
Subgroups aren't relevant to the question. And if they are—if we're allowed to probe the environment for clues—then the answer isn't 4/9, it's 2/3, because we can choose to always pick from the 1 box that only has 1 ball.

[Anonymous No.12260967]

>>12260950
>Subgroups aren't relevant to the question.
They are, because you're picking from a box, and then picking from the balls within that box. Subgroups are naturally baked into the problem.

>because we can choose to always pick from the 1 box that only has 1 ball.
The assumption that the second pick is from a randomly chosen box such that you can't distinguish which box you already pulled from. This is common practice in probability problems; in lieu of specific qualifiers, assume uniform randomness where appropriate.

[Anonymous No.12261016]

>>12260967
>Subgroups are naturally baked into the problem.
Only if the boxes somehow interfere with you reaching out and picking any random ball.

>you can't distinguish which box you already pulled from.
You're assuming that the edges of the boxes can touch your hand but that your hand can't touch the balls inside the boxes. Doesn't make sense.

[Anonymous No.12261029]

>>12261016
>Only if the boxes somehow interfere with you reaching out and picking any random ball.
They do, because the balls are in the boxes. You pick a box, and then take a ball out of the box.

>You're assuming that the edges of the boxes can touch your hand but that your hand can't touch the balls inside the boxes. Doesn't make sense.
What are you even trying to say?
Direct from the OP:
>you pick a box at random
>you put your hand in and take a ball from that box at random
You pick a box first, and then pick a ball from that box.

[Anonymous No.12261044]

Since you don't have to keep using the box you picked you can consider as one box.

Given you already picked a gold ball, there are 2 out of 5 balls left that are gold.

Therefore, 2/5

[Anonymous No.12261047]

>>12261044
>Since you don't have to keep using the box you picked you can consider as one box.
Nope; not all balls are equally likely, since they're partitioned in different boxes. It's a subtle difference, but it does matter.

[Anonymous No.12261052]

>>12261047
It only matters if the problem were to continue beyond a second pick.

[Anonymous No.12261057]

>>12261044
Lmfao this 2/5 shit is literally the 50% version for this edit of the question.

[Anonymous No.12261058]

>>12261052
No, it matters for the second pick as well.

[Anonymous No.12261062]

>>12261029
That's the first pick. The second pick specifically leaves out the step you're trying to highlight.
>What are you even trying to say?
There are 5 balls. The subgrouping only matters if you have to choose a box before choosing a ball. If you don't, the answer is 2/5.
And if you're allowed to probe the environment for extraneous information, what stops you from feeling how many balls are in a box? In that case, the answer is 2/3.
The answer is only 4/9 if you add idiosyncratic rules to the question.

[Anonymous No.12261069]

>>12261057
t. probability 101 brainlet

[Anonymous No.12261071]

>>12261062
Not that anon.
You have to choose a box if you want to choose a ball. Your explanation is idiosyncratic considering the problem says
>Note: You can't see into any of the boxes

[Anonymous No.12261075]

>>12261062
>"what is the probability that the next ball you take from a box"
It's the exact same process, as you must first choose a box in order to choose a ball.

>And if you're allowed to probe the environment for extraneous information, what stops you from feeling how many balls are in a box?
The fact that you are randomly choosing a box, and are randomly choosing a ball from that box. The exact physical mechanics (as you are trying to bring into question) are irrelevant and do not matter.

>if you add idiosyncratic rules to the question.
Anon, that's exactly what you are doing.

[Anonymous No.12261076]

>>12261069
>hurr probability of winning lottery is 50%, you either find out you win or lose durrr

[Anonymous No.12261084]

>>12261071
>You have to choose a box if you want to choose a ball.
No, you don't.
>You can't see into any of the boxes
What does that have to do anything? If you can see into the boxes, then the answer is 1.

[Anonymous No.12261099]

>>12260254
if you've picked a gold ball then what box are you more likely to have picked it from?

[Anonymous No.12261101]

>>12261084
>No, you don't
Yes you do, because you have to pick a second ball, which mean you have to pick a box.
>What does that have to do anything? If you can see into the boxes, then the answer is 1.
It means you can only arrive at the correct answer by following the methods instructed in the question, that is, you have to pick a second ball to arrive to a conclusion, and you have to do it via putting your hands first into a chosen box, and then picking a ball. No guesswork from just observation.

[Anonymous No.12261110]

>>12261075
No, you don't have to choose a box. In fact, choosing a box is specifically omitted from the second pick instructions after being included in the first pick instructions.

The only way you'd be forced to choose a box first is if you couldn't simply reach out and pick any one of the five balls without running into the edges of one of the boxes. And if the default environment forces you to interact with the boxes, as you suggest, it makes no sense that you wouldn't also be able to interact with the balls.

[Anonymous No.12261118]

>>12261101
>which mean you have to pick a box
>putting your hands first into a chosen box
This isn't in the question—you're adding it as an idiosyncratic rule of your own.

[Anonymous No.12261120]

>>12261110
>The only way you'd be forced to choose a box first is if you couldn't simply reach out and pick any one of the five balls without running into the edges of one of the boxes
Kek, that's de facto choosing a box as all balls are contained within a box. The rest is just cope.

[Anonymous No.12261122]

>>12261110
>No, you don't have to choose a box.
Yes, you do. The balls are in the boxes, and thus to pick a ball, you must necessarily pick a box. It is impossible to choose a ball without also choosing a box.

This is because you cannot see into the boxes; you have no idea of knowing which balls are in which box. Thus, to take a ball, you must first choose a box to take a ball from.

>the choose a box instruction is specifically omitted
"what is the probability that the next ball you take FROM A BOX"
At worst, it is still implied. It in no way describes the ability or instruction to take a ball independent of a box, as you are describing.

>is if you couldn't simply reach out and pick any one of the five balls without running into the edges of one of the boxes
Firstly, stop with this. The physical mechanics of the selection are irrelevant. Secondly, no, you can't just grab a ball without opening a box first. They are boxes. The balls are contained within the boxes.

>And if the default environment forces you to interact with the boxes, as you suggest, it makes no sense that you wouldn't also be able to interact with the balls.
Nothing you typed here makes any sense. What are you trying to say?

[Anonymous No.12261123]

>>12261118
It is in the question as you would have no other way of knowing what ball you picked.

[Anonymous No.12261127]

>>12261118
>This isn't in the question
Yes it is

>you're adding it as an idiosyncratic rule of your own.
No he isn't. All balls are contained within boxes. You cannot see into the boxes. You must choose a box in order to select a ball. This is baked into the structure of the question.

[Anonymous No.12261135]

>>12261058
Youre right, im a brainlet, it is 4/9

P(G2|G1)=P(G2|B1G1) + P(G2|B2G1) + P(G2|B3G1)

=((2/3)((1)(1/3)+(1/3)(1/2))+(1/3)(1/3)+0

=4/9

[Anonymous No.12261199]

>>12261120
>>12261122
>>12261123
>>12261127
>The physical mechanics of the selection are irrelevant.
This is exactly why you're wrong. If the boxes aren't involved in the physical mechanics of your pick, it's no different than picking from 5 balls on a carpet with a random pattern of squares.

[Anonymous No.12261226]

>>12261071
You can't see into the boxes, but unless they're reshuffled, you'll know which box you pulled the first ball from and can choose to draw from it again. Obviously the question was poorly written so all we can do is make assumptions, but there's no correct answer here.

[Anonymous No.12261239]

>>12261199
>This is exactly why you're wrong. If the boxes aren't involved in the physical mechanics of your pick, it's no different than picking from 5 balls on a carpet with a random pattern of squares.
My point is that you're focusing too much on the actual physical mechanics, rather than recognizing the underlying structure of the problem, which is that there are a number of elements collected into groups, and that you must randomly choose a group first before selecting an element from within that group.

Boxes, balls, grabbing, seeing into the boxes, etc, are all narrative window dressing that are irrelevant to the underlying question,

[Anonymous No.12261250]

>>12261239
>you must randomly choose a group first
No, this is exactly why you're wrong, as the analogy of choosing a ball from a patterned carpet makes perfectly clear.

[Anonymous No.12261255]

>>12261250
>you cannot see into any of the boxes

Sure, fine, lay your balls out on a patterned carpet, but by some arbitrary mechanism you're unable to distinguish the contents of each group, and so you must first randomly choose a group, and then choose a ball from that group.

[Anonymous No.12261267]

>>12261255
>arbitrary
That's my point. You're invoking an arbitrary (idiosyncratic) mechanism. The default answer to the question, as written, is 2/5.

