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Board
/sci/ - Science & Math

You cannot reply anymore.

>Continuation thread old one aged out.
So /sci/ what is your answer to question one and why is it 1/2?
>>
>>12933011
please keep this going forever
>>
>>12933011
>and why is it 1/2?
you know damn well that's not how mathematical probability works.
>>
>>12933011
The answer to question 2 is 4/9.

The probability of a ball being picked depends on the number of balls in the same box.
Therefore, all balls don't have the same probability of being picked.
This probability is directly related to the box in which the ball is contained.
One must be rigorous in their quest to epiphany.

This is dedicated to you, wherever you are, Senior Package Hater.
>>
>>12933011
your total number of options given boxes with blue balls and picking a blue ball first are b1 then b2, b2 then b1, or b3 then r1.
2/3rds of these options involve choosing a second blue ball.
>>
>>12933503
You know damn well that it's how boxes work. You cannot select a ball without first selecting a box. Since it's guaranteed the first ball selected will be blue we can only choose from two boxes. Since it's guaranteed the FIRST ball will be blue the only two selections left are a single blue and a single red marble.

>1/2
>/
>2
>>
>>12933514
The problem is you see the boxes almost as one big entity. The reality is you select the box prior to the ball and the foreknowledge that a ball is blue means there's only 2 box selections possible. Eliminating the first blue marbles leaves a 50/50 chance blue or red mate.

>This cannot be disputed.
>>
>>12933482
It has to go on until the brainlets can understand probability.
>>
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>>12933011
this image is a much better size, I wish good luck to you and your thread
>>
>>12933011
>So /sci/ what is your answer to question one and why is it 1/2?
Presumably because in this hypothetical scenario I am retarded
>>
>>12933677
The first one is literally just Bertrand's Box and if you don't know that one then you have no business talking about probability
>>
Q1: 2/3
Q2: 2/5
Q3: 2/5
>>
>>12933508
That’s a common mistake in threads like this. Per Q2, the second box isn’t chosen. You choose the ball. The color and the box are conditional on which ball you choose, not the other way around. You’ve weighted your choice of the ball based on the box it’s in, which is the opposite of what the question asks you to do.
>>
>>12934030
What is Bertrand's box
>>
>>12934358
A boxful of Bertrand's mate. Pretty simple stuff.

The answer is 1/3
>>
>>12934264
You can't choose a ball before a box, and here's why.

You are blindfolded and the boxes are shuffled.
That means you don't know which box contains a different number of balls.
Also remember the boxes are separate and distinct.
This implies that when picking a ball, not knowing which box you took it from, a ball could have a different probability of being picked.
This is because you don't know if you're picking a ball from a box with 2 balls or a single one.
This means the box in which the ball is contains will affect the ball's probability of being picked
>>
>>12933011
I thought the two thirders had won?
>>
>>12934836
>you don't know which box contains a different number of balls.
Correct.
>a ball could have a different probability of being picked.
Doesn't follow.
>you don't know if you're picking a ball from a box with 2 balls or a single one
Correct.

It's a simple conditional probability problem. You start with 6 balls

2 are blue color and left box (bL1, bL2)
1 is blue color and middle box (bM)
1 is red color and middle box (rM)
2 are red color and right box (rR1, rR2)

You then remove 1 of the 6 balls. The color of the removed ball is given as blue. The box of the removed ball is not given. This gives you 3 equally likely states, depending on whether you removed bL1, bL2, or bM.

bL2, bM, rM, rR1, rR2
bL1, bM, rM, rR1, rR2
bL1, bL2, rM, rR1, rR2

The question then asks for the probability that your next random of choice a of ball gives you a ball of blue color.

P(b) = (2/5+2/5+2/5)/3 = 2/5

You can find the conditional probability of any other attribute, or combination of attributes, in the same way.

P(r) = (3/5+3/5+3/5)/3 = 3/5

P(L) = (1/5+1/5+2/5)/3 = 4/15
P(M) = (1/5+2/5+2/5)/3 = 1/3
P(R) = (2/5+2/5+2/5)/3 = 2/5

P(bL) = (1/5+1/5+2/5)/3 = 4/15
P(bM) = (1/5+1/5+0/5)/3 = 2/15
P(bR) = (0/5+0/5+0/5)/3 = 0/5
P(rL) = (0/5+0/5+0/5)/3 = 0/5
P(rM) = (1/5+1/5+1/5)/3 = 1/5
P(rR) = (2/5+2/5+2/5)/3 = 2/5

Note further that

P(b)+P(r) = 1
P(L)+P(M)+P(R) = 1
P(bL)+P(bM)+P(bR)+P(rL)+P(rM)+P(rR) = 1
>>
>>12934944
>Doesn't follow.
I'll show you. Read carefully.

