/sci/ - Science & Math

2/3

stage #1: 3 ways to have picked the first gold
stage #2: Next ball, 1 silver, 2 golds

So, would the solution be
You first get G1 then G2
You first get G2 then G1
You first get G3 then S1

So a 2/3 chance the second is a gold ball?

>>9462739

lol this exact question was on my 2nd year probability exam

>>9462739
2/3

>>9462739
1/2
you take 1 gold out
you remove the box with 2 silver entirely
you have 2 boxes left
that means you either have box a or box b
only a has a gold

Fifty percent.

>>9462751
>>9462755
>>9462757
There are only two balls in each box. How can the probability of grabbing a certain ball from a single box be two thirds? Please explain, I haven't taken maths since underground

>>9462739
50%
if you picked a gold ball, then your hand is either in box 1 or 2
now you can only pick a silver ball if you landed in box 2, or a gold ball if you landed in box 1
so 1/2 = 50%

>>9462739
1/3 of course. because there is only one box with 2 gold balls.

>2/3 brainlets
why haven't you cleansed this world of yourselves yet?

>>9462739
B RV for chosen box ({gg,gs,ss})
F RV for the color of first ball ({g,s})
[eqn]
P(B=gg|F=g) = \frac{P(F=g|B=gg) \cdot P(B=gg)}{P(F=g)} \\
P(B=gg|F=g) = \frac{1 \cdot \frac{2}{3}}{\frac{1}{2}} \\
P(B=gg|F=g) = \frac{1}{3}
[/eqn]

>>9462739
1/2 duh

>>9462780

>>9462769
>How can the probability of grabbing a certain ball from a single box be two thirds?

The probability of drawing a second gold ball from the leftmost box is 1; the probability of drawing a second gold ball from the middle box is 0. However, given that you drew a gold ball as the first ball, the probability that you initially chose the leftmost box is 2/3, and the probability that you initially chose the middle box is 1/3.

>>9462784
>the probability that you initially chose the leftmost box is 2/3

>>9462784
>the 3rd box with 2 silver balls is still relevant
this is really fucking with my brain

>>9462739
2/3. You're more likely to have picked the first golden ball from the box with 2 golden balls inside.
>>9462776
The problem states you've already picked a golden ball, the box with 2 silver balls is irrelevant.

>>9462739
If you picked 1st ball from 2Gold, the 2nd ball must also be gold.
If you picked 1st ball from 1Gold1Silver, the 2nd ball must be silver.
Since you don't know which box is which, the probability of getting a gold from the same box is 50%.

What is the probability that my next CAPTCHA will involve vehicles? 98.9%

>>9462793
You're ignoring the fact that you might have picked EITHER of the two gold balls in the first box. The probability of holding either gold ball is 1/3, you have a 2 in 3 chance of having picked the gold ball form the double gold box.

2/3 bc 2 gold balls over 3 balls left (in the two relevant boxes)
You don't know which box you picked so you consider both

>>9462739
>>>9462768
>>9462769
>9462755>>9462776
>>9462784


It's already stated that the ball is gold. This eliminates the the box with two silver balls. Now that we know only two options for the second Ball, it's 1/2

>>9462778
>t. humanities graduate
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

Let's look at it this way by labeling the balls, not the boxes.

G1
G2
G3
S1
S2
S3

You have the following combinations that are possible
G1 then G2
G2 then G1
G3 then S1
S1 then G3
S2 then S3
S3 then S2

Since we know the first is a gold ball
What are the chances you got G1 then G2? 1 out of 3
What are the chances you got G2 then G1? 1 out of 3
What are the chances you got G3 then S1? 1 out of 3
What are the chances you got S1 then G3? 0
What are the chances you got S2 then S3? 0
What are the chances you got S3 then S2? 0

You have a 2/3 chance that you picked from box 1 and the second is a gold ball and only a 1/3 chance you picked from box 2 and the second is a silver.

>>9462739
gotta be 1/2. You're initially picking from a box, not from a collection of all the gold balls.

>>9462791
thanks for opening my eyes, i have always failed at statistic

>>9462786
>the probability that you initially chose the leftmost box is 2/3
... given that you drew a gold ball as the first ball.

Initially, the probability of drawing a gold or silver ball is 50:50. If you drew a gold ball, the odds of the box you chose being (left, middle,right) is (2/3,1/3,0). If you drew a silver ball, the odds are (0,1/3,2/3). If you drew a "gold or silver ball" (i.e., you don't know what color it was), the odds are (2/3,1/3,0)/2+(0,1/3,2/3)/2 = (1/3,1/3,1/3), which makes sense: clearly the initial choice of the box makes each equally likely.

>>9462814
actually. i take that back. this question is fucked. Why is probability like ths.

>>9462808
Using this logic, consider the odds that you chose the middle box. Initially, those odds were 1/3. You're saying that if you drew a gold ball, the odds of it being the middle box would increase to 1/2. But consider: what if you had drawn a silver ball? By symmetry, the odds of it being the middle box must be the same as if you'd drawn a gold ball. So by your logic, the odds would be 1/2.

But you are certain to draw a gold or silver ball! In other words, we start saying that the odds of having chosen the middle box are 1/3, but then we say that no matter what color ball we draw, the odds will increase to 1/2. Does this not seem wrong to you? That we can predict a change in the odds with certainty before making a measurement? If we know that the odds of being the middle box will be 1/2 no matter what we draw, doesn't that mean the initial odds must have been 1/2?

In other words, you are assuming a contradiction: the only way the the odds of choosing the middle box could be 1/2 *regardless of the color of ball you draw* is if the initial odds were 1/2, rather than 1/3.

Why are any of you confused by this question? I am interested in why some people understand math better than others.

>>9462780
You mean:
[math]P(B=gg|F=g) = \frac{P(F=g|B=gg) \cdot P(B=gg)}{P(F=g)} \\ P(B=gg|F=g) = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} \\ P(B=gg|F=g) = \frac{2}{3}[/math]
Since:
[math]P(B = gg) = \frac{1}{3}[/math]
With no prior information on the box.

>>9462808
>It's already stated that the ball is gold. This eliminates the the box
it's doesn't matter what is stated. we still need to pick the first ball at random.
silver box cannot be eliminated with that conditions. in fact, picking the second has absolutely no impact and the problem can be simplified by completely remove that step

>>9462791
>The problem states you've already picked a golden ball, the box with 2 silver balls is irrelevant
You already determined destiny by picking the fitst ball. this is irrelevant to calculate probability after that point. This is false probability

>>9462854
Ups, what shit did I post. Yes of course. Thx for correction.

>>9462854
this

>>9462886
It's not irrelevant because it means you were more likely to have picked the box with 2 gold because you got a gold

take one:
You begin with 3 boxes and 6 balls. 3 balls are by definition left since the right most box is eliminated and the golden one you picked. There are 2 golden ones left and 1 silver so the next pick will have a 2/3 chance to be gold

I just saw this same puzzle on /b/.
Fifty-percenters: go back to /b/, you'll probably feel more at home there.

>>9463315
but you're not picking from those 3 remaining balls, your box is decided and there's 1 ball remaining, it's a question of which is it

>>9463474
You do because it's basically a permutation of the three remaining balls - {gold, gold, silver}.

Let's take the moment you know you have picked a golden ball as the starting moment, anything before that does not affect any probability.

So there are 3 balls left:
- one in the box you just picked from
- two in the other box

(the one in your hand and the two in the right-most box are excluded from the starting six: 6 - 2 - 1 = 3 balls left.

The 3 balls are {gold, gold, silver} by definition, thus gold has a 2/3 chance to be picked.

>your box is decided and there's 1 ball remaining, it's a question of which is it
Yes the box is decided, but the question is if the next ball will be gold.

Just work it out on paper. Assume the balls are numbered from left to right so g1, g2, g3, s1, s2, s3.

pick g1: next picks are: g2.
pick g2: next picks are: g1.
pick g3: next pick is s1.

These are the only picks possible. Whenever you pick up a golden ball (g1,g2, or g3), two out of three choices result in a golden one -> 2/3 chance.

To those that believe that the answer is either 1/2 or 2/3, both are technically correct in their own right. On one hand, one may assume that the person that removes the gold ball replaced it (i.e. the image stays constant), or the other may assume that the person removing the ball does NOT replace the ball, and there is 3 balls left instead of 4. The latter is logically correct, and is the preferred answer, but 1/2 is technically correct due to the vagueness of the problem at hand.

