HOW SMART IS THE AVERAGE ROBOT?

We already had another thread about this faggot. It's 2/3. Fucking kikes.

File: 1482936396098.jpg (194KB, 1000x750px) Image search: [Google]

194KB, 1000x750px

>>34180071
100% because I will it too be so with my mind

>>34180071
40% chance 2 out of 5 balls left 2/5 x 2 = 4/10 which means 40%

50/50 nigga, there's no chance it's the silver/silver box.

>>34180071
3/4

Orginaliopastonoliopleaseacceptthisasoriginalcomment

>>34180071
2/5
assuming you are taking the gold ball out and not placing it back in once you are finished with it

>>34180071
2/3rds

Let's break it down, call the balls from left to right balls A, B, C and D (and E and F but we don't care about those). A, B and C are the golden balls.

Now you just picked the golden ball. If you picked A, then you will pick B and get the golden ball. If you picked B, then you will pick A and get the golden ball. Finally, if you picked C, you will pick D and get a silver ball.

There are thus 3 possibilities, with 2 giving you a golden ball.

Hence 2/3rds.

>>34180184
WRONG originaloo

>>34180071

50% the boxes are irrelevant and a red herring.

3 gold balls. 3 silver.

>>34180099

This. It's just 50/50. It's not complicated.

You guys are overthinking it.

>>34180083
>>34180085
>>34180097
>>34180099
>>34180106
>>34180152
>>34180212
>>34180235
brainlets

>>34180166
High IQ ubermensch

>>34180166
But what if the original golden ball that was take out was taken out of the middle box containing one gold and silver ball.
That would make the odds 1/3

>>34180212

Woops. Misread the question. Nevermind. It's 2/3rds.

50%.
Start with boxes A, B, and C, where A has 2 gold and C has 2 silver. Because I have one gold ball out of the box, I know the box I'm picking from must be box A or box B.
If it is box A, there is one gold ball left, so I will pick the ball. If it is box B, there is only a silver ball, so I will not pick another gold ball. There is a 50% chance based on the given information that it is box A, so the odds of pulling a gold ball is 50%.

>>34180258
Irrelevant, you don't know that.

>>34180266
But there are two different ways of having picked a ball in box A (either one of them), hence 2/3rds.

>>34180272
No one does. The question never specified which box the golden ball was taken out of. So there are multiple answers depending on which box the golden ball was taken out of. Right?

>>34180307
Nope. Reread my solution carefully.

>>34180071
it's 50/50
the prompt says you pull a ball out and it's gold, therefore you know which 2 boxes you chose from, the chances the next one is gold is 50/50 it's either silver if you chose box 2 or gold if you chose box 1
those who say 2/3 learned about the door puzzle and think that it's applicable here

>>34180502
No it's 2/3. The flaw is that you think it's equally likely to be either box 1 or 2 given that it was a gold coin. But in reality it's twice as likely to be from box 1 (since there are 2 gold coins there).

it's definitely 5/8ths right? how is this even a question?

>>34180292
>there are two different ways of having picked a ball in box A

irrelevant. this isn't like the goat-door problem. if i have two identical balls in a box and i've picked one, it doesn't matter which was picked first. i know that i'm going to pick the same ball next.

>>34180166
False, there's FIRST the bottleneck of picking a box before picking a ball. Even though there are 3 gold balls to select, it is only possible to select the gold ball that will lead to another gold ball by FIRST selecting the BOX with two gold balls, which is a separate decision.

It really looks like this:
50% chance to select the box with 2 gold balls (outcome A)
50% chance to select the box with a gold ball and silver ball (outcome B)
even though there are 3 outcomes, and two of them lead to selecting a second gold ball, both of those outcomes have a 100% chance of being picked if you select Outcome A, and a negative result, the silver ball, has a 100% chance if Outcome B is selected first.

File: 1484479632382.jpg (134KB, 807x935px) Image search: [Google]

134KB, 807x935px

>>34180563
>irrelevant. this isn't like the goat-door problem
It is indeed similar

>if i have two identical balls in a box and i've picked one, it doesn't matter which was picked first.
Exactly. Which is why there are TWO possibilities of picking another golden ball from the first box. Because you could've picked either one of them during the first pick.
>>34180576
Wrong. There are three golden balls. You could've picked either one of the three in your first pick, hence your premise:
>50% chance to select the box with 2 gold balls (outcome A)
>50% chance to select the box with a gold ball and silver ball (outcome B)

Is already false. It is false because you have a higher chance of having picked the ball from box A, by virtue of there being two golden balls there (as opposed to one in box B).

I would say 100% because you are certain to not take from the double silver box at random, if the ball you pick up is gold. But which of the two balls you take is also random, so how would you simulate randomly choosing the gold ball every time from the middle box? If both box choice and ball choice is random, but the ball is guaranteed gold, then only the 1st box is possible.

File: untiitled.jpg (2MB, 2964x1896px) Image search: [Google]

2MB, 2964x1896px

I don't need to justify myself to you.

I've had ample validation already and have concluded that I'm easily the smartest person on this board.

>>34180071
50%

If you already have a gold ball, you are either picking from the double gold box or the gold/silver box.

Therefore the correct answer is 50/50.

I can absolutely believe the people who use this board are so devastatingly stupid as to not understand this.

>>34180731
It's 2/3 you brainlet.

>>34180540
the moment you pull 1 gold, the fact there are 3 is irrelevant, at that point you either pulled from box 1 or 2
50/50
>But in reality it's twice as likely to be from box 1
that is a logical flaw, this question is not about how many gold/silver balls there are in total, it's about what are the chances you pull another gold ball in the box you just chose, it's either box 1 (2 gold) or box 2 (gold/silver)

Approx 66%

>>34180744
>that is a logical flaw
No it's not.

If you have a box with 1000 golden balls, and another box with 1 golden ball and 999 silver balls, if you pick a ball from one of the boxes at random and it's golden then that means it's pretty fucking likely you picked it from the box with 1000 golden balls.

The problem is underspecified, and Generation Z thinks "I read a Wikipedia page once" constitutes understanding. The answer is 1/2 or 2/3, depending on what you think it means to choose a gold ball from a box at random.

>>34180790
we aren't talking about 1000 balls per box we are talking about 2 box's, don't try to change the parameters, in this scenario it's 50/50

>>34180864
>we aren't talking about 1000 balls per box we are talking about 2 box's,
It's the same idea you dingus

>>34180071
Could anyone actually solve this shit without looking at the posts in the thread first?

2/3*1+0.5=0.9=90%
>tfw 0% content
And not original

>>34180864
>don't try to change the parameters

Are you unironically autistic? The same math holds whether it's 2, 100, or 1000. It's just easier to intuit when you imagine it with more balls.

>>34180876
it's not, not by a long shot
a correct alteration of this would be there are
box 1 1000 gold balls
box 2 500 gold 500 Silver
because in the original it was
box 1 100% gold
box 2 50% gold
you are changing it to
box 1 100% gold
box 2 1% gold
>>34180935
yes if the percentages were EQUAL again he changes it to
box 1 100% gold
box 2 1% gold
as opposed to
box 1 100% gold
box 2 50% gold

Is this what Magic gatherings are like?

>>34180966
All right Einstein.

So in the situation where one box has 1000 golden balls, and the other one 500 golden/500 silver, you still think that if you pick a ball at random it is EQUALLY LIKELY that you picked from the box with 500 MORE GOLDEN BALLS than the other?

>>34180966

Just read the wikipedia page:

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

>>34180071
easy
there are 3 gold balls and 3 silver
after you take a gold there are 2 gold balls left
the answer is 2/5

File: 1390174415791.jpg (17KB, 208x294px) Image search: [Google]

17KB, 208x294px

Picking the left box, regardless of which of the two balls you pick from the left box, will always yield a second gold ball.

Picking the Middle box will always yield a silver ball.

There is a 50% chance you'll pick the left box, and a 50% chance you'll pick the middle box, after which there is no more variance. The left has a 100% chance of yield two gold balls, the middle a 100% chance of yielding a silver ball second.

People saying that it's 2/3 are only looking at the number of gold balls and don't understand the relevance that it happens in stages.

>50% chance you pull from box 1
>50% chance you pull from box 2
>if you pull from box 1, 100% chance to pull 2 gold
>if you pull from box 2, 100% chance you pull silver
>(1*0.5) + (1*0.5) = 1 = 100% all outcomes accounted for

Alternatively, you can do it like this:
>50% chance of pulling box 1
>50% chance of pulling box 2
>Option A, if you pull box one, there is a 50% chance you'll pull gold ball 1 (and therefore gold ball 2 second) or Option B you'll pull gold ball 2 (and therefore gold ball 1 second)
>Option C, if you pull box the second box you'll silver second 100% of the time

Option A, 0.5*0.5 = 0.25
Option B, 05.*0.5 = 0.25
Option C, 0.5*1 = 0.5
Options A AND B both yield a second gold ball, Options A + B = 50%
0.25+0.25+0.5 = 1 = 100%, all outcomes accounted for.

ALL THE MATH IS THERE.

>>34181035
>>50% chance you pull from box 1
>>50% chance you pull from box 2
Wrong, faggot

KYS

50/50 look at the balls, not the boxes

>>34180998

Read carefully. After you pick a ball the second ball you pick comes from the same box. And you are given the prior info that you first pick a gold ball, so you would never have chosen box 3

>>34181046
You have to pick a box BEFORE you pick a ball, and you CANNOT pick box 3 because you always draw a gold ball first.

You are NOT more likely to pick box 1 because it has 2 gold balls.