>It's a simple conditional probability problem.
It is.

> [...] probability that your next random of choice a of ball gives you a ball of blue color.
So far so good.

>P(b) = (2/5+2/5+2/5)/3 = 2/5
There's the problem. After the first pick, the balls don't assume uniform probability distribution.

Hear me out on this.
3 equally likely states after the first pick, i.e. the ones you stated.

Again, you don't know if you're picking a ball from a box with 2 balls or a single one.
You can't see inside the boxes.
Remember the boxes are shuffled.
Therefore, even if you don't want to "choose a box", when you go and pick a ball, the ball you pick comes from a random box which you don't know the content of.
Since the boxes are random, distinct and separate:
1/3 of the time you will pick a ball from the left box.
1/3 middle.
1/3 right.

If you pick a ball at random from a box in which there is only one ball, that very ball will be picked everytime.
If you pick a ball at random from a box in which there are two balls, each ball will be picked 50% of the time.

Then you do the math for all 3 scenarios.
>>
>>12933677
>Since it's guaranteed the first ball selected will be blue
It's not guaranteed, it's just something that happened.

>Since it's guaranteed the FIRST ball will be blue the only two selections left are a single blue and a single red marble.
The former is twice as likely as the latter since the former is from a box that was all blue while the latter is from a box which was only half blue. So the answer is 2/3.
>>
>>12933688
Wrong. The 2 selections are not equally likely one you know you chose a blue ball.
>>
>>12934258
>Q2: 2/5
Wrong.
>>
>>12935068
>3 equally likely states after the first pick, i.e. the ones you stated.
Let S1, S2, S3 denote the three 5-ball states conditional to removing 1 random blue ball from the initial 6-ball state given by OP.

S1 = {bL2, bM, rM, rR1, rR2}
S2 = {bL1, bM, rM, rR1, rR2}
S3 = {bL1, bL2, rM, rR1, rR2}
P(S1) = P(S2) = P(S3) = 1/3
P(b|S1) = P(b|S2) = P(b|S3) = 2/5
P(b) = P(b|S1)P(S1) + P(b|S2)P(S2) + P(b|S3)P(S3) = 2/15 + 2/15 + 2/15 = 2/5

>Again, you don't know if you're picking a ball from a box with 2 balls or a single one.
Yes.
>You can't see inside the boxes.
Yes.
>Remember the boxes are shuffled.
Yes.
>Therefore, even if you don't want to "choose a box", when you go and pick a ball, the ball you pick comes from a random box which you don't know the content of.
Yes.
>Since the boxes are random, distinct and separate:
Yes.
>1/3 of the time you will pick a ball from the left box.
>1/3 middle.
>1/3 right.
>If you pick a ball at random from a box in which there is only one ball, that very ball will be picked everytime.
>If you pick a ball at random from a box in which there are two balls, each ball will be picked 50% of the time.
No, that math assumes the ball you pick is conditional on its box. If Q2 said you picked a box and then picked a ball from the box you picked, you'd be right.
>>
>>12935094
See >>12935127
>>
>>12934264
How do you choose a ball such that each has an equal probability of being chosen when there is a box, unknown to you, which has a lone ball?
>>
>>12935127
>P(b|S1) = P(b|S2) = P(b|S3) = 2/5
I'd be curious to see you demonstrate a single one of these.

>If Q2 said you picked a box and then picked a ball from the box you picked, you'd be right.
"the next ball you pick from any one of the distinct boxes"

I'd like to know your answer to these questions:

A. If the boxes are separate and random, and you can't see their content, how can you assure all of their combined elements assume the same probability distribution as if they were pooled?