[Anonymous No.12261274]

>>12261267
No, the mechanism is arbitrary, but the effect (you cannot see the balls within the boxes) is clearly defined. It does not matter what the mechanism is, only that you cannot distinguish the contents of a given group. Whether its boxes, blankets, or selective blindness and amnesia, whatever it is, you cannot determine what elements compose a given group. Because of this, you MUST select a group first as an inherent requirement in the process of selecting a ball, as every ball is part of a group, and as you cannot observe the balls directly.

The answer is, always has been, and will continue to be 4/9.

[Anonymous No.12261414]

>>12261274
No, arbitrary groups don't matter at all. If you're picking from five pool balls—1, 2, 3, 4, 5—the fact that the balls can be arbitrarily grouped into odds and evens, or primes and nonprimes, has no relevance at all to the probability of picking any one ball over any other.

Stop trying to redefine the question itself in order to make your answer fit. 4/9 is the wrong answer, and you simply can't change that without changing the question itself.

[Anonymous No.12261420]

>>12261274
No, arbitrary groups don't matter at all. If you're picking from five pool balls—1, 2, 3, 4, 5—the fact that the balls can be arbitrarily grouped into odds and evens, or primes and nonprimes, has no relevance at all to the probability of picking any one ball over any other.

Stop trying to redefine the question in order to make your answer fit. 4/9 is the wrong answer, and you simply can't change that without changing the question itself.

[Anonymous No.12261426]

>>12261414
>No, arbitrary groups don't matter at all. If you're picking from five pool balls—1, 2, 3, 4, 5—the fact that the balls can be arbitrarily grouped into odds and evens, or primes and nonprimes, has no relevance at all to the probability of picking any one ball over any other.
Except the groups aren't arbitrary, they're explicitly defined in the question.

One box (a group) has two gold balls (elements); one box has a gold and a silver ball, and one box has two silver balls. Those are explicitly defined groups.

>Stop trying to redefine the question itself in order to make your answer fit. 4/9 is the wrong answer, and you simply can't change that without changing the question itself.
I am reading the answer exactly as it is written. You are the one trying to redefine it by ignoring the explicitly stated, and important consequences of, the balls being part of groups.

The answer is 4/9. You are wrong.

[Anonymous No.12261429]

>>12261420
>>12261426

[Anonymous No.12261437]

>>12261426
No, you're pretending that the second pick instructions are the same as the first. You're wrong, and you couldn't be any more clearly wrong if you tried. The exact words of question are right there for everyone to read.

[Anonymous No.12261449]

>>12261437
lol you dumb, it clearly states
>what is the probability that the next ball you take from a box will also be gold
>next ball you take from a box
>from a box
idk wtf you even trying to do, just admit to being a dumbass and move on.

[Anonymous No.12261451]

>>12261437
The second pick instructions are the same as the first, there is no difference between them. Even if they are not explicitly defined to be equal, there is no other way to select a ball other than the already stated process (for the first pick) that makes sense in the context of the problem.

I am sorry that you are functionally illiterate and are in the throes of dunning kuger, but you are wrong. The answer is 4/9, as has been repeatedly explained and demonstrated (in multiple ways) throughout the thread, along with explanation of how the question unambiguously leads to that construction of the problem.

[Anonymous No.12261464]

>>12246941
im pretty sure thats a high-school textbook question and im pretty sure the answer was more complicated than i believed after a fast glance, since i always sucked at probability

[Anonymous No.12261478]

>>12261420
Are you trolling? It clearly states you need to pick from a box the second time in the question.

[Anonymous No.12261484]

>>12261451
>there is no other way to select a ball
The only way that conforms to the question as written is that you select directly from the single group of five balls without interference from the three subgroups of boxes.

>I am sorry that you are functionally illiterate and are in the throes of dunning kuger
I'm sorry you ran out of logic and into insults, but it doesn't really surprise me.

[Anonymous No.12261493]

>>12261478
The balls are all in boxes, just as five pools all have numbers written on them. It's irrelevant. The first pick instructions are that you pick a box, then take a ball from that box. The second pick instructions are that you take a ball from a box. I don't see how the difference could be highlighted any more clearly.

[Anonymous No.12261495]

>>12261478
The balls are all in boxes, just as five pool balls all have numbers written on them. It's irrelevant. The first pick instructions are that you pick a box, then take a ball from that box. The second pick instructions are that you take a ball from a box. I don't see how the difference could be highlighted any more clearly.

[Anonymous No.12261501]

>>12261484
>The only way that conforms to the question as written is that you select directly from the single group of five balls without interference from the three subgroups of boxes.
There is no single group of five balls. They are all contained within boxes.

>the next ball you take
>FROM A BOX
The next ball you take is from a box, meaning the box must be selected first, as you cannot see into the boxes.

>insults
Statements of fact, because there is no other explanation to your willful ignorance aside from trolling.

[Anonymous No.12261506]

Why do idiots think CS isn't legitimate? It's built with pure math and stats. It's the biggest gateway for pure math into the real world relative to other studies.

[Anonymous No.12261508]

>>12261495
>The second pick instructions are that you take a ball from a box. I don't see how the difference could be highlighted any more clearly.
There is no difference. It's lexical shorthand that implies the previous process, as opposed to writing it out explicitly longform, again.

[Anonymous No.12261512]

>>12261495
>The first pick instructions are that you pick a box, then take a ball from that box. The second pick instructions are that you take a ball from a box
What exactly is the difference here? Isn't your entire point hinged on the basis that the boxes are irrelevant, whereas it states you're still picking from a box, and therefore still under the limitations of the balls being located in separate boxes, thus needing separate probability states that include the boxes in the second pick?
Logically it does not follow that you skip the boxes and make them irrelevant, as the question demands you take from a box. The "a" used is for ambiguity in the selection of boxes unlike the original unedited question which states "take from the same box", I don't see how you are extrapolating box ephemerality from it.

[Anonymous No.12261518]

>>12261501
>There is no single group of five balls.
>willful ignorance aside from trolling
At this point, it's pretty obvious that you're the one who's trolling.

[Anonymous No.12261535]

>>12261518
Anon, there are three boxes of two balls each.
After the first pick, you are left with two boxes with two balls each, and one box with one ball.

There has never been a group with six balls, nor a group with five balls. The question is entirely predicated on the balls being contained within a box, and unobservable, until after a box has been chosen and a ball is chosen from that box.

Since logic, reason, mathematics, and rhetoric have not worked, I am forced to resort to just outright assertion.

You are wrong, you do not understand the question, and you are too prideful (on an anonymous imageboard, no less) to accept that you are wrong.

[Anonymous No.12261538]

>>12246962
>>12246969
learn Bayes' Theorem, you fucking brainlets

[Anonymous No.12261546]

Ahem, OP here, can you anons actually post some other unsolvable mathematical problems like thread title stated?

[Anonymous No.12261571]

>>12261546
Post 3 more and we'll think about it, faggot.

[Anonymous No.12261582]

>>12261546
If you start a shitposting thread, don't be surprised when people shitpost.

[Anonymous No.12261588]

>>12261535
>You are wrong, you do not understand the question, and you are too prideful (on an anonymous imageboard, no less) to accept that you are wrong.
This describes your posts quite well. I’d simply add that you’re also now doubling down on the idea that a group of three subgroups of two balls each isn’t a group of six balls. Come on, even you have to realize that’s literally “schizo” tier.

[Anonymous No.12261589]

>>12261571
r-rude...
>>12261582
I didn't mean too.. it's not my edit, I just noticed it and couldn't figure it out

[Anonymous No.12261597]

>>12261589
God why are OPs always like anime MCs in being spineless weak faggots. Why can't we get a chad OP for once.

[Anonymous No.12261604]

>>12261588
No, because you're not choosing between the six balls. You are choosing between the three groups, and then choosing between the balls within that group.

This is explicitly defined and stated.

Crying like a kindergartner and impotently throwing back insults doesn't change the fact that the answer is 4/9, and that you are wrong.

[Anonymous No.12261612]

>>12261597
Ok I'll be leaving then...

[Anonymous No.12261663]

To anyone not being willfully obtuse, it's 4/9 because the boxes are uneven once you take the first gold ball out and now just one of your boxes gives you as probability of 1 of getting silver once you pick it.

Consider the following.

You pick a random ball out of a random selection these 3 boxes. What is the probability of picking a silver ball?

Hint: it's not 1/2

[Anonymous No.12261669]

>>12261663
(1/3)(1/2) + (1/3)(1) + (1/3)(1/4)
1/6 + 1/3 + 1/12
7/12

[Anonymous No.12261689]

>>12252554
This.

