This should be easy, but i still predict that a great number of you will get it wrong

50/50

how many balls are left?
after picking a gold ball you have 2 gold balls left, by that logic its 50/50

Fucking come on, this thread again?

I'm the lab analyst from a few weeks ago that actually trialled this for real, using gratnells trays and coloured ping pong balls, the result was around 68% of cases in my 400-some trials which fits with the math stating the probability in the exact scenario listed above is 0.66 recurring. That is all. Reality has spoken, please don't sperg your wrongness all over this thread folks.

Attached: IMG_20200113_121350-1728x2304.jpg (1728x2304, 642K)

1

And one silver ball. Three balls are left. It's not rocket science, user.

oh whats up dude how you been? im the guy that made this thread the last time.
Do you never leave Cred Forums or what?

oh i know exactly how it works, you just worded it poorly, allowing people to still draw the 50/50 conclusion from your logic.

The trick is in the "also" - so the answer is 1/3 since you can only choose the first box.

Probably 2/3

Idiot

Uh huh

then what is the answer in the last third?

how can you only choose the first box?
2 boxes have a gold ball in them.

Depends on how you "pick a box at random." It might not be a uniform distribution, so the problem is underspecified.

I turn 360 degrees and walk away with my gold ball that I'll be selling for profit. don't have time for that riddles.

its a given that the first try is a gold ball so you have to leave the 2 silver balls box out of the starting possibilities..therefore it should be 50/50

but its not 50/50, try again.

Simple. This comes down to facts more than chance. Fact 1, Whether you were able to see And pick a box or have to pick one “at random”, it’s established that you’ve picked one of TWO boxes that contain at least one gold ball. The third box is irrelevant. You have two boxes in this equation right off the bat, not three.

Fact 2, we know you select a gold ball (there is no chance to this), yet you still don’t know which of the two boxes you’ve selected the gold ball from because both boxes contain at least one gold ball. What you do know is, there’s one remaining ball in that same box. If it’s box A, you will get another gold ball since that is all that remains in box A. If it’s box B, you will get a silver ball since that is all that remains in box B.

The probability you will pull a gold ball as ball two from the same box you pulled the first gold ball from is 1/2 or 50%.

Oh good, this question again

Attached: 75EEBB0D-CF41-4112-A88D-16451C4A4CB9.jpg (400x400, 19K)

It’s 33%

1/3 since only 1 box has both gold balls

By taking out a gold ball, you eliminate possibility of the box being 2 silver balls, so 50/50 that second ball is either gold or silver.

50%

Genius

Also genius

You have a 50% chance.

Im having a really rough poop this morning b/ros

Right now there are more wrong answers than right answers, i called it!

its a little depressing how science gets overpowered by ignorance.

Statistics are some strong shit, i tell you man.

Attached: 27d.jpg (680x378, 70K)

the real question is what are the chances of picking a gold ball out of the 2 boxes left untouched?

Also for jews it would be 100% lmao

Fuck now you’ve done it.

1/2
All other answers are incorrect.

We know that he didn't pick a ball from the box of silver balls, so that is out of the game.

If he picked from the mixed box, it's a 100% chance the next ball is silver and if he picked from the all gold box there's a 100% chance it's gold.

Mathematically that's a 50-50 chance.

Some people will try to say that "because he picked a gold ball it's more likely he chose the gold box" but this is incorrect and shouldn't be counted into the equation.

You have two possible outcomes. The probability is therefore divided into two.

>All other answers are incorrect.
>gives an incorrect answer
ishygddt

Genius #3.

Stupid people keep forgetting or missing the fact that it’s established you select a gold ball to begin with

its called the Betrand's Box paradox, its a well known probles used in teaching conditional probability, you sir (like all the so called "geniuses") are actually retards and proud of it.

The answer is 2/3

>posts simple math problem on Cred Forums
>100% chance to hear all kinds of retarded anwers
>"i predict..."
easy there nostradamus, you a time traveller or something? seriously - neck yourself retard. this shit aint original, it aint fun either. its equivalent to expecting a noble prize for sayin " when i wake up at 12 pm the sun will be up".

did i even say i expected a reward, respect or anything else? i knew ppl would get it wrong, im just trying to rub it in their faces.
You arent taking this personally are you?