B. Different but similar problem.
Box 1 contains one single green ball. Box 2 contains 1000 yellow balls.
Boxes have the same properties as in OP's. Can't see inside them. Shuffled => random.
Boxes are in front of you, shuffled, contents hidden.
You don't know which box contains which set of balls, and you can't see the balls.
You pick a ball from one of the boxes.
What is the probability that it's a green ball? A yellow ball?
>>
>>12935131
See >>12935199
>>
>>12935127
>P(b|S1) = P(b|S2) = P(b|S3) = 2/5
This is incorrect. How are you choosing balls with equal probability when it's unknown which box has the lone ball?
>>
You have a box with \$1M and a box with a lottery ticket for a lottery in which the prize is \$1M. According to the halftards, if you make a choice and end up with \$1M, there is a 1/2 chance you chose the box with the lottery ticket. But the chance of getting \$1M was 1/2 from the beginning (assuming chance of winning the lottery is negligible). This the chance of choosing box 2 AND winning the lottery is (1/2)(1/2) = 1/4.

So next time you want to win the lottery, just put your ticket in a box and you can increase your chance of winning the lottery to 1/4.
>>
>>12935408
wow i fucking love science now
>>
>>12935199
>>12935341
>>12935363
>How do you choose a ball such that each has an equal probability of being chosen when there is a box, unknown to you, which has a lone ball?
You reach out and choose a ball. There's no reason to assume the box affects your choice.

>>12935204
>>P(b|S1) = P(b|S2) = P(b|S3) = 2/5
>I'd be curious to see you demonstrate a single one of these.
By definition, P(A|B) = P(A∩B)/P(B)

>>If Q2 said you picked a box and then picked a ball from the box you picked, you'd be right.
>"the next ball you pick from any one of the distinct boxes"
Yes, every ball is in 1 of 3 distinct boxes. That doesn't imply that picking a ball is conditional on picking a box.

>I'd like to know your answer to these questions:
>A. If the boxes are separate and random, and you can't see their content, how can you assure all of their combined elements assume the same probability distribution as if they were pooled?
Q2 gives no reason to assume they don't.

>B. Different but similar problem.
>Box 1 contains one single green ball. Box 2 contains 1000 yellow balls.
>Boxes have the same properties as in OP's. Can't see inside them. Shuffled => random.
>Boxes are in front of you, shuffled, contents hidden.
>You don't know which box contains which set of balls, and you can't see the balls.

>You pick a ball from one of the boxes.
Box pick conditional on ball pick.

>What is the probability that it's a green ball?
1/1001

>A yellow ball?
1000/1001

>You pick a box and then pick a ball from the box you picked.
Ball pick conditional on box pick.

>What is the probability that it's a green ball?
1/2

>A yellow ball?
1/2

By conditioning the ball pick on a box pick which isn't given, you've ended up with the right answer to the wrong question. The answer to Q2 is 2/5 because your ball's box was determined by your pick of a ball, not by your pick of a box.
>>
>>12935491
>You reach out and choose a ball.
That doesn't answer the question. I'm asking you how you do that such that each ball is equally likely to be chosen.

>There's no reason to assume the box affects your choice.
It's not an assumption, it must affect your choice since you don't know which box has the lone ball.
>>
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>>12935491
>By definition, P(A|B) = P(A∩B)/P(B)
Everybody knows this anon. I meant I wanted to see how you got P(b|S3) = 2/5, when in this case clearly only one box has blue balls.

So even if you don't see inside the box, you'll just happen to pick from the same box 1000/1001 of the time?

Remember you can't see the balls, so you can't single them out and pick them equally.

>Box pick conditional on ball pick.
You can't choose/pick a ball just like that, because the ball is contained in a random box which you don't know the contents of.

Picking a ball implies picking it from a single, random box.
>>
>>12933011
Q1: 2/3
Q2: 4/9
Q3: 2/5
>>12925627 was a much better thread.
>>
>>12933011
1/2
1/4
2/5
>>
>>12933688
>The reality is you select the box prior to the ball and the foreknowledge that a ball is blue means there's only 2 box selections possible.
Except you still had to randomly select a box, and in this scenario, that selection resulted in the first draw being a blue ball, but 2 blue balls being in one box compared to 1 blue ball in another box weights the odds that you pulled from the box with 2 blue balls. If you had to randomly draw from two boxes, one had 100 blue balls, and the other 1 blue ball and 99 red balls, and you pull out a blue ball, do you really think it's a 50% chance you managed to pull the 1 blue ball from the box of 99 red balls?
>>
>>12935491
>By definition, P(A|B) = P(A∩B)/P(B)
P(b|S1) = P(b∩S1)/P(S1) = 3 P(b∩S1)

Now how do you calculate P(b∩S1)?

