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/sci/ - Science & Math


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7979124 No.7979124 [Reply] [Original]

This basic probability question was posted on the board /b/.
Since so many retards and autists struggled with this, I'm taking this to the maths board to see if the stupidity lessens.

>> No.7979150

>>7979124
It's 50-50, since you can rule out the grey box. It's also a feint of the Monty Hall problem, which is probably where the stupidity comes from. The difference is so slight that it's almost unnoticeable.

>> No.7979161

If there is a gold ball in the box, what is the probability of the other one being gold.

[math]P(G|G) = \frac{P(G \wedge G)}{P(S \wedge G) + P(G \wedge G)} = \frac{1/3}{1/3 + 1/3} = \frac{1}{2}[/math]

Babby's first intro to prob and stats.

>> No.7979174

May I add

https://en.m.wikipedia.org/wiki/Bertrand%27s_box_paradox

>> No.7979179

>>7979174
Yeah, just use Bayes'

>> No.7979193

The frequency of 50/50, 50% and 1/2 answers were so high on /b/ that it was almost painful.

The answer is 2/3.

>> No.7979206

>>7979124
You already know the box you picked has at least one gold ball, so the third box is null.

Now, you picked either the 2 gold ball box or the 1 gold ball box.

You had a 2/3 chance of doing so but the question is asked after the fact, making this probability null.

Now it is a clear 1/2. Think of it this way, it is like the box with 2 silver balls never existed in the first place.

>> No.7979209

>>7979193

How do you figure, sports fan?

>> No.7979212

>>7979193
It would be 2/3 if you picked the ball at random, but you pick the box at random before you pick the ball at random.

>> No.7979232

There are 3 boxes. The third box with SS is eliminated since a gold ball can't be grabbed from that one. Now we're left with 2 boxes. Box 1 contains GG and box 2 contains GS. Collect like terms and you get 3G S. It states that you've grabbed one gold ball which leaves you will 2G S. Thus, the probability of grabbing a gold ball next is 2/3.

>> No.7979235

>>7979209
Sports fan?
Rather random false assumption

>> No.7979260

P(G from 1st box) = 2/3
P(G from 2nd box) =1/3

so probability of getting gold again is 2/3, silly kids

>> No.7979283

>>7979232
It says from the same box though.

With a gold ball on the first pick there are two possible scenarios: you picked the first box so the next one will be gold or you picked the second box so the next ball will be silver. Both scenarios are equally likely under the condition that the first ball you picked is gold so the probability of the second ball from a box that contains at least one gold ball being gold is 1/2.

>> No.7979312

>>7979283
Approaching this from the perspective of the chances of each box you could grab a gold ball from, the probability would be 1/2. But, we're finding the probability of grabbing a gold ball from the same box. We don't know which box the gold ball is grabbed from so it's pointless finding box 1's and box 2's individual probability. You then collect like terms from the two boxes since we don't know which box the gold ball was grabbed from. So that's 3G and S. The first ball you grabbed was a gold ball, so now it's 2G and S. The total number of balls is 3 of which 2 are gold. Thus, the probability of grabbing a gold ball from the same box would be 2/3.

>> No.7979366

>>7979312
nah

>> No.7979372

>>7979312
It is not 2/3 in any way shape or form

You don't understand conditional probability

>> No.7979388
File: 293 KB, 1000x750, 2HneMZQ.jpg [View same] [iqdb] [saucenao] [google]
7979388

>>7979193
Came here to post this.

Clearly 2/3, 50-50 fags can suck my cock.

second gold ball.

>>7979212
>but you pick the box at random before you pick the ball at random.
Doesn't matter, two of three gold balls come from a box with a second gold ball.
You aren't more likely to get the gold with silver ball than either of the gold/gold balls.

>>7979283
>Both scenarios are equally likely
Nope, you're twice as likely to pick one of the gold/gold box balls since there are two of them.
Any of the six balls are equally likely, we've just been told that three of them (the silvers) didn't get picked.
Just imagine you knew the original layout, and that you pick a gold ball.
What retardation would convince you "hey, this is probably the gold ball from the mixed box!"?

>> No.7979402

50/50 obviously.

Greyball box is never relevant to this, since it's already known you chose a box with at least 1 gold ball. So there's a 50% chance you chose a box with 2 gold balls, and a 50% chance you chose the box with only 1 gold ball.

If its anything else I'm retarded

Is this supposed to be hard or something?

>> No.7979429
File: 28 KB, 450x521, 1456518676394.jpg [View same] [iqdb] [saucenao] [google]
7979429

>>7979402
>Is this supposed to be hard or something?
For you (apparently).

>> No.7979441

>>7979150
>recognizes monty hall similarity
>still answers 1/2
Just another stupid post from our resident retarded tripfag.

>> No.7979445

>>7979312
No, you're over complicating

The question is basically asking for the probability you chose box 1 P(A), given that you chose either box 1 or 2 P(B). By conditional probability that's the joint probability of both events occurring divided by the probability of event B, i.e. (1/3)/(2/3).

>>7979388
>you're twice as likely to pick one of the gold/gold box ball
In a general case sure but in this case that's irrelevant because we KNOW that we've picked one of them because of our gold ball. See above.

>> No.7979448
File: 54 KB, 610x288, 1439267451846.jpg [View same] [iqdb] [saucenao] [google]
7979448

>>7979429
>he thinks it's 2/3

>> No.7979455

>>7979441
It's not the month hall problem though numb nuts, that's their point.

>> No.7979456

>>7979455
*monty

Autocorrect

>> No.7979461

>>7979455

I don't get why people think the total number of balls factor into it. It's about the boxes themselves.

Correct me if I'm wrong, but here's how I see it

You know there's two boxes with 3 gold balls out of 4 in them.

If you pick up a gold ball you've put your hand in 1 of 2 boxes with at least 1 gold ball. That leaves a 50% chance you put your hand in the box with 2 gold balls and a 50% chance you put your hand in the box with 1 gold ball.

If you'd picked up a silver ball first things would be different.

>> No.7979462

the people that are saying 1/2 are not takingn into account that you could've picked the S from the 2nd box. Therefore it's twice as likely that the G you picked was from the 1st box. This problem is way easier than the monty hall problem and you retards still can't get it lmao.

>> No.7979474

>>7979455
Yes it obviously isn't the monty hall problem "numb nuts" because if it was there would be goats. But the answer is still clearly 2/3, if you can't figure this you should review basic probability and statistics.

>> No.7979475

>>7979461
Yeah pretty much

>>7979462
The probability that you picked a gold ball first time is irrelevant because it's already happened.

>> No.7979487

>>7979475
No it isn't because we're trying to determine the chances of which of the gold balls you picked are. This is basic probability. It's a 2/3 chance you picked a gold ball from the 1st box.

>> No.7979502

>>7979474
Let's lay out every possible scenario then.

Info: the box we chose contains at least one gold ball

A) we have box 1
B) we have box 2

Neither is more likely than the other and they are the only two options so each of these events gets prob of 1/2

Next level

C) get gold again in scenario A
D) get gold again in scenario B

Prob of C is 1, prob if D is 0.

Our total probability is the probability of getting A AND C OR B AND D. That is (1/2*1)+(1/2*0)=1/2.

>> No.7979506

>>7979461
Let me dumb this down for you. Before you pick a ball the chance you will take one from the box with one silver and one gold is 1/3 right? This is box HALFER

Ok now you take one ball out of the box. According to you if the ball is gold the chance is now 1/2 it came from HALFER. On the flip side if the ball is silver the chance is also now 1/2 it came from box HALFER. So now matter what color the ball is the chance is 1/2 it came from HALFER.

Do you see the problem now? Before you even look at the ball you know it must be gold or silver. So according to you the chance you take a ball from the HALFER box is 1/2 even though the problem says you pick a random box from 3 of them.

>> No.7979507

>>7979502
Slight correction, info should really be "box contained at least one gold ball that has now been removed"

>> No.7979515

>>7979474
It's not equivalent to the Monty Hall problem, either.

The Monty Hall problem involves picking a wrong door and calculating the remaining probability. This is just basic probability where you already know you didn't pick the grey box.

>> No.7979519

>>7979506
The half is a conditional probability

>> No.7979524

>>7979487
>we're trying to determine the chances of which of the gold balls you picked are
Wrong. The questions asks if it's gold or not. The configuration of boxes doesn't factor in because the ball is never identified. This isn't a question of whether or not you should "switch" boxes, it's a simple question of which of two boxes you've already selected.

>> No.7979527

>>7979161
This is a good example of Dunning-Kruger. Smart enough to go through the motions to fool yourself into thinking you know what you're doing but too dumb to actually do it right.

Let A = other coin in box being gold
Let B = the first coin chosen is gold

P(A|B) = P(A) P(B|A) / P(B) = (1/2) (2/3) / (1/2) = 2/3

>> No.7979529

>>7979502
A is twice as likely as B given we've pulled a gold ball.

>> No.7979530

>>7979524
Oh wait no I'm wrong. If you picked a gold ball it's more likely that you've wound up in the double gold box than the grey-gold box. 2/3 is probably about right.

Still not the Monty Hall problem.

>> No.7979537

>>7979502
>>7979515
This problem is fucking trivial and your fucking wrong you stupid faggots go take baby's first probability and statistics. I know what I'm talking about and Im telling you ignorant fucks what the answer is so I already did half the work for you now go study it until you have the right answer. Don't reply to this post unless its to suck my dick for helping you.