Well if you want a "mathematical" answer
P(A knowing B) =P(A∩B)/P(B)
A: probability of the second draw being gold
B: probability of the first draw being gold
P(A) = 1/2
P(A∩B) = 1/3
==> P(A knowing B ) = 2/3

Conclusion: most of those anons are tards

Clearly, but you're not as smart as you think you are. They always told me I was smart too. I dropped out of hs in my senior year after knocking up a girl I couldn't get along with for more than a day or two. I continued thinking I was the smartest person in my city until I realized that this belief came from 1. a lack of familiarity with other people; 2. narcissism developed out of pride and shame; and 3. actually being surrounded by dumb people. You sound like you need to grow up. I'm starting to suspect you are just fishing for attention/something to do here because you're anonymous and there's no consequences. I hope I'm right about that, because you seem like you could be pretty cool after you grow up a lot more.
You sound like a spoiled, spiteful little boy. "Let me show everyone why they're wrong because I'm so smart and clever tee hee! Nobody can out-wit me I'm one of the intelligent kids."

You’re a complete moron and you’re not understanding the question.

The question is what is the probability of pulling a gold ball as a second ball to the gold ball you already pulled. The problem is specifically telling you that you have one of two boxes. You can forget everything previously stated to that including the “random” selecting of 1 of 3 boxes, and the “random” reach in to pull a gold ball as the first ball.

damn nigga you are going next level with that reading between the lines thing, i never said im smart, you think i "sound like" i should grow up and even think you know my motives.

Well I think that you should seek professional help.

you didnt look up Betrand's Box paradox did you?
go do that and then come back.

it can only be box 1 or 2 if you managed to pick at least one gold ball out of the box. there's a 50% chance that the remaining ball is gold.

Attached: asdasdd.png (328x79, 22K)

50/50 and anyone who says otherwise is a fucktard autist.

Attached: 20-bucks.jpg (450x358, 27K)

oh the irony.

Just did and went through multiple explanations of the 2/3 math, and they all use math explaining the probability of getting a gold ball as a second ball when calculating that probability from the start including randomness of which of the 3 boxes selected, and also randomness of whether you pull a gold or silver ball first out of the box you selected. 2/3 is correct in that case, but not in the case of this problem. This isn’t exactly the same problem.

In this problem we establish very clearly that you have ALREADY selected a box containing a gold ball, and that you have ALREADY selected a gold ball first out of that box. In this case the answer is 1/2.

The balls were mad of iron? Well then that changes everything. The answer is: 6

so you are either too stupid or too stubborn to get it.
The whole thing is that you dont know which gold ball you picked, it could be any of the 3, 2 of which are in the box with 2 golds, this is where the 2/3 comes from.

If you still dont get it i cant help you and feel sorry for you.

the boxes come into play. the boxes contain 2 balls each. the question asks the probability of containing 2 gold balls if you already picked a box containing 1 gold ball. does the probability matter across all 3 boxes or across this one box? if it's within the 1 box, it would be 50%. if it's across all 3 boxes, minus 1 ball, it would no longer be 50%.

that's not what the question is asking, fucktard.

see this guy Last time i made this thread several weeks ago, he did this experiment, you can do it too if you dont believe me or him, throwing words at me wont change the answer to 50/50 nor will it convince me its 50/50.

the only thing that can change at this point is what YOU think the answer is, are you willing to find the answer or would you rather stay ignorant? its your choice.

There are 3 boxes. Each box contains 2 balls. There are 3 gold and 3 silver balls within these boxes. You choose a box and it contains at least one gold ball. What is the probability that this remaining ball in the box is gold?

You obviously didn’t pass 3rd grade math. You realize that the 2/3 probability is derived from factoring the chance of selecting a gold ball #2 BEFORE you even put your hands on any of boxes 1, 2 or 3 right? Once again, in that case, yes, it’s 2/3.

This is not the same question. This question is asking the probability of getting gold ball #2 AFTER you already have a box with a gold ball picked out. It is already established that the silver/silver box has no relevance in this particular question.