Tl:dr, both 1/2 and 2/3 are correct due to the vagueness of OP's question.

>>9462768

i agree

>>9462780

lmao over complicated much

>>9462768
no, assume box A is gg and box B is gs,
- initial pick is from A (first g), then again from A results in g.
- initial pick is from A (2nd g), then again from A results in g.
- initial pick is from B then the next one is silver by def.

of these possibilities, 2/3 are gold, not 1/2.

>>9462739

>>9462739
100%, because I put my hand with the gold ball back in, then pull it out again.

>>9463550
It asks for the probability of getting gold from the *same box*. If you replaced the gold ball you found then the probability would be 5/6.

>>9462787
It's not. The question rules out the possibility of the third box being chosen and thus it's entirely irrelevant to the question. Anyone considering the third box as relevant to the probability is just as mistaken as anyone adding a hypothetical fourth box. As far as the question is concerned the box with two silvers doesn't exist because there is no reality in which it could have been chosen

>>9462739
50%.

If you take a gold ball out of the box you know you chose one of the boxes with the gold balls. If you chose the box with the 2 gold balls, the next one you pull out will be gold. If you chose the box with the gold and silver, the next one you pull out will be silver. There are 2 possibilities and 1 of those possibilities is a gold ball. 1/2 is 50%.

The people saying 2/3 are retarded. They are assuming that it's 2/3 because there are 2 golden balls in any of the boxes left with golden balls in them and one silver ball in the boxes left that had any golden balls in them, but the boxes didn't magically merge into one box when you pulled the first one out.

It's ambiguous and depends on how you read the problem. That is, if you consider the question to be asking the random chance to incur from the first selection (of the three boxes), or after you know you've picked gold.

>>9462784
No.

The probability that you chose the leftmost box is 50% and the probability that you chose the middle box is 50% because you have a golden ball in your hand.

>>9463701
No it doesn't, it outright says that it's after you know you've picked gold the first

>>9462811
90% of /sci/ confirmed brainlets and BTFO'd

>>9463572
The first two cases are identical. The first pick is between boxes, not balls. You pick between Box A or Box B, not between Ball A, B and C. You're coming to an erroneous result because you're mistakenly making the choice between individual balls and not the boxes that contain them.

>>9463711
If you choose a box and then you choose a ball from that box, and it's gold, you're twice as likely to have chosen the box with two gold balls as you were to have chosen the box with only one gold ball. What's in the box matters.

>>9463720
Only if you assume that there were 3 boxes to choose from, but we've established there weren't. The box with 2 silvers is ruled out.

You have a choice between two boxes. You make a choice, and once you make that choice your decision is fixed. You pulled out a gold ball what is the probability the other is gold? 0.5

The third box is not a valid choice. There are two boxes, you pick one and the probability is that the OTHER ball will be gold.

You're calculating the probability from a point further back than the question asks. It states you have already pulled out a gold ball, the probability of having chosen box 2 and pulled the silver first is irrelevant to the question, it has NO probability of occurring because the decision has already been made.

>>9463733
>Only if you assume that there were 3 boxes to choose from
It says right in the problem that there were. But this is not even relevant to my point. There could only be the two boxes and my argument would be exactly the same. If you have a gold ball in your hand, it's twice as likely it came from the box with two gold balls as it did from the box with only one.

>You're calculating the probability from a point further back than the question asks.
No, I'm calculating it from the point after you have chosen the gold ball. The fact that I chose a gold ball and not a silver ball tells me I probably choose from the box with two gold balls. Imagine that the box with only one gold ball had 999999 silver balls. If you chose that box initially it would be highly unlikely you found the one in a million gold ball. It's much more likely any gold ball you have is from the box with all gold balls. But you're arguing that it's equally likely to have gotten a gold ball from either box, ridiculous.

For the brainlets

>>9463753
>The fact that I chose a gold ball and not a silver ball tells me I probably choose from the box with two gold balls
The box you probably chose it from has already been determined. Going back in time and calculating the probability you chose that box in the first place is retarded and beyond the scope of the question. You have the ball in your hand, the decision has been made.

its 50%
if you picked a box and pulled a gold ball the other one is either gold or silver.

all of you dumb 2/3 niggers need to kill yoursleves for trying to equate this to thats stupid goats and cars problem

>>9463707
It's ambiguous in what it's looking for as the solution.

>>9463711
>The first two cases are identical
thats why it's 2/3. It's about the permutation of B = {g1,g2,g3,s1} that favors your pick.

Again, the balls are numbered g1,g2,g3,s1.Wich is {3g,1s} pick one gold and you get {2g, 1s}. Just go over the possibilities of the next step (there are only 3):
say you pick g1: g2 is the next pick
say you pick g2: g1 is the next pick
say you pick g3: s1 is the next pick

Whatever you do, 2/3 of the possible picks after gold (g1,g2, or s1) is another golden one (g_c \ B). You dont pick a ball, you pick from a box, but the ratio of gold and silver applies

>>9463754
OK so let's play a game. I have a box with two gold balls and a box with one gold and one silver. Every time a gold ball is taken from the first box, I get $1 from you. Every time a gold ball is taken from the second box you get $1.50 from me. According to you, you should get more money since you will win half the time and get $0.50 more than me, correct?

First box-> left ball -> gold, I win $1
First box-> right ball -> gold, I win $1
Second box-> left ball -> gold, you win $1.50
Second box-> right ball -> silver, no money

So on average I win $2/4 = $0.50 cents each game and give you $1.50/4 = $0.375

Meaning I make $0.125 each game. So how much money would you have to lose before you admit you're wrong?

>>9462811
Bertrand was wrong then. Holy shit, if probability is built on such flawed thinking then we better discard the entire field NOW

>>9463754
>balls disappear
>you always pick the silver-gold box
brainlet

>>9463767
That's an irrelevant situation to the question though? You're starting from the point where I choose a box, when that is already determined. If your situation was the same as the question where I have already selected a ball, I know it's gold and the only possible outcomes are box A holding a gold and box B holding a silver you're absolutely right I'd take you up on it.

Why do you keep trying to solve a problem that isn't asked by the question? The situation is very clear. The ball has been picked.Stop going back and calculating probabilities of events that have occurred. All that matters is that I have selected a ball and the other ball depends solely on whether I picked box A or B initially. There are no other factors involved, period.

can a 1/2-nigger solve pic related?

For what it's worth, "random" as it applies to the chosen box seem to merely be there as a means of suggesting that you don't know what the other ball is, rather than being included as part of the calculation requested.

What it asks is "What is the probability that the next ball you take from the same box will also be gold?" which implies that it is limited to that predefined scope rather than including the probability of the person picking a box then a gold ball in the first place.

>>9463759
>rying to equate this to thats stupid goats and cars problem
(congratz on your spelling skills btw)

Actually it's the opposite.
In Monty, it looks like there are no groups, but the doors fall in two groups - the one you chose, and the ones you didn't.
In Bertrand, it looks like there are two groups, but the odds are the same even if the 3 golds and 1 silver were all in one box.

1/2

Since its the same box, there are only two boxes with gold balls, if it is the one that has one golden ball (already taken then ) then it is a 0% chance

If it is the one that has two golden balls, then it has a 100% chance.

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

>>9463756
>The box you probably chose it from has already been determined.
Yes, but it's unknown to you. A probability is simply a measure of how much information about an event you have. If the answer was "determined" from your perspective, then you would already know with certainty which box it is. But you don't. All you know is that it's twice as likely you chose a gold ball from the first box as it is you choose it from the second.

>Going back in time and calculating the probability you chose that box in the first place is retarded and beyond the scope of the question.
No one went back in time you insufferable retard. I simply looked at the possible events which led to the current outcome and saw that one is more likely than the other.

Let's say a person with a winning lottery tickets come into your bodega. You put their ticket into a box. You tell someone to scratch off a regular lottery ticket and put it in a second box. Now you choose a random box and pull out a winning ticket. So you're telling me it's just as likely you won the lottery with a random ticket in the second box as it is you simply chose the first box that array had a winning ticket in it? Well this changes everything. Now to win the lottery you just need a winning lottery ticket and two boxes. There's a 50% chance of winning the lottery each time you do this and you can do it over and over again. Wow!

Jesus Christ, sci. It's a simple probability problem...

>>9463777
>each box contains 2 balls
>the final box contains 4 gold balls and one silver ball
ebin

Anyway, we are given as a matter of fact that the first ball picked was gold, so there's an 80% chance that the outcome of the second ball will also be gold.