>> No.7979544

>>7979537
Okay, but you'll have to help me first.

>> No.7979545

>>7979124
P(G) = 1/2 (prior probability of drawing a first gold ball: half of all balls are gold)
P(GG) = 1/3 (prior probability of choosing the double gold box: it is one of the three)
P(G|GG) = 1 (if the box is double gold, you're sure to get a second gold ball from it)

P(GG|G) = P(G|GG)*P(GG)/P(G) =1*(1/3)/(1/2) = 2/3

>> No.7979555

>>7979530

>2/3 probably about right

How probably? 2/3 chance? its 3/3 exactly right you cheeky fuck


>it's not the Monty Hall problem.

What who said it was the monty hall problem

>It's also a feint of the Monty Hall problem. The difference is so slight that it's almost unnoticeable.(you)

Ok buddy you showed us

>> No.7979559

>>7979555
>you showed us
No I didn't. I don't even have Bayes' rule memorized.

>> No.7979560

Guys the probability is definitely 5/6

>> No.7979574

>>7979150
But on the first draw, it's more likely that you drew a ball from the box with two golds than the box with one gold, so it should be higher than 50%.

If you drew a gold ball, there is a 2/3 chance it came from the first box, and a 1/3 chance it came from second box.

If it came from the first box, the chance of drawing another gold is 100%. if it came from the second box, the chance is 0%, so the answer to the question is 2/3.

>> No.7979604

-1/12

>> No.7979652

>>7979388
>twice as likely to get your first ball from...
>>7979506
>Before you pick a ball...
>>7979574
>but on the first draw....

Do none of you fucking morons know what CONDITIONAL probability means?

God damn you autist circlejerking dickfucks, stop fucking trolling and go back to /b/

>> No.7979789

>>7979559
if you can't intuitively derive it by visualizing intersections of sets why are you posting?

>> No.7979790

>>7979652
I like that you're calling people morons for not understanding conditional probability when it's clear that you are the one who doesn't understand.

>> No.7979797
File: 12 KB, 579x368, chanball.gif [View same] [iqdb] [saucenao] [google]
7979797

#include <stdlib.h>
#include <stdio.h>
#include <math.h>

int main(void)
{
int N=1000000,n,randbox,randball,otherball,m=0,M=0;
int boxes[3][2]={{1,1},{1,0},{0,0}}; /* 1=gold, 0=silver */

for(n=0;n<N;n++)
{
randbox=rand()%3; /* pick a random box */
randball=rand()%2; /* pick a random ball */
if (boxes[randbox][randball]==0) continue;
M++; /* count times we got a gold ball */
otherball=1-randball;
if (boxes[randbox][otherball]==1) m++;
}
printf("m:%d M:%d prob:%f\n",m,M,m/(double)M);
return(0);
}
gives
m:333143 M:499928 prob:0.666382

So in a million tests, we get a gold ball on the first draw about 50% of the time. Makes sense since 3 of the 6 balls are gold. Of those times when we first draw a gold ball, about 2/3rd of the time our second draw is gold.

The answer: 2/3.

>> No.7979810
File: 146 KB, 701x576, 1459811109203.jpg [View same] [iqdb] [saucenao] [google]
7979810

For anyone who has trouble understanding why the original problem is 2/3.
This version makes it more obvious to understand that if you've picked a box at random and then randomly pick a gold ball out of it, it is most likely that you picked from the first box, because if you had picked from the second box, getting the gold ball would have been very unlikely.

>> No.7979814

>>7979652
watch this video on conditional probability https://www.youtube.com/watch?v=cpwSGsb-rTs
don't call people morons when you are wrong

>> No.7979828

>>7979652
Sounds like you're the one who doesn't understand conditional probability.

We're working under the assumption that a gold ball has already being drawn. That means it was either ball 1, ball 2, or ball 3. If it was ball 1 or 2, the next ball will be gold. If it was ball 3, the next ball will be silver.

2 of the three possibilities yield a favorable result, so 2/3.

>> No.7979835

>>7979810
Good job if the 1/2 retards can't understand this I don't know what will

>> No.7979840

There are 6 possible results of this experiment, one for each ball. Balls will be numbered 1-6 from left to right

ball 1: Next ball is ball 2. Result is favorable.
ball 2: Next ball is ball 1. Result is favorable.
ball 3 Next ball is ball 4. Result is not favorable.
ball 4: Not gold so result can be discarded.
ball 5: Not gold so result can be discarded,
ball 6: Not gold so result can be discarded.

That leaves us with three possible results where the first ball is gold. Two gold balls are drawn in two of the three results, so the probability is 2/3.

>> No.7979894

>>7979828
drawing ball 1 or 2 are equivalent outcomes, it's downright stupid to consider them independent drawings.

>> No.7979909

>>7979894
:^)

>> No.7979925
File: 2.49 MB, 1083x1079, 88530921.png [View same] [iqdb] [saucenao] [google]
7979925

>>7979814
>>7979828
This is a classic example of completely ignoring logic in favor of completely irrelevant mathematical calculations.

I'll say it again, the specific gold ball you chose is completely irrelevant in the context of this problem.The condition that you've already drawn a gold ball eliminates every single other relevant possibility except for
A. You picked box 1
B. You picked box 2

For the reference frame this problem sets up, the chance of getting a second gold ball is only dependent on which box you have (read: NOT which ball you have) and therefore there is one positive outcome and one negative. a 50/50 chance.

>>7979810
This is still a 50% probability for the exact same reason jesus christ


You can sit there and show me the math all day long, but any calculation that comes out to 2/3 is including practically redundant outcomes. There is no reason whatsoever to argue this problem except to circlejerk semantics and vaguely defined frames of references that have plagued probability since the dawn of time.

>> No.7979934

>>7979894
am i being memed?

>> No.7979937

another vote for 1/2.

... someone go get some marbles or coins, and do it a couple hundred times already.

>> No.7979939

>>7979925
the probability of having chosen box is not the same as the probability of having chosen box two since box 1 has more gold balls.
There are three possible first draws that result in a gold ball. two of them are from box 1.
So there are three possibilities, we chose ball 1, we chose ball 2, or we chose ball 3. each ball is equally probable to have been picked. there is a 2/3 chance that you picked with ball 1 or 2 and a 1/3 chance that you picked ball 3. refer to >>7979810

>> No.7979964

>>7979925
This has to be bait, no one is actually this retarded

>> No.7979973

>>7979925
Maybe this example will help clear things up for you.
You notice that your bike has been stolen. You know that it must have been a nigger, since white people don't steal bikes.

>> No.7979977

>>7979964
/b/ are all this retarded.

This is the problem with probability theory - you have to think and not just plug numbers into a formula.

>> No.7979980

>>7979977
but this is /sci/ not /b/
would have thought that most people here would know how to think

>> No.7979984

The goat is behind the door you didn't initially choose

>> No.7979992

>>7979441
He definitely is retarded, but you appear moreso.

>> No.7979994

>>7979992
why?

>> No.7979997

>>7979124
I always sucked at these probability problems, 4/9? I got this from 3 cases. Case 1: you pull from the middle leaving 2 boxes with silver balls and 1 with 2 gold balls -> 1/3 of getting gold. Cases 2 and 3 (one for each ball in the first case): you have 1/3 of picking the case containing only gold balls, and 1/2*1/3 = 1/6 of picking the gold ball from the middle (1/3 of picking the case, then 1/2 for picking the gold instead of silver) -> 1/2. Then you average the three together and get 1/3+1/2+1/2=4/9. I could be wrong, though.

>> No.7980000

>>7979124
Enumerate the gold balls from the left and split into three cases where we pick that gold ball : 1) we get ball nr 2 ( gold) , 2) we get ball nr 1 (gold) 3) we get ball nr 4(silver). So 2 out of 3 times that other ball should be gold.

>> No.7980014

>>7979997
You have to pick from the same box twice
>>7980000
nice quads

>> No.7980017

>>7980000
These quads don't lie

>> No.7980018

You pick a specific ball you don't pick a box. You are doing conditioning on the wrong entity.

>> No.7980019

>>7980014
Oh wow, I missed that. Then I get 2/3. If you get a gold ball you most likely (2/3) picked from the right box, so repicking will give you 100% chance of getting a gold ball. You could also have picked from the middle box (1/3) in which case your chance of getting gold would be 0. So, 2/3*(1)+1/3*(0)=2/3. Ok, now that makes sense.

>> No.7980024

>>7979124
It's 2/3.

Effectively you are picking the ball instead of the box, and 2/3 of them are in the box which guarantees the second pick will also be gold.

I should really study probability theory, since I'm not well versed in this stuff.

>> No.7980028 [DELETED] 

I can't stop laughing.