[Anonymous No.12261692]

>>12261669
ye

[Anonymous No.12261694]

>>12261663
50%

[Anonymous No.12261747]

>>12261604
>No, because you're not choosing between the six balls.
That’s what you say, not what the question says.
>Crying like a kindergartner and impotently throwing back insults doesn't change the fact that the answer is 4/9, and that you are wrong.
Wow, it’s starting to sound like you are, in fact, literally a “schizo.” Be well friend.

[Anonymous No.12262174]

>>12261663
Based retard

[Anonymous No.12264225]

>>12261747
>That’s what you say, not what the question says.
That is literally what the question says. I am sorry that you are too uneducated to understand that the groupings are a critical part of the question, and too prideful to listen to people who know better than you.

>Wow, it’s starting to sound like you are, in fact, literally a “schizo.” Be well friend.
Thank you for proving my point.

[Anonymous No.12264270]

>>12252846
It's bait considering OP's responses.

[Anonymous No.12264277]

>>12262174
Sauce on that

[Anonymous No.12264287]

>>12264225
>Crying like a kindergartner and impotently throwing back insults
This is what you wrote. Clearly you’re not well.
>That is literally what the question says
It literally is not. You’re wrong, and all you’re doing now is digging yourself into a deeper and deeper hole.

[Anonymous No.12264311]

>>12264287
>This is what you wrote.
Yes, because it is what you are doing.

>Clearly you’re not well.
Says the anon unable to understand a simple probability problem.

>It literally is not.
I literally is, and the question explicitly says that it is. Are you ESL, by chance? That's the third possible explanation for you continued failure to get this right, besides a mental disability or just trolling.

[Anonymous No.12264523]

>>12264311
>Crying like a kindergartner and impotently throwing back insults
>Yes, because it is what you are doing.
The fact that you not only imagine this sort of thing, but that you've also convinced yourself that your imagination is real, is indicative of actual schizophrenia. If you're not trolling, you should seek professional medical help.

>I literally is, and the question explicitly says that it is.
Watch how all I have to do, to prove that I'm right and that you're wrong, is to transcribe the exact words from the two pick instructions.
1. "You pick a box at random. You put your hand in and take a ball from that box at random."
2. "If it's a gold ball, what is the probability that the next ball you take from a box will also be gold?"

Now watch how you won't be able to do this. The only way you'll be able to support your own idiosyncratic reading is by adding extra words and descriptions that don't actually exist in the original text. Watch how this works. It's a form of cognitive dissonance.
It's because you've convinced yourself that your imagination is real, just as you did with the bizarre kindergarten scene you hallucinated.

[Anonymous No.12264602]

>>12264523
>2. "If it's a gold ball, what is the probability that the next ball you take from a box will also be gold?"
Yes, the next ball you take, FROM A BOX. Y

We know that, in order to take a ball from a box, you must first choose a box. We know this because 1) that's what it tells us to do in the first, explicit pick instructions (which carry over to the second pick, especially given the absence of any explicit instructions to the contrary which you are inventing), and 2) because it is explicitly said that we "cannot see into the boxes". For anyone with even a marginal understanding of English, in the context of this problem, this means that the balls are indistinguishable and un-pickable until we "open" a box to pick one.

You CANNOT choose a ball without simultaneously picking a box, and as you cannot know which balls are where, or how many balls are in any given box for the second pick, logically, you must first choose a box to open before you can choose a ball from it.

This hinges on the fact that you cannot see into the boxes; the balls are unobservable, you will only know which ball you have taken once you have taken a ball out of the chosen box. This matters because, after the first pick, one of the boxes only has one ball. The balls are no longer evenly distributed between boxes, and so you can't just randomly choose between balls.

This is not "adding words", this is not adding idiosyncrasies, this is logical deduction from how the problem is stated to be constructed. The slightly different wording in the second part is lexical shorthand that does not change the underlying described process.

>litterally everything else in your post
You're not schizophrenic, but your continued inability to understand reason, simile, analogy, or hyperbole convinces me that not only are you mentally disabled, you're also highly autistic.

The answer is 4/9

[Anonymous No.12264628]

>>12264602
>[insert several paragraphs of extra words and descriptions]
qed

[Anonymous No.12264632]

>>12264628
>nooo you can't explain why I'm wrong
okay retard

[Anonymous No.12264656]

>>12264632
I already explained (cf. pool ball analogy, carpet pattern analogy) why you're wrong.

Then for fun, I predicted that you wouldn't be able to validate your reading by simply transcribing the exact words of the question as written—that you'd be forced to add extra words and descriptions—because if you didn't, then your reading wouldn't make sense, not even to you. And I was right again. I'm actually quite impressed by my own predictive power. Thank you for playing along.

[Anonymous No.12264672]

>>12264656
>cf. pool ball analogy, carpet pattern analogy
No, you didn't, because both of your analogies are incorrect and do not conform to the underlying problem.

>Then for fun, I predicted that you wouldn't be able to validate your reading by simply transcribing the exact words of the question as written—that you'd be forced to add extra words and descriptions—because if you didn't, then your reading wouldn't make sense, not even to you.
Because I see what game you're trying to play, and simply transcribing what was already said doesn't let me explain why your reading is incorrect.

Using what has been explicitly given by the problem, I have logically show the correct interpretation that allows one to reach the correct answer.

The fact that you think explaining a logical deduction is "adding extra words" just further proves my assertion that you're a dimwitted autist.

[Anonymous No.12264851]

>>12246941
Okay I'll put it in a way /sci/ will understand: You have three neighbours: a lesbian couple, a straight couple, and a gay couple. You knock on a door at random, and a girl answers the door. She asks "would you like to come in and fuck my partner?". This should make the probabilities clear.

[Anonymous No.12264866]

>>12264851
>yes i-i'm here to fuck?

[Anonymous No.12264881]

>>12264851
50%

[Anonymous No.12264962]

>>12264672
>do not conform to the underlying problem
My logic conforms perfectly to the exact problem as written.
Your logic conforms to a different problem that you've adapted from the original by including arbitrary elements of your own imagination.

What's more, your own post history already proves beyond a shadow of a doubt that even you know that your answer is wrong.

[Anonymous No.12264980]

>>12264962
>My logic conforms perfectly to the exact problem as written.
No, it doesn't. It conforms to an imagined problem where the balls are, magically, no longer in boxes or where the subgroupings no longer matter.

>>12264962
>What's more, your own post history already proves beyond a shadow of a doubt that even you know that your answer is wrong.
Put up or shut up. Links and quotes.

[Anonymous No.12264993]

>guys it's totally 50% lmao

[Anonymous No.12265008]

This question is probably a decent psychometric test but a much better psychological test for personality disorder.

[Anonymous No.12265149]

>>12264980
>the subgroupings no longer matter.
Correct, the subgrouping is irrelevant to the second pick. You’ve now explicitly typed out your own mistake. It’s a good first step.

>Links and quotes.
>>12264602
You already proved that you are unable to transcribe the actual, unaltered text of the problem, because you know as well as I do that it doesn’t fit your adaptation.

[Anonymous No.12265231]

>>12265149
No, the subgrouping is relevant. It is as relevant to the second pick as it is to the first pick.

I will repeat, explicitly:

It is IMPOSSIBLE to select a ball without first picking a box. This is the fundamental structure of the problem.
You're not dumping the balls out of the box
You're not opening up all the boxes
Your hand isn't phasing through the boxes to randomly grab a ball.

You pick a random box, and then take a random ball from that box, for the first pick
>"If it's a gold ball, what is the probability that the next ball you take from a box will also be gold?"
The operative part of this sentence is "the next ball you take from a box"

In this clause, the ball is explicitly being taken from a box. Whichever ball is chosen is coming from its respective box, which means it is IN the box prior to be taken out.
As it is still in the box (which is a direct precedent for it to be taken from the box), there is the question of which box(es) the ball(s) are in.

The first state of the question is the one as described in the original question; two balls per box, as colored.
After the first pick, the state has changed; one ball has been taken out of one box (of course, we cannot know for sure what box in specific it was, as the boxes are identical and we cannot see into them)
After this singular state change (one ball being taken out of one box), we are now at the second pick. Nothing else has occured. No balls have been shuffled around, mixed together, poured out, or anything. All balls are still in their original box, except for the first ball that was taken. The second pick is the same as the first.

>You already proved that you are unable to transcribe the actual, unaltered text of the problem, because you know as well as I do that it doesn’t fit your adaptation.
I explained why my interpretation logically follows from the unaltered text, and why you are wrong.