Yes, it says you selected a box at random (so what?) and that you selected a gold ball at random (so what?), and now that it’s established you have a box that has gold ball #1, what is NOW the probability of ball #2 being gold.

Try to read and understand what I’m explaining.

This thread is for talking about this question, im not interested in talking about other questions, make your own damn thread for that.

i see. everyone here has already explained it, but i see.

mentally, i view it as a 50% chance of picking the first box, and a 50% chance of picking the second box.
but there are three golds, so you either pick gold 1, gold 2, or gold 3. and if you pick gold 1 or gold 2, you always get another gold. and if you get gold 3, you never get another gold.

therefore, each ball has a 1/3 chance of being picked, and 2/3 of the time you'll get another gold.

you are cutting out important parts.

> box at random (so what?)
this is important

> you selected a gold ball at random (so what?)
this is also important

the silver-silver box is meaningless. ignore it.

the chances for the first two boxes are as follows.

box 1:
G(L), G(R)
G(R), G(L)

box 2:
G(L), S(R)
S(R), G(L)

since you don't get a silver first, ignore the box 2 situation where you get a silver first. that means there are three ways to pick a ball.

then, since you know each one is random, you have an equal likelihood of getting each possibility.

2 of the 3 times, you get another gold ball. 1 of the 3 times, you get a silver ball.

does this make sense to you?

Attached: Gold!.gif (640x360, 1.88M)

your calculations are wrong. your terminology is wrong.

if you're pulling a gold, you're twice as likely to pull that first gold from the first box.

sorry, i see what you mean.
the
>mentally, i view it as a 50% chance of picking the first box, and a 50% chance of picking the second box.
was my initial wrong conclusion. i meant that i initially viewed it as a 50/50 chance, but later understood the thirds.

3.14....duh

This one isnt easy as pie, most ppl get it wrong and you were one of them.

Attached: troll_math.png (292x74, 14K)

50%

the wikipedia page on this is much better for understanding what's going on than the retards trying to explain it in this thread

en.wikipedia.org/wiki/Bertrand's_box_paradox

That makes absolute sense if you have a single box that begins with 4 balls (3g 1s), where you already have one gold one in your hand. There would be 3 remaining balls in it, then you’re reaching your hand in. Out of those 3 balls, yes, two are gold, one is silver. Your chances of taking a gold ball out is 2/3.

That is not the case here. In this case, once you already have a gold ball in your hand, you are only dealing with one remaining ball in that specific box. That ball is silver, or gold. One of the two.

>That ball is silver, or gold. One of the two.
true, but 1 of them is twice as likely as the other one.

2/3
Three gold balls. If you pick either one in the "only gold" box you will get another gold. If you pick the one in the "one of each"-box you will not get a gold.

In the case of it being a single box that housed 4 balls. Yes, the remaining 3 have a twice as likely chance of it being gold.

In this case we have established the scenario of you selecting ball 1 as gold is 100%. There’s no going around that.

If there are only two boxes that are relevant, and you remove one gold ball from the GG box, (if that’s the box you happened to select), that would leave a gold ball remaining. It does NOT matter which of the two gold balls you pulled out of the GG box, it still leaves a single ball remaining. Same with the GS box. If you remove the gold ball, that leaves a single silver ball.

Since you don’t know which box you removed the first gold ball from, that leaves a single ball left which could be either the remaining gold ball from box 1, or the remaining silver ball from box 2.

you just listed two options, one of which involved picking the gold ball.

how fucking retarded are you to then say 2/3?

>pick Box 1 or Box 2
>pick out a golf ball
>if you've picked Box 1, you get a gold ball
>if you've picked Box 2, you get a silver ball
>50/50
50%

Also, what is that gif of? That’s pretty funny shit.

no there are 3 options, because the first box has 2 SEPARATE gold balls, it could be either one of those and you dont know because you picked it at random.

idk i didnt post it.

Attached: 1580471643506s.jpg (125x125, 2K)

if you pick up a ball and it's gold, then you can ignore the box on the right because its not that one. that leaves two options. one still has a silver ball in it, the other has another gold ball in it. therefore, there are two options and one of them will be the right one.