All of you still saying 1/2 please understand why the answer is 2/3. I looked at that question and had the answer within a matter of seconds because it's that simple.
Dear those who say the probability is 1/2 or 50%, this means half, which means your answer tells us that you're equally as likely to grab a gold ball than a silver. Are you retarded? Clearly there's not an even amount of gold and silver balls to grab from, so how on earth can it be a 50% chance? There are 3 balls you could grab and 2 of them are gold. Firstly, forget all of this method shit and already done for you methods you lazy fucks. Look at it. Just look. There is not an even amount of gold and silver balls to choose from. Just because a method pops up on Wikipedia which is correct if you use it properly, but you autists can't so it's better for you to be thorough with your basic logic and hope that it pushes you to realise that a 50% chance isn't possible.
I'm really not the best at explaining the things because when doing math because I don't use words in my head, thus it comes out as a bit of a jumble when I try to put it into words, but I hope even the most autistic of you will understand why it's 2/3. There are many other clearer explanations on this thread so learn from them too.
2 gold balls
1 silver ball
3 balls
2/3
I really do hope the ones answering with 1/2 or 50% (they're the same just in case you're confused, in the same way that you're confused with this question) are 18- because I'm definitely 18- and if anyone 18+ on the maths board can't see why the probability is 2/3 then I'm quite worried. Supposedly math majors worked their brains (well, what's left of them) to the probability 1/2. Help.

"Two things are infinite: the universe and human stupidity; and I'm not so sure about the universe."
-Albert Einstein

Dear those who still persist that the answer is 1/2, thank you for keeping one of Eintein's quotes applicable to use to this day.

>> No.7980030

>>7979810
Thank you for making me slightly less retarded, I appreciate it greatly.

>> No.7980031

How do you calculate the odds of one of two mutually exclusive events occurring at a given time?

Say there is a 50% change the earth will explode and a 50% chance it will freeze on a given day. What are the odds of one of the two happening on a given day?

>> No.7980034

>>7980028
Hurr durr human stupidity is only infinite if humans last forever and not everyone has taken statistics so someone could be confused. Fortunately I remembered that 2>1 so your explanation is enlightening.

>> No.7980036

>>7980028
You know... Some people sometimes pretend to be stupid to make you either angry, sad or overconfident. You obviously seem to be in the third category there.

>> No.7980037

>>7980034
2>1
What has that got to do with what I explained?

>> No.7980039

>>7980031
It depends if you know wether they can have overlap or not. Can the earth both freeze and explode on the same day? Make a venn-diagram.

>> No.7980042

>>7980036
'some people pretend to be stupid'

And there lies the problem.

Also, if I know an answer is right, why would I agree with a wrong answer just to seem that I'm not overly confident? What's wrong with you? So you most likely couldn't get the answer correct so you've decided to comment on how I approached the explanation? Bravo.

>> No.7980043

>>7980039
They're mutually exclusive here.

>> No.7980044
File: 10 KB, 302x329, 1363829229715.jpg [View same] [iqdb] [saucenao] [google]
7980044

>>7980028
lol look at this autist

>> No.7980046

>>7980044
Take a good look at yourself first, m8

>> No.7980049
File: 152 KB, 312x333, 1363826195985.png [View same] [iqdb] [saucenao] [google]
7980049

>>7979527
You're really smart for having the right answer after about 20 others did it before you. Respect man, yuo vrey smrt.

>> No.7980051
File: 5 KB, 431x338, earth_cats_venn_diagram.png [View same] [iqdb] [saucenao] [google]
7980051

>>7980039
>>7980031
Green is that none happens, red is explode only, blue is freeze only, yellow is both explode and freeze. Sum of all possibilities give us Green+Red+Yellow+Blue = 1 and we know red+yellow = 50 percent, blue+yellow = 50 percent. Now you can use linear algebra to figure out how much and what kind of information is missing.

>> No.7980052

>>7979124
Assuming you're not factoring in the chances of picking a golden ball on the first pick, 50% or 1/2.
Assuming you are factoring that in, I don't know, maybe 2/3.

Most of the arguing in this thread would be gone if the question just stated whether or not you're factoring in the chance of the initial pick.

>> No.7980054

>>7980044
Feynman's rolling in his grave. There's still some hope for you though - Paul erdos didn't think monty Hall was 2/3 until he saw a computer simulation of it running.

>> No.7980055
File: 16 KB, 160x160, 1361964689282.jpg [View same] [iqdb] [saucenao] [google]
7980055

>>7980046
>N-no u!
Yeah you have autism faggot.

>> No.7980056

>>7979260
The probabilty of getting a box with gold would be 1/3, but this is dismissed as the box is chosen and ss drops out. You take one out of the box you are currently in, making either box G or box S. Since you have alreafy determined the box, the probability is 1/2

>> No.7980060

>>7979527

How do you calculate P(B|A) =2/3?

>> No.7980061

>>7980042
That's not the problem. The problem is that you don't know you need to adjust your behaviour to make those types of people stop wanting to fuck with you.

>> No.7980062

>>7980055
It's nice to know that I'm one of your kind.

>> No.7980064

>>7980061
I don't need to adjust my behaviour because I simply couldn't care if they fuck with me or not.

If there are some people who genuinely think the answer is 1/2, they could possibly learn from what I've explained even if it was a response to someone pretending to be stupid.

>> No.7980066

>>7980056
:'( draw it out mate. It's very helpful for first year probability students.

>> No.7980068

>>7980064
Well you should care as If they are just fucking with you then you are wasting your time trying to explain to them. Time you could instead go hone your skills to get good enough at something useful to back that self-image up.

>> No.7980069

>>7980062
You really are autistic with your wall of text and edgy Einstein quotes you know. I'm trying to help you here.

>> No.7980070

>>7979124

1/2*0 + 1/2*1 = 1/2

>> No.7980072
File: 145 KB, 701x576, boxes_experiment_numbered.jpg [View same] [iqdb] [saucenao] [google]
7980072

>>7980066
I'm not him but here's a numbered example that goes with >>7980000 this explanation.

>> No.7980073

>>7980068
What self-image exactly?
Shall I just agree with the people who say 1/2 just because they believe that someone knowing the answer is 2/3 is someone who thinks they have a huge self-image?

>> No.7980076

>>7979925
Right but choice A is 2/3, choice B is 1/3.

Rephase the problem.

Suppose you have three boxes. Box A has 99 Gold, 1 black.

Box B has 1 gold 1 black,,
Box C has 2 black.

You pick Gold.

What is probability that the next ball is gold?

I am going to say its more than a half because I am not a fuckwit fuck you all fuck you all and your stupid fucking arguments.

>> No.7980079

>>7980073
You should accept that some are stupid and you can't help them because they don't want to learn and that some are just wasting your time on purpose. If you really are so smart you think you are... Use your time to get real good at something that matters and not just inflate your ego by arguing about silly toy examples.

>> No.7980080

>>7980076
Actually I reconsider it. Box A and B are both 50/50, it doesnt matter how many gold balls are in A because its still a 50/50 chance of reaching into either box and picking the ball.

>> No.7980083

>>7980079
It's not to 'inflate my ego', it's to try and help them learn.

But whatever, I'm leaving this thread to run without my input now.

And for the last time, the answer is 2/3.

>> No.7980089

>>7980080
It's not a 50/50 chance of reaching into either box. Remember that we are assuming you have already picked a gold ball, so there's a 2/3 chance you picked from box 1.

>> No.7980091

>>7980083
But its not.

Consider this problem.

There are 2 boxes.

Box A has 99 Gold balls, 1 Black.
Box B has 1 Gold, 1 Black.

You put your hand in a box at random and pull out a gold ball. What is the probability you put your hand in box A?

For anyone still following the thread, this is essentially what the problem boils down to. From a likelihood perspective I suppose it is more likely that it came from Box A.

So I am wrong.

>> No.7980092

>>7980089
Yup I changed my answer again.

Its been a while since I have done stats. I should stop posting.

>> No.7980095

>>7980000
Quads of truth

>> No.7980099

>>7980083
Yes I know the answer but you aren't helping anyone. That attitude scares away people who might actually want to learn and it encourages people who have no intent on anything but to waste your time and make you angry / overconfident.

>> No.7980107
File: 1.10 MB, 2448x3264, IMG_20160405_095546.jpg [View same] [iqdb] [saucenao] [google]
7980107

Contradiction for all you 1/2 sausages

>> No.7980110

There are three possible gold balls I can select. Regardless of which gold ball I randomly select, I will always be left with 3 other balls - 2 out of 3 them will be gold.

>> No.7980116

>>7980110
right but once you pick a box you are stuck with it, so it does matter if you choose box B. your argument needs work.

>> No.7980132

>>7979124
Do you take the ball out of the box and keep it out, or you put it back inside? Because I'm starting to think that's where most confusion comes from.

>> No.7980133

>>7980000
you are not getting the probability of when you get gold at first
just what is the probability of the next ball being gold AFTER YOU ALREADY GET GOLD THE FIRST TIME

meaning the problem doesn't count if you get silver as a first ball

>> No.7980135

>>7980107
>getting baited this hard

>> No.7980137

Intuition says 1/2 wikipedia says 2/3, I'll side with wikipedia on this one.

>> No.7980170
File: 6 KB, 220x251, 1455882361091.jpg [View same] [iqdb] [saucenao] [google]
7980170

Jesus Christ, you people are retarded. READ THE QUESTION AGAIN. It says:

"It's a gold ball". So you have already picked one of the two possible boxes containing gold balls. At that moment the box with the two grey balls is IRRELEVANT. So, clearly, without any doubt, the answer must be 50%. You idiots.

>> No.7980173

>>7980170
We aren't saying 2/3 probability of getting a golden ball from boxes with a gold ball, we are saying 2/3 probability of the next ball you grab being a gold ball because there's 2 balls which are golden that you could possibly grab next and 1 ball which is silver that you could possibly grab next

>> No.7980177

>>7979448
>meme faces make me right

>> No.7980179
File: 3.11 MB, 345x184, MySides.gif [View same] [iqdb] [saucenao] [google]
7980179

>>7980177
>parroting wikipedia makes me right
wew lad

>> No.7980180

>>7980173

how could you possible grab a silver ball next once you grabbed the first gold ball??