[Anonymous No.12265406]

>>12246941
Fuck you and your ambiguous problem. It's just like the coin flip problem. Back when those were getting posted all the time, they said "One coin comes up heads" but after people started pointing out the fact that in English, "One coin" refers to a specific coin (i.e. the first coin flipped), the wording was changed to be unambiguous: "At least one coin comes up heads." "If it's a gold ball" is equivalent to "One coin comes up heads," i.e. it implies a specific coin/ball, the first, is known.
>>12246971
WRONG.
>and a gold coin is found inside it
That's the info given in the problem on wikipedia's article. "A coin" is not the same as "If it's a gold ball" because "A coin" could be either coin while "If it's a gold ball" refers specifically to the first ball picked.


The answer is 50%.

[Anonymous No.12265418]

>>12265406
retard

[Anonymous No.12265440]

>>12246969
Yes, yes it is. It is exactly like the ambiguous coin flip problem which was rewritten from "One coin comes up heads" to "At least one coin comes up heads." It's not that people don't understand probability. It's that people don't understand English.

[Anonymous No.12265498]

>>12253562
Without intent being specified you have to assume each selection is made at random

[Anonymous No.12265522]

>>12265406
>they said "One coin comes up heads" but after people started pointing out the fact that in English, "One coin" refers to a specific coin (i.e. the first coin flipped), the wording was changed to be unambiguous
Actually, neither is ambiguous, they just mean different things. "one coin" is not a specific coin. It's just exactly one coin, no more, no less. "At least one" is one or more coins. There is absolutely no rule in English that says if you mention "one coin" it's the first one. Same with the balls. Don't know where the fuck you got this. Also, of course the coin, or ball, you take from the box is the one being referred to in both instances. You say it's ambiguous but you are simply semantically challenged.

[Anonymous No.12265555]

Does the change in OP's edit from the original
>It's a gold ball
to
>If it's a gold ball
Mean anything? Does that additional "if" affect the problem in any way? I can't wrap my head around it. And yes I know it also changed "same box" to "a box".

[Anonymous No.12265569]

>>12265555
It's the same thing. You're dealing with probability. It's all a big "if," implicitly.

[Anonymous No.12265572]

>>12265569
Ok thanks. I guess it was just thrown in as a red herring then.

[Anonymous No.12265575]

>>12246941
Am I hallucinating or does this bait problem read different than usual?

[Anonymous No.12267089]

>>12265231
>the subgrouping is relevant
Nope.
>I will repeat
Lol, why? We both already know your mistake. You can repeat that mistake as many times as you'd like, but the text of the original problem will never change.

>dumping the balls out of the box
>opening up all the boxes
>phasing through the boxes
You keep imagining that the boxes are somehow physically involved in the pick. I already explained your mistake to you two days ago.

The problem says nothing at all about the boxes being closed, or having lids, or having magic superwalls that somehow interfere with your hand when you reach out to pick a ball. You are literally imagining an arbitrary physical constraint, and then treating your own imagination as if it were a real element of the problem. The only real constraint in the real problem is visual: "Note: You can't see into any of the boxes."
This is why you're unable to simply quote the actual text of the problem, without adding extra caveats. Quoting the text forces you to append the text with a description of your own imaginary element.

Furthermore, as I also already explained two days ago, 4/9 isn't even the correct answer to your own adaption of the problem.
If we allow the walls of the boxes to be used as a physical constraint—your imaginary element—then the probability of picking a gold ball is 2/3, because we can then use the position of the walls to feel how many balls are inside each box and always pick from the box containing a single ball.
You're not only wrong about the logic of the original problem, you're wrong about the logic of your own imaginary problem! That's what amuses me the most.

[Anonymous No.12267134]

>>12261694
Holy Based

[Anonymous No.12267145]

>>12246941
Assuming gold has different density all you have to do is lift box a little and you know which box is which.

[Anonymous No.12267703]

>>12267089
>Nope
Yep

>The only real constraint in the real problem is visual: "Note: You can't see into any of the boxes."
Are you seriously this dumb? Are you genuinely, truly, this mentally challenged?

Firstly, The balls are IN the boxes. They are contained WITHIN the boxes, as is stated in the first line of the problem. This tells us that the balls are grouped by box. This is the "physical restraint" that corresponds to the underlying logic.

As has been stated, secondly, you reach INTO a box to grab a ball. This is stated for the first pick, and as nothing has changed between the first pick and the second pick, it remains true.

Thirdly, you cannot see into the box. This tells use that you cannot distinguish the boxes, nor can you choose a specific ball from a box.

>then the probability of picking a gold ball is 2/3, because we can then use the position of the walls to feel how many balls are inside each box and always pick from the box containing a single ball.
Firstly, you're autistic for trying to exploit a physical loophole. What you just said was on par with "measure the weight of the boxes to find out what they contain"

Secondly, no, because I have been repeatedly saying, you first choose a box, then choose a ball from that box. You can't "unchoose" a box if you discover the number of balls inside before selecting a ball.

You are wrong.
You are illiterate.
I have explained both my reasoning, and my math.
You have only asserted your own incorrect interpretation.

The answer is 4/9

[Anonymous No.12267771]

I read through 220 posts and can't tell who's baiting who
The answer is that OP lied, it's not a problem. Where is the problem? It's just a question.

[Anonymous No.12267935]

>>12267703
Like I said, it doesn't matter how many times you explain your mistake, or how many different words you use to describe your mistake. It's always the same mistake, and 4/9 is always going to be the wrong answer to the question.
>my reasoning, and my math
Your reasoning and your math are both wrong, as I've proven over and over again for two days now.

>nothing has changed between the first pick and the second pick
You either aren't reading the problem, or you're lying. I assume your cognitive dissonance is so strong that you're afraid to go back and read the original problem, and are instead relying on your own false memory of words that don't actually exist. It's okay. False memories happen all the time, to everyone. Try reading the words out loud to yourself. That can sometimes help.

"You pick a box at random. You put your hand in and take a ball from that box at random."
"If it's a gold ball, what is the probability that the next ball you take from a box will also be gold?"

By the way, this is particularly funny:
>Thirdly, you cannot see into the box.
Yes, I pointed that out myself. It tells us that you can't use your eyes to distinguish the balls.
>This tells use that you cannot distinguish the boxes
Lol, what? This is a non sequitur/ nonsense.
Do you really not see that your conclusion doesn't follow from your premise?

By the way, I could easily adapt the original problem into a new problem whose correct answer is 4/9. It blows my mind that you can't.
The answer to the real problem is 2/5, and the answer to your adaptation (so far) is 2/3. Nothing you've described so far can be correctly answered by 4/9.

[Anonymous No.12268142]

The wording makes the question ambiguous. That said:
If we pick the 2nd ball from the same box, the answer is 2/3.
If we pick the 2nd ball from any box, the answer is 4/9.

[Anonymous No.12268317]

>>12267935
>Your reasoning and your math are both wrong, as I've proven over and over again for two days now.
You haven't proven anything, not one word you've posted has even approached anything resembling an argument.

>"You pick a box at random. You put your hand in and take a ball from that box at random."
Yes, you randomly choose a box, then randomly choose a ball from that box.
>"If it's a gold ball, what is the probability that the next ball you take from a box will also be gold?"
Yes, the next ball you take from a box. As has been stated, and proven (despite your continued inability to read English), in order to take a ball from a box, you must first choose a box, leading to the answer of 4/9.

The answer is 4/9 if the second pick is from a random box, which it is, by OP's formulation.

In no formulation is it 2/5, aside from your own deranged imagination.

>Lol, what? This is a non sequitur/ nonsense.
>Do you really not see that your conclusion doesn't follow from your premise?
No, it does. You're the one getting hung up on the exact, physical, material narrative of the problem, instead of understanding and analyzing the underlying problem. This is because you are autistic and focus on irrelevant and pointless non-issues, to the exhaustion of everyone around you.

Respond if you want, but I'm done lecturing you.

[Anonymous No.12268351]

>>12267935
Not the guy you keep arguing with, but I'm here to defend 4/9. This will require a certain amount of brain cells.

We are looking for the probability that the 2nd ball is gold (B) given that the 1st ball is also gold (A). Without replacement. I'm gonna assume we can pick the 2nd ball from ANY box. The probability of picking a box at random {I,J,K} is 1/3. I have chosen to not incorporate this explicitely, to make for a lighter demonstration.

Given A, there are 3 gold balls we could have picked in the first round, so there are three different outcomes {X,Y,Z} after A, given that we don't replace the ball. The first two outcomes are actually identical. Given A, {X,Y,Z} represent the total set of events with each having an equal probability of 1/3.