1 out of 2. 1/2. 50%.

yes we all know we ignore the third box.
now you chose 1 of the 3 golden balls at random, what part of that is so hard to understand?

[G1, G2] [G3, S1] [S2, S3]
How many possible ways can you choose 2 balls from the same box in this setup?
G1, G2
G2, G1
G3, S1
S1, G3
S2, S3
S3, S2
We only consider the first 3 since it's given we picked a gold ball first.
G1, G2
G2, G1
G3, S1

As you can see, 2 out of 3 are from the double gold box. 1 out of 3 is from the Gold/silver mixed box.
Does your small brain comprehend?

youve already chosen one golden one, it says that in the image. that leaves only 2 left, in two different boxes

How stupid are you? You are NOT choosing 1 of the 3 remaining balls. If you were choosing 1 of 3 remaining balls, that means you’d be selecting from a box that is currently holding all 3 balls remaining balls. You are not doing that here.

You’re selecting the last ball from the box you currently have in front of you. There is only ONE ball remaining. It’s gold or silver.

That likeliness was already decided when you chose a box, from then on in theory you have
>the box with 1 silver in it
>the box with 1 gold in it
choose one, and tell me how this choice is 2/3

>2/3
You're all a bunch of morons, it's not Bertrand's Box Paradox, that is the probability of the entire scenario. *This* scenario specifically asks for the probability after the first gold ball has been picked, the answer to this scenario is 1/2

[G1, G2] [G3, S1] [S2, S3]
How many possible ways can you choose 2 balls from the same box in this setup?
G1, G2
G2, G1
G3, S1
S1, G3
S2, S3
S3, S2
We only consider the first 3 since it's given we picked a gold ball first.
G1, G2
G2, G1
G3, S1

As you can see, 2 out of 3 are from the double gold box. 1 out of 3 is from the Gold/silver mixed box.
Does your small brain comprehend?

>it's not Bertrand's Box Paradox
Do you have Down's syndrome?

Attached: BBP.png (1568x201, 34K)

it doesn't matter if we pick up G1 or G2, the outcome is the same. those two outcomes are interchangeable

Does anything change if the gold balls don’t have fixed positions/identities in the boxes?
i.e., if I grab a gold ball, which could either be R or L in your explanation, but the box is shaken up after (or while) I grab it, am I still performing the same experiment with a 2 out of 3 chance of grabbing another gold ball?
I feel like a part of the quantum/observation piece is dependent on each ball being unique—if you take that away, you’re back to “probability you picked the box with 2 gold balls” instead of “probability you grab one specific ball out of 3.”
If you have 4 balls labeled 1 through 4, and you grab ball #1, you now have a 2/3 chance of grabbing either ball #2 or 3. But if someone else grabs the first ball and can only tell you that they grabbed an odd numbered ball, and you know the box it came from had exactly one odd and one even ball in it, you can’t pull another odd ball, so you must be about to pull an even one—which has a 50% chance of being #2 and a 50% chance of being #4.

Choosing either box is 50-50, but if you draw a silver ball first, that trial is not recorded due to the condition of the problem being that you draw a gold ball first. So if you consider the number of times each box was chosen when a gold ball was drawn first, it is obvious that the middle box has a smaller probability of being chosen since the trials where the silver ball was chosen first were disregarded.

50% bc if you pick a gold one the first time, it's impossible that you picked the 3rd box. So you picked either the 1st or the 2nd box. The 1st one contains another gold ball and the 2nd one a silver ball. So the chanses are 50/50

Stupid question: Is the gold ball returned to the box after its selected or is it gone?

Wait, hang on, that’s not quite right; if someone picks an odd ball for you, the equivalent scenario is that you want to pick another which is

One in two. You've grabbed a ball from either the box with two golds or the box with a gold and a silver. Depending on which box you've selected either there is a silver ball remaining or a gold ball. The two silvers box is completely irrelevant.

Can you read?

