100% balls to the wall I LOVE GOOOOOOLLLD

2/3 There are 3 ways to initially pick a gold ball. If you pick one of the 2 in the 2 gold box you pass, if you pick the 1 in the 1 gold 1 silver you will fail. Therefore you have a 2/3 chance of getting a gold.

Well it's either 100 or 0 percent depending on which of the boxes you got.

2/5 ez

>>677922989
50%

>>677923278
But you don't know which box you get anon so it's the overall probability

>>677922989
It's 50% because if you got gold on first try we can eliminate the all gray box. Now, the next ball we get will define which box we are picking from. If it's gold, box one, grey, box two. 50/50

>>677922989
Unless we are a statistician, have seen similar problems explained before, or are currently taking a mathematics class discussing this very topic, we won't get the answer. This information is useless.

2/3 should be right. I'm guaranteed a gold ball. Meaning 2/3 times I've picked the double gold box and 1/3 times I've picked the single gold box

>>677923230
That didn't take long at all

>>677923517
In order to have gotten the gold ball initially you are more likely to have picked the double gold box though. You can still pick the silver gold box and fail on your first try. So if I'm guaranteed a gold ball. It means there's a 66% chance I've picked the double gold box

>>677923230
No. Imagine there are 1 million boxes with two silver balls instead, but the same problem statement: I pick a box at random, take a ball at random, and it's gold. By your reasoning the chance of getting another gold ball would be incredibly tiny, but all the boxes that don't contain any gold balls don't matter anymore once we've established that I've already drawn a gold ball. All that matters is the chance that of the remaining possible boxes, I'll draw a gold ball.

>>677923962
However, and please correct me if I'm wrong, we don't need to factor the initial take from our probability, as the question is asking what our second ball will be. The question is not in effect until we draw a gold ball. At least that's how I read it.

>>677923962
I was wondering why the answer wouldn't be as simple as just saying 50%, seems like there was something behind this cause it's just too easy.
Thanks for clearing that up.

This is just a version of the Monty Hall problem, right? So 50%.

>>677923490
Fair enough.

Wouldn't it.be 50/50 because you pulled one gold ball so you either have the box with one gold one silver. Or two gold

>>677924493
This

>>677924589
No because there are are a total 3 remaining balls in the 2 remaining boxes. Of those 3 balls 2 are gold and one is silver

Based on how the example is worded, the answer is 50%. Word it slightly differently and its 66%.

Just imagine the middle box is all gold since the split doesn't matter.
You choose silver - 100% of gold next
You choose gold A - 50% chance of gold next
You choose gold B - 50% chance of gold next
(100 + 50 + 50) / 3 = 66.6...
So you have a 2/3 chance.

>>677924705
Yea but we aren't going by how many balls there are in both boxes we are going by whats the chance that we got box 1 or box 2 if we got one gold ball

>>677924705
But you cant use the gold balls to total balls as your numerator and denominator in 2/3, as the two golds are in one box. This makes the problem more complex than a basic probability question

>>677922989
50/50
The fact the first was gold shows that you have one of the two boxes containing gold balls.
Each option is equally likely, so your next ball is also equally likely to be gold or silver.

>>677924493

it seems like its meant to resemble that question but is actually different

>>677924320
Why would you not factor in the first box? Just because you are given a gold ball doesn't mean that you could have picked either box 50 50. The fact that you got a gold ball means you've probably (66% likely) picked the double gold box.

>>67792396
No even with a million silver ball boxes and then a gold gold and gold silver box still equates to 2/3.
Imagine you have a box with a million gold balls. Then a box with a gold ball and a million silvers. Then a box with a million silvers. If on your first draw you draw a gold ball. It's 99.999 percent sure you've picked the all gold box. Not 50 50

>>677924866
Yea but we aren't picking out of the all silver box

>>677925102
How is it 99.99%???
There are only 3 boxes

>>677925102
Because the question is asking for the second ball, and we know that it is impossible to have the third box, as decided by the first draw, limiting it to a simple binary 50/50 option.

>>677923339
>>677923517
>>677924589
>>677924493
>>677925037

These guys get it.
To all those saying it's 2/3:
You are factoring in the initial odds here.
But the initial scenario is already set up.

The all silver box is part of the problem, but by saying you grabbed a gold ball first, you have effectively removed the silver box from the problem.
The problem then becomes the all gold and half gold/silver boxes.
If all gold, you get gold next.
If half silver, you get silver next.
These are the ONLY two options left, so it is 50%

>>677925251

because the odds that you happened to pick the only gold ball in the 1 gold 99 silver box are so low that seeing a gold ball tells you that theres a very high chance you took a ball from the 100 gold box

>>677922989

Definition of conditional probability:
P(A|B) = P(A&B)/P(B)

This means: the probability of event A occurring, given that event B has already occurred.

This problem is asking us the probability of the 2nd ball being gold given that the 1st ball is gold. The 2nd ball has to come from the same box the 1st ball came from.

Let A = the 2nd ball you pick is gold
Let B = the 1st ball you pick is gold

Then, P(A&B) = 1/3, since only one box has two gold balls
P(B) = 1/2, since half of the balls are gold and you have an equal chance of picking each ball

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

Final Answer: 2/3

>>677924964
No it's the probability of the second ball being gold given that the first ball is
>>677924974
You don't know which box you have selected, so theoretically the probability is 2/3

>>677925496
See >>677925604

>>677925604
>Then, P(A&B) = 1/3, since only one box has two gold balls

That 1/3 isn't correct though. You are counting the all silver box, but basic reading comprehension tells us the silver box became irrelevant before the math even starts.

>>677925701
I do know it's not silver, making it 1/2

>>677925792
see>>677925802

>>677925251
Its 99.99 because I've just drawn a gold ball. There's no fucking way I drew a gold ball from the one in a million box. I'm sure to have chosen the all gold box.
>>677925322
You are right in the fact you can eliminate the all silver box. It has absolutely nothing to do with this question. But like I've stated a million times. It's not equally probable that we've drawn from either box. Since weve just drawn a gold ball. It's actually more probable that we have just chosen the double gold box

>>677925604
You can't have 1/3, as we discontinue the relevancy of the silver box.

>>677925802

i cant tell if this is a troll

i remember the last time we had one of these threads there were people saying stuff like this, it almost sounds reasonable, but its definitely wrong

>>677925604

this poster, for instance, has taken into account every parameter set forth in the OP in his equations and calculations, but you are trying to confuse people by taking one line of his analysis out of context

>>677925802
>>677925963

See this: >>677923765

>>677926223
Let me put it this way:

Since we KNOW the first ball is gold, we can without any shadow of a doubt, state that we did not pull from the all silver box.