>>34180989
it's the same you have a 50/50, well more like 49/51, chance of grabbing a gold
>>34180993
well theoretically yes it would be 2/3 but in an applicable sense it's 1/2, the 2 silver box is irrelevant the moment you pull a gold
you have 2/3 chance of pulling a gold, then you have 1/2 chance of grabbing another gold

>>34181097
>You are NOT more likely to pick box 1 because it has 2 gold balls.
Yes you are dumbdumb

Or rather, rephrased properly :
>If you get a golden ball during your first pick, it is more likely to have come from box A than from box B.

Don't even try to argue with me. I have a phd in mathematics and make 300k/y

>>34181123
No idiot, because chronologically you FIRST pick a box and THEN you pick a ball. The fact that the first ball you pick is always gold does not influence the probability of which box you took.

100% chance I've got BALLS OF STEEL!

https://www.youtube.com/watch?v=eckoYQqdk28

>>34181136
>The fact that the first ball you pick is always gold does not influence the probability of which box you took.
Oh? So you still have a non-negative probability of having picked from the third box? Tell me more.

Checkmate, brainlet.

https://www.youtube.com/watch?v=6cS18yG8DJI

>>34181102

> theoretically yes it would be 2/3 but in an applicable sense it's 1/2

Wtf?? The fact is that since you are given the info you picked a gold ball, it's more likely to be from box 1 than from box 2. That makes it not 50-50 but rather 2/3 - 1/3

>>34181174
Having picked a gold ball excludes the possibility of having picked the third box, but the problem STILL necessitates you pick the box before you pick the ball. Thus, you are not twice as likely to pick the first box, rather the conditions ONLY exclude picking the third box which means that you could have only picked box 1 or 2, a 50/50 chance.

>>34181199
You are not more likely to have picked box 1 just because you picked a gold ball first. Rather the question simply excludes the possibility of picking the silver ball in box 2 first.

>>34180071
The answer is infinity.

Shroedinger's balls. You don't know what's going to occur until it's observed. The possibilities are infinite.

>>34181210
>Thus, you are not twice as likely to pick the first box
Yes you are since there are two golden balls in the first box, but only one in the second box.

Think about it.

>>34180071
>3 gold balls
>3 silver
I would say 50% chance?

File: Goliath-chan.jpg (138KB, 577x1024px) Image search: [Google]

138KB, 577x1024px

It's technically 1/3, but the actual chace of pulling the second gold ball is 1/2 concidering the two silver ball box becomes a null factor after you pull the first golden ball.

>>34181199
>it's more likely to be from box 1 than from box 2. That makes it not 50-50 but rather 2/3 - 1/3
the chance of getting a gold ball is 2/3, the chance there is 2 gold balls in there is irrelevant, once you pick a gold ball it becomes 50-50

>>34180166
The question was, "what are the odds that your next pick will also be gold"? That is the only criteria it counts, that the next ball is golden. Picking A then B and picking B then A is the same fucking shit. The question does not take into account the balls' "identity"

>>34181235
No, because that's not how the question works. You first select the box, this is the FIRST decision, and it's probability impacts the second decision.

You can pick 3 boxes, 1, 2 or 3. Because 3 has no gold balls and we know the subject inevitably picks a gold ball first, box 3 is a null factor with a 0% chance of happening, thus it leaves us with 2 boxes, and thus a 50% chance to pick box 1 or 2.

Second, the person always draws a gold ball first. Therefore the possibility that the player picks box 2 and draws the silver ball is null, 0%.

We're left with 3 outcomes.
>player picks box 1, and then picks gold ball 1 (25% chance, successfully picks 2 gold balls)
>player picks box 1 and then picks gold ball 2 (25% chance, successfully picks 2 gold balls
>player picks box 2, and then picks the gold and then silver ball (50% chance)

There is a 50% chance that 2 gold balls are picked. The number of gold balls CANNOT influence the probability box 1 is picked because chronologically boxes are chosen before the balls are chosen and which box is chosen directly influences the possibility of picking up 2 gold balls.

students are failing this math class because of you all

>>34180071
There were 3 boxes so wouldn't the chance remain at 1/3?

You knowing the 2silver box is no longer a chance doesn't affect the actual probability of the universe. To think otherwise would be arrogance. You're still working on the original 1/3 chance.

>>34180071
If he put the ball back in the box it's
>50/50
otherwise it's
>60/40

>>34181123
There's no need to go on the internet and tell lies. That said, there are two ways of formulating this problem, which provide different results. This is the reason that people get it "wrong" and people who think Wikipedia (or, for that matter, the literature) is a reliable source get it "right."

Formulation one: how to get 2/3
Imagine there are 6 rooms as indicated. A person goes into each room and chooses a box, then removes a ball. Eliminate all the rooms where a person chose a silver ball. At this point, the expected number of rooms remaining is 3: two rooms with (G, G) and one room with (G, S). The (G, S) room can only choose a silver coin, while the (G, G) rooms can only choose a gold coin. Therefore, two of the three remaining rooms contain a second gold coin, so the answer is 2/3.

Formulation two: how to get 1/2
We take the problem at face value: I've picked a box, and it contains a gold coin. Therefore, it had to have been chosen from either (G, S) or (G, G). It doesn't matter that it's more likely to have been from (G, G), because the past is the past, we might as well ask something idiotic like, "I just flipped a coin and it came up heads. What's the probability it actually came up tails?" In this situation, there's two possible boxes, so I have equal odds of it being either one.

>>34180071
its 50 percent.

i wrote a simulation to determine

~
import random

def test():
dictionary = {'1': [0, 0], '2': [0, 1], '3': [1, 1]}
checks = ['1', '2', '3']

x = random.choice(checks)
first = random.choice(dictionary[x])
dictionary[x].remove(first)
y = random.choice(checks)
second = random.choice(dictionary[y])
if second == 1:
return 1
else:
return 0

number = 0
hmm = 1000000

for i in range(hmm):
number += test()

number = number/hmm

print(hmm * number)

>>34181427

nvm, it's still 50/50 I'm fucking retarded.

>implying all probability questions aren't merely about buggering people around with misdirection and ambiguity rather than genius intellect

The question is essentially asking "what are the chances you pick which of these 3 boxes has 2 gold balls, and also since we assume you've picked 1 gold ball it presupposes you never pick the box with 2 silver balls"

The question is "of these 2 boxes, what are the chances you pick the box with 2 gold balls rather than 1 of each"?

>>34181411
Is it a maths question? Or is it a philosophy question?

There is no ball.

>>34181433
actually that is coded wrong i didn't read it had to be a random choice from teh same box that you picked -

def test():
dictionary = {'1': [0, 0], '2': [0, 1], '3': [1, 1]}
checks = ['1', '2', '3']

x = random.choice(checks)
first = random.choice(dictionary[x])
dictionary[x].remove(first)
second = random.choice(dictionary[x])
if second == 1:
return 1
else:
return 0

so the function should pass x to check the same list.

if CASIO calculators had mouths they would talk like this >>34181505
>>34181483
>>34181448
>>34181433
>>34181430
>>34181427
>>34181419


good thing they don't.

Good lord the IQ of the people in this thread

>>34181609
it's okay to not understand, anon

>>34181621
It's 2/3s
https://en.wikipedia.org/wiki/Bertrand's_box_paradox

>>34181433
Your simulation doesn't account for the fact that the first ball chosen has to be gold, which is why you got 1/2. A correctly-implemented simulation that's based on the "eliminate cases where a silver ball was chosen" will give you 2/3.

Here's an example, though you'll have to bear with the fact that I only have the Glasgow Haskell Compiler installed on this computer:

import Control.Monad
import System.Random

data Result = Gold | Silver | Neither deriving Eq
boxes :: [(Result, Result)]
boxes = [(Gold, Gold), (Gold, Gold), (Gold, Silver), (Silver, Gold), (Silver, Silver), (Silver, Silver)]

choose :: IO Result
choose =
randomRIO (0, 5) >>= \first ->
case (boxes!!first) of
(Gold, Gold) -> return Gold
(Gold, Silver) -> return Silver
(Silver, Gold) -> return Neither
(Silver, Silver) -> return Neither

main :: IO ()
main =
replicateM 10000 choose >>= \results ->
print (fromIntegral (length (filter ((==)Gold) results)) / fromIntegral (length (filter ((/=)Neither) results)))

Which prints, e.g., 0.6681478489194103

>>34181700
>boxes = [(Gold, Gold), (Gold, Gold), (Gold, Silver), (Silver, Gold), (Silver, Silver), (Silver, Silver)]
You're doing it wrong, you're simply using the possible combinations without regard for the fact that first a selection determining the box is made, and then a selection determining the balls is made.

the idea of six pairs, GG, GG, GS, SG, SS, SS, is incorrect.

>>34181700
x = random.choice(checks)
first = random.choice(dictionary[x])
if first == 1:
dictionary[x].remove(first)
second = random.choice(dictionary[x])
if second == 1:
return 1
else:
return 0
else:
return 0

im pretty sure the simulation is coorect now.

>>34181700
>>34181505

Lol the haskellfag is right.

You have to re-pick if you don't get a gold coin on the first try:

import random

def test():
dictionary = {'1': [0, 0], '2': [0, 1], '3': [1, 1]}
checks = ['1', '2', '3']

x = random.choice(checks)
first = random.choice(dictionary[x])

if first != 1:
return test()

dictionary[x].remove(first)
second = random.choice(dictionary[x])
if second == 1:
return 1
else:
return 0

num = 0
hmm = 1000000

for i in range(hmm):
num += test()
print(num/hmm)

>>34180071
I'm pretty sure you only have a 33 percent chance to get another gold ball if you reach into the same one

>>34181825

No it's not just a matter of returning 0 if the first coin isn't gold. Picking gold on the first try doesn't count as a failure. Look at my code.