>That doesn't imply that picking a ball is conditional on picking a box.
It does though. Which box you pick from decides which balls are avaliable to you. And when the number of balls in each box is unequal, you can't choose balls with equal probability.

>Q2 gives no reason to assume they don't.
The fact that they are in boxes is the reason they don't, no assumption needed.

>Box pick conditional on ball pick.
Gibberish.

>1/1001
>1000/1001
How would you know to choose from the yellow box 1000 times more than the green box when you don't know which is which?

Let's play a game where you give me \$2 and I let you choose a ball from either of two boxes, one with 1 green ball and the other with 1000 yellow balls. If you choose a yellow ball I'll give you \$3. If you choose a green ball you get nothing. Would you play this game?
>>
>>12935545
>was a much better thread.
>>12925627
>Q1. You pick a random box, then pick two random balls from it.
>Q2. You pick a random box, pick a random ball from it. Remove it from the box. Then again, pick a random box, pick a random ball from it.
Of course it was. These questions are a model of pedagogical clarity, stated in a simple language that forestalls the (I)(D) ambiguity. Very professional, very well done.
>>
>>12933011
Q1: 0.5
Q2: 2/3
Q3: 0.2
>>
>>12935777
Wait no. I wasn't sure wether distinct meant all three or the remaining two.
For all three it's 2/3
For remaining two it's 0.375
>>
>>12935728
> These questions are a model of pedagogical clarity
i.e. pedantry

OP's problem is a test of word problem translation, hence the title, "Autism test".
>>
>>12935831
Oh, and here I thought it was a conditional probability problem. No matter, the comparative results of that thread and the other threads speak for themselves.
>>
>>12935537
>Everybody knows this anon.
No, not everybody does. Your original question indicated that you didn’t. I can’t read your mind.

>I meant I wanted to see how you got P(b|S3) = 2/5, when in this case clearly only one box has blue balls.
You asked how I got P(b|S) in general, not P(b|S3) in particular. Yes, in S3 both of the remaining blue balls are in the same box. How are you saying this does or doesn't impact P(b|S3)?

>So even if you don't see inside the box, you'll just happen to pick from the same box 1000/1001 of the time?
There’s no reason to assume otherwise.

>Remember you can't see the balls, so you can't single them out and pick them equally.
Now this is something everybody does know, because it’s given in the OP.

>You can't choose/pick a ball just like that, because the ball is contained in a random box which you don't know the contents of.
You’re assuming the box influences which ball you can choose. It doesn’t, unless you’re first choosing a box and then choosing a ball.

>Picking a ball implies picking it from a single, random box.
Correct. We agree on that.
>>
>>12935510
>That doesn't answer the question. I'm asking you how you do that such that each ball is equally likely to be chosen.
I'm sorry, maybe I don't understand your question. You're asking for a description of how you reach out and choose any 1 of 5 balls?

>It's not an assumption, it must affect your choice since you don't know which box has the lone ball.
It only affects your choice if you assume picking a box is a precondition to picking a ball. There's no reason to assume that. Q2 says that you pick a ball from a box, not that you pick a box and then pick a ball.
>>
>>12935724
>P(b|S1) = P(b∩S1)/P(S1) = 3 P(b∩S1)
>Now how do you calculate P(b∩S1)?
P(A∩B) = P(A)P(B) <-> A, B independent
P(b∩S1) = 2/5*1/3 = 2/15

>It does though. Which box you pick from decides which balls are avaliable to you.
No, nothing like that is happening. Every ball is available to you until you pick one.

>And when the number of balls in each box is unequal, you can't choose balls with equal probability.
Yes, that's precisely what you're doing. The number of balls in a box doesn't weight your ball pick unless you predicate it on a prior box pick.

>The fact that they are in boxes is the reason they don't, no assumption needed.
Gibberish.

>>1/1001
>>1000/1001
>How would you know to choose from the yellow box 1000 times more than the green box when you don't know which is which?
You don't "know to choose" from any box. The box is as meaningless to your choice of a ball as its color is.

>Let's play a game where you give me \$2 and I let you choose a ball from either of two boxes, one with 1 green ball and the other with 1000 yellow balls. If you choose a yellow ball I'll give you \$3. If you choose a green ball you get nothing. Would you play this game?
As a simple conditional probability problem the way you described it? Sure. But I assume you'd try to add in some arbitrary physical design feature that forces me to choose a box before choosing a ball. I'd have to see your real life design before deciding.
>>
>>12933011
This isn't an autism test it's just an unintuitive probability problem. People have been posting this for months. Get better b8.
>>
>>12935892
>Oh, and here I thought it was a conditional probability problem.
It is, you translate a world problem into a conditional probability problem.