This means we drew from ONE of TWO boxes.
THIS is the initial state of the question. The silver box might as well have never been posed, and the question can be rewritten to:

You have two boxes, one all gold, one half gold half silver. You just pulled a gold ball out of a box, what is the probability the next ball is gold?

The two possible out comes are:
1)You chose the all gold box, in which case the next is gold.
2)You chose the half silver box, in which case the next is silver.

Given that there are only two possible outcomes, then your odds are essentially a coin flip (not counting that the weight of coin sides affects the odds of flipping)

>>677925909
The odds of drawing the second ball would still be 50/50, because if we remeoved a gold one already, that means there's one box that has one gold, or one box that has one silver. We have to treat it as though we removed a ball from a super stae

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I'd say 1/3

I'm finishing a math degree this semester but only cucks like statistics so meh, give me some ODEs and systems to model and I'll do it.

who fuckin cares

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>>677922989
50%

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>>677925802
Ah yes, the basic reading comprehension theorem. The most famous theorem in all of mathematics. I see now that I am wrong and you are right.

>>677926664

We know this: the first ball picked was gold.

How many ways can this possibly happen?

1. It happens when I pick the leftmost gold ball in the double gold box as my first pick.

2. It happens when I pick the rightmost gold ball in the double gold box as my first pick.

3. It happens when I pick the gold ball in the gold-silver box.

There are three scenarios possible, given the fact that my first pick was a gold ball.

In how many of these scenarios will I also get a gold ball on my second pick?

In the first two.

2/3

The first probability is picking a gold ball which is 3 of 6. Then you have the probability that the next ball will be gold. As there are 2 gold and one silver the chance is 2/3 the next will be gold.

>>677927049
>word problem
>thinks you don't need reading comprehension
Go eat a dick, good sir.

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>>677926698
To help explain

>>677925604
Based on the wording, p(a&b) is 1/2. You have to completely ignore the third box.

>>677926698
Huuurr durrr there's one box gone means 50 50 next ball I'm so fucking stupid and can't read.

50/50 is actually wrong. Got a gold ball first try? Run the simulation a million times. Every time you got a gold ball first try, you will draw a gold ball again 66% of the time

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

>>677927138
THIS

75%

50%

>>677927250
This is a quantum representation of the problem. If superstates were taken into account then yes you would be correct.

>>677926664

okay lets do it your way

there is no silver box

there are only two boxes, box A has 2 gold balls, box B has 1 silver ball and 1 gold ball

think of it like this, half the time you reach into box B, you will get a silver ball

so every time you do NOT pull a silver ball on the first try, you have eliminated exactly half of the likelihood that you picked box B, but leaving the likelihood that you picked box A the same

Box A and Box B had equal likelihood before, so removing half leaves you with a ratio of 1 to 1/2 or 2 to 1, which equates to 2/3 probability

>>677927138
That's... actually something I did not consider.
Granted, I am currently a bit sleep deprived.
I concede defeat. I forgot to factor in the chances of the starting box and only considered the static chance of the second ball.

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>>677922989
Its 50% dumbasses

>>677927439
Die

>>677927328

Nope, see my follow up here: >>677927138

>>677922989
How are you fuckers not getting 50%? Its 50%.

>>677927732
I hope you don't wake up

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>>677927250
I'm laughing in real life at this absurd picture. Please draw us some more, anon

Wow I feel so dumb. It has to be a 2/3 chance of a gold ball.

OP here, everyone who has said 2/3 is correct. The question is a variation of this: https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

>>677922989
1/2

>>677927732

Yeah it's past 2AM for me and I've decided that it's more important to help people understand this than to get sleep before work tomorrow.

>>677927958
If this were quantum mechanics he'd be correct, but sadly these are not.

Fuck you're all retarded its a 33% chance

>>677927923
Not who you are replying to, but:
>don't wake up
>not asleep
k

Anyways, why be a dick to him in particular?
In this entire thread, that guy is the only one to actually admit he was wrong and gracefully bow out.

>>677922989
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

ITS 2/3

GET OVER IT

https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
>>677922989
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

ITS 2/3

GET OVER IT

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

>>677928071
How the fuck is this even possible? Wouldn't pulling a gold coin on the first try remove the third box from the problem?

50%, easy math. Only two boxes have gold, so you are equally likely to have the box with one gold as two.

>>677927138
This is incorrect, there are only two scenarios, 'left box' and 'right box' both are guaranteed to give one color, right will give silver no matter what and left will give gold no matter what.

>>677927923
That's a bit harsh...

>>677928122
I'm currently depriving myself of sleep to reset my sleep schedule.
I've always been good at things like physics and chemistry, but probability is never something I really shined at. Dunno why.

>>677928362
No. A good way to consider that is if you where doing it like a hundred times and then seeing how many times you got two gold balls. Reducing that to 2/3.

>>677928475
You are forgetting that to factor in the uneven odds of the first box.
It's more likely that you selected the all gold box than you did the half and half box.
This gives slightly boosted odds (2/3)

>>677928362
It does you fuckin retard. Doesn't mean it's 50 50 cause there are 2 boxes are left in the equation. Read the fucking thread. Itso explained 20 times 20 different ways

50/50

>>677928280
You do realize that the "paradox" is literally wrong right?
>>677928639
Yes, it actually does, theoretical probability theory is almost all wrong.
>>677928733
The question only concerns whether or not the final ball will be gold, previous parts dont affect the REAL probability.

>>677928956
>all the experts are wrong and i am right

>>677927699

What if there were 100 gold balls in a and 1 gold ball and 99 silver balls in b?

We picked a gold ball so it's only a 1 in 100 chance that we're in box b. So it's a 99% chance that we're in box a and the next ball we pick is gold.

By the way op, that image doesnt even say the fucking paradox correctly, the way its written the probability is 50%.

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>>677922989

50%. You have a 50% chance of having picked the double gold box. The double silver doesn't factor in at all

50%, assholes. Fucking idiots.

>>677929125
The "experts" are actually asking a different question, the way that this question is written makes the answer 50%.

>>677928956
Dude, I came into this thread assuming the same thing.
And you are right that the silver box IS eliminated before the problem begins, but that doesn't change the odds of which of the other two you picked.

Read this:
>>677927138

>>677929484
I read that, its wrong, re-read the question instead of acting like its actually Bertrands box, its written significantly differently.

>>677928956
Tell me more about this REAL probability

>>677929606
I actually did exactly what you said at first, and assumed this was a troll problem based on the wording, but it's not.

The main issue with the 50/50 answer is that you are assuming the starting state is this:

You have selected ONE of TWO boxes.
This SOUNDS correct, but in reality it is:
You have selected ONE out of THREE gold balls.
Two of those three gold balls are in the box.
Since the initial box is unknown, we can factor in the chances of which box we have.

Since there is a higher likelihood that the initial state is the all gold box, it can't be 50/50.