>>34181788
The six pairs of GG, GG, GS, SG, SS, SS is just to simplify the programming. They represent the fact that you could choose balls from the box in either order, so we need a GS and a SG case. Since we're doubling up the heterogeneous boxes, we need to double up the homogeneous boxes. If this is really a sticking point, the program can be rewritten as,

import Control.Monad
import System.Random

data Result = Gold | Silver | Neither deriving Eq
boxes :: [(Result, Result)]
boxes = [(Gold, Gold), (Gold, Silver), (Silver, Silver)]

choose :: IO Result
choose =
randomRIO (0, 2) >>= \box ->
randomRIO (0::Int, 1) >>= \order ->
case ((boxes!!box), order) of
((Gold, Gold), _) -> return Gold
((Gold, Silver), 0) -> return Silver
((Gold, Silver), 1) -> return Neither
((Silver, Silver), _) -> return Neither

main :: IO ()
main =
replicateM 10000 choose >>= \results ->
print (fromIntegral (length (filter ((==)Gold) results)) / fromIntegral (length (filter ((/=)Neither) results)))

but it produces the same results.

>>34181825
No, if first == 1 then you need to return nothing (or start the selection process over), because we're not in one of the cases we're talking about. This is the reason the canonical result is "unintuitive" and/or wrong.

>>34181887

Haskellfag strikes again.

For those who want a saner language without monads in your gonads here's the corrected python script

https://ideone.com/KuHgpt

>>34181640
>I read the wikipedia post about this equation so I'm too smart to read any comment that might be against what I read

>>34181887
>if first == 1 then you need to return nothing
Or rather, if first == 0, I lost track of the integer encoding there. >>34181899 provided an ironically functional solution (and that's not a good idea in Python!), a simpler version would wrap the action of test in a while True:, where if first != 1: continue, and you break the infinite loop by ultimately returning if first == 1.

>>34181958

That's too much work for a throwaway script. And the stack is not exactly going to overflow since it's on average < 2 rolls to the first gold coin

>>34180071
There was a 50/50 chance to pull a gold ball. If you picked up a gold ball it's more likely you pulled it from the box with 2 gold balls

>>34181916
A friend of mine thinks this is the big problem with Gen-Z, that they were raised in an environment of Wikipedia and standardized tests, so they expect to have an authority there to tell them the answer, and intelligence is the ability to repeat that answer back.

>>34182051
It's not really any more work than anything else, it adds one line of code (while True:), and replaces return test() with continue. Obviously you're not going to have a stack overflow in this problem, and yes the recursion depth will tend to be small. I just think even throwaway scripts should be written well, so that you develop good habits.

>>34181322
>>34181353
>>34181430
I give up, you're retards.

It's 2/3

P(pick 2 gold ball box | pick a gold ball) = P(pick 2 gold ball box and pick a gold ball) / P(pick a gold ball)
= P(pick 2 gold ball box) / P(pick a gold ball)
= (1/3) / (1/2) = 2/3
simple conditional probability

Either you pick a golden ball or not.

These things work in paper, but in reality, it either happens or not. You can have a 99% chance of success in something and still fail, not because of the 1% of chance, but because you either succeeded or failed, and you failed.

>>34182199
anyone that argues against this is wrong canonically
https://en.wikipedia.org/wiki/Conditional_probability

>>34180071
50 percent anyone else is braindead

Sheldon's Math Club

Level = Highschool / Local CC

>>34182227
The thing you're gesturing at is actually really important, which is, probability is fiction used for papering over dynamics. If we take mathematical physics for granted, then a coin flip isn't "random," it's just complicated. Since faithfully modeling a coin flip isn't generally possible, probability is the band-aid that's applied. This problem is an example of what happens when mathematicians take the band-aid as a fundamental abstraction and proceed to get lost up their own asses.

>>34182199
>>34182228
i guess i'll explain this in laymen's terms
So you're given that you have already picked a gold ball and you're trying to find the probability that you have picked the box with 2 gold balls, which is contingent on the fact that you have already picked a gold ball.
So let A = the event that you have picked the box with 2 gold balls
let B = the event you pick a gold ball
Denote P(X) the probability of event X occurring, with P(A|B) meaning the probability of event A given that event B has already occurred, and P(A & B) meaning the probability of event A and B both happening.
The formula to compute conditional probability is
P(A|B) = P(A & B) / P(B).
Now you just compute P(A & B) and P(B).
P(A & B) = P(A) = 1/3, since if you pick the box with 2 gold balls, you must pick a gold ball.
P(B) = 3/6 = 1/2
Hence, P(A|B) = (1/3) / (1/2) = 2/3

>>34182257
>2 gold balls in one box
>1 gold ball in other box
>still thinks it's 50/50

https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
everyone in this ITT thread: kys

>>34182228
except he did the calculation wrong

you're picking a box randomly, not a ball randomly, so P(pick a gold ball) is the probability that you pick a box with a gold ball in it, not the proportion of gold balls across all 3 boxes. so P(gold ball)=2/3, not 1/2

File: mqdefault.jpg (7KB, 320x180px) Image search: [Google]

7KB, 320x180px

I'll put this in terms /r9k/ can handle.

There are 3 waifus
Eliza, Marky and Agatha. You get the Instagram of one of them and after 20 minutes they send you lewd pictures of their feet.
Marky sends feet half the time since she's manic/depressive
Eliza sends feet 100% of the time because she needs attention all the time.
Now it's 24 hours later and you need more feet. What's the chance they send you more pictures?

By virtue of the fact that you received any lewd pictures yesterday means that you are more likely contacting Eliza. Get it?

>>34182343

While this is wholly right, I think the bayesian solution (in the wiki page) gives a bit more intuition as to why the naive answer doesn't work.

File: CWY17gQUYAAThoh.jpg (71KB, 600x652px) Image search: [Google]

71KB, 600x652px

>>34182377
All of these are whores

CRISPY FOREVER

>>34182375
the probability that you pick a gold ball is 1/2, it's a very simple counting problem. you're overcomplicating it, there are only 6 balls, 3 of which are gold, it doesn't matter what box you pick since you're gonna end up picking one ball anyway. choosing a box beforehand just narrows your choices before you pick just one ball.

File: 1480802421074.jpg (41KB, 381x539px) Image search: [Google]

41KB, 381x539px

>>34182377
>tfw to intelligent to understand basic probability

>>34182377

Someone should create an entire book on probability with anime and waifus

>>34182429
oh yeah, you're right. silly mistake on my part

ITT: Dunning Kruger in action. The ones who have the least understanding of statistics are the most vocally loud. And wrong.

Think of it like this. There are six different universes when you pick different ball each. Universe 1,2 you reached into the left box. Universe 3,4 you reached into the middle box. Universe 5,6 you reached into the right box.

Since you have drawn a gold ball, universes 4, 5 and 6 are disqualified.

Now you have 3 universe in hand. 1, 2 and 3. Two of those, you have reached in the left box and will draw a second gold ball if you reach into it again.

It's 2/3.

>>34182578

But when you try to observe which ball you drew the universes collapse

>>34180891
My first instinct was that it was 50/50 but I was pretty sure that was wrong because I correctly identified that it was a similar question to the Monty Hall problem. I didn't actually know the answer though, I had to look up how the Monty Hall problem went first.

>>34182793
This isn't the Monty Hall problem. Stop overcomplicating it. You're picking a box, not a ball.

>>34182429
no, this is wrong.

imagine a machine with 3 arms. each arm is attached to 1 gold ball. 1 arm is attached to gold ball a in box 1, another arm is attached to gold ball b in box 1, the last arm is attached to gold ball c in box 2. if the machine can only draw a single gold ball randomly, then there is a 2/3 chance of pulling a "winning" gold ball from box 1 and only 1/3 chance of the "loser" gold ball c.

>>34180166
no. You pick from the same box.
it can only by one of the first two boxes. it's 50/50 that you picked one of the gold balls from box 1, or the gold one from box 2.

>>34182936
You're wrong. In your scenario you are giving twice the chance for the two-gold box to be picked. The chances for picking the box are equal.

>>34182978
>it's 50/50 that you picked one of the gold balls from box 1, or the gold one from box 2.
>one of the gold balls

this changes the probability. It's no longer 50-50.

>>34183032
The scenario basically states only gold balls can be chosen. All gold balls have an equal chance to be chosen. 2/3 gold balls are in box 1. You need 2 gold balls to win. You have a 2/3 chance.

50%
There's only six balls, and there is an equal amount of each color. I don't understand why everyone is over thinking the question.

>>34183035
How? If you pick up a gold ball from one of the boxes, there's a 50% chance that you picked one of the balls from box 1 (doesn't matter which one) or the gold ball from box 2.

>>34183119
Because you can ONLY choose a gold ball, and since 2 of 3 of the gold balls are in one box, the distribution is unequal.

File: FKIN TREE DIAGRAMS NIGGER.png (51KB, 1288x1922px) Image search: [Google]

51KB, 1288x1922px

>>34180071
Maybe this will be of some help.

>>34183184
or if you simplify it, imagine there are no boxes or silvers, and only 3 gold balls : a, b, & c.
If you draw a or b, you win. if you draw c you lose. what's the chance of getting a or b?

the boxes and silvers are red herrings.

>>34183248
Yeah that simplifies it.

if picked box A probability is 1
if picked box B probability is 0

therefore average probability is 0.5

File: 1484548732126.png (13KB, 184x184px) Image search: [Google]

13KB, 184x184px

>be dumbfag
>get answer instantly
>see most people get wrong answer
ah thanks

>>34182436
>my imagine
SICK
funny comment, too!

File: 123081923.jpg (582KB, 1920x1080px) Image search: [Google]

582KB, 1920x1080px

50%. Final answer.

You know your box is 1 of 2.