>>9463775
>You're starting from the point where I choose a box, when that is already determined.
No, you only win money after a box is chosen and after getting a gold ball, it's the same.

>>9463787
>I simply looked at the possible events which led to the current outcome and saw that one is more likely than the other.
Except it doesn't matter which one was more likely, it's already happened and the question begins from the state of already having picked a box and a ball. Probabilities are irrelevant to events that have occurred. Events that occur have a probability of 1. "Which box is more likely to pull a gold out of" ceases to be a relevant question the moment you pull a gold out.

>>9463799
>it's the same.
How can it be the same when your scenario still has the potential to pull out a silver, a situation that is completely ruled out in the question? You start from the state of already having a gold. The selection has been made so how is a situation where you can pull out a silver even remotely comparable

>>9463777

It might help these tard turds if you put a ton of gold balls in the first bin, but only one gold ball in the second bin with a bunch of silver balls.

They'll see that, since there are more gold balls in the first, the condition that you got a gold ball in the first draw skews you toward considering the first box.

>>9463803
>Except it doesn't matter which one was more likely,
Exactly, so if you put a scratched off winning lottery ticket into a box and a random scratched off ticket into a second box, then you have magically increased the chance of that ticket winning, since half the time you find a winning lottery ticket, it must have been the one in the second box. So go and buy a ton of lottery tickets, find someone with a winning ticket, and explainto them how you can create infinite winning tickets with this strategy. Go ahead and do it retard.

>>9463812
The probability of lottery tickets winning isn't connected to which box you find them in though

>>9463803
>Probabilities are irrelevant to events that have occurred. Events that occur have a probability of 1.
Ah so then the chance of you having picked from the 1st box is not 1/2 as you previously argued. It's 1. Or is it 0? You tell me side you understand probabilities so well. And no it can't be 1/2 dive you already chose the box. It already handed meaning it can't have probability 1/2.

Jesus Christ you are retarded. Again, probabilities are a measurement of your knowledge of events. Your knowledge can be incomplete even if an event already occurred.

>>9463803

I haven't invested much time in figured out what you are saying, because it probably isn't very important/intelligent, but let me try to help anyway.

You say that events that occur have a probability of 1. Let's pretend that's true. After you pick out the gold ball, the prob that you got a gold ball is 1, so that means after performing an action that picking out a ball, every probability changes (for example, the prob. you get a gold ball just changed from 1/2 to 1). Do you have a rule for how the probabilities change? I mean a general rule. If you come up with a consistent rule, I think you'll end up finding the correct answer.

>>9463820

Hint: p(a|b) = p(a and b)/p(b)

You can consider the conditional probability p(x|b) for any x in the measure space as a full-fledged probability in the sense of those axioms a probability has to satisfy. That's basically what you're doing by setting that probability to 1. These other fruitcakes in the thread can't grasp that.

>>9463806
>How can it be the same when your scenario still has the potential to pull out a silver, a situation that is completely ruled out in the question?
The question states that you had the potential to choose a silver ball, but didn't. Learn how to read. Nothing was stopping you from choosing the box with only silver or the silver in the mixed box. You just happened to not do it.

>>9463815
You just argued it did. When you get a winning lottery ticket there is a 50% chance it came from the second box. Yes or no? If yes you think that putting a lottery ticket into a box magically makes it likely to win.

>>9463824
>The question states that you had the potential to choose a silver ball
No you didn't. The question states you pulled out a gold ball. The possibility of pulling out silver is not relevant at all, it didn't happen and within the framework of the question, can't happen. Why can't it happen? Because you ALREADY HAVE A GOLD. There is no reality within the question where you ever get a silver. It's as if everything popped into existence the moment you pull a gold out. What occurred before has no relevance to the question

ITT /sci/ shits itself over which interpretation of a vaguely-worded problem is the right one.

>>9463826
>You just argued it did
No I didn't. You're pushing a retarded analogy that has no relevance to the question.

>>9463828

Suppose there are two boxes. One has 100 gold balls, and the other has 1 gold ball and 99 silver balls. You randomly pick a box. From that box, you pick a ball randomly. It is gold. What's the prob. you picked the first box? The second?

Now answer the same Q in the first problem.

>>9463828
>The question states you pulled out a gold ball.
This is a non-sequitur. You had the potential to pull out a silver but you didn't. You had the potential to learn basic probability theory but you didn't. That something didn't happened has fuck all to do with whether it had the potential to happen. Retard.

>>9463832

u are getting tolled hard bro

>>9463830
So why can't you answer? Yes or no? It shouldn't be that hard for a genius like you.

>>9462739
For anyone who hasn't learned Bayes theorem yet:
Pr(Box 1|Gold) = (Pr(G|B1)×Pr(B1))/Pr(G) = (1×0.5)/(0.75) = 2/3.

You're more likely to draw a gold ball from box one than box two, so given that you drew a gold ball it's more likely that you're in box one than box two.

>>9463831
Has no relevance to the question because you're talking about events that occur before pulling a gold out.

Lets just simplify this. Existence starts at the point you pull a gold out. There is no past that exists, only the facts you have before you. You pulled out a gold. What is the probability that you will pull out another gold from the same box. Is there ANY way you can claim it's not 0.5 without appealing to the past that does not exist beyond the point where you pulled the ball out of the box?

>>9463832
>You had the potential to pull out a silver but you didn't.
No you didn't. The question states you pulled a gold out. There was no possibility of ever pulling a silver out. See >>9463838

>>9463838

You should just answer my question bro. Even if it isn't relevant

>>9463839
>you pick box at random
>you pick a ball at random
You lose. Thanks for playing.

>>9463842
Of course it's relevant. If he answers no then he admits that what's in the box matters and his shitty troll fails.

the box you just picked out is obviously not the one with 2 silver balls, so you only have the scenario of the ball being gold or silver with 1/2 chance due to it being the box having two gold, or one gold one silver.

>>9463845
>It's a gold ball

>>9463851
See >>9463832
>>9463833

>>9463849

I know it's relevant, but you can't argue so many things at once on the internet. Just insist he answer that question.

>>9463854
He left, he's a pussy.

>>9463849
>If he answers no then he admits that what's in the box matters
Only if you're calculating from the point from before you pick a ball. After you pick a ball the facts of the question change and there are less possible results. It's not my fault you think the probability of picking a gold before you pick a ball has anything to do with what is inside the box you pick after it's already determined you have a gold.

Yes it would matter if the question was asked at the point before you pick a ball. But it isn't. We know what ball we picked. It's gold. It can never be silver because we've picked it's gold, it's fact. The present situation is we have a gold ball that was selected from one box and the ONLY remaining variable is whether we picked box A or B. 50/50.

>>9463845
Where does it ask you about the probability of the first ball being gold? You're making the major assumption that it's asking that when all it's really asking is what the probability of the second ball is.

>>9463858

Suppose there are two boxes. One has 100 gold balls, and the other has 1 gold ball and 99 silver balls. You randomly pick a box. From that box, you pick a ball randomly. It is gold. What's the prob. you picked the first box? The second?

>>9463858
>bla bla bla
Still no answer to >>9463826 because you know you lost. Pussy.

>>9463862

Fucking puuusssssssssss

Pussaayyyyyy muthafucka

PUSSY BITCH

BABY BACK BITCH

DONT TELL ME HOW TO FUCK YOUR BITCH, PUSSY

>>9463859
>Where does it ask you about the probability of the first ball being gold?
It doesn't. It says that you choose randomly, meaning you had a 1/3 chance of choosing the silver box. You could have potentially chosen silver, but you didn't. You lose.

>>9463860
Has no relevance to the question asked.

>>9463868

Please dude. Just help me out. I need your probability expertise, I can tell you know your stuff

>>9463867
>It says that you choose randomly
It also says that you picked gold. 100%. The random chance stops being meaningful when it tells you outright what the result was. You have gold.

>>9463868
how is it irrelevant?

>>9463870
>It also says that you picked gold.
Yes, thank you for agreeing with me and admitting defeat.

>>9463867
That actually doesn't matter, you can, in fact, ignore the silver balls altogether.

The relevant question is, "which box did the ball I'm holding come from?" And there a 2/3 chance it came from box one, and a 1/3 chance it came from box 2. Therefore, there's a 2/3 chance the next ball you draw will be gold, and a 1/3 chance the next ball you draw will be silver.