>> No.7980182
File: 283 KB, 714x335, probability.png [View same] [iqdb] [saucenao] [google]
7980182

>>7979461
>If you pick up a gold ball you've put your hand in 1 of 2 boxes with at least 1 gold ball. That leaves a 50% chance you put your hand in the box with 2 gold balls and a 50% chance you put your hand in the box with 1 gold ball.
pic related

Like with monty hall, things will probably get more intuitive if you consider larger numbers.

All boxes have a million balls
1 box has a million gold balls
1 box has 999999 silver balls and one gold ball
1 box has a million silver balls

Now if you pick up a gold ball, do you still think there's a 50% chance you picked the one with a million gold balls and 50% chance you picked the one with only 1 gold ball?

>> No.7980183
File: 1.65 MB, 300x196, 1411091196226.gif [View same] [iqdb] [saucenao] [google]
7980183

>>7980173
But that's wrong. If you grab gold, you are not allowed to switch boxes. It says from the same box. So once you grab gold, you're either in box 1 or box 2. After that it'd just 50/50. Chance is 1/2. You guys really need to learn how to read.

>> No.7980186

>>7980183
>So once you grab gold, you're either in box 1 or box 2.
>After that it'd just 50/50.
This does not follow.

>> No.7980187

>>7980180
Because one box has one gold ball and one silver ball. If the ball you grabbed was from that box then the next ball you would grab would be silver

>> No.7980188

>>7980186
Yes it does.

>> No.7980189

>>7980137
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

>> No.7980192

Fuck this shit.
I'm sending this to NASA.

>> No.7980193

>>7980188
Just because there are 2 possibilities doesn't imply they happen equally likely.
>you are either a faggot or not, so there's 50/50 chance you're a faggot
in reality it is 100%

>> No.7980194

>>7980183
Yes I know you'll be either in box 1 or box 2, but you have to take both boxes contents' into consideration because we don't know which of the two boxes the golden ball was first grabbed from

>> No.7980195

>>7979402
You realised there is 0% chance of box 3 being the one that was picked
Now do that again for 1 & 2

Or to make it even clearer, imagine 10 boxes with 10 balls, 9 boxes have 1 gold and 1 has 10 golds.

>> No.7980197

>>7980193
I'm this case it is, fagbait.

Scenario 1: I've picked gold from 1, chance of grabbing gold is 1.
Scenario 2: I've picked gold from 2, chance of grabbing gold is 0.

So you see, either it happens of it doesn't chance is 1/2.

>> No.7980198 [DELETED] 

>>7980197
>>>7980193 (You)
>I'm this case it is, fagbait.
No you're not, that's me.

>> No.7980199

>>7980197
>Scenario 1: I've picked gold from 1, chance of grabbing gold is 1.
>Scenario 2: I've picked gold from 2, chance of grabbing gold is 0.
Still, no 50/50 follows from this.
Draw a tree or something you'll see.

>> No.7980200

Wtf I'm not 193 #
Tf 4chan

>> No.7980203

>>7980197
You still have to take into consideration the contents of the box you didn't grab the gold box from since we have no idea which box we could've grabbed the gold ball from

>> No.7980214

>>7980203
But that doesn't make sense. Sorry if I sound retarded... But you're either in one of the two remaining boxes. So then you either get the silver ball or the remaining gold ball. I just don't see where the contents of the box you're not in comes into play.

>> No.7980217

The people who say 1/2 are the kind of people to see from a distance that an object is a book since it has a cover on which makes it look as if it is a book and then they'll swear blind that the object is a book and nothing else.

The people who say 2/3 are the kind of people who went up to the object and explored what was inside. They opened up the object and it turns out that it was a foldable map with a cover on stating that it's a map.

I've approached the question with the mentality of landing on the answer 1/2 and with the mentality of landing on the answer 2/3 and this is how I can best describe the thought processes.
2/3 is correct btw.

>> No.7980220

>>7980217
You mean you just read the wiki article and now you're all full of delusional grandeur.

>> No.7980221

>>7980214
The contents of the box you're not in needs to be taken into consideration because you're either in box 1 or box 2, right? There's no point in arguing the 1/2 answer because that's with the assumption that you know what box you're going to grab the ball from. But we do not know which ball we grabbed the gold ball from since there's a 1/2 probability of the first ball being grabbed from either box 1 or box 2. Since there's a probability of both boxes being the box that the gold ball was firstly picked from, we then have to take both boxes contents' into consideration.

Thus, the answer is 2/3

>> No.7980223

>>7980220
I haven't even bothered to read the article because the answer is so straight forward.

>> No.7980239

>>7979124
The box with the 2 silver balls can be disregarded. The question states that you have already limited the options to the two boxes containing one or two gold balls, i.e. what is the probability AFTER you’ve already picked one gold ball.

So the options are
Gold – Gold
Gold – Silver

The answer is 0.5

>> No.7980243

>>7980217
1/2 is correct. Read the question.

>> No.7980261

>>7979574
this. Nice to see it translated into English. This is the second time I've come across the Monty Hall problem ( first time was Mythbusters ) and the second time I've been fooled. ( this time I even recognized it like guy you replied to ). I hope I'll be able to avoid being fooled the third time.

>> No.7980282

>>7980214
Consider this exactly equivalent problem:

You have a big bag with six balls in it: three gold, three silver. Two gold balls are marked with the letter A; one each of the gold and silver are marked B; and two of the silver balls are marked C.

You reach into the bag and pull out a gold ball. What is the probability it has an A on it?

Bonus points: You now rummage through the bag to find the other ball with the same letter as the one you drew. What is the probability it is gold if you drew A? B?

>> No.7980297

I'll ask again because I think most people are missing this.
Do you put the ball you pick back on the box, or not?

>> No.7980312

1/3

Monty hall problem

>> No.7980315

>>7980243
The question was read and answered.

You pick a box at random. Then you pick a ball at random. This means essentially the exact same thing as "you pick a ball at random, regardless of boxes".

Next, it says that you picked a gold ball. That means essentially the same as "you repeat this process until you pick a gold ball"

The chance of picking one of the two gold balls which are in the same box is 2/3, therefore the probability of the other ball being gold is 2/3.

>> No.7980319

>>7980239
I'm almost tempted to run a simulation of this, the amount of 1/2's is astounding.

>> No.7980321

You can eliminate the box with no gold.

Then there are 3 balls remaining after you remove the gold you picked, 2 of which are gold. Therefore the answer is 2/3

>> No.7980328

>>7979124
The probability of picking the box with two golds was 1/3 since the beginning.
Looking at the contents won't change that.

>> No.7980337

>>7980328
you have the possibility of picking the no gold box at the beginning, that has been ruled out once the ball is seen. The probability has changed to 2/3

>> No.7980341

>>7980328
It was 1/2 since the question removes the silver-silver box from the experiment.

>> No.7980344

>>7980319
see >>7979797

>> No.7980363

>>7979124
It's simple, you have to count all the possible outcomes and eliminate the ones that need to be eliminated

6 original equally likely possible outcomes
only 3 are valid (picking a gold ball)
2 of those three would give you a second gold ball

so 2/3

>> No.7980365

>>7980337
But that was ruled out after selecting a box at random.
The outcome would have been different if that information was given to you beforehand.

>> No.7980373

I can't believe the stupidity didn't lessen.
Infact, the stupidity worsened.
It's infuriating.
/b/ 's thread of this had less stupidity.
So basically, on /b/ 's thread everyone was going with 1/2 at the start and very few of us were going with 2/3. Their brains finally started to function properly and most of the people who went with 1/2 at the start, then went with 2/3 because there was so much proof or they figured out that it was the correct answer.
You don't even need to over complicate this with formulas and methods because if you use basic logic, you'll get the correct answer. Of course you can use formulas and methods, but make sure you're using the right ones correctly. It's not just a matter of plugging numbers into a formula; you have to actually use your brains and think for yourselves.
Congratulations to all who say 2/3, you are correct and I'm sorry about the high frequency of 1/2 people you've had to get angry about.

>> No.7980374

>>7979797
nice digits, but you can simply figure it out by calculating every possibility once

but of course that's asking too much from half of /sci/ posters

at least the naive bruteforce approach you chose is pretty foolproof.

>> No.7980378

>>7980373
Welcome to the collective, friend.

>> No.7980379

>>7979925
u r stupdi

you have trouble understanding that after picking box 2 there's a half chance you'll pick an invalid ball and it will stop there. You're confusing yourself because you're thinking with the boxes and not with the individual balls

>> No.7980383

>>7980378
Thank you, H...kek

>> No.7980385

>>7979797
Every probability thread results in someone posting some fucking brute force code because they can't follow a basic argument.

>> No.7980386

>>7980197
Scenario 1 is twice more likely than Scenario 2, that's your mistake

>> No.7980388

great thread I learnt lots of different things.

p.s
2/3

>> No.7980389

>>7980239
you don't understand that those options have weights, the first option is more likely than the second

>> No.7980394

Next time make a strawpoll OP

>> No.7980397

>>7979283
>>7979445
>>7979475
>>7979502
Yeah I'm this guy and I just drew out the probability tree to check and it turns out it is 2/3.