You're reading comprehension is fine, but you are assuming the wrong initial definition of the problem.

>>677928280
>paradox

what the fuck

>>677930121
*your

>>677929606

explain the difference, it seems like its pretty much copied from the wikipedia page on bertrand's box

>>677927756
U mad cuz I smater then u

>>677930175

sometimes vaguely "counter-intuitive" facts are called "paradoxes" even when they arent logical paradoxes at all

It doesn't fucking matter which of the two golden balls you pick from the all gold ball box, faggots. On pull two, you could get either the second golden ball or a silver ball, making it 50%

>>677929404

explain how its different

OP's pic is exactly like the wikipedia page on bertrand's box

When you choose the FIRST ball, there is a 50% chance of it being gold:

>GG, first or second ball
>GS, first ball
>SS, neither ball

>3/6 = 50%

But your chances of choosing a BOX with a gold ball in it is 66.66...%:

>GG, has a gold ball
>GS, has a gold ball
>SS, has no gold ball

>2/3 = 66.66...%

If you reach into a box at random and get a gold ball, it is certainly the GG or GS box. But the chances are HIGHER that it is the GG box, because there are two of the three gold balls in it.

>GG, two gold balls, 2 of 3 possible
>66.66...% that this is the box you chose from

>GS, one gold ball, 1 of 3 possible
>33.33...% that this is the box you chose from

This is intro to statistics level probability. It's a question that is meant to fool people that have no experience with probability, who are still using 'common sense,' and make them realize that they don't really understand the basics of statistics. Good job, /b/, you've proven that most of you are lacking in education.

If you don't believe me, go read up on "Bertrand's box paradox," it's a very well known problem in probability theory that's been fooling plebs for over a hundred years.

>inb4, ermehgerd, it's 50/50 even though I don't have mathematically sound reasons to support my beliefs!
Go look it up, retard.

>>677930121
This is a good explanation. There might be some ambiguity in the statement of the problem, but I believe that the way you explained it is the intended meaning of the problem.

>>677930350

this guy is definitely trolling with some of the same troll rhetoric from the "crit chance" thread a while back

>doesnt matter which of the two golden balls you pick

sounds a lot like

>the order doesnt matter

>>677928956
It's a random choice to begin with and failed attempts still count when working out the probability.

>>677926769
100%

>>677929606
>>677930218
>>677929404
Gold balls versus gold coins.
Completely different.

3/6 50%

>>677930431
thank this guy:
>>677927138

I was in the 50/50 camp at first, but realized my error when he put it this way.

The problem is definitely tricky when you first read it, as it seems pretty clear cut on the surface.
Plus, I've seen a ton of these types of problems on /b/ where the answer IS 50/50
A lot of people will post problems that seem to be compounding odds, but are basically just obscured coin flips, purposely worded to confuse the reader.
This was not one of those, though.

>>677926769

is this question nonsensical?

im confused

There's a fifty-fifty chance

Winrar bitches

>>677930429
>But your chances of choosing a BOX with a gold ball in it is 66.66...%:

It is stated that you take a gold ball, so SS is to be ignored entirely.

>>677926769

1/100[ (1/100) + (2/100) + ... + (100/100)]

=(1/100)(5050/100)
=5050/10000

=50.5%

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>>677923230
>>677923336
>>677923753
>>677923765
>>677923962
>>677924705
>>677924866
>>677925604
>>677926706
>>677927049
>>677927138
>>677927210
>>677928071
>>677928280
>>677930429

>Just kidding I was only pretending to be retarded

>>677930814

it sounds like you're one of those people who is always wrong in these threads

>A lot of people will post problems that seem to be compounding odds, but are basically just obscured coin flips

the coin flip interpretation of the monty hall or boy girl paradox involves multiplying odds, but this statement makes me think youre one of those subhumans who thinks the gambler's fallacy is relevant here when its not

40% chance

>>677930905
Yes, but given that you pulled a gold all, its more likely that your starting box is GG than GS.

Thus, it isn't 50%, its 66.666...%

>>677930905

there you go with the trolling again

taking a line of the proof out of context and ingoring that he addressed your concern in the proof as a whole

its really quite deft, im impressed with how subtle it is

>>677930905
Yep, only those two boxes matter. Which is why I wrote...
>But your chances of choosing a BOX with a gold ball in it is 66.66...%
...and then followed up by looking at just those two boxes, and the odds of the entire contents of each of those boxes.

>>677922989
OP is a lying asshole.
The last box actually has two silver balls.

>>677922989
NOICE YOU ALMOST GOT ME

1/2
because given that i pick a gold ball first the chance of picking the 3 box is 0%
leaving 2 boxes with equal chances

I will now amuse my self by reading the shitstrom of trolls and reatrads and frustated math grades

>>677931095
Wut
Nigga, did you even read what I said?
Seriously, go back and read what I said and you will realize that your response doesn't make any fucking sense at all.

I merely stated that my mind immediately jumped to assuming it was 50/50 due to previous threads.

Nowhere in there did state jack shit about it being relevant to this; in fact, I stated exactly the opposite of that.

lrn2read

>>677926769

1/10000
+
2/10000
+
3/10000
.....

=
5050/10000 or 50.5% chance of death

all these fucking retards lol holy shit good job OP
>50%
for those who really want to know

>>677931554
Yes, my brotha. High five.

This was me: >>677931062

>>677931214
That was my first post though desu

Your chances of picking a box with a gold ball are 100%, not 66.66 because doing so is a requirement.

P(G|G) = 1/2

>>677931385

imagine there is no box with 2 silver balls, it doesnt even exist

box A has 2 gold balls
box B has 1 gold ball and 1 silver ball

that makes 4 balls, 3 gold 1 silver
the odds of the first you grab being gold are 3/4
that leaves 2/3 remaining are gold

2/3 is the answer

>>677930429
>When you choose the FIRST ball, there is a 50% chance of it being gold:
Nope 100% percent bc narrator said so

>>677931815

okay then go ahead and read this post here

>>677931851

to explain why youre wrong, the answer is 2/3

>>677931820
No, if you want a conditional probability explanation, see this: >>677925604

>>677931851
lol you cant be serious? please dont be serious

>>677931948
>the point
...
...
...
...
>you

>>677932038

let me rephrase cuz i misspoke one part:

imagine there is no box with 2 silver balls, it doesnt even exist

box A has 2 gold balls
box B has 1 gold ball and 1 silver ball

that makes 4 balls, 3 gold 1 silver
you grab 1 and its gold, that leaves 2/3 remaining are gold

2/3 is the answer

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>>677922989
This isn't hard. It's a matter of how you read it. It tells you there's three boxes that you pick a potential box with 2 gold balls, a box with 1 silver and 1 gold ball and a third box with 2 silver balls. It also tells you that you cannot see into them. In which case, it says you get a gold ball. Now again..just pretend this is real and you cannot see into the box. You're left knowing that you have a shot at one gold ball, or silver all the way. If you draw one gold ball, the chances of getting another is 1/3 only because you're aware one box has two gold balls and another has one. So if you draw from a double gold box. That was still a 1/3 chance of it happening. But if it was the mixed ball box, then you're fucked. Only answer what the question specifically asks you for. 33.x% of success with a 66.x% margin of failure.