>>9463869
When you answer >>9463838
If existence starts at the point where you pull a gold out and history before that point does not exist is there any way to justify any other probability than 50/50 without appealing to a non-existent history? Because the question is exactly like that. You start existence at the point you pull gold out, there are no prior probabilities before that point and every single time you've slunk back and said n"B-But if you assume we can go back in time and calculate from before you pick the ball!" no.

>>9463867
It also says that you picked a gold ball first. You're treating the probability of doing so as part of the calculation when it was in fact postulated as part of the premises before the question was even asked.

In other words, it didn't ask "What is the probability of randomly picking a gold, then another gold ball?"

It asked "What is the probability that the next ball will be gold?"

>>9463874
So you know it's 50/50 now?

>>9463882
This guy gets it

>>9463878
>That actually doesn't matter, you can, in fact, ignore the silver balls altogether.
It matters insofar as this retard says that you had no potential to choose silver balls.

>>9463882
Nothing you posted responds to a weird I said. Try again.

>>9463879
Okay, I'll answer it, but you better answer my question in your next post! If you don't... god have mercy on your soul.

Okay, so suppose you pulled a gold ball from some box. There is no "history" before that. That means you pulled a gold ball from either the first box or the second box, but you won't know which. Let's label the balls. So, the first box has balls A and B (both gold), and the second box has balls C and D (C is gold, D is silver). You have an equal chance of holding B, C, or A in your hand. Suppose you are holding B. That means A is the other ball and the other ball is gold. Suppose you are holding A. Then the other ball is B, and the other ball is gold. Suppose you are holding C. Then the other ball is D, so the other ball is silver. Since you had equal chance of holding A, B, or C (1/3), that means 2/3 of the time you will get a gold ball. 1/3 of the time you will get a silver ball. Never talked about the "past", only "the facts I have before me".

Okay, now you better reply to both this reply AND the Q I asked you.

>>9463879
See >>9463878

At the beginning of existence you're holding a ball, that did that ball come from box one or box two? There's a 2/3 chance it came from box one, as two out of three gold balls are in box one. There a 1/3 chance it came from box two, as one out of three gold balls are in box two.

Consequently, if you draw another ball from the same box as the initial ball, there's a 2/3 chance you draw another gold, and a 1/3 chance you draw a silver.

>>9463883
No, you just agreed with me that you had the potential to choose a silver ball. Because nothing you said contradicted that.

>>9463890
You didn't. And it doesn't matter that you didn't, it's still 2/3.

>>9463894
You never had the potential to do anything before you picked the gold ball because that's where the question begins. You're fabricating a history that has no relevance to the question. You pick a gold ball out. That is the defined starting point. To assume there was any possibility of picking a silver ball out assumes anything happened before you picked the gold, it didn't.

>>9463898
see
>>9463892

oh my god, I have so much adrenaline right now

my heart is pumping the serotonin is surging

>>9463898
>You never had the potential to do anything before you picked the gold ball because that's where the question begins.
The question begins by describing the setup prior to that. Ignoring information in the question is why you dropped out of elementary school Cletus.

>>9463892
Pulling out balls A or B are identical cases and are not considered statistically different. Pulling out a gold ball from box A is the same result whether it be either ball.

>>9463891
It's postulated that you didn't choose silver first.

It only requests information about the probability of the second ball given a predefined state of the first, not the probabilities of the first and second together.

If it wanted the probability of a gold then another gold, then it would have specifically asked that instead of asking only for the probability of the second.

Your answer is correct for an entirely different question from the one being asked.

I'm not certain how much simpler I can make this.

>>9463908
But one of them has a big fat "A" on it and the other one has a big fat "B" on it, you just can't see it because you are a fucking retard

you never answered my other question you little bitch

you'll trolled me hard, friend

have a good night

>>9463909
>I'm not certain how much simpler I can make this.
I don't think they're going to understand even if you break it down far enough that a 3 year old could understand it. They're obsessed with answering a question that was never asked.

>>9463921

your ass is blasted by this guy

someone get this man an ass transplant

>>9463909
Yeah again nothing you says responded to what I said. The answer is 2/3. You have a 2/3 chance of the next ball in the box you chose a gold from being gold. That's what the question asked. T be reason this is true is because it's twice as likely you chose from the box with two gold balls. Note that your attempt to characterize this reasoning has no bearing on whether it answers the question.

>>9463937
>T be reason this is true is because it's twice as likely you chose from the box with two gold balls
Like he said that's pushing it back and calculating the joint probability of picking gold > gold, which is not what the question asked. The probability of picking the first gold has absolutely no relevance to the question and if you're including the probability of picking the first gold in your answer that's an instant give away that your answer is wrong.

Again the question does NOT ask what is the probability of picking gold then another gold, which is the question you just answered. It is presuming you have selected a gold (the probability of selecting which has no relevance) then asks what is the probability of the other ball being gold. This is not a question about joint probability, it's a simple question about what color the other ball is. And the answer is it's either silver or gold. 0.5

>>9463953
Ok let's try this again retard.

Thereis a known winning lottery ticket in one box and a random lottery ticket in a second box. You pick a box randomly and you find a winning lottery ticket. What is the probability it came from the second box?

>>9463953
>Like he said that's pushing it back and calculating the joint probability of picking gold > gold
No. The probability of picking gold then gold is 1/3. The probability of picking gold given the events in the problem is 2/3. Try again, retard.

>>9463937
Okay, let me explain it in the form of two different problems.

There is a jar with 50 pennies and 50 nickels.

#1: If coins are chosen randomly and removed, and the first coin is a penny, what is the probability that the second coin picked will be a penny?

(This is my interpretation of the problem: a predefined state for the first element, while the question asks about the resulting independent probability for the second, given the changes to the scenario induced by the predefinition of the first element.)

#2: If coins are chosen randomly and removed, what is the probability that you will choose a penny followed by another penny?

(This is your interpretation of the problem, where it asks about the probability of the second as conditionally dependent on the probability of the first. Here the probability of selecting a penny first would matter because it's part of what's being asked.)

>>9463964
>The probability of picking gold given the events in the problem is 2/3
It's 0.5. The only variable is which box you picked. A or B. 0.5 you picked the gold in A, 0.5 that you picked the gold in B. Thus which box you picked is the variable that determines the color of the second ball you draw.

>If coins are chosen randomly and removed, and the first coin is a penny, what is the probability that the second coin picked will be a penny?

100%, given that the problem stipulates you have to draw from the same box.

>>9463970
Read the question again smart guy. One jar with 100 coins. 50 pennies, 50 nickels.

>>9463970
...I think you might have replied too fast.

Slow down and read it again.

>>9463975
>>9463972
Yeah, I assumed they were making a comparable analog.

It doesn't even get them what they want, though, since if you draw a penny the first time and throw it away, the probability you draw a nickel the second time = 50/99 != 0.5.

So all in all, just a horrible example.

The question has you picking boxes, not balls. If we rephrase the question as

"If you select a box that has one gold what is the probability the second ball is gold?"

The answer then becomes trivially obvious.

>>9463966
>#1: If coins are chosen randomly and removed, and the first coin is a penny, what is the probability that the second coin picked will be a penny?
49/99

>#2: If coins are chosen randomly and removed, what is the probability that you will choose a penny followed by another penny?
49/198

>(This is your interpretation of the problem
No, retard, it is not. The correct interpretation is like #1 and the answer is 2/3. You have illuminated nothing, just shown your total lack of understanding of anything said in this thread.

>>9463968
So answer this >>946396 oh wise one.

>>9463979
You missed the point again. He was comparing two separate questions and using them to illustrate how you're answering the wrong question.

>If coins are chosen randomly and removed, and the first coin is a penny, what is the probability that the second coin picked will be a penny?

Is distinct and a separate question from

>If coins are chosen randomly and removed, what is the probability that you will choose a penny followed by another penny?

The answer for one will be incorrect for the other. Just as answering the probability of getting gold > gold is different to the probability of getting gold given you already have one gold.

>>9463968
So answer this >>9463962 oh wise one.

>>9463987
No one is answering the second you fool. That would be 1/3, not 2/3.

>>9463982
>If you select a box that has one gold what is the probability the second ball is gold?
This is essentially what the question is asking

It doesn't ask you which box you chose, it asks you "If you drew a gold from a randomly chosen box, then what is the probability that the other ball is gold?"

>>9463999
Where does the other ball come from?

>>9464001
The same box. Yes you can redefine it that way, but "which box did you pick?" isn't what's in front of the question mark.

If you do rephrase it that way, as other anons have mentioned, the question basically becomes "Between box 1 and 2, what is the probability that you picked box 1?"