Feeling p silly now tbqh

>> No.7980402
File: 349 KB, 350x233, 1e0.gif [View same] [iqdb] [saucenao] [google]
7980402

>>7980373
You're so fucking autistic you can't even tell when you're being trolled. Just get off 4chan and come back when your chromosomes have dropped.

>> No.7980407

>>7980402
I don't think chromosomes can drop.
Would you like me to make a thread where we can debate whether chromosomes have the capability to drop or not?

>> No.7980410

>>7980407
Sure, then you can get trolled in two threads at the same time, sounds like a winning strategy.

>> No.7980411

>>7980402

>autism
>chromosomes

murican edumucashun

>> No.7980413

>>7980410
Okay. I'll also mention that mr Anonymous thinks that down syndrome is autism

>> No.7980417

>>7980413
You can be both you know. An autistic downie. Which is you, such a coincidence.

>> No.7980427

>>7980182
Event gold ball has been pulled has already occurred. It doesn't matter what probability the event has to happen, because it's already happened.

You're looking at weather the next object will be gold or silver.

Even if there are 9 million silver, it doesn't matter, because the question states that you've already chosen a gold.
Basic reading comprehension.

50/50

>> No.7980435

>>7980427


you got it.

>> No.7980437
File: 59 KB, 390x463, Cloning_Cartoon.jpg [View same] [iqdb] [saucenao] [google]
7980437

These threads should be removed and op banned

>> No.7980439
File: 76 KB, 867x761, box.png [View same] [iqdb] [saucenao] [google]
7980439

Gold ratio: 0.5
Silver ratio: 0.5

>> No.7980462

Please forgive that it's Javascript and poorly written I had poor resources and little time at work.

var boxes = [[1,
1],
[1,
0],
[0,
0]];

var numTrials = 10000;
var num2ndIsGold = 0;

function getRandomIntInclusive (min, max) {
return Math.floor(Math.random() * (max - min + 1)) + min;
}

var trial = 0;
while (trial < numTrials) {
var chosenBox = getRandomIntInclusive(0, 2);
var chosenBallSlot = getRandomIntInclusive(0, 1);
var unchosenBallSlot = chosenBallSlot === 1 ? 0 : 1;
var chosenBall = boxes[chosenBox][chosenBallSlot];
if (chosenBall === 0) {
continue;
} else {
var unchosenBall = boxes[chosenBox][unchosenBallSlot];
if (unchosenBall === 1) {
num2ndIsGold++;

}
trial++;
}
}
console.log(num2ndIsGold);

First trial: 6693
Second trial: 6690

It's 2/3

>> No.7980465

>>7980462
brute force LMAO

>> No.7980466

Just kill this thread someone.

>> No.7980477

>>7980465
I figure it's the best method to settle it once and for all, if the 50%ers can't agree with real math then they should be able to agree with 10000 simulated trials. Plus I had a couple minutes at work.

>> No.7980482

>>7980477

var chosenBox = getRandomIntInclusive(0, 2);
would be right... If the question was worded that way.

>> No.7980490

>>7980482
>would be right... If the question was worded that way.
It is.

Like Monty Hall you (the contestant) knows the general rules, but not what's where.
You reach into a box, and draw a gold ball.
Knowing what you do about the contents of the three boxes, it's clearly twice as likely you drew one of the two from the "both gold" box.

>> No.7980495

Since the chance is twice as high that you selected the gold ball from the box with two gold balls, the chance is twice as high that the ball in the same box will be gold, hence the answer is 2/3.

>> No.7980499

>>7980495
Do you also assume the chance that you were the one sperm who reached your mothers egg every time you calculate a probability as well ? The timeframe is set when you have already drawn a gold.

>> No.7980502 [DELETED] 
File: 58 KB, 368x367, 1427323250266.jpg [View same] [iqdb] [saucenao] [google]
7980502

>>7980439
>correct code
>misleading post

>> No.7980507 [DELETED] 

>>7980502
I just realised this is porn. Ffs.

>> No.7980511

>>7980389
You are right.
I'm correcting my post >>7980239

>The options are
>Gold1 - Gold2
>Gold2 - Gold1
>Gold - Silver

Giving an answer of 2/3

>> No.7980514
File: 102 KB, 212x204, 1459582125998.png [View same] [iqdb] [saucenao] [google]
7980514

>>7980439
>correct code
>misleading post

>> No.7980533

>>7980239
If you only see that the ball you picked is gold it could be any of those three gold balls. 3 cases.

>> No.7980544

>>7980499
Doesn't change the fact that if you draw a gold ball and pick another from the same box, 2/3rds of the time it'll be gold.

"Only two options" doesn't imply 50/50.

>> No.7980548
File: 29 KB, 697x399, Untitled.png [View same] [iqdb] [saucenao] [google]
7980548

In: experiment(1000)
Out: 'Gold prob: 0.659184 Silver prob: 0.340816'

In: experiment(10000)
Out: 'Gold prob: 0.670671 Silver prob: 0.329329'

In: experiment(100000)
Out: 'Gold prob: 0.665354 Silver prob: 0.334646'

>> No.7980558 [DELETED] 

NASA responded.

The answer is 2/3.

Case closed.

>> No.7980563

>>7979810
Wouldn't it be 10 balls in the secount box?

>> No.7980564

>>7979124
2/3 no question.

>> No.7980568

>>7980563
Honey, the second box can have whatever you want it to have :')

>> No.7980576

>>7980563
you could do that
then you would be back to 2/3 I believe

>> No.7980586

>>7980568
So even 100% gold?
>inb4 it must have less than 100%


>honey
Cringed
Thx for the answer

>> No.7980590

>>7980586
Omf read the thread dear

>> No.7980592
File: 24 KB, 883x874, are you fucking sorry.png [View same] [iqdb] [saucenao] [google]
7980592

>> No.7980597

>>7980590
Learn from him he gives good answers
>>7980576

>> No.7980644

>>7980590

>dear
>honey
>condescending false sense of superiority

Fat bitch detected.

>> No.7980645 [DELETED] 

>>7979193

You already picked a box, you melon. The probability that you choose a box with a gold ball in it is 2/3, but it's already given that you picked one of those, so it's 1. Now, given that you chose a box with a gold ball in it, what is the probability that you chose the one with two gold balls?

SPOILER: You're a fucking mong.

>> No.7980651

>>7980644
Dealing with the trollies like yourself

>> No.7980655

Btw all, if you're thorough with Bayes' theorem probability would be solved with ease. If you struggled with this then I recommend you to learn Bayes' theorem.

>> No.7980658

Here it is
https://en.m.wikipedia.org/wiki/Bayes%27_theorem

>> No.7980660

>>7980651

>moo

>> No.7980662

>>7980660
Great onomatopoeia...
That should remind you of yourself

>> No.7980678
File: 73 KB, 596x539, funny-fat-women-girls-people-obese-images-pics-photos-world-17.jpg [View same] [iqdb] [saucenao] [google]
7980678

>>7980662

>let me explain relatively minor statistical methods with a sense of superiority

>> No.7980736

ITT: autism central. The question is not "how many gold balls are there in total"; it is "which box did the gold ball come from".

>> No.7980741

It's 2/3 right?

>> No.7980768

>>7980049
The point is not to show I have the right answer, nimrod. It's to show the correct way to use Bayes' Theorem.

>> No.7980772

>>7980060
There are 3 possible ways to have the other coin in the box gold. 2 of them result in a gold coin.

>> No.7980823

>>7979124
Obviously 50-50. It's either the GG or the GS. 50-50 you fucking scrubs.

>> No.7980829

>>7980741
No, it's 1/2. You know it's either the GG or the GS box. There are only two more outcomes now, you pick out a gold, or a silver.

>> No.7980834

>>7980655
How could you possibly struggle with this question? Fucking Koko the Gorilla could answer it.

Somehow the autists on here are fucking retarded.

>ITT: Engineers, Biologists and Physics majors.

>> No.7980839

>>7979937
I'm going to have to wake Caprice up in a minute and write something I locomotive basic.
See what I'm reduced to.
Bah.

>> No.7980849

>>7980829
When you pick the second box, half of the outcomes will be you picking the silver balls, thus 2/3 of the gold balls you pick will be in the left box.

>> No.7980852

subtracting the gold ball you are holding you are left with 3 silver and 2 gold for a total of 5. 2 of the 5 are gold.

you have a 2 in 5 chance of getting a gold. this is an easy one.

>> No.7980854
File: 59 KB, 541x531, image.jpg [View same] [iqdb] [saucenao] [google]
7980854

All of the 1/2 people are bait right?

People here can't be that stupid can they?

>> No.7980860

>>7980854
Probability is unintuitive.

>> No.7980867

>>7980854
They just don't take into account that, if you choose the mixed box, you only have a 50% chance of picking the gold ball, as opposed to a 100% chance of picking the gold ball if you choose the all-gold box.

>> No.7980876

>>7979124
Just use mindless probability equations
P(gold2|gold1) = P(gold2&gold1)/P(gold1)
=P(gold2&gold1box1 or gold2&gold1box2)/(P(gold1box1 or gold1box2))
=(P(gold2|gold1box1)P(gold1box1) + P(gold2|gold1box2)p(gold1box2)) / (1/3 +1/6)
=((1/3 + 1/6) * 1/3 + 1/3 * 1/6) / (1/2)
=2(3/18 + 1/18)
=4/9

Now wreck me if I'm wrong :^) (I'me probably wrong, geddit?)