P(B|A) is equal to the sum of P(B|X)P(X), P(B|Y)P(Y) and P(B|Z)P(Z)

P(B|X)P(X): Picked 1st gold ball in box 1
Box 1: P(I) = 1/3
One ball left in box 1, a gold ball, Probability of 1 given I and X
1/3 * 1 = 1/3
Box 2: P(J) = 1/3
One gold ball and one silver ball, Probability of 1/2 given J and X
1/3 * 1/2 = 1/6
Box 3: P(K) = 1/3
Two silver balls, Probability of 0 given K and X
P(B|X) = 1/3 + 1/6 + 0 = 1/2
P(B|X)P(X) = 1/2 * 1/3 = 1/6

P(B|Y)P(Y): Picked 1st gold ball in box 1
Box 1: P(I) = 1/3
One ball left in box 1, a gold ball, Probability of 1 given I and Y
1/3 * 1 = 1/3
Box 2: P(J) = 1/3
One gold ball and one silver ball, Probability of 1/2 given J and Y
1/3 * 1/2 = 1/6
Box 3: P(K) = 1/3
Two silver balls, Probability of 0 given K and Y
P(B|Y) = 1/3 + 1/6 + 0 = 1/2
P(B|Y)P(Y) = 1/2 * 1/3 = 1/6

P(B|Z)P(Z): Picked 1st gold ball in box 2
Box 1: P(I) = 1/3
Two gold balls. Probability of 1 given I and Z
1/3 * 1 = 1/3
Box 2: P(J) = 1/3
One ball left in box 2, a silver ball, Probability of 0 given J and Z
Box 3: P(K) = 1/3
Two silver balls, Probability of 0 given K and Z
P(B|Z) = 1/3
P(B|Z)P(Z) = 1/3 * 1/3 = 1/9

P(B|A) = P(B|X)P(X) + P(B|Y)P(Y) + P(B|Z)P(Z) =1/6 + 1/6 + 1/9 = 4/9

Seethe

[Anonymous No.12268433]

So what I got from this is, for example if there are no boxes, you have infinitely many golden balls and say some countable number of silvers, then the chance of picking a golden is 1.
But say you have infinitely many boxes that either contain one of those silvers or they are empty, and one box that contains infinitely many goldens, then the chance of picking a golden is 0.

[Anonymous No.12268457]

>>12268433
Yes

[Anonymous No.12268465]

>>12268457
Thanks, I didn't have probability yet, so this problem was cool and useful to remember. I almost said 1/2 at first as well, before understanding that you first pick the box and then the ball

[Anonymous No.12268476]

>>12265149
>>12267089
>>12267935
Base rate fallacy

[Anonymous No.12268500]

>>12268317
>As has been stated, and proven
Nope.
>you must first choose a box
Nope.

That's the whole point of the carpet pattern analogy. It highlights your exact mistake. You don't seem to have any intention of learning from your mistakes, but others might.

If choosing a box isn't a specific step—it is for the first pick, but not for the second pick—then the fact that the balls all happen to be laying in various boxes is no more relevant to the probability of which ball you pick than if those same balls all happened to be laying in various polygons on a carpet of polygonal patterns.

Asking you to pick one of the balls from a polygon isn't the same as asking you to first pick a specific polygon, then pick a ball from that polygon. And asking you to pick one of the balls from a box isn't the same as asking you to first pick a specific box, then pick a ball from that box.

>No, it does.
This is fascinating. You're also now doubling down on the idea that not being able to see what's inside a box is the same as not being able to see the box itself. I hope you eventually realize how silly this sounds.

>The answer is 4/9 if the second pick is from a random box
Nope.
>, which it is, by OP's formulation.
Yep.
>In no formulation is it 2/5, aside from your own deranged imagination.
The exact text of the problem is right here for all to see.
Feel free to point out exactly which part you think I'm imagining.
I've been pointing out the exact part that you're imagining for two days now.

[Anonymous No.12268504]

>>12268476
>Base rate fallacy
Nope. The text is perfectly clear.

[Anonymous No.12268521]

>>12268351
>The probability of picking a box at random {I,J,K} is 1/3.
This is relevant to the first pick but not to the second pick.
As I've said, it would be quite easy to adapt the OP into a new problem whose correct answer is 4/9. But the answer to the actual problem as written is 2/5

[Anonymous No.12268657]

>>12268504
>>12268521
>This is relevant to the first pick but not to the second pick.
How is it not? Given that you picked the 1st ball from a certain box WITHOUT REPLACEMENT, the probability of getting a second gold ball depends on which box you took the first one. The text clearly says you pick a box at random. It's part of the process of picking a ball.

I understand how you view the problem. You say that once a golden ball is taken, there are two golden balls left and three silver balls. Probability of picking 2 out of 5. The problem here is that you are indivduating the probability of picking each ball and disregarding any prior conditions. I already explained all of this but I will do it again. Baby steps.

Lets say you picked the first gold ball in box 2.
Box 1 has two gold balls.
Box 2 is left with one silver ball.
Box 3 has two silver balls.
The balls are still in the boxes. You pick a box at random (1/3 chance). You pick ONE ball from that box.
If you pick from box 1, the ball is 100% gold.
If you pick from box 2, the ball is 100% silver.
If you pick from box 3, the ball is 100% silver.
The point here is that the probability of picking each ball differs because you didn't replace the first one you picked. Given the same example, if you put your hand in box 1, each ball in box 1 has a probability of 50% of being picked, whereas if you put your hand in box 2, the one ball that is left in box 2 has 100% chance of being picked. Each ball doesn't have the same probability of being picked, because that probability depends on how many balls are in the same box. You CAN'T individuate the probability of each ball because you are not choosing the ball from the whole set, they are distributed unevenly in the boxes and you CAN'T SEE where the balls are.

[Anonymous No.12268732]

41.6%
If you originally chose gold from the left box, you have a 50% chance of picking gold again. If you chose gold from the center box, you now have a 33% chance of picking gold. Averages to 41.6%

[Anonymous No.12268739]

>>12268732
no

[Anonymous No.12268799]

>>12268657
>I understand how you view the problem.
And I understand how you view the problem.
You're imagining the same arbitrary constraint that the other poster imagined. You're imagining that the last sentence of the problem is something like:

"If it's a gold ball, what is the probability that [you will get another gold ball if you again pick a box at random and take a ball from that box]."

The answer to your imaginary problem is 4/9.

But in the real problem, the step where you pick a box before picking a ball is specifically included in the first pick instructions, and specifically omitted from the second pick instructions.
The fact that the 5 balls are all in 1 of 3 boxes doesn't matter at all.
It's no different than if you were asked to pick from 5 numbered pool balls—1, 2, 3, 4, 5—and blindfolded so that you couldn't see the colors or the numbers of any of the balls. The fact that the 5 balls happen to also form subgroups of odds and evens, or primes and nonprimes, has nothing at all to do with the probability of you picking any particular ball.

The only correct answer to the real problem is 2/5.

[Anonymous No.12268809]

What? If you pick out a gold ball, the third box is irrelevant.
Your chances of picking another gold depending on the gold ball you pick first are 1, 1, and 0.5. Average them out and it's 2/3.

[Anonymous No.12268881]

>>12268799
As per the original question:
" [...] what is the probability that the next ball you take from a box will also be gold?"
This implies that the balls are still in the boxes.

The balls are still in the boxes.
Each box doesn't have the same number of balls.
Therefore, each ball doesn't necessarily have the same probability of being picked as a ball from another box.
You don't know where the balls are, because you can't see.
You don't know which box contains two balls and which box only contains 1.
You can't just select a ball from the set of five balls because you can't see them, you don't know where they are.
To pick a ball you have to put your hand in a box, not knowing how many balls are in it.

To make this more easy to understand, I will change the problem a bit:

Box 1 has 1000 gold balls and 1 silver ball.
Box 2 has 1 silver ball.
Box 3 has 1 gold ball.
You can't see in the boxes, so you don't know which box has what.
You pick a ball from a box, what are the chances of it being gold?

You could say there are 1001 gold balls out of 1003 balls in total, making for a probability of 1001/1003.
But all balls don't have the same probability of being picked. Again, you don't know where the balls are, so you have to put your hand in a box and pick a ball. If you put your hand in box 1, each ball hasa probability of 1/1001 of being picked. Where as the ball in box 2 has 100%, same for the one ball in box 3.

If we were to iterate this, approaching infinity, with replacement, 2 times out of 3 the ball you would pick would be the only one from its box.
And 1 time out of 3, you would pick from the box with 1001 balls, each ball having a 1/1001 probability of being picked.
More simply:
1 time out of 3, the ball is gold
1 time out of 3, the ball is silver
1 time out of 3, the ball is either 1000/1001 gold or 1/1001 silver.
Because you can't see where the balls are. You can't individuate their probabilities.