>>9464008
The facts we know are.
1. You pick a box
2. That box has a gold ball in it
From there it's quite simple to see that since we have those two facts the two boxes with a gold ball are A and B and it could be either of them because the only fact we have is that there is at least one gold ball in the box we picked.

>>9464010
The fact that you found the box contains a gold ball makes it more likely that it's the box with two gold, since you would have had to pick a specific ball to find the gold ball in the mixed box. If you disagree, you must think that the answer to >>9463962 is 50%. Ignoring useful information will give you the wrong answer.

>>9464021
What is the answer to "If you select a box that has one gold what is the probability the second ball is gold?"

Once you understand that's the question in the OP just rephrased it becomes obvious it's 0.5

I'm no stadistics expert, but I think the answer is 2/5, because you only have left 2 gold balls + 3 silver balls so after the first pick you 2 gold balls out of 5 balls, then (#of gold balls)/(#of remaining balls).

The probability for the third pick have to be lower and so on.

50%

>>9464023
>What is the answer to "If you select a box that has one gold what is the probability the second ball is gold?"
How do you know you selected a box that has one gold? Your question is ambiguous since it does not tell me what information I have/don't have.

>>9464023
Why would you need to rephrase the question when it can be easily answered as is. Do you not know basic probability theory?

>>9464023
Referring to a "second ball" implies that the "one gold" refers to a specific ball that has been identified. Without telling us how it was identified, the question is ambiguous.

>>9464027
Logically how you obtain the knowledge doesn't have any affect on the probability of whats in the box so we should ignore that to avoid muddling the question. If we just stick to the known facts it's fairly straightforward.

1. We picked a box
2. That box has at least one gold ball
3. Both A and B have at least one gold ball
4. Our box could be either A or B
5. The probability of the second ball being gold or silver depends on if we have box A or B
6. Therefore the probability of the second ball is 0.5 for it to be Gold or Silver since it is dependent on having box A or B

I don't see any problem with this logic.

>>9463829
It's not vaguely worded. There are two types of people in this thread: people who have taken prob&stats, and people who are arguing from intuition. 2/3s is correct.

>>9464032
>2/3s is correct
Only if you interpret the question incorrectly. See >>9463909

>>9464035
You're mistaken.

>>9464031
Logically how you obtain the knowledge has a large affect on the probability of whats in the box, since the different outcomes can affect the probability of that knowledge being obtained. Again, ignoring information won't help you, it just makes you look like a fool.

>>9464031
Wouldn't the simplest solution to this be to set up a simulation, and look at the answer?

Make sure you set it up exactly like the problem specifies though, if you rephrase the simulation, you'll get the wrong answer.

Or do both. The initial problem, and then how you rephrased it, so you can see why you're getting the wrong answer. That would probably be best for learning.

>>9464031
>Our box could be either A or B

Yes, it could, but not with equal probability.

>>9464031
So still no answer to >>9463962

Yeah you lost and you know it.

>>9464042
The answer becomes trivial when you realize it's basically "If you select a box that has one gold what is the probability the second ball is gold?"

If there are 1 googol gold balls in 1 box and 1 googol silver balls + 1 gold ball in another box. What is the chance that the next ball you choose is gold after you just chose a gold ball.

Protip: It's 50%. The number of balls doesn't matter because the outcome is decided as soon as you choose a box. The only thing that changes is the likely hood of choosing a gold ball initially.

So many fucking brainlets in /sci/ these days.

>>9464048
Shitty troll is on a loop.

>>9464050
This guy is correct

>>9464048
Should be a boring simulation then. Give it a shot.

>>9464050
Yo bro, your likely hood is harlem nigga, cuz its clear you aint got no private education

>>9462739
It makes a lot more sense that /sci/ doesn't understand bayes theorem when you remember that this place is full of high schoolers

>>9464008
>Between box 1 and 2, what is the probability that you picked box 1?
That's not question. The question is: if you pick a ball at random and it is a gold ball, what is the probability that second ball from the same box is also gold. This can be simplified to the "what is the probability that you chosen a box with 2 gold balls".

You determined the second ball by picking the first ball. You cannot count the probability of deterministic system by definition. You can't count the probability to pick up the second gold ball, because you don't have the options to choose between balls. You can count that 50%, but this is just artificial meaningless numbers, they do not say anything about your actual chances, you can't use this numbers to make better decision.

Another way of solving it, for the brainlets benefit.

If you choose box 1 there is a 100% chance of it being gold. If you choose the second box there is a 0% chance of it being gold. The average of these two is 50%.

QED.

It's pretty obvious that anyone who disputes this is a troll or actually retarded. I hope for the latter for hilarity.

>>9464063
What's the probability you choose box 1, given that you drew a gold?

I'll help you start: Pr(Box 1| Draw Gold) = (Pr(Draw Gold | Box 1) × Pr(Box 1)) ÷ Pr(Gold)

You just have to plug in the numbers now.

>>9464063
Jesus Christ kid.

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

>>9464077
>What's the probability you choose box 1, given that you drew a gold
0.5

Both boxes have a gold and me drawing one has absolutely no impact on the other ball inside the box I picked.

>>9464077
The probability that you chose box 1 is irrelevant to the question asked. You attempts at trolling are low energy.

>>9464082
No.

>>9464082
>Pr(Box 1| Draw Gold) = Pr(Box 1|At least one gold)

Why do you think this is true? Drawing a gold ball from your box tells you something about your box.

Think of it this way. I draw a ball from my box. It's gold. I put it back. I draw again. It's gold. I put it back. I draw again. It's gold. I put it back. I draw again, it's gold, I put it back. I draw again, it's gold I put it to the side. What's the probability that I draw a silver on my next draw? All we know for sure, is that there's at least one gold ball in the box, but that series of draws is much more probable if we have a box with two golds, than if we have a box with one gold and one silver.

>>9462776
You read the problem wrong boy, read it again. You probably never did probability in school and learned how these problems work.

We are told the ball drawn IS gold. Therefore all events that result in a silver ball being drawn are irrelevant.

You choose box 1.
Two possibilities.
You draw a gold1 and then gold2 or
you draw gold2 and then gold1.

You choose box 2.
One possibility.
You draw gold3 and then silver1.
You cannot drawn silver1 and then gold3.

You choose box 3.
No possibility.
You cannot draw silver2 and then silver3.
You cannot draw silver3 and then silver2.

You choose a box RANDOMLY.
It cannot be box 3 as there is no possibility.
Therefore it is either box 1 or box 2. Both have an equal likelihood of being chosen. If you chose box 1 you have two possibilities that both result in gold being the second draw. If you chose box 2 you have one possibility that results in a silver being drawn. Since the chance you chose either box is equal (50%) the chance that either a gold or silver ball is chosen is by extension, also 50%.

>>9464099
https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox
https://en.wikipedia.org/wiki/Bayes%27_theorem

>>9464102
That considers that there was ever a possibility that silver was drawn. Which, based on the question asked, is not possible. Because gold IS drawn.

>>9464102
If you physically do this and discard and choices that result in a silver being drawn you will see that the chance is exactly 50%.

>>9464106
>If a gold coin was withdrawn in the first step, it is more likely that this happens because the chosen box is number 1 (the box with two gold coins) than box number 3 (the box with one coin of each type).

Have you taken a course on probability theory?

>>9464108
You're mistaken. If you physically do this, according to the procedure described in the image, and then discard all of the cases where the first ball drawn was silver, you will will see that the chance is exactly 2/3.

>>9464099
>Both have an equal likelihood of being chosen.
The only have an equal likelihood BEFORE you discover you're holding gold. The knowledge that you're holding a gold ball tells you about which box you picked.

>>9462739
If you picked the mixed box, you know it's 1/3 if you pick randomly.
If you picked the all-gold, it's still 1/2 if you pick randomly.

You can know which box you picked from based on weight. The chance of you picking all-gold and getting a gold is higher than picking mixed and getting a gold, so you know the one you picked from is more likely to have another gold. If I get to keep the balls, I'm going to pick the same one as last time.

You can also know EXACTLY what's in each box based on weight, since gold weighs almost twice as much as silver, so you can actively choose the box with two gold balls twice making the chance of you getting gold both times 100%.

since the problem is solvable with drawing only, I took this route.

>>9462739
If you were to draw a gold ball from any of the boxes, there's a 2/3 chance that it came from the leftmost box. If it came from the leftmost box, there's a 100% chance that the next ball will be gold.