You know one of the two has either 2 gold 1 gold left, which are equivalent here.

Or the other one of the two has 1 silver.

So it's one or the other, hence 50%.

Such Bayesian, lad.

>>34184692
It's more likely the box had 2 gold coins in it, than 1 gold and 1 silver, because you pulled a gold out.

The answer is 2/3

>>34184692
There's a higher chance you picked the box containing both golds though. Imagine if you diluted the second box with a million grays for example. Same principle

50/50?

>Scenario A
>You took a gold ball from box 1, and the other ball is gold
>Scenario B
>You took a gold ball from box 2, and the other ball is silver

Box 3 doesn't matter because there's no gold ball there to begin with.

>>34184735
that's not bayesian, bayesian is improving upon prior probability with experimented data

It's 2/3, it's really not that difficult, holy shit.

You have a gold ball, you have chosen a box, which box is it? Well, it's definitely not the silver one. Is it the Gold/Gold one, or the Gold/Silver one? Well, since you took a gold ball out, odds are you chose the Gold/Gold box.

The answer is 2/3

You guys it's stupid, it's 50%.
The question asks what the likelihood that you have the box that contains 2 golds is. The silver-silver is disqualified.
You picked one box, there's only a 50% chance between the two boxes.
The identity of the balls does not matter at all, what matters is the identity of the box.

>people incorrectly assuming that drawing a gold ball retroactively determines which which box you took
>people simply stating there's 3 outcomes, GG, GG, and GS, with no regards for dependency or causality

You have three outcomes, but each individual GG pair is only 25% likely to be chosen while the GS pair is 50% likely to be chosen, because the possibility to choose the silver ball first from the middle box is impossible.

You have a tree diagram that looks like this:
Choice 1:
box A (50%) because we intrinsically know there are only two possible boxes to pick, it is impossible to have picked the all silver box)
Choice 2:
-Gold Ball 1 (50%, positive result) 50%*50%=25%
--Gold Ball 2 (50%, positive result) 50%*50%=25%
Choice 1:
Box B (50%)
Choice 2:
-Gold Ball (100%, as we know it is impossible we pick the silver ball first)
50%*100%=50%

both 1/2 and 2/3 are correct as the question is open to interpretation and whether or not there's an extra layer of chance/choice.

If you pick a ball from the box at random then the answer is 2/3 as per
>>34183248

Otherwise if the box is designed in a way where you always get one particular ball first (like a tube containing tennis balls) OR if you purposely pick out a ball in a specific way (e.g. you pick the one on the left first always) then the answer will be 1/2.

TLDR the question isn't specific enough to have one true answer.

Why in the name of Greek buggery are you spastics arguing this?

Inb4 300+ replies

>>34180071
It's 50/50. Since either a silver ball will remain or a gold one will remain.

2/5 or 40%
Originale comment

>>34184865
R u mad? Prior probability of picking one with 2 golds is 1/3. Posterior is 2/3 = ((1)*(1/3))/(1/2)

>>34180636
It's not like the fucking Monty Hall. There are two boxes with gold balls and you picked one of them. The third is irrelevant because the problem doesn't allow you to pick it.

The problem attempts to mislead the reader by presenting false outcomes which have a 0% chance of being selected based on the assumptions and limitations of what's already happened partway through the problem.

Here's the problem reworded removing all the impossible outcomes:
>there are two boxes in front of you
>one box has 2 gold balls, one box has a gold ball on the top layer and a silver ball on the bottom layer which can only be picked up after the gold ball on top has been picked up
>you first select one of the two boxes at random
>you then pull out a gold ball
>what is the probability the other ball is also gold?

There are 3 gold balls, each of them equally likely to have been chosen. The ball in your hand is either:

Gold Ball A, comes from Box #1
Gold Ball B, comes from Box #1
Gold Ball C, comes from Box #2

3 equally likely scenarios. If you picked Gold Ball A or B you will pick a second gold ball. If you picked Gold Ball C, the second ball you pick will be silver. 2 out of 3 scenarios = Another gold ball.

Pretty cut and dry 66.67% chance that your next ball will be gold.

>>34186000
>3 equally likely scenarios
False, because these 3 scenarios are first dependent on the box that's chosen. See: >>34181035

>>34186017
You have a gold ball in your hand. There's only 3 ways it got there: You picked one of the three gold balls.

Stop worrying about box #3. It's impossible that you picked a ball out of it, so it's irrelevant.

>>34186000
There are 3 scenarios:
You pick box 1(50%) then ball 1(50%) 25% chance total of this scenario

You pick box 1(50%) then ball 2(50%) 25% chance total of this scenario

You pick box 2(50%) then the gold ball(100%) 50% chance total of this scenario

Do you see how first having to select the box before picking a ball out of it modifies the probability of each individual outcome? That despite there being 3 outcomes they are not equally likely?

The third box is a lie the answer is 1/2

>>34186049
Anon, that solution already excludes the ability to pick box 3.

2/3
Tfw high IQ robot

File: Autism.jpg (7KB, 192x185px) Image search: [Google]

7KB, 192x185px

>>34180071
Well, if you've picked up a gold ball, there are three ways that might have happened. You might have picked the first box and the first ball, or the first box and the second ball, or the third box and the third ball. There are only two outcomes; the next ball being gold or silver, but there are three ways that outcome might be reached and all are just as likely. So it's a 2/3 chance that you picked the first box and will pick up a gold ball next.

You might say that there's a 50/50 chance, because there are two boxes you could have picked from (since you know you haven't picked the third). But since you've picked up a golden ball, you're already past that 50/50 point, and a new development has happened that has given you new information which allows you to predict what is more likely to happen next.

There's a 50/50 chance you picked the first or second box. But if you picked the first one, then you had a 100% chance of picking a gold ball, while if you picked the second you had a 50/50 chance of getting the silver one. To pick the first box and a gold ball is just one 50/50 chance, but picking the second box and also a gold ball is two separate 50/50 chances; picking the box and also the gold ball. So if you've picked up one gold ball, it's twice as likely that you got it through a 50/50 chance than though two 50/50 chances, and thus it's a 2/3 chance you picked from the first box and you're next ball will be a gold one.

I don't know if this is coherent or rambling too much since I'm really just trying to explain it to myself.

Did I get it right?

>>34186099
>if you've picked the second box there is a 50/50 chance you pick the silver one
No there isn't, because the problem prohibits the possibility of picking the second box and then picking the silver ball first.

2/3 of the gold balls are paired with gold balls. The answer is 2/3

>>34186128
You're looking at it purely through the lense that there are three outcomes, GG, GG, and GS, but falsely making the assumption that these are equally likely and neglecting the fact that the probability of each of these pairs is influenced by which box you pick first.

>>34186122
It's not that you 'have' a 50/50 chance but you 'had' a 50/50 chance. This was before you picked the golden ball. Now that you have the golden ball though, you can look back and say that it is more likely you got it from the box where you were guaranteed a gold one, rather than the box that only had a chance of giving you a golden one (and you can say for certain that you didn't pick the box that had no golden balls). Once you've picked the golden ball you can say; there's a 2/3 chance I got it from the first box, a 1/3 chance I got it from the second, and a 0/3 chance I got it from the first.

The possibility of picking the second box and then picking the silver ball first only became 'prohibited' after you picked out the golden ball, before that it was entirely an option so you have to take it into account if you're going to consider the most likely way you got the golden ball.

>>34186274
You have to select a box before you select the ball. You're saying that because we always draw the gold ball, were retroactively more likely to have picked the first box.

Can we agree on this in principle anon, that before any balls are drawn, we first pick either box 1 or box 2, which is a 50/50 decision, and then pick a ball after?

BEST ANSWER HERE
________________________________________
it's 2/3
the first ball had to either come from the first or second box
however knowing only that you picked a gold ball, you can see that you had a 75% chance of doing so between both of those boxes
since the first box has twice the gold ball opportunities as the second box, you can conclude that the probabilities of choosing the two boxes respectively are 2/3 and 1/3
this means that, since you can't see which box you chose but it had to be one of the two, you have a 2/3 probability of drawing the second ball from the first box and a 1/3 probability of drawing the second ball from the second box
this means that the two outcomes are weighted thusly:
% to draw SILVER = 1/3
% to draw GOLD = 2/3
since the problem is asking what the probability of drawing the second ball as a gold ball is, you can tell from the information in the first draw that the answer is 2/3 or 66% chance of drawing another gold ball rather than silver (it's implicit that previous chance of selecting box 1 was higher than selecting box 2 if in fact, as the question tells you, your first pick for sure had a gold ball)
the thread has been closed since before the first comment
this and all threads like it are highly effective brain bait and the government is forcing all you lazy people to do math in the name of trivial problems that you don't realize subject you to the subtle intellectual brainwashing of the world
wake up

>>34180071
50%
I hope everyone who says 2/3 is trolling, otherwise kys

>>34186324
>it's implicit you're more likely to pick box 1 than box 2
This assumption is false. You're equally likely to pick box 1 or box 2, the problem merely prohibits picking the silver ball from box 2 first.

Alright so you take one ball and it's gold, that has already happened. Now you take a second ball, and it has to be gold also. If you took a gold ball from box 1, it's gonna be gold again. If you took it from box 2, it won't be. So in one out of two instances you will draw gold. 50%.

Learn to read you idiots.

To everyone answering 2/3, what is your response to the Monty Hall problem?

>>34186362
let me try to give you an example to help you understand how the first box is more likely
imagine you had a choice between drawing a single grain of RED sand from either box 1 or box 2
however, box 1 is entirely filled with millions of grains of red sand and box 2 is filled with millions of grains of normal yellow sand with only a single grain of red sand
do you understand why it's safe to assume, given that you drew a red grain of sand, that the box you drew from was probably the one filled entirely with red sand?