>>677924292
Are you a retard? By his reasoning there would be a high chance of getting another gold ball.

>>677932196
LOL still wrong
this thread is the funniest ive seen all year

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>>677932004
How does that prove I'm 'wrong' anon?

All I ever said was that there was no need to mention the silver box at all, that post backed my statement.

>>677932350

good trolling sir, excellent work

really showing your stuff

maybe if you had a fake demonstration to go with your fake conclusion it might be even more impressive

>>677923339
Jup

>>677932380

you thought, and still believe the answer to OP's question is 50% do you not?

you are wrong if you do

if not i apologize for putting you through this rigmarole

金玉です。

>>677932438
there is 2 different boxes why the fuck would add the remainder of balls together
you dont know how to read the question do you
your hand grabs a gold ball
the next is either gold or silver, 50/50
where the fuck do you get the rest

>>677932730
see: >>677927138

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>>677932552
>do you not?

>>677932820
are you fucking stupid? scenario 1 and 2 are the same, you still end up with a 50% chance to pull out a gold ball you fucking tard
>combinations/permutations?

So... Your answer is 1/2.

It is asking what is the probability that the NEXT ball will be Gold. Yeah, you can argue that 2/3 is, because regardless of the ball you pick, you run a 2/3 chance that your second ball will be Gold. BUT the question has already provided that your first balls HAS to be Gold. So, the box without a Gold can't be the box you picked... Leaving TWO. Only ONE of those boxes has two Gold in it.

The answer is 1/2, or 50%

For these problems it makes it easier to understand if you make the problem more extreme. Read the example above where it's 100 balls in each box. (100 gold in a, 1gold and 99 silver in b)

Then you see that since you picked a gold ball already, then there is a 2/3 chance that you are in box a, and a 1/3 chance you are in box b. That's all that matters.

>>677932730

>there is 2 different boxes why the fuck would add the remainder of balls together

because revealing a single ball does not tell us which box we've chosen, until we've established which box is which, we are working with the probability of all possible options added together which includes both boxes

think about it

3/4 of the balls are gold. its more likely than not that you will grab a gold one first. why is that? because one of the boxes has 0% of having a silver ball, that box has a greater chance of having a gold ball on your first grab than the other box, therefore the fact that you grabbed a gold ball means the odds that you picked the double gold box are higher than that you picked the one gold one silver box

You have all been trolled by a question that seems familiar to the paradox but is actually quite easy since it adds a crucial limitation:

>What is the probability that the next ball you take from the same box will also be gold?
>same box
>same

It's either the left or middle box you initially begin with:

a) left box: you got one gold left
b) right box: you got one silver left
=> 50%

All the ones saying that it's 2/3 are missing the part "same box". Go die faggots trying to reason with a paradox that does not even apply here.

And whoever tries to even reason with the previous probability of getting to the state the question starts at should die even more.

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>>677932552
>rigmarole

>>677933438
you dont like reading whats given to you lol
see >>677933346
bet you were that retard that took the bait on tests when more than enough info was given
>because revealing a single ball does not tell us which box we've chosen
LOL i think it tells me im either in box A or B
and that the next ball is either gonna be gold or silver. 50/50

>>677933662
You keep making yourself look dumber and dumber. You must be trolling.

>>677933346

>the question has already provided that your first balls HAS to be Gold.

what it says is that the first ball IS gold, not that it HAS to be gold, whatever you think that means

>So, the box without a Gold can't be the box you picked

thats true

>Leaving TWO. Only ONE of those boxes has two Gold in it.

also true

>The answer is 1/2, or 50%

this part is wrong though

if your first ball was gold, and it is, that scenario is most likely given that you chose the double gold box. less likely given that you chose the 1 gold 1 silver, and 0% chance you chose the double silver box.

>>677933346
The fact that you already have a good ball pulled out of the box makes all the difference, and eliminates the all silver box from the probability. If it just said "what would be the probability that the second ball would be Gold?", 2/3 would be right. But it establishes that a Gold was the first ball. Since it is already established, then we know you couldn't have the all silver box. It is asking the probability of the two remaining boxes.

>>677933180
See
>>677925604

Now imagine you exclude the 2 silver box, as some are stating you have to do.

P(A|B) = P(A&B)/P(B) - as before

P(A&B) = 1/2 (2 boxes and only one is all gold)
P(B) = 3/4 (100% in one box 50% in the other, or just 3 gold balls and 1 silver)

P(A|B) = (1/2)*(4/3) flipped the fraction for clarity
= 4/6
=2/3

It does seem counter intuitive and that's why it is a paradox, but it makes sense if you think about it as a chance of outcomes.
GG GS SS
GG SG SS
Of these outcomes only 3 are a gold ball draw first. 2/3 of those a gold is drawn next.

>>677933799
this is the same as:
I flip a coin and get heads 9 times in a row
what is the probability that next will be heads?

>>677933662

>you dont like reading whats given to you lol

obviously you dont either since you only bothered coming up with a trolly response to 1 line of my post

>>677933346
Yet there is a 100% chance that you're a retard...

Even if there were two boxes to start with, GG, GS, if you chose a gold ball to start there is still a 2/3 chance that the other ball in your box is gold. Go look it up, rather than rely on your obviously inept reasoning skills.

>>677933979
No it's not. Are retarded?

>>677934071
same answer, different question being asked

>>677927138
Wrong. You cannot factor in the same box and say left ball right ball. The fact is you already have a gold ball. So that means the 2 silver ball box is irrelevant. But you could of drawn from the 2 gold box or the mixed box. You can't say a 2 out of 3 chance because that's basically saying two boxes have two gold balls. No. The only box that scenario could go down like the question asked in the first place. The two gold ball box is the only way this could work. So the answer is def. 1/3. The question gives you process of elimination. You fools.

>>677933555

the "crucial limitation" is also part of the bertrand's box paradox

if that limitation were not there, the question would be quite different

in that case you would just say 3 of the 6 balls are gold if you pick a gold one, that leaves 2 of 5 remaining are gold.

the answer is 2/3

>>677934071
Ok, you got me. I was just trolling you guys. Lol.

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>>677932552
I have yet to make up my mind, actually.

What I'm wondering is whether or not to factor the individual balls. Since the third box is factored out, you have 2 boxes. Since picking a gold ball is a requirement, you do...