But you probably didn't get that either. Congratulations, you passed the test. A for Autist.

>No matter, the comparative results of that thread and the other threads speak for themselves.
Yes, this thread does a better job at revealing how many people on /sci/ are just equations monkeys that can't think for themselves without pedantic handholding.
>>
>>12935953
>But you probably didn't get that either. Congratulations, you passed the test. A for Autist.
Lots of insinuations about what other anonymous posters do or don’t get. Sure you’re not just getting your own reflection back here?
>Yes, this thread does a better job at revealing how many people on /sci/ are just equations monkeys that can't think for themselves without pedantic handholding.
Oh, is that what’s happening now?
>>
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>>12935919
>Your original question indicated that you didn’t.
You literally need this equation to get the 4/9 answer. An equation you're not using carefully. Garbage in garbage out.

>You asked how I got P(b|S) in general
"a single one of these"
I then insisted on P(b|S3), because i think it is a more revealing scenario.

>Yes, in S3 both of the remaining blue balls are in the same box. How are you saying this does or doesn't impact P(b|S3)?
See below.

->can't see the balls, so you can't single them out and pick them equally.
>[agrees]

->Picking a ball implies picking it from a single, random box.
>[agrees]

ok cool.
Now if you pick from a distinct, separate, random box, and you can't single out a ball, how can you make sure they will get picked equally?
Picking a ball implies picking a box because you can't see the balls and they are contained in separate in distinct separate boxes.
>>
>>12935934
>People have been posting this for months.
And yet they still fall for it everytime.

>>12935980
>Sure you’re not just getting your own reflection back here?
Yeah and I like what I see. Thanks anon, I want to fuck myself now.

>Oh, is that what’s happening now?
I don't know anon, I can't read.
>>
I haven't read the thread but here are my solutions.
>Q1
P(Both balls blue) = P(Box A was picked) = 1/3

P(Second ball blue | First ball blue)
= P(Both balls blue) / P(First ball blue)
= (1/3)/(1/2)
= 2/3
>Q2
P(Both balls blue)
= P(Both balls blue | Both balls came from box A) * P(Both balls came from box A) + P(Both balls blue | One ball came from box A and the other from box B) * P(One ball came from box A and the other from box B)
= 1 * (1/9) + (1/2) * (2/9)
= 2/9

P(Second ball blue | First ball blue)
= P(Both balls blue) / P(First ball blue)
= (2/9)/(1/2)
= 4/9
>Q3
P(Both balls blue) = (3 choose 2) / (6 choose 2) = 3/15

P(Second ball blue | First ball blue)
= P(Both balls blue) / P(First ball blue)
= (3/15)/(1/2)
= 2/5
>>
>>12933011
the real answer is testicals.
>>
>>12934258
For Q1
There are only two boxes that even have blue balls. If the first ball you picked is blue - it must have been from one of those two. Of those two boxes, whichever one it is one will have a blue ball (box A) the other one (box B in the middle) would only have a red ball remaining.

So 1/2.
>>
>>12936092
it's true, i've been feeling pretty blue balled since this covid shit started
>>
>>12933011
Forget Q1, why is 1/2 the answer to Q2?
>>
>>12936103
>2/3
Agreed.
>>
>>12933011
>Q1
50%
>Q2
33%
>Q3
Unknown value. Am I selecting from a box? If so, 0%. Am I selecting from the floor? If so, it depends on whether I prefer red or blue.
>>
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>>12936006
>You literally need this equation to get the 4/9 answer.
You never said what your answer was. There are 19 posters in this thread. I can't read your mind or know who you are.

>An equation you're not using carefully. Garbage in garbage out.
If you're getting 4/9, then you're the one not using it carefully. Or yes as I said last thread garbage in garbage out.

>"a single one of these"
Yes, that's a generalized question.