If you chose the middle box, there would be a 0% chance. Thus, the odds of drawing another gold ball are (2/3)*1 + (1/3)*0 = 2/3.

If there were one additional silver ball in each box, then the odds would be 1/3.

If there were one additional gold ball in each box the odds would be (1/2*1) + (1/3*.5) + (1/6*0) = 2/3.

If there were 2 additional gold balls in each box
(4/9 * 1) + (3/9 * 2/3) + (2/9 *1/3) = 20/27.

>>9464150
The first two scenarios are identical, there is no reason to split them. When you remove one you get the correct 1/2 result

>>9464167
>The first two scenarios are identical
Just like your brain and ass are identical

>>9464167
since the scenarios are identical, according to you, we could add 1000 golden balls to each box and it wouldn't change a thing.
weak troll or simply stupid / 10

>>9462739
>Take a ball from that box at random. It's a gold ball.
Therefore the box with two silver balls gets discarded.
1/2

>>9464348
>he thinks you're just as likely to pick the box with 1 gold ball as the box with 2 gold balls if you picked a gold ball

You pick a gold ball.
There are 3 possible balls left, 2 gold, 1 silver.

2/3

2/3

>>9465112
that implies both the remaining gold balls are both relevant when only one is, if doesn't matter if you got ball 1 or ball 2 in the double gold box, there's 1 remaining gold there, and there isn't any remaining gold in the 2nd box, 50/50

>>9462739
Ita not 50% fucking brainlets it depends on witch box and witch gold ball you tale.

I have picked a golden ball,
This means, another ball is gold or silver.

It depends on how balls was placed in the box, I picked from middle box, thus ball is silver.

inb4 brainlets> There is robotic arm, that always toss the coin same.

>>9465112
>You pick a gold ball.
>There are 3 possible balls left
wrong there is 1 possible ball left, there aren't 3 more balls left in the box

Knife has two edges, what is the probability, that if I slide it on somebody's throat in the fight for live it will be sharp side?

You are retards.

>>9465214
If you get a gold ball, you know your box is either the one that has two gold balls or the one that has a gold and a silver ball. Therefor, there are three balls remaining that you could possibly pick. Two of them are gold, one of them is silver. 2/3 chance.

>>9465217
You take a ball, not pick a ball, try harder.

Also if boxes are closed, we can simply weight boxes if balls are same sizes.

>>9465217
No because you are picking from the same box, there aren't three balls remaining to pick from, there's 1, it's either silver or gold as a consequence.

>>9465221
If all balls are golden, and somebody is trying to shill you a lot, what happened?

>>9465220
>>9465221
This problem is called Bertrand's box paradox. Like the Monty Hall problem, the result is counterintuitive, but the probability really is 2/3, not 1/2. See mathematical and logical proof here: https://en.wikipedia.org/wiki/Bertrand's_box_paradox

>>9464150
so if you pick the gold ball first, there is higher chance that the second ball is also gold, which implies that you have a higher chance to pick the box with 2gold balls. how is that possible?

>>9465247
I proved that 2+2 is 5 but this is hard to understand.

>>9465247
>https://en.wikipedia.org/wiki/Bertrand's_box_paradox
It's called paying somebody money and time so he tells you a lot of bullshit.

>>9462739
at first i thought it was 1/2, but now i understand why 2/3 is the right answer.

imagine you're doing the experiment multiple times. we take out the golden ball, which rules out the container with 2 silver balls. now it's either container with two or one golden ball. let's not choose the answer and repeat the experiment. we take out the golden ball again. then we repeat the experiment again and the same thing happens. we do it a lot of times and every time we take out the golden ball first. but if the chances of us picking from both containers are equal, then why haven't we picked a silver ball even once?

>>9464032
>be me
>take stats
>still have to manually draw out this shit to understand basic stats problems
feels brainlet

>>9465269
>at first i thought it was 1/2, but now i understand why 2/3 is the right answer
but the only right answer is 1/3
how the problem is formulated is a sophism. you cannot count the probability of deterministic system. you have one box with one ball left and states of other boxes are completely unrelated

>>9462739
>itt people who took basic probability class vs people who didn't and are trying to use logic

Don't know if this link works, but here is a small python simulation:

https://trinket.io/python/08a466b111

>>9465158
It's 2/3

P(A|B) = P(AnB)/P(B)

Let B = picking first gold
therefore A = 1/3 + 1/2 * 1/3 = 3/6 = 1/2

AnB = 1/3 as only one box will contain both

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

>>9465542
meant to be therefore B = 1/2 not A

>>9462739
1/2

If you take one gold ball out, then the next ball out then that means that either the gold ball has been taken out and there is only one ball left. THE SILVER ONE

Or you put your hand in the double gold one and took out the gold ball and now there is one gold left.

So there are only two golds left if you took out the one gold ball and you are either in the box with the one gold ball left or the silver ball left which means that there 1/2 a chance of getting a gold ball.

>>9465518
>pick a box at random
Third box should never be allowed to be picked, as it basically doesn't exist for the problem anymore

>>9463534
There aren't. if you take one ball out then you are either in the gold silver box or the double gold box, which gives two possible out comes, you take a ball out and it is gold or you take a ball out and it is silver.
Therefore the probability of getting a gold ball is 1/2.

>>9465765
I have a known winning lottery ticket in one hand and a random lottery ticket in the other. You choose randomly and find a winning lottery ticket. Is there a 50% chance you chose the random lottery ticket?

>>9465759
Yes and that's why it does nothing when box_nr==3:
>37: # nothing to be done for box 3
it is just picked to not confuse people. Change line 19, it will just decrease std deviation.

>>9465821
I don't know why you keep posting this retarded analogy when it's been repeatedly pointed out it has no relevance to the actual question.

>>9466022
Obviously it does or you wouldn't keep avoiding answering it.

>>9465249
Its possible because you already know that your first ball is a gold ball.

okay guys I just did a computer simulation and the answer is 2/3 so shut the fuck up

It's 1/2.
Part of the reason why people keep getting the answer wrong is because the problem is posed in a way that makes you think the box with only silver is significant in calculating the percentage.

Because a gold ball is picked out initially, the sample space is reduced to only boxes with a gold ball. Thus you either remove a silver ball next or a gold ball (1/2 probability).

>>9466107
I did a physical simulation and got 1/2

>>9464150
there is absolutely no difference between 1 and 2 you brainlet.
the fallacy is that you assume you can pick first or second ball from the first box, but you can't do that.
if we add another 10 in the first box this doesn't change the probability at all, because we pick THE BOX with balls, not the ball
correct answer is 1/2

>>9466074
If a tree falls in the woods and no-one is around to hear it does it make a sound? Huh? Huh?Answer me! Obviously you keep dodging my question because it's so relevant and proves my argument correct, clearly not because it's dumb, irrelevant and will cause an argument that is completely unrelated to the question posed in the OP!

>>9466113
>because we pick THE BOX with balls, not the ball
Exactly.

>>9462739
>>9462751
>>9462811
>>9463769

can someone explain my flawed thinking

here's the simulation, you can run it in your javascript console

new Array(10000000).fill(1).map(()=>Math.floor(Math.random()*3)).filter(b => b>0).map(b => b - 1).reduce((c,b,i,a)=>c + b / a.length,0)

result: 0.49985474513964967

explained:
1) Take an array of 1000000
2) fill it with random integers values [0,2]
3) remove all boxes with 0 gold
4) remove 1 gold from each remaining box
5) sum all boxes with 1 gold, divide by number of remaining boxes

>>9466109
I have a known winning lottery ticket in one hand and a random lottery ticket in the other. You choose randomly and find a winning lottery ticket. There are only two options, the winning ticket was the known one or the random one. It's a 50% chance that it was the random one, right?

You magically increase the chance of a random lottery ticket to 25% simply by putting it in one hand with a winning ticket in the other.

We can think about this problem in a different way: instead of boxes with balls, we have two boxes with buttons. When you press the button, the box will output the stream of random numbers with certain probabilities (picking balls can be that stream of events).
1st box produces 1 with 100% chance, and the 2nd box produces 1 with 50% chance and 0 with 50% chance.
You pressed the random button and get 1. What is the probability that when you press the same button second time you will also get 1?

>>9466116
Yes it obviously makes a sound since sound is defined as vibrations in air. Now answer my question retard.

>>9466162
im a new guy, thats missing the point of the question,

The tree in the woods question is trying to ask "if something cant be observed by anything, then does it happen"

>>9466163
The answer is still yes.