[Anonymous No.12268883]

>>12268809
the question is ambiguous and there are two answers depending on how you interpret the question. 2/3 is correct, yes.
see >>12268142

[Anonymous No.12268900]

>>12246941
Wouldn't this require a permutation/combination probability? It's easy to solve but my brain stopped working years ago.

[Anonymous No.12269123]

>>12268881
>The balls are still in the boxes.
The fact that the balls are in boxes only matters if you imagine some sort of physical constraint that doesn't exist in the real problem.
Maybe the boxes have lids that you have to open.
Maybe the boxes have high walls that prevent you from simply reaching out and picking a ball.
Maybe.
If you add a new, arbitrary element to the problem that forces you to pick a box before picking a ball, sure, it would have (mostly*) the same effect as adding "first you pick a box" to the second pick instructions. But again, that would be a different problem. No such element exists in the actual problem.

(*mostly. If we're allowed to interact with our physical environment, you'd also have to add a caveat that we're not allowed to feel inside each box before picking a ball, because we could then always pick the box with the single ball, giving a probability of 2/3.)

>you can't see them, you don't know where they are
Try to explain the difference between the actual problem as written, and the analogous blindfolded pool ball pick I described in the last post, without adding any new, arbitrary elements.

[Anonymous No.12269186]

>>12269123
>Maybe
>Maybe
>Maybe.
It doesn't matter if there is a lid or not, it doesn't matter whatever fuckin loophole you're trying to exploit.

>Try to explain the difference between the actual problem as written, and the analogous blindfolded pool ball pick I described in the last post, without adding any new, arbitrary elements.
That's what I've been doing all along.
The difference is that in your problem, every ball has the same probability of being picked.
The numbers on the balls are irrelevant because they don't affect the pick.
You can individuate the probabilities because they're independant.

The original problem IS different, because the box you happen to put your hand in WILL affect the pick.
The balls are in the boxes, which means you need to reach into a box to pick a ball.
I'm not adding this, this is just the problem as it is written.
" [...] probability that the next ball you take from a box [...]?"

You can't see what's in the box.
You don't need to choose the box, you could do it blindfolded.
But to pick a ball, you have to put your hand in a box.
You can't see what's in the box, so you can't just select 1 ball out of the 5, because you don't know where they are.
You just have to pick a ball from some box, not knowing how many other balls there are in it.
The probability of a ball being picked depends on how many other balls are in the same box.
The box you happen to put your hand in WILL affect the pick.
Again try to think about problem if we were to iterate it infinitely.

[Anonymous No.12269368]

>>12268500
Impressive if bait, impressively retarded if not

[Anonymous No.12269421]

>>12269368
truly a sight to behold

[Anonymous No.12271100]

>>12269186
>fuckin loophole you're trying to exploit
I'm not the one trying to alter the problem with an imaginary loophole where the boxes have magic walls, or a lid, or are somehow physically intrusive on your hand as you reach out to pick any one of the 5 balls. That's what you're doing. That's your mistake. I'm explaining your mistake.

>The numbers on the balls are irrelevant because they don't affect the pick.
Neither do the boxes. Nowhere in the actual problem does it say anything about magic walls, a lid, or whatever other arbitrary physical constraint you're imagining. There is no difference.

>You can't see what's in the box.
Correct.
>You don't need to choose the box, you could do it blindfolded.
Correct.
But to pick a ball, you have to put your hand in a box.
Correct.
>You can't see what's in the box,
Correct.
>so you can't just select 1 ball out of the 5,
Wrong. That's exactly what you're doing. Reaching out and picking 1 of 5 balls.
The fact that your hand also crosses into some intangible orthotope of air—which may or may not be shared by any number of other balls—is completely irrelevant. The only way the box is relevant is if you're either specifically instructed to choose a box first, or if the contour of the box somehow acts as a physical impediment to simply reaching out your hand and picking any 1 of 5 balls.
You know as well as I do that neither of those constraints exist in the real problem, otherwise you'd surely be able to quote that part of the text.
>because you don't know where they are.
Lol, what? So when your hand crosses into the airspace of a box, there's a magic radar that suddenly starts pinging the location of the balls in that box? That's even sillier than the lid or whatever you were previously imagining.

[Anonymous No.12271102]

>>12269186
>You just have to pick a ball from some box
Correct. The balls are all in boxes, just as 5 balls on a polygonally patterned carpet are all in polygons. Just as 5 pool balls are all in parity groups.
>, not knowing how many other balls there are in it.
Correct.
>The probability of a ball being picked depends on how many other balls are in the same box.
Wrong.
>The box you happen to put your hand in WILL affect the pick.
Wrong.
Again, you keep making the same unsubstantiated leap of logic: that these boxes have some sort of magical or physical property that prevents you from simply reaching out and picking a ball.
That's your own imaginary rule you're adding.
We both know that you can't support your imaginary rule with a quote from the actual problem—otherwise you'd have done so already.
>Again try to think about problem if we were to iterate it infinitely.
Again, I understand perfectly the imaginary problem you've imagined, and the imaginary rule you're trying to shoehorn into the real problem.
I understand it even better than you do, apparently, since I keep highlighting your exact mistake, and yet you keep rewriting your same mistake over and again without correcting it or addressing it.

[Anonymous No.12271499]

>>12271100
>>12271102
I will do this one last time

> boxes have magic walls, or a lid, or are somehow physically intrusive
It doesn't matter. The boxes could be ethereal. The physical context is irrelevant.
The box only acts as a container for a sub group of the set, and hides its content.
This information only defines the parameters of the problem, nothing else.

>So when your hand crosses into the airspace of a box, there's a magic radar that suddenly starts pinging the location of the balls in that box?
No. You don't know where the balls are, because you can't see them.

The boxes could have walls or not, they could be completely open.
The boxes could be juxtaposed or miles away from each other.
It doesn't matter.
The problem perfectly states that the balls are DISTRIBUTED in a particular way in SEPARATE boxes.

" [...] probability that the next ball you take from a box [...]?"
"next ball you take from a box"
from A box.

You can see the box, but you can't see inside it.
You pick a ball from A box, without knowing what's inside it.
Again, the physical context is irrelevant.
But while your hand is in a box, you can't pick a ball from another box.
It's not about physical constraint.
If this were to be written as a program it would behave the same way.
The containers are separate and don't share their content.

Separate boxes. You can't see their content.
You need to pick a ball from A box.
What's in the fucking box?
No idea. The balls were distributed unevenly.
The fact the you need to pick something from a separate container without being able to see what's inside it means you have to pick from A box, a random box.

The point of the boxes is simply to distribute the balls in separate subgroups.
This is just a normal conditional probability problem.

You're the one who puts it in a physical context and ignores the boxes because it wasn't specified that they had material walls separating them.
You ARE exploiting an imaginary loophole.

[Anonymous No.12271546]

>>12246941
Since we chose a gold ball, that implies that we chose from either the first box or the second box. We can ignore the third box completely.
What is the probability of the gold ball being from the first box?
2/3.
What is the probability of the gold ball being from the second box?
1/3.
If we choose from the second box, then grabbing the next ball in that box will result in a silver ball.
If we chose from the first box, then the second ball is certainly a gold ball.
Choosing from the first box has a probability of 2/3, and swapping results in a probability of 1 (certain) of getting the next ball.
Thus 2/3(1) + 1/3(0) = 2/3

[Anonymous No.12271684]

>>12271499
>No. You don't know where the balls are, because you can't see them.
Then how do you pick one, ever?
Your argument is that not being able to "see" the balls somehow prevents you from moving your hand toward a ball when your hand is outside a box, but doesn't prevent you from moving your hand toward a ball when your hand is inside a box.
This only makes sense if you're imagining some sort of magic radar that only turns on if your hand is inside a box. In other words, it's nonsense.

>But while your hand is in a box, you can't pick a ball from another box.
Sure, and while your hand is hovering over a polygon, you can't pick a ball in another polygon.
And while your hand is closer to an even pool ball, you can't pick an odd pool ball.
Yes, whenever your hand is closer to any arbitrary group of balls than it is to any other arbitrary group of balls, you can only pick a ball from the closest arbitrary group.
You're arguing that the airspace of the boxes act as some sort of Zenonian paradox, without realizing that the parameters of your own argument would disallow any sort of subgroup probability altogether.
Yes, the argument is vacuously true, but it's also trivial and degenerate.

