My brain is full of fuck

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]

1/2 duh

...

>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.

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

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

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

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%

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

>
>9462755
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

>t. humanities graduate
en.wikipedia.org/wiki/Bertrand's_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.

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

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

>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.

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

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.

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.

>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

>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

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

this

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.

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

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.

lmao over complicated much

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.

...

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

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.

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

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.

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.

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

90% of Veeky Forums confirmed brainlets and BTFO'd

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.

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.

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.

>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

>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

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

>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

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?

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

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

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.

>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

>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...

>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.

>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.

>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.

>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

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.

>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.

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

>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.

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.

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.

>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.

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.

>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 Veeky Forums shits itself over which interpretation of a vaguely-worded problem is the right one.

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

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.

>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.

u are getting tolled hard bro

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

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.

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?

>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

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

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

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

>It's a gold ball

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.

See

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

He left, he's a pussy.

>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.