Thing is, the whole thing doesn't account for the possibility of taking a silver ball at all. It simply isn't possible in the universe of this question because you always pick a gold ball. Going from there, the chance that the other ball in the box is also gold is p much just 50%

>>677934267
>I was just pretending to be retarded!
waht a twist

i just asked 30 math majors and they all said 50%
congrats /b/ you're fucking retarded, stay on this board and rot

It's 50%?

i am on the virge of blackin out fdrunk and could still tell that the answer is 50 percent. jesus fuckin christ this forum is confrimed autist

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>>677922989
50% having pulled one gold ball out you know that it has to be one of the two boxes that contained gold balls. If one had two gold balls and the other a gold and a silver ball, then you have equal chance of pulling either gold or silver on the next ball.

>>677933898

>think about it as a chance of outcomes
>GG GS SS
>GG SG SS
>Of these outcomes only 3 are a gold ball draw first. 2/3 of those a gold is drawn next.

This is a good way of explaining it as well.

>>677922989
The sequential point of vew: it's 2/3 chance
Quantum physics say 50/50

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>>677934451
I'm smarter than the 30 math majors you asked. Pic related is proof. The answer is 2/3.

>>677934344

>Thing is, the whole thing doesn't account for the possibility of taking a silver ball at all. It simply isn't possible in the universe of this question because you always pick a gold ball. Going from there, the chance that the other ball in the box is also gold is p much just 50%

i dont know where you got that nonsense but the OP implies nothing of the sort

you are reading OP's question as if its some sort of inviolable prophecy when its a play by play account of a hypothetical scenario that could, for instance, have already occurred in the past

the 3 boxes all exist. the one with two silver balls exists. the odds that you would put your hand in that box are not 0%, it JUST SO HAPPENS that you did not put your hand in that box.

all OP said is this (in parentheses i show past tense because i think it makes it more intuitive):

>You pick(ed) a box at random. You put your hand in and take(took) a ball from that box at random. It is(was) a gold ball.

notice it doesnt say that it couldnt possibly have been silver, it just says that it was, in fact, gold, thats just how it turned out, this one time.

and given that it did turn out that way, what is the probability that the other ball in that box is also gold

the answer to that question is 2/3 as demonstrated by all the intelligent posters in this thread

>>677933894
I meant Gold ball. Too tired at this point. But versed enough in statistics to know that any who said 2/3 should never breed.

The point is, the question doesn't say probability off all of the boxes, it is implying "of the remaining boxes". After you eliminate the possibility of 1 of the boxes, that leaves two... Because 3-1=2.

Have I lost you yet? Sorry is you didn't know that 3-1=2. Math is hard, I know.

And yes, since the problem has said the first ball IS Gold, then the first balls HAS to be Gold. Why? Because logic says if that first ball is Gold, then guess what, it can't be fucking silver... Very very very simple. Sorry... I forgot who I was talking to.

Anyone who thinks that it is anything other than 1/2 did not read where it said the first ball is Gold, therefore eliminating any and all chance that the all silver is a factor in the probability of the asked question.

Please... Keep working with those flash cards until you get to subtraction, I'm rooting for you.

My god...

Think of it if you had 100 gold balls in a and 1 gold ball in b with 99 silver.

Since you picked a gold ball what is the chance that you are in b? It's 1%. And 99% that you are in a.

In ops question, what is the chance you are in b? 50%. And so 50% a.

Jk. It's 2/3 chance your in a..

>>677932650

kek

>>677934898
See
>>677933898

>>677934898

i think you must be actually so tired that its destroyed your statistics skills

get some sleep and try again

>>677934016
That either proves you are either a troll or somebody needs to give you coloring books back. Very simple logic here. If you only have two possibilities, then you can never have anything higher than 2 as the denominator in the probability...

2/3 MASTER RACE

>>677935310
ding ding ding we have a winner
>3 outcomes when its only 2 balls?
you can all go to sleep now
2/3 people please make sure to spade or neuter yourselves

You will pick the gold/gold box 1/3 of the time. This skews your chances of picking a gold ball again. 1/2 of the time you iterate through this scenario, you will always pick gold (100% chance). The other half, you have a 50% chance of picking gold.

If the boxes looked like this:

gs gs ss

You would have a 50% chance of picking a gold ball every time

But they look like this

gg gs ss

So it's more like a 66% chance. 1/2 is only the truth if you selected the gs box.

50%

>>677933555
>two gold balls in case A
>two outcomes as I can pick gold ball 1 or gold ball 2 first in case A
Was that so hard to understand`?

>>677922989
then who was gold? eh?

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>>677922989
Nuclear engineer here.

It is actually 2/3 you fucking degenerates.

>>677935310

>If you only have two possibilities, then you can never have anything higher than 2 as the denominator in the probability...

the denominator of probability is always defined relative to the numerator

you can say that you wont budge on the denominator but that doesnt fucking matter lol, the numerator will change accordingly

for any given fraction there are an infinite number of equal or essentially equal (cant account for irrational numbers) fractions with a different denominator

if you demand 2 in the denominator, then the probability is 1.33333/2

which is the same as 2/3

2 / 3
_

3

>>677922989
50%, either it is the other gold ball or it's the box that has the silver in it which would have had gold.

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>>677935473

>spade or neuter
>spade

pic related is a spade anon

The number of people saying 50% makes me want to kill myself.

>>677936054

trolls make you want to kill yourself?

maybe you should take a break from the internet

>>677934898
It also could be 1/3.

It says you drew a gold ball out. So you're left with knowing ss is useless. You have gs and gg.

What if you picked gold from the mixed box. The same box it asks, well it's zero. But you wouldn't be aware of that. You can't see inside it.

What if you picked gold from the gg box. You also don't know that. But it would mean a 1/3 chance since it was at random when you picked. The question just gave you the default first ball you picked which is gold.

But consider this.

Six balls to start, 3 gold 3 silver. Three boxes have two balls. Two g two silver two mixed. Six. You draw gold first. So then they ask what are the chances of getting another gold from the same box.

You know one box has two gold. So in order for that to happen. It's 1/3 only one fucking box has a pair of gold before factoring out the other boxes. Prove me wrong pls.

>>677935622
There is never anything about the first gold ball you picked up being put back in.

>Was that so hard to understand`?

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So, lets walk thru it. I'm naming boxes A, B and C (left to right) and balls 1, 2, 3, 4, 5, 6.

You have picked a ball out of box A or B since those are the only boxes with gold balls.

You have picked a gold ball, which means that if you picked B, you've picked the only gold ball in that one.

The odds that its A or B is 50/50.

If the box you picked from is A, then there is a 100 % chance you'll pick a gold ball again.

If the box you picked from is B, then there is a 0% chance you'll pick a gold ball again.