>> No.7980877

>>7980829
So you have G1 G2 G3 and S

If you picked G, you still have two golds and one silver left.
Two favorable results out of three total.
The chance of picking a second gold is 2/3

>> No.7980880

>>7979150
See? This is what I'm talking about when I say you lower the quality of this board. You claim to have significant knowledge of AI but you can't even solve a basic probability question.

>> No.7980891

>>7980877
But you only have TWO BOXES.

>> No.7980906

>>7980891

Yes, one box has G1 and G2,
The second box has G3 and S.

So?

>> No.7980931

>>7979530
>>7979828

1/2 to 2/3 convert here. People listen. I have a peculiar theory that they listen more when not being called retards straight away, but thats just the soft science of psychology, probably a meaningless observation.

Anyhow, here's another way to think about it, . 0 chance were in box 3 so forget it.
Since you know you did not reach into box 3, the chances of getting any of the 4 remaining balls is 1/4. The probability you'd get the gold ball from box two is 1 out of four. The chance that you'd pick one of the two gold balls from box one is 2/4 = 1/2.
1/2 is twice as likely as 1/4 so you get 2/3.

not the simplest explanation here, but if I help just one child...
One tricky thing is that we are tempted to visualize picking the first ball in each box as it appears in the picture because we are conditioned to think linearly, left to right. Another tricky thing is when people post a problem that is designed to exploit a kind of rational, intuitive thinking that generally is pretty solid, with the apparent aim of just feeling superior and special.
The likelihood of someone getting this right has much more to do with one's exposure to problems of this nature than it does with any innate intelligence or specialness, and anyone who has implied otherwise here is ... no not a retard... just kind of an ass

>> No.7980951

>>7979206
>Now, you picked either the 2 gold ball box or the 1 gold ball box.

Nope, you picked a ball, not a box.
You picked one of the three remaining gold balls.
Next you will pick from 2 golds and 1 silver

2/3

>> No.7980975

>>7980379
>>7980385

I wrote the code specifically because people weren't buying the basic argument. It's obviously 2/3 to me, but others have doubt.

>> No.7981010
File: 60 KB, 487x578, hot meme injection t. bayes.png [View same] [iqdb] [saucenao] [google]
7981010

>> No.7981013

>>7980891
You need to forget about the boxes and just focus on the balls

>> No.7981023

2/3, its a Monty Hall problem

>> No.7981031

>>7979124
2/3, now fuck off.

>> No.7981039

I thought it was 1/2 but now I think it's 2/3 but that raises a new problem, that I'm a retard and should kill myself.

>> No.7981049
File: 111 KB, 523x720, image.jpg [View same] [iqdb] [saucenao] [google]
7981049

>>7979487
>160 IQ here

>> No.7981052

>>7981039
You're not a retard for not seeing things that many people have problems seeing. These problems are very unintuitive and immensely helpful to have students work with to understand Probability and Statistics.

>> No.7981053
File: 257 KB, 701x576, explanation.jpg [View same] [iqdb] [saucenao] [google]
7981053

>> No.7981055

>>7981013
You need to forget about the balls and focus on the boxes.

>> No.7981072

I see what people are saying about 2/3. But if you're picking the ball from the same box doesn't it mean you've only made one decision and that is basically just pickig a box. It seems to me that it's asking what are the chances you pick the box with 2 gold balls.

I haven't studied probability at all though so tell me why I'm wrong

>> No.7981083

It's a 50 50.
It's either gold or silver.
What's so hard to understand about it?

>> No.7981084

>>7981072
Yes, that is what it is asking. You are seeking the probability of picking box 1, given that you have picked a gold ball. That probability is 2/3.

>>7981010
Third paragraph onwards.

>> No.7981086

>>7980482
If you read the whole program, and knew how to program, you would realize that it selects a box (0,1, or 2) and discards results where the first ball it pulls is silver (0).

>> No.7981092

>>7981053
this is fundamentally different from the OP though

>> No.7981095

>>7979810
If the first ball was picked from the left box, the second ball will be gold 100% of the time.

If the first ball was picked from the middle box, the second ball will be gold 0% of the time.

The average will be 50%

*evil cackling*

>> No.7981096
File: 57 KB, 548x546, 1457417094886.png [View same] [iqdb] [saucenao] [google]
7981096

>>7981072

Define the box with two gold balls as box A, one silver one gold as box B, two silver as box C.

Box A contains balls 1 (gold) and 2 (gold), box B 3 (gold) and 4 (silver), box C 5 (silver) and 6 (silver).

There are 6 possibilities at the start.

You select:
1. Box A, then ball 1. You are holding a gold ball, there is another gold ball in your box.
2. Box A, then ball 2. You are holding a gold ball, there is another gold ball in your box.
3. Box B, then ball 3. You are holding a gold ball, there is a silver ball in your box.
4. Box B, then ball 4. You are holding a silver ball, there is a gold ball in your box.
5. Box C, then ball 5. You are holding a silver ball, there is another silver ball in your box.
6. Box c, then ball 6. You are holding a silver ball, there is another silver ball in your box.
Now, because we've limited you to only selecting a gold ball, we can eliminate all outcomes that result you drawing a silver ball.

So now the possible outcomes are:
1. Box A, then ball 1. You are holding a gold ball, there is another gold ball in your box.
2. Box A, then ball 2. You are holding a gold ball, there is another gold ball in your box.
3. Box B, then ball 3. You are holding a gold ball, there is a silver ball in your box.

Three outcomes, two of which are form box A. 2/3 chance that you've drawn a box A ball.

>> No.7981107

>>7981092

It's not.

>> No.7981134

>>7981107
it is, there are 3 boxes, and you know the outcome of event 1, which removes one of the boxes, and reduces the remaining balls to 3. You have 2 remaining gold balls, and 1 remaining silver ball that could potentially be in your specific box, so that the odds are 2/3 of you picking another gold one.

>> No.7981145

>>7981134

Yes and the altered image gives the same results. The point of that modified problem is that the silver balls and the boxes ultimate don't matter, everything hinges on what gold ball you select from the lot.

>> No.7981154

In all seriousness, what is the answer to this? Can someone answer it and post proof of credibility, like a mathematics diploma or something?

>> No.7981158

>>7981154
1/2

>> No.7981169

>>7981154
You can run this code which brute forces it and returns a rate of about 2/3

here:
>>7980548

>> No.7981181

Just kill me already.

What Bayers is about her is that you could find yourself in 3 situations where you are fondling a gold ball. You picked number one from first box, number 2 from the same one, or the only gold ball in the second box. 2 of those end up with your second ball being gold.

You let 6 people in and tell them to choose the box. They now pick their first ball from there. You throw out guys who picked grey ones. Let's say every one of them picked a different ball. You need to kick the two guys who chose the third box, and one guy who chose the grey ball from the second one. You are left with 2 box one people and 1 guy from box two.

>> No.7981189

>>7981181
But there are two boxes, and you have already chosen one of them. So its 50/50.

>> No.7981202

>>7981154
Its 2/3

Think of them instead as balls labeled (1,2) (3,4), and you chose at the beginning a ball that isnt 4.
If you chose 1, then you have box (1,2) =G

If you chose 2, you have box (1,2) = G

If you chose 3, you have box (3,4) = S

Thus 2/3

Its pretty unintuitive though. When in doubt number things uniquely.

>> No.7981205
File: 26 KB, 655x385, gold balls.png [View same] [iqdb] [saucenao] [google]
7981205

This is pretty clear, right?

>> No.7981208

>>7981189
There are 3 boxes nigger, you choose one at random here. You just know that you are a point where you couldn't have gotten 3rd one because you have a gold ball. Seeing how 2 gold balls out of 3 come out of box number 1 it's more likely to be box number 1.

>> No.7981217

>>7981208
I'm not a nigger.

>> No.7981219

>>7981217
Yes you are until proven otherwise

>> No.7981220

>>7981154
There are multiple pieces of code written in this thread that brute force the problem. All of them get 2/3rds. Satisfied?

>> No.7981231

How can such a simple question have this much discussion attached to it.

It is so painfully obvious 2/3. What is fucking wrong wth all these inbred mother fuckers.

Just look at the damn picture, and use your fucking imagination is you must.

I`m not a smart guy, but holy shit how can some of you fucks be this retarded?

>> No.7981233

>>7981202
you chose the box at the beginning. finding a gold ball doesn't change the probability that you chose box 1 or box 2, which determine the color of the second ball. it only excludes box 3.

>> No.7981235

>>7981217

This is just like people that say "I'M IN CHARGE HERE."

>spoiler alert, they're never in charge.

>> No.7981236

>>7979402
you are retarded
https://en.m.wikipedia.org/wiki/Bertrand%27s_box_paradox

>> No.7981237

>>7981205
There's plenty of clear explanations, but without putting a gun to their head you can't stop trolls from creating a 300 post thread.

>> No.7981246

>>7979124
1/3

>> No.7981249

Try to think about it this way: replace the GG box with one that contains G and a platinum coin P, and rephrase the problem adding 'or platinum' next to every 'gold' you find in the problem formulation. You can see that these manupulations don't change the solution. Now consider what you draw: if you drew a gold ball you can be in either the first or the second box, if you drew a platinum one you're in the first, so we have 3 scenarios. out of those 3, the first and the third lead to being able to draw a gold/platinum coin, the second no, so 2 out of 3 scenarios.