>>34186385
the second ball doesn't have to be gold, it's only asking you to find the probability that is will be gold

>>34180071

>>since the first box has twice the gold ball opportunities as the second box, you can conclude that the probabilities of choosing the two boxes respectively are 2/3 and 1/3

This is where you are wrong. The text states you already picked a gold ball. There are no possibilities of picking a gold ball. The text guarantees it. The probability regards wether you picked it from two choices of boxes. So 50%.

>>34186393
But that logic is not conducive to the logic used in the problem. To rectify your recreation of a similar problem, it would be impossible to draw anything but the red grain of sand first from the box with one grain of red sand and 1000 grains of yellow sand.

>>34186393
>misrepresenting the problem
You first pick one of two boxes. You then pick a red grain of sand out of the box. The probability you pick a second one red grain of sand from the box is 100% I'd you picked the all red box, and 0% if you picked the mixed box.

The contents cannot impact the probability of which box is selected. Rather the contents is dependent on the box selected.

>>34186439
see
>>34186393

>>34186442
actually in the case of two boxes of sand, even though it's assumed you pick a grain of red sand, you can measure the probability of each red sand as a separate equal outcome
so in other words you have a 1million/1 chance of picking a red sand grain from the red sand box vs the normal box with only 1 grain of red sand

it's like if you select a random white person from the world you can assume they're from a country that is majority white rather than from ethiopia

>>34186393
Like I already said previously. It is indeed harder to get the red grain from box 2, but that doesn't matter because the sentence guarantees that you already took a red grain from either box 1 or 2. Even if in an example you manage to take the red grain from your box 2, the chances to get another red grain is impossible for box 2, and a fact from box 1. Two possibilities... 50%

>>34186480
even if you pick a white person that happens to come from ethiopia, before you know exactly where they come from it's safer to assume they don't come from ethiopia, hence >50% chance from not ethiopia

>>34186487
You're incorrectly modeling the problem because you're analyzing the probabilities out of sequence. Given the parameters of the problem, there is a 100% chance you always pick a red grain of sand first, all that matters then is the colour of all the remaining grains, which is entirely dependent on which box you selected, which is a 50% chance.

>>34186500
yes and outcome of two possibilities indeed (either box 1 or 2) but the chance of winning the lottery is also an outcome of 2 possibilities, wherefore one outcome obviously does not equal the second (not 50/50)

>>34186503
That's not the same internal logic that the problem uses however.

Refer to >>34185998

>>34186528
Are you joking? The chances of winning lottery are not 50 50 lol you have millions of possibilities, but in the proposed exercise there are only two options... Remember, they don't ask about the ball, the ask about the box (between 2)

>>34186313
>You're saying that because we always draw the gold ball, were retroactively more likely to have picked the first box.
You weren't more likely to pick either box. It was a 50/50 chance when you picked. At that point there was still a chance you would get a silver ball. However, when you have the golden ball in your hand, then you can say it is more likely that one thing happened than the other. You can say that you either picked the first gold ball, the second, or the third, and none of these are more or less likely than the other. It's a 1/3 chance that you got either of the balls. But two of those balls cause one outcome, and only one causes the other, so you can say the first outcome is more likely.

>Can we agree on this in principle anon, that before any balls are drawn, we first pick either box 1 or box 2, which is a 50/50 decision, and then pick a ball after?
No.

Before any balls are drawn, there's no reason you can't pick the third box. It's only when you've picked a gold ball that you can dismiss that possibility. When you've picked the golden ball, only then can you say 'I've either chosen from the first box or the second'.

Okay anon, let's rephrase the sand example.

You have two boxes. A box with only red sand in it, and a box with one grain of red sand and the rest is yellow. You're guaranteed to always pick the red grain of sand regardless of box. What is the probability the remaining grains are red or yellow?

In actuality, the question is merely "what is the chance you pick the all red box or the mixed box?". The first grain always being red is purely misdirection.

>>34186516
you have to reconcile your line of reasoning with >>34186503

>>34186530
it is the same, because if you try to apply >>34185998 to a situation where you have a box full of gold balls and a box with all silver besides 1 gold, you wouldn't still try to assume a 50/50 split
the reason for that is because even though you assume a gold ball, you will need to try a disproportionate amount of times to get that 1 out of a million gold ball NOT vs the chance of getting a silver ball but RATHER vs the chance of getting a gold ball that comes from the other box

>>34186581
No anon, because the problem PROHIBITS the possibility you pick the mixed box but then pick the silver ball from it. If you pick box 2, there is a 100% chance you pick the gold ball from it first, regardless of the ratio of gold balls to silver balls.

>>34186581
For the third time... The sentence guarantees you get the red ball... Doesn't matter how many there are in either of the two boxes

>>34186562
yeah thats the point
and yes, they ask about the box but one box had a higher chance of being drawn because of the ball chosen
you can tell from how the 3rd irrelevant box is not weighed equally to the others
because that box is not weighed equally you can conclude that boxes of different values of gold ball amount may likewise not be weighed equally
if someone wins the lottery and there are two lotteries, one where everyone wins, you can assume it was more likely for a winner to have drawn from the lottery where everyone wins

>>34186609
yes
>you will need to try a disproportionate amount of times to get that 1 out of a million gold ball NOT vs the chance of getting a silver ball but RATHER vs the chance of getting a gold ball that comes from the other box
>NOT vs the chance of getting a silver ball but RATHER vs the chance of getting a gold ball that comes from the other box
just considering the gold balls and nothing else

>>34186614
You're modeling the problem wrong. You're assuming that you're more likely to pick box 1, rather than one of the supposed outcomes from picking box 2 being in actuality impossible.

>>34186480
>The probability you pick a second one red grain of sand from the box is 100% I'd you picked the all red box, and 0% if you picked the mixed box.
No it's not. It's 99.999% the first box and 0.001% the second box. This means both boxes are options, but one is far more likely. It's a 99.999/0.001 chance. It's not a 100% chance, but it's also not a 50% chance. The golden balls aren't so extreme, but they're also not a 100% or 50% chance, it's a 66.666...% chance.

>>34186614
There are no chances of getting red ball... The sentence makes it a fact. Wether it's from box 1 or 2, it says you DO get a red ball, regardless of how many are there in each box.

>>34186613
>>34186642
>>34186655

https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
>Solution...the probability is actually
2
/
3
wikipedia explains it best read that if you're actually confused

>>34186645
See
>>34186577

You're not working through the problem correctly.

>>34186669
They are not asking for the same thing... You seriously need to read the exercise nice and slowly

It is 50 percent.

Really all the example tells us is that we picked either the first or the second box.

>>34186682
>>34186681
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

>>34186669
>The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the GG box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin

This is not conducive to the problem, because the problem in the OP outlines there is a 100% chance you drew the gold coin from the mox box.

>>34186562
There are millions of possibilities in the forms of the numbers you choose, but only two outcomes; winning or losing. There are less possibilities with the balls and boxes, but still more than two. The order in which you can pick out the balls is equivalent to the numbers drawn in a lottery. You can pick the first and then the second ball (gold), or the second and then the first (gold), or the third and then the fourth (silver). There are two outcomes, getting a gold or silver ball, but there are three possible ways to get there; two leading to the first outcome and one leading to the second. It's thus more likely that you get the first outcome.

The entire problem comes down to whether you assume that the wording means picking the silver ball first from the mixed box is impossible or not.

File: paradox.png (36KB, 947x426px) Image search: [Google]

36KB, 947x426px

can't you faggots just shitpost on /sci/?

Here's the answer, brainlet mode. If you don't understand this consider suicide.

>>34186710
I don't care what wikipedia says.

You either pick box 1 or box 2

If you pick box 1 the order will always be gold gold and if you pick box 2 the order will be gold silver or silver gold

since we are told that the first one was gold, the orders we are left with are gold gold(box 1) and gold silver(box 2).

>>34186386
it is the same principle of gained information but inverted to infer about the doors you instead didn't choose

>>34186712
>>34186734
end with
>>34186738

>>34186742
lol

>>34186734

that is exactly what the wording says. this isn't pick between 4 balls. it is pick between 2 boxes.

>>34186712
This. The "flaw" that indicates it's 2/3 isn't concrete, and lies entirely in the problem's ambiguity. However when modeled under the assumption that SG is an impossibility we're left with a 25% chance to draw GG, a 25% chance to draw the second GG, and a 50% chance to draw GS

>>34186722
I think you aren't even trying to understand another explanations besides yours. Let me tell you and please try to understand. The sentence guarantees you get a red ball... There are no chances here... How many red balls are in each box is irrelevant. You already get one wether it's from box 1 or 2. Read the exercise carefully.

Can someone explain to me why this anon's model:
>>34185998
Is not accurate and cannot be used to conduct a practical experiment to evaluate the probability?

6 possible outcomes
Gold, Gold
Gold, Gold
Gold, Silver
Silver, Gold
Silver, Silver
Silver, Silver

>Gold ball was picked
3 possible outcomes
Gold, Gold
Gold, Gold
Gold, Silver
2 lead to another gold ball

2/3

>>34186765
you should physically try the experiment
fill a jar with jizz then jizz into a jar of water
run your hand through and see how many tries it takes you to scoop jizz from the jar of water
then kill yourself for being a faggot

>>34186796
wrong

3 possible outcomes

box 1
box 2
box 3

we learn that it cannot be box 3

what remains

box 1
box 2

1/2.

>>34186796
>gold ball was picked
The gold ball being picked occurs after the box is picked.
>box 1 50%
Gold ball 1 50%(25% total)
Gold ball 2 50%(25% total)
>box 2 50%
Gold ball 100% (50% total)

>>34186742
>You either pick box 1 or box 2
Yes. You could have picked box 3, but since you've picked a gold ball you know that you didn't.