This means, as the probability you've picked A (100% chance for a gold ball) or B (0% chance for a gold ball) is 50/50... the chance is...

50/50.

Prove me wrong.

the first ball your grabbed was gold

what does that tell you

all of you mongoloids have managed to grasp that it tells you the box you reached into was not the one with two silver balls

but it also tells you something else: there is a higher chance you reached into the double gold box than the chance you reached into the single gold box

so yeah there are two options, you chose doulbe gold or you chose 1 gold 1 silver

but the odds are not 50% because its more likely you chose the double gold box

>>677935310
You really are very retarded. There are 6 balls in the original and only choice. Since OP has revealed that the choice was made and resulted in a gold ball, that choice could only have been one of the three gold balls. Since the choice was already made, and there is no other choice to be made (you are simply trying to determine the odds of what the remaining ball in your chosen box happens to be), you can determine the chances easily:

1. Your box must either be the GG or GS box
2. The ball you chose could be any of the three gold balls:
>The first of GG
>The second of GG
>The G in GS
3. Since two of those options result in another gold ball, and one results in a silver ball, the odds of drawing another gold ball is 2/3.

Since you are the type to throw around unrelated mathematical nonsense, I don't expect you to follow this rational argument. I expect you to hur and dur a bit more, and stick to your guns even though this is a well known problem with a well known solution. Any normal retard could just look up the correct answer with a 15 second google search, but you are a special kind of retard.

>>677936442
the question is not:
>What are the odds of you picking something after something?

it is:
>You are either in A or B, what is left for you to pickup?
-> It's eiteher gold or silver -> 50/50

>>677935086
I get that. But there is one flaw you are missing. Yes, all 3 boxes exist. But once you pull out the first ball and it is Gold, then the all silver box is no longer a factor in your probability. The question doesn't say "of all of the boxes".

Okay, here... Say you pull the first ball out of the box and it is silver, it is automatically disqualified you from this probability because it says the first ball pulled is Gold. Because the first ball pulled is Gold, the all Silver box is no longer important. At this point, you must consider that there are only two boxes the one you have COULD be, either GG or GS. And only one of those boxes can the second ball be Gold...

The points are, as soon as "the first ball is gold" is said, the all silver box is no longer important to the situation or the probability. Because it isn't saying probability of the all the boxes, it is checking for probability of the boxes that had at least 1 Gold in it.

My statistics isn't bad, your reading comprehension... That is bad.

Need a summary. Gold is pulled. All silver is eliminated. Leaving two boxes. Only 1 has 2 Gold balls. 1 out of 2 boxes that had at least A gold ball in it. 1/2.

Sorry, but you needed educated

>>677922989
66%

>>677936278

okay so before you reach into any of the boxes, the odds of you reaching into the double gold box first are 1 in 3

but after you've reached in once and found a gold ball, we have more information about which box you reached into, and we can update the odds.

we know the double gold box has more golds therefore a gold is more likely to come from that box than the other. we know it is exactly twice as likely because the ratio is 2 gold balls to 1.

and since the double silver box is now definitely not an option, 2:1 is our final odds

2:1 can be rewritten as 2/3 chance

>>677936742
hahahahaha
faggot

>>677936054

You have six guns, and they are placed among three boxes.

Box 1 has two guns. The first gun has a blank in chamber 1. The second gun has a real bullet in chamber two.

Box 2 has two guns. The first gun has a real bullet in chamber 3. The second gun has a blank in chamber four.

Box 3 has two guns. The first gun has a blank in chamber 5. The second gun has a real bullet in chamber 6.

You pick a box and pick one of the guns. You point it at your head and pull the trigger 3 times. You're still alive.

Next, you will pick the other gun in the same box and fire it at your head 3 times.

What is the chance you kill yourself. (Note: you can't hear the blanks)

>>677936757
2:1 = 2/3
MOM GET THE CAMERA

You're all retarded it's 2/3 or 1/2 depending on how you interpret the first part of the question. Do you already have a gold ball before working out the probabilities? Or does the chances of picking a gold ball first need to be factored in? It's a retarded Internet math question that gets shit posted on pol and b because you all take the bait and endlessly argue whether it's 50% or 2/3. Enjoy faggots

>>677936657

perhaps you should reread the OP

im pretty sure the question IS in fact:

>what are the odds of you picking something after something

that other nonsense you wrote is just your unfounded assumptions about OP's question

>>677936386
>If the box you picked from is A, then there is a 100 % chance you'll pick a gold ball again.
There is no second choice. There is only one event in which there is a choice, which is the initial draw of a ball. The question is, and has always been, what are the odds that your initial choice was the box with two gold balls? And the answer is 2/3:

>Box A, GG, you picked the first gold
>Box A, GG, you picked the second gold
>Box B, GS, you picked the only gold ball

Those are your only possibilities from your only draw. And two out of the three possibilities result in your draw coming from box A. Which would mean the box you drew from has another gold ball 2/3 of the time. Anything past this is NOT a choice, it is simply a reveal of the only remaining ball left in the box you already chose. There are no new events that determine any new odds.

>Prove me wrong.
Done.

>>677923339
Statistics and probability are lies. Everything is 50 /50. It either happens or it doesn't.

>>677937066
Yeah but can you see why kids love cinnamon toast crunch

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>>677936657
>the question is not:
>What are the odds of you picking something after something?

dude. that's literally the question.

>>677936977

if the odds of something occurring are 2 to 1, that means that it is twice as likely to happen as to not happen

if the chance of something occuring is 2/3, that means the chance of it not occuring is 1/3.

2/3 to 1/3 is the same ratio as 2 to 1

>>677922989
50% 3/6 balls are gold

>>677937220
>>677937075

read this:
>>677933555

I think you drooled a little on your keyboard when trying to use critical thinking skills.

Based on:
Definition of conditional probability:
P(A|B) = P(A&B)/P(B)

and using your definitions as is:
Let A = the 2nd ball you pick is gold
""""""Let B = the 1st ball you pick is gold""""""""
B = 100% since it is a given constant based on the setup of the question:

"You pick box at random. You put your hand in and take a ball from that box at random. It's a gold ball"

So based on the same logic you were using with a little critical thinking skills to help set the variables....

P(A&B) = 1/2 (2 boxes and only one is all gold)
P(B) = 100%, 1/1 = 1

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

>>677937354
forgot to tag you

>>677933898
>>677933898
>>677933898
>>677933898
>>677933898
>>677925604
>>677925604
>>677925604
>>677925604

>>677937334

now you read this

>>677934258

>>677937354

>>677933898
>>677933898
>>677933898
>>677933898
>>677933898
>>677925604
>>677925604
>>677925604
>>677925604
I think you drooled a little on your keyboard when trying to use critical thinking skills.