>> No.7981255
File: 23 KB, 655x385, 1459886212969.png [View same] [iqdb] [saucenao] [google]
7981255

>>7981205

>> No.7981257

>>7980876
jesus fucking christ

>> No.7981258

>>7981055
You can simplify the problem: you have 1 box with 4 balls. 3 gold and 1 silver.
You pick two balls, the first one is a gold ball.

>What are the odds that the second ball is also gold?

>> No.7981260

>>7980170
But if you got a gold ball it is more likely (2x as likely, so 2/3 or 66.7%) that you picked from the rightmost box. In this case it's a 100% chance your next pick will be gold. Likewise in the less likely case (you selected from the middle box which accounts for 1/3 of the gold balls, so 1/3 or 33.3%) you won't be able to get another gold ball (0%) so 1*(2/3) + 0*(1/3) = 2/3.

>> No.7981268

Ahhhh, this is bertrands box paradox,
https://en.m.wikipedia.org/wiki/Bertrand%27s_box_paradox

>> No.7981274

>>7980183
You're right, you can't switch. If you grabbed a gold ball, there's a 66.7% chance you picked from the leftmost box which means you'll definitely get a gold ball in the next grab. There's a 33.3% chance you picked from the middle in which case you won't get a gold ball. That's how it's 2/3.

>> No.7981276

#include <iostream>
#include <string>
#include <ctime>
#include <cstdlib>

using namespace std;

int main() {
string boxes[3] = {"gg", "gs", "ss"};
srand(time(NULL));
int temp, temp2;
double counter = 0;
double counter2 = 0;
for (int i=0; i < 1000; i++) {
temp = rand()%3;
temp2 = rand()%2;
if (boxes[temp][temp2] == 'g') {
counter++;
if (temp2 == 0) {
if (boxes[temp][1] == 'g')
counter2++;
}
if (temp2 == 1) {
if (boxes[temp][0] == 'g')
counter2++;
}
}
}
cout << counter << " " << counter2 << " " << counter2/counter << endl;
return 0;
}

>> No.7981287

I sent this to a group of guys at NASA and they said the answer is 2/3.
If you think NASA has the incorrect answer then please take your time to bring Einstein back to life so we can ask him.

>> No.7981289

>>7981287
great job anon, wasting nasas time, rocket launch explosions imminent

>> No.7981291

>>7981289
I said a group of guys at NASA. I know them very well, so I asked them this whilst we were talking in general

>> No.7981298

>>7981287
>>7981289
>>7981291

Hey guys, it's me. The answer is 2/3.

>> No.7981304
File: 2.03 MB, 1767x1981, H.jpg [View same] [iqdb] [saucenao] [google]
7981304

>>7981287

>I sent this to a group of guys at NASA and they said the answer is 2/3.

I doubt you even sent it to a single one. Name someone at NASA you can't google and what department he works in.

>> No.7981306

>>7981298
Omg I told you guys to stop stalking me.

>> No.7981310

>>7981304
No.
I have no right to give out their personal information.

>> No.7981315

>>7981310

Another internet victory for you. Good game.

>> No.7981350 [DELETED] 

>>7979124
P(GG|G) = \frac{P(G|GG)}{P(G|GG) + P(G|GS) + P(G|SS)} = \frac{1}{1 + 0 + 1/2} = \frac{2}{3}

>> No.7981353 [DELETED] 

>>7979124
[math]P(GG|G) = \frac{P(G|GG)}{P(G|GG) + P(G|GS) + P(G|SS)} = \frac{1}{1 + 0 + 1/2} = \frac{2}{3}[/math]

>> No.7981360

Come on guys I am not proud of the people who say the answer is 1/2. Let me walk it through for you.

I explain it with ascii art, but because /sci/ is too retarded to support boxes like a proper board I can't do ascii art in this post.

if you want to know why the answer is 2/3, I explain it here on pastebin: http://pastebin.com/uQ5J4RpK

>> No.7981366 [DELETED] 

>>7979124
[math]P(GG|G) = \frac{P(G|GG)P(GG)}{P(G)} = \frac{1/3}{1/2} = \frac{2}{3}[/math]

>> No.7981372

to the OP problem and not the wikipedia one (I didn't look at it) it's impossible to figure out the probability because the problem excludes the word "only" when talking about the balls. if a box contains 2 gold balls it can also contain more than 2 gold balls, silver balls, a purple box or some worms. autist logic but whatever

>> No.7981373

>>7979124
[math]P(GG|G) = \frac{P(G|GG)P(GG)}{P(G)} = \frac{1*1/3}{1/2} = \frac{2}{3}[/math]

>> No.7981393

>>7981372
why did you even bother posting

>> No.7981422

>>7979124
1/2
Say otherwise and you cannot into maths

>> No.7981424

>>7981372
God please kill yourself

>> No.7981428

Maybe the paradox is 2/3 but the OP picture fucked that up. All it is in the OP is a question of "Which of these two boxes did you pick?"

>> No.7981457

>>7981053
1/6, Danke

>> No.7981461

Question for 1/2-fags: how do you explain the brute force simulations that come up with 2/3?

>> No.7981507

The answer is 50% and not 2/3 and let me explain why:

The question is what is the probability to pick another gold ball from the bag.

You have 3 bags at the beggining. You pick a gold bar. This means, that even though it is more likely that you picked the gold bar from the bag with 2 gold balls, it still is possible that you picked the gold ball from the bag with one gold ball. This mean that you have two possibilities:
1) you picked the gold bar from the bag with G/S and therefore your chances of another gold ball is 0% (the next one is silver)
2) you picked the gold bar form the bag with with g/g and your next gold ball is 100% (the next one is gold)
this leaves you with the following 2 possibilities:
50% gold
50% silver
or a 50% chance of getting the gold bar

Now if the question was what are the chances that the bag you currently took the gold ball from is the G/G ball, it would be 2/3 since the chances of getting a gold ball from the G/G bag is 100% and the chances of getting a gold ball from the G/S bag is only 50% this means that if you picked a gold ball from a bag, there is 2/3 chances that the gold ball is from the bag with 2 gold balls.

TL;DR the question is as it is asked leads to the chances being 50% and not 2/3.

>> No.7981534

>>7981507
Apply for NASA

>> No.7981546

>>7981461
>how do you explain the brute force simulations that come up with 2/3?
Arbitrary distinction assigned between the two balls in box one. The math treats them like two distinct outcomes, dependent on the first draw, when the problem is not at all concerned with this.

>>7981507
>Now if the question was what are the chances that the bag you currently took the gold ball from is the G/G ball, it would be 2/3 since the chances of getting a gold ball from the G/G bag is 100% and the chances of getting a gold ball from the G/S bag is only 50% this means that if you picked a gold ball from a bag, there is 2/3 chances that the gold ball is from the bag with 2 gold balls.

This guy has exactly the right logic. The conditional probability of 2/3rd is not wrong, but it is not what the problem is asking!

>> No.7981581

>>7979814
One thing I don't understand about that video: why are m/f and f/m treated as two different cases? By that logic, if there are 100 frogs and you here 99 croaks, then there's a 99.1% chance one of them is a female, which doesn't seem right. Can someone explain?

>> No.7981589

I saw this on r9k first and thought that this might come from /sci/.

I wanted to check out what you guys said and jesus christ, the amount of people who said 1/2 is horrifying.

look up Bayes' Rule, turn on your brains and try to compute all the information you get from anything, retards.

>> No.7981606

>>7981507
>there are two possibilities
>THIS MEANS THAT BOTH MUST BE EQUALLY LIKELY

I dont know this board's culture and I assume this is bait, but if it isnt: how likely is it that you got the first box and therefore have a 100% chance of getting another golden ball? how likely is it that you got the second box and therefore have a 0% chance of getting another golden ball?

its much more likely that you are at box 1 than at box 2, because 2 out of 3 golden balls are in box 1. the likelihood of being in box 1, given that you got a golden ball in the first place, is 67%.

0,67 * 1 = 0,67

>> No.7981652

>>7981606
Read the whole thing before bitching. It doesn't matter how likely it is to get the box with 2 gold balls vs the box with 1 gold ball because it ain't part of the fucking question. The question is basically given two boxes, one with a gold ball and one with a silver ball, what is the odds of getting the gold ball. It's 50%. The question is not what are the odds of getting a gold ball from the box with 2 gold balls.

>> No.7981657

>>7981652
You should apply for NASA too

>> No.7981660

>>7981652
the question is how likely it is to get another golden ball after getting a golden ball.

jesus christ, just off yourself. if you are still convinced that you are right, literally just do an experiment in real life. get 3 boxes, put something in there that can serve as golden and silver balls, and do the experiment. randomly take a ball out of a box, check if its a golden ball (or whatever you took as a substitute), and IF IT IS A GOLDEN BALL, check if the other thing is also a golden ball and write it down.

do this 60 times and tell me how often the second ball was also golden after the first ball was a golden ball.


when you realize that you were wrong, get educated or kill yourself

>> No.7982064

>https://trinket.io/python/0ef57e8ef9

Simple python program demonstrating that it is indeed 2/3.

>> No.7982103

I don`t know anything about probability. But if you took a gold ball out, then you're only possible options from the same box are gold, gold, and silver.

isnt it 2/3 chance?