>>9466148

OOOOH I got it

nevermind

It's really simple

you have

1000 boxes with 2 silver
1000 boxes with 1 silver 1 gold
1000 boxes with 2 gold

you discard all boxes with 2 silver
and you discard HALF the boxes with 1 silver 1 gold

so you're left with

500 1 silver 1 gold => gets you a silver
1000 2 gold => gets you a gold

probability of getting a gold on the second grab, given the first grab, is 1000/1500

>>9466175
Quantum mechanics says otherwise brainlet

>>9466175
no its not that simple. If something becomes unobservable then we cant say that it happens. theres this thing called superposition

>>9466190
An observer is not "someone" in quantum mechanics.

>>9466191
>If something becomes unobservable then we cant say that it happens.
Nothing became unobservable, it simply wasn't observed when it happened.

>>9464050
this is true
people cant understand independent events i guess. pretty disappointing. /pol/ is hemorrhaging retards into every corner of 4chan.

>>9466208
I think it's more that there's a bunch of pseuds who validate themselves by trying to propose an illogical answer without really understanding why the question in the OP is fundamentally different than the one they're asking. The question in the OP can be phrased as:

>If you select a box that has one gold what is the probability the second ball is gold?

Once you do that it immediately reveals every person who thinks it's 2/3 is wrong.

>>9465286
your answer is wrong. events are independent. i cant help but wonder id this is the most elaborate troll thread of all time.

>>9464032

if you can't explain it with intuition then you don't understand it.

>>9466180
What was the point of your posts?

>>9466274

I didn't get it, then I got it.

maybe it'll help someone. idk & idc.

this thread is a proof of concept of some way to take advantage of all you morons who think the answer is 2/3

>>9462739
There are three pairs of balls. If you pick the first in a pair, you must then pick the second.
There are six balls to choose from. We assume we've picked one of the three gold ones. Two of them will result in the second pick being another gold one, while the third will result in us picking a grey one. Answer is 2/3.

There's also the Bayesian statistics route if you like plugging and chugging:
P(box = gold,gold | first = gold) = P(first = gold | box = gold,gold) * P(box = gold,gold) / P(first = gold)
where
P(first = gold | box = gold,gold) = 1,
P(box = gold,gold) = 1/3,
P(first = gold) = 1/2,
so
P(box = gold,gold | first = gold) = 2/3

>>9466268
I tested this my self
what I recognized is that the second box has a lower chance of picking a ball from it, so It is more likely that you chose the gold box from the get go if you picked up a gold ball.

>>9466334
independent events. youre wrong. the problem starts when you are in the universe where you have already picked a gold ball. and asks what the next probability was when you pick the next ball. stop taking the past into account. independent events. you have a formula but you are misusing it.

>>9466313
lol. this may actually be true.

>>9462739
When first presented with the Monty Hall problem, an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter (Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995:713). In her book The Power of Logical Thinking, vos Savant (1996, p. 15) quotes cognitive psychologist Massimo Piattelli-Palmarini as saying that "no other statistical puzzle comes so close to fooling all the people all the time," and "even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans (Herbranson and Schroeder, 2010).

>>9466364
too bad you had 2/3 chance of picking the box with 2 balls in it in the first place to get in that universe lol

>>9462848
Humans are notoriously bad at statistical probability. Some people just aren't locked into the practicality of it and can think in terms of genuine probability without effort.

Box 1: G,G
Box 2: G,S
Box 3: S,S
Since our first ball is gold, it's either from 1 or 2.

If 1:
Choose again from same box: 100%
Choose again from new box: 25%

If 2:
Choose again from same box: 0%
Choose again from new box: 50%

Which gives us a probability of 2/3rd

There are [math]n[/math]boxes.
There are [math]m[/math]colors. Each box [math]i[/math]contains [math]c_{i,j}[/math] balls of color [math]j[/math]
You pick [math]x[/math]boxes at random. You pour these boxes out onto the floor.
You pick up [math]y[/math]balls at random. They are balls of colors [math]j_1, j_2, \ldots, j_y[/math], respectively.
What is the probability that the next ball you pick up will be of color [math]z[/math]?

For those of you looking for a challenge:

Let [math]\mathcal{N}[/math]be a set that indexes a set of boxes.
Let [math]\mathcal{M}[/math] be a set that indexes a set of colors.
For the box with index [math]\eta[/math] and the color with index [math]\mu[/math], let [math]\mathcal{C}_{\eta, \mu}[/math] be a set that indexes the balls in that box of that color.
Let [math]\mathcal{X}[/math] be a set that indexes the boxes you pick at random. You pour these boxes out onto the floor.
Let [math]\mathcal{Y}[/math] be a set that indexes the balls you pick up at random.
What is the probability that the color of the next ball you pick up will have index [math]\xi[/math]? Note that any of the aforementioned sets may be uncountable.

>>9466364
The problem describes the exact procedure, starting with randomly choosing a box, and then randomly selecting a ball from it. This procedure is what guarantees the 2/3 odds. The 1/2 answer violates the "choosing a box at random" stipulation. (To see why, consider the symmetry of the problem with respect to ball color.)

>>9466313
>>9466369
Please take a single course on probability.

>>9462739
Assuming you grabbed a yellow it means you have a higher chance of being in the double yellow box than the 1 yellow 1 white box.

Hence the % chance of having a double yellow box is higher.

>>9466729
Just reformulate this question

You have the 3 boxes. What are the odds you are in the double yellow box if you grab a yellow first?

Hence the chance of it being white or yellow is not 50/50.

>>9462739
Wow one nice paradox.

t. brainlet

>>9466160
>>9466160
This one doesn’t get any replies because the answer is fifty. But that’s because it’s a different question. You can’t make your second push of the button before the first one, but you can draw the second ball in a bag before the first one.

>>9462848
The problem is people are assuming you are removing the ball from the box, instead of putting it back in.

>>9466772
wrong

>>9466224
Is the probability that you draw one gold equal to the probability that the box contains at least one gold? Prove it mathematically.

Otherwise, the problem says you drew a gold ball, not that the box contains a gold ball.

>>9462769
>>9462757
>>9462755
>>9462751
T. sophists

The meme logic isn't actually valid. Fuck off.

I can't believe this thread is still going on, what a gold mine.

Interestingly, no one has responded to any of the proofs that the answer is 2/3:
>>9463836
>>9465542

Or the simulation showing the answer:
>>9465518

Which seems to indicate that the 1/2 side knows that they don't know what they're talking about. Otherwise they'd prove it. After all, this is the science and math board, right? Not the vague analogies and intuition board? So use science or math to come up with an answer, and show your work, otherwise yield to the people who are smarter than you.

>>9466785
look, an idiot grunted

>>9462739
It's 1/2. If you took out a gold ball, then you have either box 1 and the second ball will be gold, or box 2 and the second ball will not be gold. If it was randomly pickled, then those options are both 50% likely.

>>9466891
Actually, I've figured out where the 2/3rds people are getting their number from: you don't have to have picked a gold ball if you picked from box 2. Once the ball is revealed to be gold, then the odds are 50/50 that the next ball will be the same, but if you choose the ball as well, then there's a 2/6 chance you picked from box one and a 1/6 chance you picked from box two, which fills up to add 1 as 2/3 and 1/3. I'll amend my answer.

Guys, guys I got it. I finally understand where is 2/3 error comes from.
Listen, guys, the idea here is that picking the second ball doesn't matter at all, and *the only thing that actually probabilistic is the chance to pick the gold ball from the first box*.
The first box will output gold balls more often than other boxes. That's it, *the probability to pick the gold ball from the first box is 2/3*. And indeed, if we run simulation over long time we will get the answer 2/3.
That's from where this 2/3 is come. BUT THAT'S NOT WHAT WAS ASKED. We are not asked "what is the probability to pick the gold ball from the 1st box?" We are asked "what is the probability that the next is also gold"? We shouldn't care about probability of picking the gold ball from the first box at this point.

This is a tricky sophism, but the correct answer is 2/3

>>9466904
>but the correct answer is 2/3
Fuck, I mean 1/2 of course

>>9466904
>What is the probability that the next ball you take from the same box will also be gold?
So basically what is the probability that the first ball was taken out of the first box.

>>9466885
>psychology
Lefty pol is over there >>>/his/

>>9466914
No, this is 2 different questions.
People who answer 2/3 and 1/2 actually answers two different questions.

>>9466924
And yet the answer to the question asked by OP is 1/2.