Based on:
Definition of conditional probability:
P(A|B) = P(A&B)/P(B)

and using your definitions as is:
Let A = the 2nd ball you pick is gold
""""""Let B = the 1st ball you pick is gold""""""""
B = 100% since it is a given constant based on the setup of the question:

"You pick box at random. You put your hand in and take a ball from that box at random. It's a gold ball"

So based on the same logic you were using with a little critical thinking skills to help set the variables....

P(A&B) = 1/2 (2 boxes and only one is all gold)
P(B) = 100%, 1/1 = 1

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

>>677936541
You are missing one BIG piece to the whole question. It says the second ball is picked from the SAME box.

That seems to be the big thing most of you are missing.

Step 1: pick a box, pull a ball. Find out it is a gold one.
Step 2: realize you can't have SS, because you pulled a gold ball.
Step 3: that leaves two boxes that you could have, because only 2 boxes had a good ball in it.
Step 4: you must pull the other ball from the SAME box, what is the probability THAT ball is gold?

Only 1 box had 2 gold, and only 2 boxes had a gold in it.

If you were allowed to pick you second ball from a different box, then your answer would be 2/3, but you must pick from the SAME box.

That is part that is throwing everyone off. It says the SAME box.

guys stfu
NEW NERD EPISODE
https://youtu.be/Q3iEn5rzMnw

>>677937066
>depending on how you interpret the first part of the question.
There is only one question:
What is the probability that the next ball you take from the same box will also be gold?

>Do you already have a gold ball before working out the probabilities?
Yes. There is a clearly stated order of operations:
>You pick a box at random
>You pick one of the balls inside at random
>The choice results in a gold ball
>What is the probability of the other ball in the same box being gold

There is no ambiguity here. You choose, see the result, and determine the resulting probability. The only correct answer is then 2/3, as anyone that actually understands the probability as applied to this problem has already pointed out.

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>>677937461
I think you shat yourself trying to seem intellimagent.

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

>>677937418
>>677934258

keep getting trolled by OPs question, don't care anymore

>>677937066
Btw it's 2/3 look up bertrands box, however that paradox specifies you need to consider the chances of picking a gold ball first, ops version is ambiguously worded so as to enrage autists

>>677936757
GG SS GS
3 box

Pick gold ss eliminated

You already know that two boxes have either...1 silver ball if you take the gold out of that one.

Or...

One gold taken out of the double G box. Since you're told from the git go that one is a mixed and you picked gold. You don't know which box you picked from.

The chances of that happening are 1/3 if you picked the gg box. It did say the same box too. But it's not implied you ultimately get the gg box in the question. So it would make it 2/3.

>>677937775

>if that happens to be a gold coin, of the next coin also being a gold coin.
the scenario OP posted is NOT the same as this. The OP's scenario guarantees an initial gold ball pull, Bertrand's box paradox does not.

>>677937492
>It says the second ball is picked from the SAME box.
The problem as posed does not include a choice after drawing the first ball. It asks you to determine the probability based on that draw, if that draw is a gold ball. That's what you are missing.

>If you were allowed to pick you second ball from a different box, then your answer would be 2/3
>but you must pick from the SAME box.
Doesn't matter. Here are the possibilities, if the first ball is gold:

>Box GG, you picked the first gold
>Box GG, you picked the second gold
>Box GS, you picked the only gold ball

2/3 of the time, if you choose a gold ball to start, that ball will be in the GG box. Which means that the second ball, from the same box, will be another gold ball 2/3 of the time. Pretty simple, and I did address that point. You just seem to have missed it.

>>677937868

you are mistaken, OP's question is worded exactly how the problem is always worded, here's an excerpt from the wikipedia page that should help clear up things for all the "50%" faggots

>There are three boxes, each with one drawer on each of two sides. Each drawer contains a coin. One box has a gold coin on each side (GG), one a silver coin on each side (SS), and the other a gold coin on one side and a silver coin on the other (GS). A box is chosen at random, a random drawer is opened, and a gold coin is found inside it. What is the chance of the coin on the other side being gold?

>The following reasoning appears to give a probability of 1⁄2:

>Originally, all three boxes were equally likely to be chosen.
>The chosen box cannot be box SS.
>So it must be box GG or GS.
>The two remaining possibilities are equally likely. So the probability that the box is GG, and the other coin is also gold, is 1⁄2.
>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. >Specifically:

>The probability that GG would produce a gold coin is 1.
>The probability that SS would produce a gold coin is 0.
>The probability that GS would produce a gold coin is 1⁄2.

3 out of 5 loosers

there are 3 cards

1 is black on both sides
1 is white on both sides
1 is black on one side, white on the other

the cards go in a hat

you pull one at random and put it on the table flat so you can only see one side of the card

you can see that side of the card on the table is black

what is the probability that the other side of that card is also black?

2/3

>>677937129
You are totally right, in almost everything. The one thing missed was this: the first gold ball being pulled has ALREADY happened. That event doesn't factor in because it already took place.

You established this:
You either picked: the first gold ball of GG, the second of GG, or the gold of GS.

The problem here is that whether or not you picked the first or second ball of GG doesn't matter. The reason is because in this probability, the first pull has already happened. It is asking for the probability of the second pull and the second pull ONLY. The only event occurring in this probability is the pulling of the second ball.

The question wasn't asking from the original 3 boxes, it is asking from the point AFTER you already pulled the first Gold ball. From that point, you know that you pulled a Gold ball, the variable in this is whether that ball that is left is either G or S.

I really do like how you supposed that, but because the first ball was already pulled, that event has already taken place.

You asked me to prove you wrong, and I did.

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>All these fucking faggots trying to prove their worth on a brazilian postage stamp forum

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>>677923339 >>677923517
>>677924493 >>677924589
>>677925322 >>677925496
>>677925802 >>677925856
>>677925865 >>677926664
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>>677937130 >>677937324
>>677937334 >>677937354
>>677937461 >>677938244

>>677936323
Are you fucking daft?
Two balls, two picks you fucking mongoloid.
You can pick gold ball 1 or gold ball 2 and still have gold 2 or gold 1 in the box depending of outcome.
Fucking hell kid have you even taken a statistics course in your life?

>>677937129
>Done.

... no. The question "starts" when I stand with one gold ball in my hand.

The initial pick has already been made. This is where this problem differs from Bertrands.

File: xcJb5zm.gif (1 MB, 288x250) Image search: [Google]

1 MB, 288x250

>>677922989
I 100% do not care.

If you want both the gold balls then picking one to begin with means you're more likely to have picked the box with two in it.

>>677938244
>The OP's scenario guarantees an initial gold ball pull
No it doesn't.

>You pick a box at RANDOM. You put your hand in and take a ball at RANDOM. It's a gold ball.