>If you pick box 1 the order will always be gold gold and if you pick box 2 the order will be gold silver or silver gold
Yes, but you're ignoring something. If you pick box one, the order can be either gold1 gold2 or gold2 gold1. These result in the same outcome for you, but they are different ways of reaching that outcome.

>since we are told that the first one was gold, the orders we are left with are gold gold(box 1) and gold silver(box 2).
The orders you're left with are gold1 gold2 (box 1), gold2 gold1 (box 1), or gold3 silver1 (box 30)

You cannot treat gold1 and gold2 as if they were the same object. There are three gold balls, each is a distinct entity. There are two outcomes from picking one of these gold balls, so two of them seem identical, but there are still three of them.

>>34186807
>run your hand through and see how many tries it takes you to scoop jizz from the jar of water
The problem assumes you always scoop the jizz first try. The ratio is purely misdirection.

>>34186818
>If you pick box one, the order can be either gold1 gold2 or gold2 gold1.

That is irrelevant. The question is not 'what is the chance you picked box 1'. but 'what is the chance of drawing another golden ball'.

>>34186818
>Yes, but you're ignoring something. If you pick box one, the order can be either gold1 gold2 or gold2 gold1. These result in the same outcome for you, but they are different ways of reaching that outcome.
Yes but also by the same principle each of these are individually less likely, and in total these both together have an equal chance based on picking the mixed box.

>>34186813
>>34186816
See
>>34186710
>tfw too intelligent to understand this paradox

>>34186790
because the separation of the boxes is key to the likelihood of each outcome and when you combine them all it changes the answer

>>34186843
If you pick box 1, you will ALWAYS pick two gold balls. If you pick box 2, you will NEVER pick two gold balls. The question is purely what is the chance you pick box 1 or 2, and the fact you pick the gold ball first is merely misdirection.

>>34186816
Exactly. And how many red balls has each box is irrelevant. The exercise only asks wether you get it from box 1 or box 2. Two choices, 50%

>>34186862
But that anon's model does separate the boxes?

There are two boxes, and the second box is rigged to always give the gold ball first.

Why do these "riddle" threads always max out? The answer is 1/2, but people keep posting contrived explanations for some other wrong answers. Please tell me honestly, are you just trolling to keep the thread going or ate you genuine?

Here's the assumption the entire problem hinges on which is open to interpretation:

Is the box with one gold and one silver ball rigged to always have the player pick the gold one first and the silver one second, based on the assumption we never pick the silver ball first?

File: maxresdefault.jpg (37KB, 1280x720px) Image search: [Google]

37KB, 1280x720px

>>34186886
It's proven to be 2/3 but some people just dont get it

>>34186863
yeah but picking a silver ball from box 2 is not a possibility.

therefore you cannot say that getting a gold ball shows a 2/3 chance of it being box 1.

>>34186874
Exactly. And how many lottery winners in each lottery is irrelevant. The exercise only asks wheteher you get it from lottery 1 or lottery 2. Two choices, 50%

>>34186886
https://en.wikipedia.org/wiki/Monty_Hall_problem

>>34186883
he changes it by assuming the second box has the gold ball on a top layer, when the original 50% chance of drawing that ball rather than silver would prohibit that change

>>34186917
>When the original 50% chance of drawing that ball rather than silver would prohibit that change
There isn't a 50% chancel of drawing that ball, there's a 100% chance of drawing that ball, because the problem prohibits the possibility of drawing the silver ball first, just like it prohibits picking the all silver box.

>>34186917
Wrong paradox anon
>>34186710

>>34186913
picking a silver ball is a possibility and it has to be removed from the end probability since a gold ball needs to be picked in order to satisfy the problem

>>34180071
50%. How are you guys this stupid?

>>34186937
>picking a silver ball is a possibility

no.

if you were to set the example up in practice. then yes. every 2/3 you would get a gold ball it would be box 1.

as it is given in the OP, no.

>>34186917
Kek. Then by your lottery example, you're given a winning ticket and a losing ticket. Of course you have 50% chance. But that's not how lottery works doesn't it

>>34186937
>picking a silver ball is a possibility and it has to be removed
False, picking a silver ball is not a possibility, for the same reason picking the double islver box is not a possibility.

>>34180071
fifty-fifty, you either win or don't win

>>34186765
You have to consider HOW you got that red/gold ball. The question is a hypothetical situation in which you have already gone through the 50/50 chance of picking one of three gold balls from six gold and silver balls. You now have a gold ball, which could have been any of the three. You know that one box has two gold balls and another only has one. You can't say for certain which box you picked from, since you have yet to pick the next ball, but you can say that it is more likely you picked from the box that had two.

>you aren't even trying to understand another explanations
The other explanations cannot be both understood and believed, because if you understand them then you understand why they're wrong; if you think they're true, then you don't really understand. You need to know that you are objectively wrong, since the 2/3 chance is confirmed by experiment. You can try it yourself. If you don't understand something, try to understand it. It's fine not to understand it, so long as you recognize it. Don't just reject reality.

>>34186904
This. It's based on interpretation of the question's assumptions.

(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)(link to wikipedia here)

>>34186990
>The question is a hypothetical situation in which you have already gone through the 50/50 chance of picking one of three gold balls from six gold and silver balls
False, it's the hypothetical situation you have already gone through the 50/50 chance of picking the 2 gold box or the mixed box.

The first decision is which box you want itself.

Then picking a ball.

THEN picking the second ball.

Your model only works if the person does not first pick a box, and instead merely picks from one of six balls which merely has a pair attributed to it.

>>34186990
And yet you're still ignoring the FACT that it doesn't matter how you got the red ball... All it matters for this exercise is that you already have one. It's not even implied... It's literally there. The FACT is that you already got a red ball from two boxes. Now the question is... Will you get another red ball from two choices of boxes? You have two choices... In one you get a red ball, in the other, you don't. So, 50%

If you reach into my anus and pull out a brown ball, the probability that the next ball is also brown is 100%.

>>34186904
>>34186904
>>34186904
>>34186904
>>34186904

You stupid bitch Bois

I have thought about it again and figured out a way to make it easily understandable.

Imagine box 1 with 100 gold balls instead of 2
Imagine box 2 with 99 silver balls and 1 gold ball.

Now if you picked a golden ball from one of the boxes, chances are that 100 out of a 101 times you picked box 1.

>you can't even use a model to settle this using practical probability because the two schools of thought dictate the use of two different models

>>34187096
No anon. You first pick one of two boxes.

If you picked the box with 1 gold ball and 99 silver balls, you'll ALWAYS draw the gold ball from it. Which box is first selected is misdirection and the question is ACTUALLY "what is the probability you picked the purely gold box versus the mixed box?"

>>34187117
This.
Two options.
1. If you selected the box 1, you WILL get another gold ball.
2. If you selected box 2, you WILL NOT get another gold ball
50%
Some people don't know how to read what is being asked

sometimes i feel pride for the amount of irony that can exist on this board and yet other times i feel nothing

>>34187117
>>34187166
I think the example is designed to make you backtrack and think what is the possibility that I picked from box 1 or box 2.

That is how you get the 2/3 answer.

>>34187224
But that's incorrect. The box selection occurs first. It only appears that the box selection's probability occurs retroactively, influenced by observing the gold ball, because of the order the question is worded.

>>34187166
>>34187224
The answer is 2/3. This is the same as Bertrand's box paradox

>>34187259
Tell me with your own words why I'm wrong

>>34187259
Bertrand's box paradox can yield both results dependent on the valid assumptions the reader does or does not make and the model they employ.

Either the GG, GG, GS model, or the B1/B2 model under the assumption it is inherently less likely to draw an individual GG pair, and the combined probability of drawing either GG pair is the same as drawing GS.

>>34187275
Ok. Which post is yours?

File: 1481467000879.jpg (88KB, 920x960px) Image search: [Google]

88KB, 920x960px

It's 1/2 guys

Two boxes you'll get another ball of the same color
One box you won't.

If you got a silver ball from the box you chose, you either got a box with two silver balls or just one. You can forget about the box with two gold entirely. Why are you autists over-complicating this shit?

>>34187304
> and the combined probability of drawing either GG pair is the same as drawing GS.

but wouldn't this go against how it would be if you were to do it in actuality?

>>34187306
>>34187166

Most original quote

>>34187166
Those are two outcomes, not two options. There are three options leading to two outcomes.

Outcome 1: You selected box 1 and get another gold
Outcome 2: You selected box 2 and get silver

Option 1: You picked the first gold ball, leading to Outcome 1
Option 2: You picked the second gold ball, leading to Outcome 1
Option 3: You picked the third gold ball, leading to Outcome 2

You know that only Outcomes 1 and 2 are possible, since you have a gold ball and therefore couldn't have selected box 3. Outcomes 1 and 2 were both equally likely when you made the choice, but now that you have made the choice and have a gold ball you can predict which choice you probably made. Outcome 1 is more likely because, of the three equally likely Options, two lead to it.

>>34187166
1. You WILL suck on my glistening golden balls.
2. You WILL NOT forget to put my meaty hog in your anus and mouth.
100%
Some people forget to put my fat meaty hog in their anus and mouth

>>34187331
It depends entirely on how you model the question, which is based on assumptions made.

>>34187354
No, because having a gold ball doesn't alter the probability of picking box 1 or 2. It's the opposite.

>>34187359
Well it does say random in the OP, while your model would be two boxes programmed to spit out a golden ball first.