>>9466917
grunt grunt grunt

>>9466924
People who answer 1/2, answer what isn't questioned

>>9465518
I fixed your code https://trinket.io/python/436a1be804
Now it produces the correct answer

>>9466968
They are abstract from the probability of picking the first box, because that's not what is questioned

>>9466977
>if you pull a gold
(meaning silver silver doesn't exist for this problem)
>what are the odds that you will pull another gold from the same box?

... Explain how the answer isn't half

>>9466983
call the balls G,S and G1,G2

"if you pull a gold" means you have just taken
G or G1 or G2

"pull next ball" the ones available are
if you just took ----- the next ball pulled will be
G ----- S
G1 ----- G2
G2 ----- G1

happy endings: 2
total routes available: 3

presto, 2/3

these are the same brainlets that shit on compsci or other studies>>9466973
>int i = random(1,2)
>if(i==1) return gold
>if(i==2)return silver

you must be a genius lmao

>>9467036
you count the probability of picking the gold ball from the box with 2 gold balls. this is wrong. here is right code https://trinket.io/python/436a1be804

>>9466983
the answer is 1/2.
but when after some tries you will finally pick the gold ball it is more likely (2/3) the ball from the box with 2 gold balls. but that's not what is questioned. the question is stated that you are picked the gold ball. this is not probabilistic anymore, but the determined fact

>You pick at random
>It's a gold ball
Here is the core part of that sophism. Either you pick at random and you don't know what the ball you picked, or you picked not random and you know the ball is gold

This thread is hilarious.
1/2-ers, you ignored bayes, ignored computer simulation, probably you will ignore this. But anyways:

>>9467059
nope. I explicitly did not do that. I only counted the total amounts of within gold picks.

g1 = at least 1 gold pick (either gg or gs). so totalGG/g1 and totalGS/g1 lead to 2/3.

you code is just random(1,2) lmao good troll tho

I just did it with the boxes {GG} {GS}

this still gives 2/3 because in OPs experiment the {SS} box is neglected yway.

>>9462768
/thread

2/3 fags are retarded.

>>9467072
>I only counted the total amounts of within gold picks
>I explicitly did not do that
that's exactly what you did.

>>9467070
same thing. why are you count the probability of picking the ball, if it is explicitly stated that we already picked the gold ball. are you retarded?

>>9462739
>pick box at random 1/3 1/3 1/3
>pull out ball and it's gold 1 1/2 0
>next ball is gold 1 0

>>946719
>same thing. why are you count the probability of picking the ball, if it is explicitly stated that we already picked the gold ball. are you retarded?
You are retarded. Please take a basic statistics class.

>/sci/ argues over high-school level conditional probability
am I even surprised any more

>>9467199
>why are you count the probability of picking the ball
he's not 'counting' anything, he's showing every way to take two balls according to rules in the question.
only the three leftmost "if you pull a gold".
of those 3, only 2 end up with a gold
==> 2/3

same thing verbally at >>9466995

>>9467317
no you, retard. i guess your neocortex is smooth as baby's ass. anyone with brain big enough will clearly tell that there is a logic contradiction in the question.
please take a basic logic class

>>9466995
That is the answer to "what are the odds of picking two gold balls in a row", that is meaningfully distinct from "what are the odds that you will pull another gold from the same box?"

The answer is half.

>>9467367
OMG WE ALREADY PHYSICALLY HOLDING THE BALL IN THE HAND. THERE IS NO PROBABILITY FOR THINGS HAPPENED IN THE PAST

>>9467371
>what are the odds of picking two gold balls in a row
That is 1/3 you brainlet when all boxes are accounted for and picking twice from the same box.

>>9467401
>my sophistry is valid
>your sophistry isn't valid
Sure thing bud.

>>9467385
/thread

>>9467385
Which gold ball are you holding? It's either G1 (in which case you'll draw G2 next), G2 (in which case you'll draw G1 next) or G3 (in which case you'll draw S1 next).

>>9467070
If the probability in question is 1/2, then from basic PT, that everyone here surely knows
[eqn]
P(A| B) P(B) = P(A\cap B)\implies P(chose box with gg | first g)P(first g) = \frac{1}{2}\frac{1}{2} = \frac{1}{4}
[/eqn]
So, probability of choosing box with two golds is 1/4.

>>9467429
But G1 and G2 are the same event.

>>9467443
Fuck off with your fallacious meme 'logic'.

>>9467476
Why won't you point out the fallacy then?

>>9467485
There are three boxes to choose from. The probability of picking any box is 1/3.
1/3rd is not 1/4th, therefore you have an error somewhere in your reasoning.

>>9467498
Yes, the error is in the assumption that probability of picking gold second is 1/2. That is called reductio ad absurdum.

>>9467385
those lines in the pic are physical, not abstract math formulas

pic related

>>9467385
>WE ALREADY PHYSICALLY HOLDING THE BALL
that's why only the three leftmost lines apply

>>9462739
>>>9462756 >>9462757 >>9462768 >>9462773 >>9462776 >>9462780 >>9462782 >>9462791 >>9462793 >>9462808 >>9462814 >>9463617 >>9463662 >>9463700 >>9463836 >>9464058 >>9464143 >>9464151 >>9464348 >>9465182 >>9465269 >>9465300 >>9465701 >>9466148 >>9466323 >>9466378 >>9466729 >>9466743 >>9466891 >>9467312
Okay, okay. I finally solved it.

Mark the golden balls with the letters A, B, C. We do not know beforehand which two of those is paired.

We extract one of those. Without loss of generality, call it A. Now, the subsequent outcomes are:

We get its pair, the golden ball B.

We get its pair, the golden ball C.

We get a silver ball. (A wasn't paired)

So two out of three cases. 2/3

>>9467551
yaaaayyy

>>9462739
>>9462739
This issue is called conditional probability. It is similar to the following issue:

You draw a card at random from a deck of cards. What's the probability of drawing a card of spades? 25%, naturally. However, while drawing it, you perceive a faint reflection on the table and notice the card is black. Knowing that you drew a black card, what's the probability of it being a card of spades? 50%.

The calculation goes as follows:
Probability of both events occuring (drawing card that is black and of spades) divided by the probability of drawing a black card (the condition). That's 1/4 divided by 1/2, which is equal to one half, hence 50%.

This exact same method can be applied to our problem:
The probability of both events occuring (picking a golden ball followed by another golden ball), which is 1/3 as the only chance of this occurring is if you pick the box with both golden balls at random, divided by the probability of picking up a golden ball (condition), which is 50% (1/3*1/2 + 1/3*1/2 + 1/3*1/2). 1/3 divided by 1/2 is 2/3.

So, the only correct answer is 2/3.

For anyone who doesnt get it. There are two boxes GG and GS. Of the three golds, two have a gold neighbour which you could pick next. 2/3.

>>9467539
more like this.
first people ignore IT'S A GOLD BALL,
second people ignore AT RANDOM
because there is a logic condradiction between these 2 premises.

>>9463754
OK, play this game in real life and check your odds. You'll find reality goes with 2/3, not 50/50 over any statistically great enough amount of attempts. 70 tries or so should be enough if you enforce randomization appropriately. I just don't get why people who doubt the math that much never go out and check the numbers. Probably, I'd guess, because they are averse to math in the first place, which really should let you know you might not be as right as you think you are.

>>9467942
in reality you need to pick the ball, but in the question you already picked the ball, thus the probability of picking the gold ball is eliminated. omg why can't you understand this difference

>>9466180
>Discarding half boxes with silver
>Being this retarded

>>9462739
1/2

>>9463769
probability is a pseudoscience

>>9463777
(50%+(75%))/2
btfo

>>9469014
you memeing me or what?

You flip a coin. the up side is heads, what is the probability that down side is tails?
According to 2/3 brainlets it's 50% lol.

>>9468052
Dude, is it really that hard to understand? Do the experiment (or make a comp simulation, you will even find the code in this thread), here is how:

1. Pick a box at random
2. Pick a ball from that box at random
3. Is the ball gold?
4. Yes: discard this round and goto 1.
5. No: pick the second ball
6. Write down the color of that 2nd ball
7. Goto 1 if you are not finished (take for example that other anons 70 tries)
8. Count the total number of colors written down
9. Count the numbers of "gold" written down
10. Divide result of 9 by result of 8.

You don't magically take everytime a box with a gold ball and than pick magically everytime a gold ball. It is: _IF_(!!) you have picked a gold ball...

>>9470002
Correction:
3. Is the ball silver?