In the scenario, where you pick a RANDOM ball, there is a higher chance you picked from the GG box, since the random ball COULD have been a silver.

imagine the boxes have two pockets inside, one pocket says A, one says B, and the balls each go into a pocket, and it just so happens that, unbeknownst to you, in the box with 1 gold ball and 1 silver, the gold ball is in pocket A and the silver is in pocket B

if you reach into a random box and then pick from a random pocket and get a gold ball, there are 3 different possibilities that could have occurred

notice im speaking in the past tense

if you did, in fact, reach into a box and you did, in fact, get a gold ball, then are 3 explanations for what could have happened

1-you took the gold ball from pocket A of the double gold box
2-you took the gold ball from pocket B of the double gold box
3-you took the gold ball from pocket A of the 1 gold 1 silver box

all of these are equally likely (you chose the box randomly and the pocket randomly, 6 total pockets, 1/6 chance for each)

2 out of those 3 equally likely options involves you having picked the double gold ball which would result in your next ball also being gold

therefore the odds of pulling another gold given that your first pull was gold are 2/3

>>677938271
I guess my point is this. Yeah, you could say that pulling the first G of GG and the second G of GG are two different occurrences... And technically you are right. 100% but which ball you have already pulled no longer matters, that event already happened. The question is asking for AFTER that. Whether you picked a Gold ball, whether you picked the first or second doesn't matter anymore, because that ball has already been picked. The only factor is the remaining ball.

Also, if you are going to consider the initial pull of either the first or second as two separate events, then the silver pull of GS must be included as well, as a fail... Because then you have 4 possibilities of what your initial pull would be...

Before you start to say, well the first ball is gold... Exactly. The first IS gold. And that already happened.

Okay, say... You pulled your first ball. It's gold. You know that it can't be the SS, so you mark the box you pulled from, and you took the SS box and put it out of the way.(I'm making a point). Now, the question is being asking. You've pulled a gold ball from this now marked box, what are the odds that the ball that is left is Gold or Silver? If you picked a G out of GS, the silver is left. If you picked a G out of GG, doesn't matter which G you pulled first, the other G is left.

The choice was made when you picked the first ball. The question asking about that remaining ball. The odds of the choice step is 2/3, easily, but the odds of the event post choice is 1/2.

Doesn't matter which of the gold balls you picked from GG, it only matter whether or not the other ball left is still a G.

If you still disagree, that's cool. I just wanted to make sure you saw the reasoning. Simply, the first pull is done, the choice has been made. What are the odds that the ball that is left is gold?

>>677922989
50/50

>>677939300
So if you had a chance to switch like in the Monty hall problem you are better off sticking to your first choice as a revealed gold ball is higher likely hood from a box with two gold balls.

>>677938693
>You are totally right, in almost everything
Nope, I'm right in all I said. Nothing you say negates any of the logic. I'm aware that the choice that happened is already done, but whether the choice is done, or the choice you make turns out to be gold, it doesn't matter in terms of determining odds based on that event... there is only ever one event that determines the odds in this problem.

>You asked me to prove you wrong, and I did.
No, I used green text to quote the guy I was disproving. And then I said:
>Done.

>>677940153
>The only factor is the remaining ball.
No. The only factor that determines further probability is the initial choice, because it is the ONLY variable in the entire problem. Since that variable is set directly by OP, that means your choice is one of three possible balls, two of which reside in the same box. The odds will always be 2/3.

>Whether you picked a Gold ball, whether you picked the first or second doesn't matter anymore
You are correct, but not for the reasons you assume. Either of the balls in the GG box will leave another gold ball to be revealed. Only the gold ball in the GS box will reveal a silver. That's two chances for gold, one for silver. So, no, it doesn't matter which of the two gold balls you choose in the GG box, both lead you to another gold ball, and a probability of 2/3.

>then the silver pull of GS must be included as well, as a fail...
Nope, OP said the ball drawn is gold. An initial silver draw is not a variable at all. There are three gold balls, one is drawn, and two of the possibilities, out of three, lead to another gold ball.

>and you took the SS box and put it out of the way.
This is irrelevant. You have chosen ONE box, so you can put BOTH of the other aside. The point is, you are determining what your CURRENT ODDS are, after the draw, of having chosen a box with TWO gold balls. And the answer, upon already having drawn one gold ball, is 2/3.

>The odds of the choice step is 2/3
> but the odds of the event post choice is 1/2
Jesus, no. The odds of getting a gold ball, when you are actually choosing, is 50%. OP has determined that the result of that choice is done, so there are NO MORE CHOICES TO BE MADE. All you can do now is determine the odds that your CHOSEN BOX has TWO GOLD BALLS. And the odds are, 2/3 of the time, you will have the box with two gold balls after having chosen a gold ball to begin with.

>If you still disagree, that's cool
Probability is math. It's 2/3 in this case. Look it up.

>>677940153

Let's include the silver balls "as a fail"

You have 6 possibilities of what your initial pull would be, 3 of those 6 are gold, 3 are silver

If your first pull is gold, we eliminate two of the silver balls from contention (the ones in the double silver box), and we also eliminate the one gold you already picked, there are therefore 2 gold balls and 1 silver ball remaining

2/3 chance the other ball in the box you chose is gold

>>677940652

Yup

50 per cent you fuckin idiots.

>>677942668

No

>>677942272
No idiot. They are not in one box. There are 2 boxes. so it's 50/50.

They question is not how probable it is to draw 3 gold balls out of 4 balls. It' about whether you pick the right box. The right box is the one with 2 golds. You have to boxes left. So it's 50/50.

>>677942813
Look at it this way, you'd be luckier to get the single gold ball from the box with only one and less lucky to get a gold ball from the box with two.

>>677942813

There are 3 boxes actually, and 6 balls

You reach in and grab a gold ball

It would be more likely that that ball was in the double gold box, than the box with only one good ball.

Do you not grasp that? I'm not sure where you've gone wrong. If there are 3 gold balls and 2 are in 1 box, 1 in another, it's more likely that the one you found was in the box with 2 gold balls, exactly 2/3, compared to the odds it was the one in the box with only 1 gold ball, 1/3

>>677943183

Think of it like this, before you revealed the gold ball, the odds you chose "the right box" we're 1/3, but the additional information of seeing a gold ball on the first grab means we can amend those odds

Originally we thought each option was equally likely but that's no longer the case, of the 3 gold balls, 2 of those 3, a majority, are in a particular box, call it box A.

the probability any given gold ball, such as the one we grabbed with our first pick, was in that particular box (the one with 2 gold balls) is 2/3

That alone is all you need to know

Two options, one has a 2/3 probability, the other has a 1/3 probability

There are many ways to think about the question, but the answer is always 2/3 and 50% will always be wrong

50% for sure

It's he obvious answer

That's why I think it must be wrong

Otherwise it would be a pointless question