>>34187354
What you propose is true IF the question was "what are the chances you get 2 gold balls?"
But that's not the case. You're not reading the proposed question carefully. Read again and reply

>>34187328
Wait no it is 2/3

You're more likely to get another ball of the same color

(same poster)

>>34187481
You're way behind of the level of the discussion. Kindly read the latest posts

>>34187491
Not really. We haven't come to a consensus yet

>>34187491
Kindly suck my smallest dick

>>34187514
>question with an objectively correct answer
>must come to a consensus
i'm american but i know for sure you're american too

>>34187531
How many do you have? I would like to suck the biggest one

>>34187442
It depends on if you assume that picking the mixed box dictates you always receive the gold ball first from it.

>>34187549
Yes, but you why would you assume that?

So sad that people tend to not further reply when they can't argue against the right answer

>>34187383
Having a golden ball means you've already picked box 1 or 2. It also means you've already picked ball 1, 2 or 3. That second part is what gives you the information you need to say box 1 is more likely.

It's equally likely to pick box 1 or box 2. So that seems like a 1/2 chance. But box 3 is still an option before you pick the golden ball, so there's actually a 1/3 chance of picking any individual box. It only becomes 1/2 in retrospect, when you have the golden ball and can say for certain that you didn't pick box 3.

You have to work with the information you have after you've picked a golden ball. You have to understand that we're not saying you were more likely to pick box 1 over box 2, but rather that after having made the choice, you can deduce what choice you probably just made.

It is equally likely that you picked ball 1, ball 2 or ball 3. So that's a 1/3 chance for each.
Ball 1 and ball 2 correlate to box 1, while ball 3 correlates to box 2.
You have a 1/3 chance of having picked ball 1, so that's a 1/3 chance you picked box 1
You have a 1/3 chance of having picked ball 2, so that's a 1/3 chance you picked box 1
You have a 1/3 chance of having picked ball 3, so that's a 1/3 chance you picked box 2

That adds up to a 1/3 + a 1/3 chance of picking box 1, which is 2/3, and a 1/3 chance you picked box 2. And thus a 2/3 chance the next ball you pick will be gold.

>>34187568
Based on the idea that we select the box BEFORE observing what the first ball we draw is, but we also need draw silver.

>>34187546
I have 3 full sets of penis and balls. My largest penis has 2 golden balls, the smallest has 2 silver balls, and the middle penis has 1 gold and 1 silver ball. If you're sucking one of my penises and reach underneath to fondle a gold ball, then you slide over to fondle the partner ball inside the same scrot, what is the probability that the scrot partner is also gold?

>>34187585
Well you do select the box first.

Then you pick a ball at random and it turns out to be gold.

The task we are given is now to see if we can extrapolate the probability of drawing another golden ball from this information.

It is a 50/50 percent chance of picking box 2, but if we do pick box 2 there is another 50/50 of getting a golden ball.

Therefore for it to be box 2 and a golden ball would be 1/2 * 1/2 = 1/4 probability.

To pick box 1 and a golden ball is 1/2 probability.

The chance of picking box 1 is then twice as big as having picked box 2. (in the example).

wait that doesn't add up. now I am confused. nerds aid me.

>>34187686
>>There is another 50/50 chance of getting gold
This is where you are wrong... There are no chances of getting gold. You get gold as a FACT

>>34187686
You did it backwards. Because we draw a gold ball, we can extrapolate we could have only drawn it from box 1 or 2. We also know that we can only have selected the gold ball from box 2 despite it being mixed, so in actuality there's two different pairs of drawing 2 gold balls, but each are half as likely as drawing gold then silver, because box 2 only has one possible outcome.

>>34187581
>But box 3 is still an option

And this is where YOU are wrong. The exercise makes a FACT you get a gold ball

No one here reads correctly

>>34187581
The greatest trick the devil ever pulled was convincing the world that there is a 2/3 chance the second ball is gold.

>>34187451
You haven't replied buddy

>>34187897
No, there is a possibility in the BEGINNING you could have picked box 3, and then only after observing we picked a gold ball it becomes evident we couldn't have picked 3.

>>34187354
You haven't replied buddy
**** Meant this post

>>34187923
You aren't reading the question completely. The proposed exercise forces you to get a gold ball. It's impossible to get one from the third box, so it's out of the equation.

File: duh[1].jpg~c200.jpg (8KB, 200x200px) Image search: [Google]

8KB, 200x200px

>>34187938
You haven't succ'd buddy
**** Meant this dicc

The entire question revolves around whether you regard SG, SS, and SS as impossibilities from the start or if they're possibilities until we observe the first gold ball.

>>34187959
Nice dodging. Not so nice rotten piece of barbage

>>34182578

that doesn't work because you only have one of two choices. If you reached into universe 1 then you can't get universe 3

>>34187897
No it doesn't. It says that you take a ball from the box at random, not that the ball you picked was predetermined. The question then asks if, in the case that you get a gold ball, what is the probability that the next ball you take will be gold.

It doesn't even matter in the end, because whether the first ball is predetermined or not, the question is about what happens next and you didn't refute my answer to that question.

>>34187938
See >>34188039

Stop being so impatient.

File: GoldSilver.png (54KB, 982x1459px) Image search: [Google]

54KB, 982x1459px

>50/50
Can these people even be considered human?

>>34180071
P(B|A) = P(A and B) /P(A)

A=1st Gold
B=2nd Gold
P(A) = 1/2 because there are six different balls and half of them are gold.
P(A and B) = 1/3 because it would have to be the double gold box

(1/3)/(1/2)= 2/3

Not even going to bother reading the thread. It astounds me how some people think they can solve problems of probability using intuition, when probability is just another field of mathematics with already proven formulas you can use to your advantage.

Let's start with the definition of conditional probability, which is a field of probability that focus on the probability of something given that something else has already occurred. Whether this agrees with your intuition or not, this is a field of mathematics that has been studies heavily and whose results can be applied to the real world. P(A|B) means the probability of A given B. The formula is P(A|B)=P(A and B)/P(B). The question is asking us the probability of the second ball being gold given that the first one is gold, so A= the probability of the second ball being gold, and B= the probability of the first ball being gold. First lets look at P(A and B), which means the probability of both balls being gold. Since there's only once box where this is possible, the probability is 1/3. Next, the probability of the first ball being gold, P(B). Now this gets a little trickier. To find this probability, we have to consider the probability of the first ball being gold for each box, multiply that by the probability of picking that box, then add them together. Since each box has an equal chance of being picked, we have (1/3)(1) + (1/3)(1/2)+ (1/3)(0). So basically we have a 1/3 chance of picking the first box, where we are guaranteed a gold ball, but also a 1/3 chance of picking a box with no gold ball, and so on. Finally, putting it all together, P(B) is 1/2. And the probability of P(A and B) is 1/3. So using our formula, P(A|B)= (1/3)/(1/2)=2/3. So the probability is 2/3.

>>34188108
Oh fuck I didn't notice those typos.

Guess I'm one of the subhumans too.

>>34188039
>Not that the ball you picked was predetermined
>From op's pic: IT'S A GOLD BALL

It must be a gold ball.
That s not being questioned
It's a fact
You're still failing to recognize that

>>34188153
This is true but that's not what is being asked! Jesus these people
Read the question

There are only 2 answers for the question

You are going to pick a second ball from one of these boxes
1. If you picked it from first box, then you WILL get another gold ball
2. If you picked it from second box, then you WILL NOT get a gold ball

It's no more complicated than that.

>>34188158
The picture is a hypothetical scenario in which you are playing a game and are already halfway through. In this case you have picked up a gold ball. Like the guy above said, this is a practical problem about 'probability of something given that something else has already occurred'. It is not about magic and predetermination. Try actually looking up the Bertrand's box paradox and seeing how it applies to the real world.

>>34188234
The question is about PROBABILITY. That is the exact word it uses. It is not asking how many outcomes there can be, it is asking about the probability of a certain outcome occurring under the conditions described.

It's so fucking simple.

>>34188234
Using this logic, imagine you have 3 boxes: one with 1000 silver balls, one with 1000 gold balls, and one with 1 gold ball and 999 silver balls. You pick a ball from a box at random and it's gold. What's the probability of the next ball you pull from that box being gold? Intuitively, it really shouldn't be 1/2 anymore, because the chance of pulling a gold ball from the 1 gold ball box is very low.
But you might say that this is a completely different problem from the OP. But, if you picked from the 1000 gold box, then you will get another gold ball. If you picked from the box with only 1 gold ball, then you won't get another one. It's the same logic you're using. So either the answer must be the same, 1/2, or your logic must be wrong. If you really think the answer is 1/2, I encourage you to raise the number from 1000 to 1 million and think if the answer is still really 1/2, or to try the experiment for yourself.

File: 20941-31845-24452.jpg (16KB, 480x360px) Image search: [Google]

16KB, 480x360px

Where the hell are you people getting your numbers from?

It's 50/50. The third box can be disregarded because you already picked the gold one, which leaves you a 50% chance of having had picked the all-gold box and a 50% chance of having picked the gold-silver box.

How can it be more complicated than that?

>>34188405
Just look at the OP's picture and disregard the right box.
You picked 1 gold one, there are 2 gold and 1 silver left, it's 2/3

>>34188514
That would be the case if they were all in one box

But there are two boxes, and you've either picked one or the other. 50/50.

>>34188338
See
>>34186577
It's 1/2 whether it's a million, a billion, or a trillion, because the question guarantees that if you've selected the second box, you'll always pull the gold ball from it first regardless of ratio.

>>34188258
I already know that but we are here arguing about the question OP proposed, not Bertrand box

My pussy is extremely, and I mean EXTREMELY, smelly. Does that help your little argument, boys?

People aren't analyzing the selection of boxes correctly. The fact you always draw a gold ball first does not mean it is a literal impossibility to draw a silver ball after selecting the mixed box, rather it's a case study halfway through the game after having selected a gold ball.