>wasn't specified that they had material walls separating them
What kind of box doesn't have material walls? That's clearly not the point.
The point is whether or not the walls somehow interfere with you reaching out and picking 1 of 5 balls you can't see, any differently than the shapes on a carpet or the numbers on a pool ball interfere with you reaching out and picking 1 of 5 balls you can't see. They don't, unless you add your own imaginary rule that they do.

[Anonymous No.12271688]

>>12271499
>You ARE exploiting an imaginary loophole.
No, I'm pointing out the imaginary loophole that you're trying to exploit.
I have no problem simply quoting the exact text as is—

"You pick a box at random. You put your hand in and take a ball from that box at random."
"If it's a gold ball, what is the probability that the next ball you take from a box will also be gold?"
"Note: You can't see into any of the boxes."

—because I'm not the one imagining rules that don't exist.
You try it.

[Anonymous No.12271788]

>>12271684
>>12271688
You are physically contextualizing the problem. It's not about walls or carpets or if you can just move your hand.
It doesn't matter if you can't really realistically move your hand towards a ball because you can't see it.
That again is you playing with the physical loopholes.
There's no magic radar.
It's about a set of distinct objects distributed in distincts subgroups.

You don't know the content of the subgroups.

" [...] probability that the next ball you take from a box [...]?"
"next ball you take from a box"
from A box.

You pick a random, distinct object, inside a random, distinct subgroup.

The point of the boxes is simply to distribute the balls in separate subgroups.
This is just a normal conditional probability problem.

The point of the boxes is simply to distribute the balls in separate subgroups.
This is just a normal conditional probability problem.

[Anonymous No.12272116]

>>12271788
>You are physically contextualizing the problem.
No, you are. What I'm doing is pointing out that that's what you're doing. If you think you're not, it should be easy for you to explain your own argument without physically contextualizing it.
You agree that
>>12269186
>The numbers on the balls are irrelevant because they don't affect the pick.
Your argument is then—to use your own idiom—that we can "individuate" numbered pool balls, but that we can't "individuate" balls in a box.
Try to explain why, without "physically contextualizing" your argument.

>This is just a normal conditional probability problem.
Yes, and the answer to the problem as written is 2/5.

[Anonymous No.12272127]

>>12271788
>You are physically contextualizing the problem.
No, you are. What I'm doing is pointing out that that's what you're doing. If you think you're not, it should be easy for you to explain your own argument without physically contextualizing it.

You agree that
>>12269186
>The numbers on the balls are irrelevant because they don't affect the pick.

Your argument then—to use your own idiom—is that we can "individuate" numbered pool balls, but that we can't "individuate" balls in a box.
Try to explain why, without "physically contextualizing" your argument.

>This is just a normal conditional probability problem.
Yes, and the answer to the problem as written is 2/5.

[Anonymous No.12272248]

>>12272127
I kept showing you how you were playing with imaginary physical constraints.
There is no magical radar, there are no physical constraints.
No box, no hand.

But you still try to tell me the subgroups don't matter because physical constraints were not defined.

It's about picking a random, distinct object, from a random, distinct subgroup.

The problem tells you about boxes to indicate the the balls are distributed in separate subgroups.
This is a typical conditional probability problem.

If you had learned conditional probability theory, maybe you would understand.
But you just choose to interpret the problem to your liking and exploit contextual loopholes.
You would fail this problem if it were on an exam.

A wise anon said:
>impressively retarded if not
I am inclined to agree.

You are helpless. Still, I tried very hard to show you.

But I don't care anymore. I am done with you.
I envy you, anon. I do.
Sometimes I wish I were an autistic brainlet like you.
It seems envy is my sin.

You don't need to pick a ball from a box.
Just let your hand flow with the dao and the balls will come flocking to you like bees to a cherry tree.

It doesn't matter what's in the box.
The box is not real.
I am the box.
I am.

[Anonymous No.12272335]

Just simulated 10000000 random picks from the boxes. The answer is 2/3 with the assumption that the second pick is from the same box as the first. The answer is 2/5 if the second pick is from any of the three boxes.

[Anonymous No.12272347]

>>12272248
>You would fail this problem if it were on an exam.
The other, more accurate, way to understand what's happening, is that if an exam asked you to compose a conditional probability problem, you would fail that part of the exam.

[Anonymous No.12272354]

>>12272335
show your code. your first answer is correct, but the second one is wrong.

[Anonymous No.12272358]

>>12272335
>composing the simulation wrong
It's 4/9, you can't directly select balls.

[Anonymous No.12272359]

>>12272358
this

[Anonymous No.12272360]

>>12272354
>>12272358
>>12272359
>4/9 truthers
Lol, nope.

[Anonymous No.12272368]

>>12272360
I am going to break your neck open like a pez dispenser and shit down your throat.

[Anonymous No.12272388]

>>12272127
The gods of empirical demonstration have spoken.
100000000 random picks.
The result approaches 4/9.
Full code and answer in pic related.

This is the end for you, anon.
You can't escape the box.
You might understand someday.
We are the balls and epiphany simply comes to pick one of us at random.

[Anonymous No.12272389]

Guys, 2/5 balls just flew over my house!

[Anonymous No.12272416]

>>12272388
>bl2 = random.choice(random.choice(boxes))
Oops, you coded your own imaginary problem instead of the real problem.

[Anonymous No.12272424]

>>12272416
>he's fuckin retarded

[Anonymous No.12272437]

>>12246941
What’s the point of putting them in boxes in groups of two? How does it change the equation if they’re all in a pile?

[Anonymous No.12272441]

>>12272424
Subject ambiguity interpreted in my favor and concurred.

[Anonymous No.12272611]

Okay guys, enough bickering over 2/3 and 4/9. The wording of the question leans towards 4/9, but it is somewhat shaky. Let me explain:

In the case where the probability is 2/3, you pick a box at random, and find that it has a gold ball. The probability calculated shows the likelihood of this box being the one with 2 gold balls (i.e. when you can take another gold ball out from the very same box you selected at the start), given that the box you selected had at least 1 gold ball in it.

In the case where the probability is 4/9, you pick a box at random, and *remove* the gold ball from it. After this, you once again pick a box at random (can be the same box as the first time, or a different one) and then see if it has a gold ball in it, given that you definitely removed a gold ball from the box you first picked.

Now, this is the exact wording of the question:
>You pick a box at random. You put your hand in and take a ball from that box at random. If it's a gold ball, what is the probability that the next ball you take from *a* box will also be gold?

Notice that the question states that the next ball must be selected from *a* box, not *the* box. This means that another box must be picked at random, rather than the same box again. Hence, 4/9 is indeed the right answer.

[Anonymous No.12272683]

>>12272611
*a* box does not imply it must be chosen randomly. The problem explicitly stated that the first pick must be random, but the second pick must be from *a* box. Nowhere does the second pick imply a random pick. The player has knowledge after the first pick, and so can act intelligently to get 2/3 instead of 4/9.

[Anonymous No.12272717]

2/3 = big brain
4/9 = average brain
2/5 = little brain
1/2 = rat brain

[Anonymous No.12272823]

>>12246941
Yeah I remember when Killworth, Young and Selfridge proved it to be indipendent of ZFC a couple of years ago in their celebrated KYS theorem.

[Anonymous No.12272959]

>>12272683
The box is not the one that is picked at random, it's the possible content inside it.
You may have chosen the box, it doesn't matter, because you didn't choose its content.
1/3 probability for every distinct subgroups to be the one you picked.
The balls are still in the boxes.
You can't see in the boxes.
You pick a ball from a box not knowing what's really inside it.
The pick is implicitely random since you don't know the content of the boxes.
The text perfectly tells you to pick something from *a* distinct subgroup.

[Anonymous No.12273010]

>>12272441
Tell me anon, have you successfully completed probability course that is at least of undergrad level?
Yes or No?

[Anonymous No.12273041]

>>12273010
Tell me, anon, do you always flirt with anonymous father figures?
Or am I special to you?

[Anonymous No.12273159]

>>12247048
This is the correct answer. You already chose a gold ball so that means the silver/silver box is off the table and you either picked from the gold/gold or the gold/silver, leaving the only two options to be you pull a gold from the gold/gold or a silver from the gold/silver so its a 50/50 chance.

[Anonymous No.12273194]

>>12273159
It doesn't Express the second ball taken is from the same box as the first

[Anonymous No.12273289]

>>12273041
Answer the question anon.
I only want to expose your autism to the world

[Anonymous No.12273421]

3/7

[Anonymous No.12273578]

>>12257051
[eqn] 2^{ \aleph_0} [/eqn]

[Anonymous No.12273806]

I may be misunderstanding the question, so I'll work through both interpretations:
1.after taking the gold ball you shuffle all of the boxes and pick out of a random box, probability 2/5
2.after taking out the gold ball you put your hand back into the same box and take out a random ball. probability 2/3

[Anonymous No.12273817]

>>12273806
>probability 2/5
probability 4/9

[Anonymous No.12273821]

>>12273159
If you are picking the second ball from the box, which is not explicitly stated.