Go do it. Find out. It is not hard. Do it with pieces of paper in cups if you must. Discard any trial that starts with a silver ball, record all trials where you draw a gold ball.

Yes, but obviously you can't. It clearly describes the same problem as the OP, with coins instead of balls.

So explain why it's wrong then rather than just being a cunt about it.

that's not even close to what the question is asking. did you drop out of school when you were 7?

the question asks "What is the probability that the NEXT ball you take from the same box will also be gold?"

it is assumed that you have already picked out a gold ball. that rules out the third box entirely from this scenario.
from the two (2) options left, one of them will have a silver ball, and the other will have a gold ball. the question is asking how many options will yield a second gold ball (1 option will) out of the remaining options (there are 2 boxes that it could be, remembering that you cannot initally pick a gold ball out of the third box)

how anyone says 2/3 is beyond me

already directed ppl (multiple times) to go look up the problem online, its name is Betrand's Box paradox.

Its conditional probability and its very counter intuitive, people cant deal with the condition part.

Just happened to pop in on my coffee break again.

In the words of Richard Feynman, if it disagrees with experiment, it's wrong. Experiment shows 0.66 recurring.

I know. They're welcome to write Gold and Silver on scraps of paper and do this experiment themselves.

Correct, because the probability of the first choice (which box) is relevant to the actual outcome.
Okay folks, this thread apparently won't die, so I made a new version that anyone can do. Get three mugs or cups, tear up paper into small squares, and write GOLD on three and SILVER on three more. Arrange them in the cups just like the OP. Then do the experiment. If you pick out a silver slip first, put it back and try again because that doesn't match the OP scenario. It has to be a randomly selected gold slip first.

Reality says 2/3. You are simply wrong, according to reality.

Of the three gold balls I could have drawn, two shareabox with another gold ball. 2/3.

Found my tally from last time in my junk paper tray. Results show that about 50% of the time the results must be discarded (i.e. there is a 50% chance of the first ball you pick being silver) and of the remaining half the results are a 2:1 (66%, 2/3) split in favour of gold.

That is all. Reality has spoken.

Attached: IMG_20200205_140457_copy_1728x2304.jpg (1728x2304, 1.05M)

Regardless of what the sperglords posting in here have to say it's in the 60% range.

0.66

To everyone saying 1/2, look at the wording again. It's asking what the chances are when picking from the SAME box.

this is the best way of explaining it.

Wow, you must have a lot of time on your hands to trial it with an infinite number of iterations

dont need infinite to start getting the picture

Easier way to think about it:

Scenario 1: Picked box 1, gold A
>Second ball is box 1, gold B

Scenario 2: Picked box 1, gold B
>Second ball is box 1, gold A

Scenario 3: Picked box 2, gold A
>Second ball is box B, silver B

Scenarios 4 through 6 all start with picking a silver ball, and are eliminated.

Try not to think of it as, “which box do I have,” because that part is already affected by box 3 having been kicked out (2/3 chance to grab a box with any gold balls; 1/3 chance to grab the silver-only box; and, if you grabbed the box with the gold and silver ball but grabbed the silver one first, you put it back, so of the 2/3 chance to grab a box with any gold in it, 1 out of 4 of THOSE times you put it back). You want to think of that as 50/50, but it wasn’t 50/50 when you grabbed the box, and it isn’t now that you have that box and are grabbing a second ball.

Try to think of it as, “which set of balls have I got.

489 trials was enough to get the picture. Another user wrote a simulation which did 100,000 trials. Same picture, about 2/3 of the time the second ball is gold. Reality has spoken.

youtu.be/LIxvQMhttq4

The problem isn't articulated well.

There's a 2/3 chance of probability /overall/ from the start of the scenario that you will end up with two gold balls. The /remaining/ probability after the scenario has been completed puts the odds at 50% (100% or 0%).

The issue here is english, not mathematics.

and tell us exactly, how did you conduct the experiment?

the issue is, since the conditions have been changed, the probabilities are not the same any more

It's the Monty Hall problem and the answer is 2/3

he had three faggots, stuck two balls in each of their anuses, and then he reached inside a random anus, withdrew a ball, and then licked it clean to check the color. he repeated this 400 times.