>>34188405
Notice how there's two boxes with two of the same kinds of ball and one that has one of each? Box 1 and 3 are actually the exact same. It doesnt matter what color they are. So if you were to choose any box, you probably chose box 1 or box 3

2/3

end
of
discussion

>>34188812
>noticing balls
lol fag

>>34188812
What the fuck are you talking about? None of that made any sense.

You picked a gold ball. There is a 50% chance it came from the mixed boxed, and a 50% chance it came from the all-gold box.

End
Of
Discussion

>>34188940
I support this post
And that's about it. Seriously guys... I've argued with around 5 replyers of this thread and everyone failed to answer. Accept this as the final answer

>>34189165
Because it's useless to keep trying to convince retards.
btw >>34188108

The answer is 50%.

t. college graduate

>>34189522
>>34188108
Again, this disregards the fact that there are two separate boxes, and pools all the balls together, leading to some terrible over-thinking.

The bottom line is, you either chose one box, or you chose the other. Simple.

>>34180071
>its another "anon tries to measure iq through shitty probability problems" episode

Seriously this isnt a show of intelligence, you dont have to be smart to get probability. People just get tripped in the wording.

If you worded it better ny dead grandma could figure out its 2/3

>>34180071
You fucking nigger.
Your thread got deleted on /pol/ and now you've come to troll the autists here.
All you retards should know it is 66.7%, which has been proven mathematically and empirically.
Piss off

>>34189733
>>34189747
It's 50/50

File: 71f7320b7953a2b0c0f9db3e7c323f95.jpg (136KB, 1024x768px) Image search: [Google]

136KB, 1024x768px

all of you have failed

Initially, I thought it's 50% as well, but after thinking about it a bit more, it seems obvious that it has to be a 2/3 chance.
Fuck all of you unenlightened 50% plebeians.

50%. The silver box just there to confuse you autistics.

>>34189894
Literally everybody has acknowledged that the silver box is irrelevant, yet it's still a 2/3 chance.

File: 1390966843628.jpg (476KB, 1600x1036px) Image search: [Google]

476KB, 1600x1036px

>tfw you've been saying 50% this whole thread and only now realize why it's 66%

Okay 50% anons, follow me on this.

First, you have to realize that the question is asking about the probability of an event occurring in a game that is already halfway completed. It is not asking you to play the game from the beginning with restrictions in place, I.E., it is not guaranteed that if you pick the second, mixed box, you'll always pull the gold ball first.

Let's start from the beginning. There's a 33% chance of choosing box 1, a 33% chance of choosing box 2, and a 33% chance of choosing box 3 correct?

Now, regarding the individuals who've selected box 2, there's a 50% chance that they select the gold ball first, and a 50% chance they select the silver ball first, yes?

Okay. So now, we have a 33% chance of selecting a gold ball from box 1, a 17.5% chance of selecting a gold ball from box 2, a 17.5% chance of selecting a silver ball from box 2, and a 33% chance of selecting a silver ball from box 3.

Now, let's isolate just the people who've pulled gold balls. You are equally as likely to have selected box 1 or box 2, but only half of the people who would have taken box 2 would have also pulled a gold ball from it, hence there is twice the chance you've taken box 1 if you have a gold ball.

The idea that there are only 2 boxes with gold balls in them, or the idea that it is an impossibility to have taken the silver ball from box 2, are incorrect assumptions. You are not playing the game from the beginning with the impossibility of selecting box 3 or selecting the silver ball from box 2 first, you are deducing the probability of a game in progress based on people who've so far, selected the gold ball first.

The possibility to have picked box 3, or the silver ball from box 2 first, were at some point a possibility, but now we analyze the game from the in-progress point of having already chosen the first gold ball.

>>34189763

If you know how to code, emulate the problem.
A Russian bro did it in the original thread and it approached 66.7%

Just google Monty Hall problem, this is just a shitty varient of it.

>>34190021
Anon will just emulate the problem incorrectly based on imposing restrictions from the start rather than acknowledging the prohibited events as retroactive impossibilities based on the decision made in the game so far.

>Choose, left box left ball
Gold
>Choose left box right ball
Gold
>Chose middle box left ball
Gold

There are 2 out of 3 scenarios in which you chose the gold ball from the left box. Thus there is a 2/3 chance you will pull a second gold ball from that box

>>34189977

Part of being smart is being able to change your mind.

I appreciate you, anon.

>>34190292
It was just a matter of realizing that it wasn't impossible to draw a silver ball first, but rather that we're only analyzing the people who drew a gold ball first, however we still use the probability values from the initial game that allows all 6 balls to be selected.

>>34189977
This is all correct, but, as lots of people here you're failing hard to read the question properly.
It doesn't matter where you took the balls, or of there are 3 million gold balls or whatever number. The exercise asks a specific thing only

>What are the chances to get a second golden ball, considering a guaranteed FACT that you already got a first gold ball from one of the boxes?

There are two possibilities.
1. you took the gold ball from left box
2. you got the gold ball from middle box

The exercise only asks the probability to take a second gold ball. Nothing else.
There's no need to ramble about the odds getting a gold ball... Because that's not the issue to be considering probability. That's a fact.

>>34180071
>>34190340

Also this is not half of a game. That's a false assumption. The game starts with the first question. The game uses as a fact that, just as you have 3 boxes, you already got a gold ball in your hands, which you took from one of the 3 boxes.

>>34190340
It's a guaranteed fact you've GOTTEN the gold ball, it is NOT a guaranteed fact you WILL get the gold ball from the second, mixed box provided you select the mixed box.

Essentially, because half the people who choose box 2 will select the silver ball first, of the people who've selected a gold ball first, 2 thirds of them will have chosen box 1.

The misdirection lies in the fact that you are not guaranteed to get the gold ball if you select the second box, rather the problem asks us to analyze from the perspective of someone who's already gotten a gold ball, and someone who picked box 2 is half as likely to have pulled the gold ball as someone who had pulled box 1.

>>34180071
50/50
It has to be either box A or B since you've taken a gold ball. If you chose box A, then you'll get gold. If you chose box B, then you'll get silver. Simple as that.

>>34190437
You're twice as likely to have box A though, because you only have half the chance of selecting a gold ball first from Box A.
>>34190395
The rub lies in whether you've assumed that you are guaranteed to select the gold ball after selecting box B, or if you're viewing the question from the perspective of someone who pulled the gold ball from box B but could've picked the silver ball.

>>34190395
>Essentially, because half the people who choose box 2 will select the silver ball first, of the people who've selected a gold ball first, 2 thirds of them will have chosen box 1.
But doesn't matter. There isn't the possibility of getting silver first. It's entirely out of the question.

>>34190395
>Not a guaranteed face you get the gold ball from the second box

Are you even reading the exercise? It is literally there. You DO GET a gold ball.

Here's how it works, dumbos.
You have a one-third chance to pick a box with two gold balls. That's your chance, period.

>>34180071
Why would I want the golden ball? What is it's value?

>>34190468
(from)>>34190437
Maybe the question isn't framed clearly enough. Whether or not you need to factor in those probabilities I don't know. The way I see it, the question doesn't imply that. Although that does make it easy, almost too easy, so that also makes me think that you're correct.

>>34190519
It says gold. It's gold. In most countries, gold is very valuable

It's 2/3 and this problem has been solved, countless times, just like Monty Hall.

50%fags are utter retards.

>>34190494
>>34190484
Okay.

You have a 1/3 chance of getting the double gold box

You have a 1/6 chance of getting the gold ball first from the mixed box, and a 1/6 chance of getting a silver ball from the mixed box.

You have a 1/3 chance of getting the double silver box.

You draw a gold ball.

THIS is where the question starts, and the perspective where we analyze the probability from.

You are twice as likely to have pulled a gold ball if you pulled the double gold box.

If it's guaranteed you get a gold ball first, and the last box has no gold balls, then the last box is out of the equation.

So, either you are left with one gold (first box) or one silver (second box). That's either 100% or 0%.

Therefore the probability is 50%.

To all the people saying 1/2, think about it this way: if the first box contained a thousand gold balls, while the second one contained one gold ball and 999 silver balls, would you pull out a gold ball and think that you are just as likely to pull out a silver ball as you are to pull out another gold one?

>>34190558
This is true but THAT IS NOT WHAT IS BEING ASKED
Seriously... Learn to read.

>>34190650
That IS what is being asked, because if you've already drawn a gold ball it is twice as likely to have come from the double gold box.

>>34190639
Yes. You either picked the first box or the second

>>34190569
Yes, it is guaranteed that you get a gold ball first, but there is a 2/3 chance that it is a ball from the first box.

>>34190639
Yes..... it doesn't matter how many balls there are. The number of balls doesn't change the outcomes, which is either you get a gold or you dont (50%).

>all these idiots saying 50%
I do wonder sometimes what it's like to be that stupid

>>34190639
No but, as you probably didn't read the whole topic where I explained why it doesn't matter how many balls there are... You already have a gold ball as a FACT, before the question at hand.
If you have taken the gold ball from.mixed balls with 1 gold ball and 3 billion silver balls, the chance to get a second gold ball (which is the question being asked) is zero.
If you took the gold ball from the left box with 9 million gold balls, the chance to get another gold ball is 100%
And right box is ignored since it doesn't fit in the premise "you get a gold ball" taken literally from the statement, BEFORE the question
You only have two options

>>34180166
There are only two possibilities, for the purpose of the question picking up A and then B or B and then A are the same.

File: balls.png (24KB, 632x515px) Image search: [Google]

24KB, 632x515px

this was posted ages ago so I made this

>>34190683
A deity did not promise you that you would get a gold ball no matter which box you chose. The situation is that you picked a box, realized that you removed a gold ball, and now must calculate the probability that you picked it from box 1.