>> No.7982119

>>>/int/57436264

some really good laughs there

>> No.7982384

>>7979124
so the real life answer is 2/3 because of what >>7982103
said, there's 3 balls to pick up afterward considering you picked up one of the gold balls (top pic)

I'm sure there's quantum physicists that can say it's 1/2, because technically there's only two situations you could have ended up in, considering you couldn't have picked a gold ball out of a box, and still have two gold balls remaining in it. (bottom pic)

red circles represent which "situation" each explanation considers yourself in.

>> No.7982386
File: 146 KB, 701x1105, Untitled.jpg [View same] [iqdb] [saucenao] [google]
7982386

>>7982384
whoops I did that thing again

>> No.7982392

>>7981606
Yeah you fucking nailed it. 2/3 everybody go home.

>> No.7982394
File: 14 KB, 728x441, bertrandBox.png [View same] [iqdb] [saucenao] [google]
7982394

>>7981606

>> No.7982408
File: 23 KB, 925x561, sciprob.png [View same] [iqdb] [saucenao] [google]
7982408

this thread

>> No.7982417

50%

>> No.7982422

>>7979797
>This code
kill yourself

randbox=rand()%3; /* pick a random box */
randball=rand()%2; /* pick a random ball */
M+=boxes[randbox][randball]; /* count times we got a gold ball */
m+=boxes[randbox][randball] * boxes[randbox][1-randball]

>> No.7982455

>>7979789
I was secretly waiting for someone to say that so I could have an excuse to actually do it. Personally, I try to avoid using probability too much.
>>7979810
>it becomes obvious at higher sample sizes
Probability <3
>>7979840
Underrated post.
>>7979925
Overrated post.
>>7979977
>you have to think and not just plug numbers into a formula
Someone should make a probability general after this. Even if I don't tend to use probability much, I love how these discussions go.
>>7979980
What's your prior on that? Actually, what's your prior on how many users /sci/ sees in an average day?
>>7980018
Nice one.
>>7980030
Does that mean you'll post a probability general?!
>>7980051
>what kind of information is missing
Have you studied probability formally? Because I like the way you reason.
>>7980052
>whether or not you're factoring in the chance of the initial pick
You have to. Anything else is subjectivism.
>>7980083
Hi namefag, namefag here. I just want to let you know that discussing intents is asinine. Being a namefag doesn't stop you from being anon, and you don't see anons discussing their intents do you? No? Then namefags shouldn't discuss their intents either.

Nice leading questions though. Would be a shame if somebody knew your end game here.

>> No.7982458

>>7980116
>once you pick a box you are stuck with it
Wrong. The boxes don't actually exist. The probability that you picked any given box is a separate measure from the ball-related measures.

>> No.7982461

I just want to say that you mathematicians need to work on your linguistics if you're wanting to convey your question to us properly. Just because you understand math, that isn't to say you understand linguistics.

https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

Here, you're stating an event that already happened, and as such you are not to include the probability of what has happened up to that point.

If you had instead asked
>What is the probability of drawing a gold ball first, and then, from the same box, a gold ball second;
or even
>You draw a gold ball first. You then draw a gold ball second from the same box. What is the probability of this happening?

instead of
>You draw gold first. What is the probability of drawing gold next?
you would get a lot more people to understand your question who would be less inclined to answer 50%.

What you are doing is despite stating that gold has already been found, you're including the likelihood of drawing gold first. In doing including this, it's easy to see that box 1 will be considered more times than box 2 because box 2 has less gold than box 1 (and consequently, you'll be restarting trials more when box 2 is picked over box 1).

However, saying that gold is picked first and by then introducing the question, "what is the probability of drawing gold next?", you're now opening a whole new can of worms, because you've segregated the two steps: one being a statement that has already happened, the other being the probability of what is to come because it happened already.

It's the simple difference between

>"Gold is picked first. What is the P of(Gold next)"
and
>"What is the P of(Gold first and then gold next)"

Before you pose your questions to us, please properly structure your question to align with what it is you're really asking.

Best regards,
Anon.

>> No.7982468

>>7982461
I'd also like to add that OP's question framing implies that if you were to conduct 1,000,000 trials, you would always have gold first. It doesn't matter how that happened because the question explicitly states that gold is found first.

>> No.7982472

>>7980261
There's never a third time with Monty Hall. By the time you've been duped twice, you already know the third door holds the non-goat.

>> No.7982477

>>7980373
That post was singularity the most angry post ITT. I know because I actually read the entire thread. It was grueling and nearly pointless, but I did it just so I could see the look on your post.

>> No.7982480

>>7980383
I'd decline any random email offers for the next 30 years if I were you.

>> No.7982484

So OP just made a poorly worded bait image? Lame.

>> No.7982485

>>7980499
>the chance that you were the one sperm who reached your mothers egg every time you calculate a probability
No. The entire universe is compressed out of the equation by the use of the term "you" in relation to drawing a ball.

>> No.7982486

>>7982461

This x100000000.

You're saying gold has already been found, so you still don't know which of the two boxes you've selected, that means what remains is either silver or gold, a 1/2 chance.

If you had reworded your question to one of those my quote said, I would totally agree that it is 2/3, but until you fix the wording, I'll ardently side with 50%.

>> No.7982487

>>7981095
You can take a gold ball from the first box in more than one way which stacks up against the only one way you can take a gold ball from the second one.
>lrn 2 statistical mechanics

>> No.7982490

>>7980586
>Cringed
>what happens when a non-fairy speaks aberrantly

>> No.7982494

>>7982490
Go to bed Cirno

>> No.7982497
File: 129 KB, 1174x539, yes.png [View same] [iqdb] [saucenao] [google]
7982497

Frame this.

>> No.7982507

>>7980880
Yes, I understand where you're coming from. Did I need to post that? No. Did I lower the quality of the board by posting it? Also no. I corrected myself here >>7979530 but let's ignore that for the sake of argument.

Exactly three people, including yourself, read and responded to that post. Out of those, one made a serious response, here: >>7979574 That post was constructive, easy to read and explained the issue well. Overall, the quality increased because I was wrong at first.

Now if you want to get serious—and I can't believe this is actually on topic for this thread—what is the probably that that anon would have made the same reply to another 50-50 poster if I hadn't been here to "first post" a wrong answer?

I'm not saying all of my posts are gold, but what is the probability that gold posts will follow from my silver posts?

>> No.7982517

>>7981134
>you know the outcome of event 1
You don't. You only know that you're holding a gold ball.
>which removes one of the boxes
No, it removes one and one half boxes.

>> No.7982531

>>7981291
>whilst
Which group of friends told you that the guys you were talking with were from NASA? Because they won't tend to tell you who people really are.

>> No.7982545

>>7982531
I know what you're on about.
I meant the guys I'm friend with are from NASA.
The people at NASA do have friends and a life...

>> No.7982547

>>7981507
>if the question was what are the chances that the bag you currently took the gold ball from is the G/G ball
That is precisely the nature of the question. It silently omits the fact that you have no idea which box you drew the gold ball from. You have to be keen on the situational reasoning to see your own ignorance. The question is about next pick, which depends on (but is not equivalent to) "which box."

>> No.7982553

Not going to read the thread. Make a simple java program and see what happens. Hint: it's not going to be 1/2

>> No.7982562

>>7982461
Shibboleth.

>> No.7982563

>>7982494
Get the name right or I can't abide the request.

>> No.7982567

>>7982545
The people you're friends with are from the internet.

If they told you they're from NASA, there's no real reason to believe them. Being from NASA wasn't a reason for you to believe them in the first place.

>> No.7982595

>>7982567
Ummm...You've got that slightly wrong there. I met them in reality, not on the web.

>> No.7982619

>>7982595
I have no reason to believe you're the same anon.

>> No.7982620

>>7982619
Ok

>> No.7982641

I've had a reply from Steven Hawkings too.
He said that it's 2/3.
I'm now taking my time to travel back in time to ask Einstein.
Brb.

>> No.7982712

>>7981205
>>7981096
>>7980592

The thing that bugs me about all these explanations is that it ends up assuming the boxes don't really matter

At the end you have 1 gold ball. Say there are no separate boxes any more, and the contents of the remaining boxes (which must be GOLD,GOLD and SILVER,GOLD) are placed into a bucket that you cannot see into.

What is the probability of getting another gold ball? 2/3. Thats what bugs me -- why should it be the same?

>> No.7982756

Fuck me?!
Who knew the late 40s could be so...bland.
It was great meeting Einstein though.
He said that the answer is 2/3.
Is there anyone else we should ask?

>> No.7982795

>>7982712
>why should it be the same?
The boxes literally don't matter. The question hides the fact that we don't know which box we're in. Try removing the grey-only box from the experiment entirely and running through all possible scenarios again.

>> No.7982805

>>7982461

Most of these probability problems are only difficult due to a vagueness of language. Monty Hall, Tuesday Boy, Sleeping Beauty etc. There is a certain fun in arguing, but it's all pretty pointless.

>> No.7982807

>>7982461
What?
>>What is the probability of drawing a gold ball first, and then, from the same box, a gold ball second;

1/3

>>You draw a gold ball first. You then draw a gold ball second from the same box. What is the probability of this happening?

1/3

>>"What is the P of(Gold first and then gold next)"

1/3

>> No.7982904

>>7982807
Linguists BTFO.