[Anonymous No.12273832]

>>12273806
Nope.

Case 1: You either picked box A or B, equally likely.

If you picked box A initially, then you have (A) 100%, (B) 50%, or (C) 0% chance of drawing gold, or 1/2. If you picked box B initially, then that box is now depleted of gold balls, so its 100%/0%/0%, or 1/3. Since each of those situations is equally likely, the total odds are 1/2×1/2+1/2×1/3=5/12, or 41.7%.

[Anonymous No.12273835]

>>12273832
Hit post too early.

Case 2: if you are drawing from the same box AND know that the box had at least 1 gold ball in it, then it's a straight up 60% chance. None of this 2/3 or 4/9 crap.

[Anonymous No.12273837]

>>12273835
Dammit. 50%, not 60%. Stupid thumbs.

[Anonymous No.12273911]

>>12246941
OK so we don't know what's in the contents of any boxes and have no ideas at first, so we pick at random.
We get a gold ball, at random. What's the probability the next ball you take from a box will be gold too?
It depends on how you proceed.
If you choose randomly again from the 3 boxes after the first draw, it is a 2/5 chance you will get gold again, I believe, because there are 5 balls left, and 2 are gold. If you are trying to get gold again, you are making it less likely by choosing randomly from the boxes a second time.
If you choose from the box you already drew from, it must be a 2/3 probability the other ball in the box you chose at first is gold too, because you know it's not the silver box, therefore it's eliminated 2 silver balls. By drawing from the same box which you know had at least 1 gold ball in it, you know there are 2 gold balls left and 1 silver that you could possibly get.
The question does not specify how you choose after the first time, so it's up to you.
Thanks for the fun thing, OP!

[Anonymous No.12273928]

>>12273911
You, like so many on here, are ignoring that before you pick the second ball you must also pick a box. This is not the same as simply finding the probability of drawing a second gold ball.

[Anonymous No.12273960]

>>12273928
Draw from the same box for higher chance of another gold ball, like I said.

[Anonymous No.12273970]

>>12273832
>>12273835
>>12273837
picking A or B isnt equally likely:
if you got a gold ball, that could have originated from you picking gold ball #1 in box A, gold ball #2 in box A, or the gold ball in box B. Two of those situations come from box A so 2/3 you pick a gold ball, it's from box A. Then you can weigh each situation: 2/3 * probability of getting another gold from box A = 2/3 * 1
then add 1/3 * probability of getting another gold from box B = 1/3 * 0

[Anonymous No.12273977]

>>12273928
if you pick a random box then it literally is the same as simply finding the probability of drawing a second gold ball

[Anonymous No.12273980]

>>12273960
Yeah, but you're just using it to weight the probability of selecting a given ball. If I am drawing from the SAME box, then I have 100% chance of drawing the only remaining ball. If I draw from a NEW box, then it's 50% per ball. That weights the probabilities in a nonuniform way. Your numbers assume a uniform weighting.

[Anonymous No.12273987]

>>12248588
No, because the probability of picking that box initially depends on how many golds there are in there

[Anonymous No.12274011]

>>12272683
The distinction is between the use of 'a' and 'the'. If it was instead written as follows:
>what is the probability that the next ball you take from *the* box will also be gold?
Then you would have to choose the same box. Since it says that you must pick from 'a' box, and this is a probability question, it implies that the next box is chosen at random.
Regarding your last sentence, this is a probability question so the implication is that every choice made by the 'player' is completely random. Also, you are assuming that the player wishes to get as many gold balls as possible when they make their decisions, which may not be the case - nowhere in the question is it also stated that the player wants to get more gold or more silver, just that they randomly choose boxes and the balls from within.

[Anonymous No.12274015]

100%. I will just buy a box of gold balls and take a ball from that box.

[Anonymous No.12274054]

>>12273980
I don't really know what you're talking about.
The question doesn't tell you what you have to do next, it is simply asking what the probability of you drawing another gold ball is. That probability changes depending on if you choose randomly from all the boxes again or if you choose from the box you already drew from.
If you draw completely randomly after the first draw it is a 2/5 chance you will get gold again. If you draw from the same box it is a 2/3 chance, because you know the box you already drew from isn't the silver one, leaving a possible 2 gold and 1 silver for the one you draw to possibly be.

[Anonymous No.12274083]

>>12273970
Good catch, dumb mistake. That makes it 4/9.

[Anonymous No.12274109]

>>12246941
>You put your hand in
This right here makes the problem a science problem not a math problem

>putting your hand in allows you to touch the metal, you don't have to pick it up.
>is the ball real solid metal?
>if so can you learn to differentiate between the conductivities of these balls?
if yes to both, as long as you're not a total brainlet, then the answer is 100%.

[Anonymous No.12274117]

>>12274109
Don't be a retard. It's a hypothetical. Assume the balls aren't made from gold or silver, assume they're painted or something. The question isn't about what the metals are, it's a question meant for math.

[Anonymous No.12274128]

>>12274117
it's actually a question meant to bati.

[Anonymous No.12274405]

4/9 if you're an autistic probability theorist
2/3 if you're a gold loving patrician
1/3 if you're a silver loving plebeian.

[Anonymous No.12274422]

another brazed tread, bumping for bazed thlead

[Anonymous No.12274431]

>>12273289
Lol, the brain of a 4/9 truther.

[Anonymous No.12274437]

>>12274405
>4/9 if you're an autistic probability theorist who can't read
Fixed.

[Anonymous No.12274440]

>>12246941
dis q is always asked poorly. its a problem with the language.

2/3 because the first box has 2 gold balls and i could have picked either ball

which means there are 2 possible outcomes in the first box and 1 possible outcome in the silver.

[Anonymous No.12274552]

4/9

[Anonymous No.12274870]

>>12272683
This isn't game theory, there's no win or loss, so there's nothing to act intelligently about. There's no reason you'd prefer two gold balls over a gold and a silver. It's simply asking the likelihood of a certain outcome. And given that you have no preference, the only thing you can do is act randomly.

[Anonymous No.12274969]

>>12274870
I don't understand why they keep thinking you're supposed to get the most gold balls
Anons be stupid

[Anonymous No.12274973]

>>12274431
shut the fuck up alex

[Anonymous No.12276517]

>>12246941
I don't know if this is a solution, but calculating a probability upon a probability might be a tentative solution.

[Anonymous No.12276615]

>>12274973
who is alex?

[Anonymous No.12277364]

>>12252846
even if this is bait its fucking faggotry . i actually wanted to know the answer.

[Anonymous No.12277375]

>>12253562
fuck wit how long did that take you. fucking ass

[Anonymous No.12277411]

>>12268500
this mother fucker really took this amount of effort to troll like this, is this climate science in action ?

2/5 because there are 5 balls and 2 are gold ?

but he emphatically ignores the fact that the balls are not free existing but a couple together in pairs, so that in the first pick you could have picked from the 2 golds

which ends with
2/3 chance of
100 percent (1gold) , 50/50(1gold 1sivler,. 0 percent ( 2 silvers)

or
1/3 change of
100 percent (2 gold), 0 percent (1 silver), 0 percent( 2 silver) .

i dont know how to multiply this many probabilities you fuckers do it for me .

[Anonymous No.12278090]

>>12277411
>effort to troll
In all seriousness, if this question were on a test, the answer would be 2/5. The 4/9 poster is basically arguing that they took a class in arithmetic and learned how to add.

2 + 3 = 5
Therefore:
2 x 3 = 5 because + and x are the same thing: a binary operator composed of two lines.

No, if you answered 4/9 on a multiple choice test, you would get no credit.
If you answered 4/9 and showed your work, you would probably get partial credit.

You don't have to worry, though, because this question would never be on any test. Because people who actually know the subject material, and the English language, fluently enough to be in a position to write an English language test, would never write that question.

If you want to learn basic probability, my first advice would be to not seek help on 4chan—and more specifically, to avoid bait from people who think that a semester of "undergrad level" math qualifies them for anything other than troll farm labor.

Read an actual textbook instead. This is a good one, and the full pdf is somehow being hosted for free on a random dot edu in Iran. Maybe they shot the DMCA inspector. In any case, take advantage while you can.

http://ce.sharif.edu/courses/97-98/1/ce181-1/resources/root/Text_Books_References/Papoulis_Pillai_Probability_RandomVariables_and_Stochastic_Processes-4th_Edition_2002.pdf