You can just ignore the existence of the third box altogether, and also the possibility of picking silver first on the second box. So there's 3 possibilities.
2 are always favorable (Picking gold from box 1) and 1 is always unfavorable (Picking gold from box 2) so it's 2/3.

I used a die to determine the box, with 1 and 2 being box 1, 3 and 4 box 2, and 5 and 6 box three because it would have been difficult to shuffle the boxes randomly. I then reached under and pulled a random ball from under that box after shaking a bit. If it was silver I put it back and rolled the die again after recording it as n/a. If the ball was gold I then pulled the second ball out and recorded whether it was g (gold) or s (silver). I repeated this nearly 500 times. This matched the criteria of the original, namely:
1. randomly select a box
2. randomly select a ball in that box
3. it is gold (discard all trials where it is not good)
4. what is the probability the second ball from that box is gold?

Arguably if you have a single die you could do the experiment with no boxes or balls.

Almost; there’s a 1/3 chance you pick a box with 2 gold balls, and a 1/3 chance you pick the box with one gold and one silver (but if you do, a 50% chance you see the silver one first and put it back).
But yes, the rules/conditions change after that, and now you are left with a different probability for a different set of conditions. That 1/3 + 1/3 chance of having the first or second box is no longer relevant to the 2/3 chance of the second ball you draw being gold, and you don’t use it to calculate that.

The conditions from that point (and the point where you start this scenario) are identical if you start with 2 boxes (one with 2 gold balls, one with gold and silver). In either case, you have drawn 1 of 3 gold balls, and are left with 2 gold balls and 1 silver one to draw from—THIS is where the 2/3 comes from, not the 1/3 + 1/3.

: P(G) = P(S) = 0.5

Op is a faggot for grabbing balls!

OP is always a faggot, tell us something we dont know OR give us the answer.

the thing is the "1 silver 1 gold box" isnt really a 1 silver 1 gold box, because every time you would catch the silver one first it doenst count anyway.

Wrong. See
And most importantly
>youtu.be/LIxvQMhttq4
>If it disagrees with experiment, it is wrong.

50%

it would be 40% because after you take one gold ball there'd be 5 balls left, 3 of which are silver. so a 2/5th chance you get a gold ball, therefore 40%. who gives a fuck that they're in boxes

Wrong.
Wrong.

See once again
And most importantly
>youtu.be/LIxvQMhttq4
>If it disagrees with experiment, it is wrong.

: wrong : see

That disagrees with experiment and is therefore wrong. Nice trips though.

Fucking hell people this shit has been on Wolfram Demonstrations for like 3 years. This is settled math. The probability is 0.66. it happens that the second ball/coin/whatever is gold 2 out of 3 times.

demonstrations.wolfram.com/BertrandsBoxParadox/

: wrong : P(G) = 2×P(S) = 0.66...

its 50%. As the probability for 2 silver is 0% and the other two have the same probability.

Wrong.

See once again
demonstrations.wolfram.com/BertrandsBoxParadox/
And most importantly
>youtu.be/LIxvQMhttq4
>If it disagrees with experiment, it is wrong.

Absolutely wrong. I won't explain to you how to get the right answer, but you should know that if you think you have 1/3 chances, then it is easily refuted because if you picked a silver ball instead, your odds of picking a gold ball next aren't 1/3, yet there are an equal number of gold and silver balls in the lot.

the next ball you take from the same box. as in the one you just drew a gold ball from. There's only one box with two gold balls.

If you picked a silver ball first then the probability of getting a gold ball next is indeed 1/3. This, again, can be proved by experiment. If your theory disagrees with experiment it is wrong.

See once again
tps://demonstrations.wolfram.com/BertrandsBoxParadox/
And most importantly
>youtu.be/LIxvQMhttq4
>If it disagrees with experiment, it is wrong.

This is just a glorified Monty Hall problem

The probability is 2/3.

It's its own problem with its own name that behaves very differently to Monty Hall, though both Monty Hall and Bertrand's Box are conditional probability.

demonstrations.wolfram.com/BertrandsBoxParadox/
And most importantly