>I then insisted on P(b|S3), because i think it is a more revealing scenario.
>See below.
>->can't see the balls, so you can't single them out and pick them equally.
>->Picking a ball implies picking it from a single, random box.
>ok cool.
>Now if you pick from a distinct, separate, random box, and you can't single out a ball, how can you make sure they will get picked equally?
>Picking a ball implies picking a box because you can't see the balls and they are contained in separate in distinct separate boxes.
You can't see the boxes either. Seeing a ball has nothing to do with picking a ball. Even working under your mistake assumption that you pick a box before picking a ball, you still don't "see" the ball until after you pick it either way.
>>
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>>12936189
>You can't see the boxes either.
Meaning you can't carefully pick the boxes to make sure each ball is picked equally.

>Seeing a ball has nothing to do with picking a ball. [...] you still don't "see" the ball until after you pick it either way.
Yes, so again, how is it that in your little world each ball is picked equally?

>I can't read your mind or know who you are.
Neither can I. But your box denialism is recognizable.
>>
>>12936189
Wrong, the answer is 4/9, and this can be plainly shown with a simple enumeration of what can happen.

One of four things can happen when you pick the first ball. You pick a blue from the blue-blue box with total probability 1/3. You pick a blue from blue-red with total probability 1/6. You pick a red from blue-red with total probability 1/6. You pick a red from red-red with total probability 1/3.

Since the condition is that your first ball chosen is blue, you can only have taken from blue-blue, or taken blue from blue-red. These probabilities are normalized, respectively, to 2/3 and 1/3.

Thus, there are two possible initial states: You took a blue ball from blue-blue with probability 2/3, or you took the blue from blue-red with probability 1/3.

Consider the first state, where you took from blue-blue. There is now one box with one blue, one box with one blue and one red, and one box with two red. By necesity, you must first (randomly) choose a box to choose a ball from it. Each box has an equal probability of 1/3.

The single blue box has a probability of 1 for you to take a blue. blue-red has a probability of 1/2. red-red has a probability of 0 to take a blue.

Thus, IFF the first state is true, your chance of drawing a second blue ball is (1/3)(1) + (1/3)(1/2) + (1/3)(0) = 1/2

---

The second state, where you drew a ball from the blue-red box, is similar. There is now one box with two blues, one box with one red (since you took blue from it), and one box with two reds.

Thus, IFF the second state is true, your chance of drawing a second blue ball is (1/3)(1) + (1/3)(0) + (1/3)(0) = 1/3

---

With the first state having a probability of 2/3 to occur, and a probability of 1/2 to draw a second blue in the first state, and the second state respectively having probabilities of 1/3 and 1/3, the final probability of drawing a second blue, given you first drew a blue, is:

(2/3)(1/2) + (1/3)(1/3) = (1/3) + (1/9) = 4/9.

QED.
>>
>1
2/3

>2
1/3

>3
2/5

I fucking hate you all
>>
>>12935777
>>12935796
Wrong.
>>
>>12935925
>You're asking for a description of how you reach out and choose any 1 of 5 balls?
Yes. With equal probability.

>It only affects your choice if you assume picking a box is a precondition to picking a ball.
It is a precondition, since the balls are in boxes and you don't know which ball is in which box. No assumption needed.

>Q2 says that you pick a ball from a box, not that you pick a box and then pick a ball.
It's the same thing. Otherwise you would be able to explain the procedure for choosing the balls with equal probability. But you can't.
>>
>>12933011
1. 2/15
2. 2/5
3. 2/5
>>
>>12934358
This isn't Bertrand's box paradox, as I explained in the previous thread.
>>
>>12933011
>blindfolded
>pick ball
>boxes shuffled
>while blindfolded
>pick from same box
How are you supposed to pick from the same box?
>>
>>12936665
t. retarded

You dont know which box is the same box.
The question asks for the probability that you do pick another blue ball given it's from the same box.
>>
>>12935927
>P(b∩S1) = 2/5*1/3 = 2/15
How is P(b) = 2/5? You just assumed what you were trying to prove. The whole point of your calculating P(b|S1) was too calculate P(b).

>No, nothing like that is happening.
Then explain how each ball has equal probability of being chosen.

>Yes, that's precisely what you're doing. The number of balls in a box doesn't weight your ball pick unless you predicate it on a prior box pick.
It is predicated on a prior box pick whether you like it or not. You keep avoiding the question because you know they're is no way to choose balls with equal probability.

>Gibberish.
What don't you understand? It's simple English.

>You don't "know to choose" from any box.
Right, therefore it's impossible for you to end up choosing from the yellow box 1000 times more than the green box. But that's exactly what your interpretation concludes. As you yourself said, the ball choice determines which box you chose from. So please explain where the information comes from when you choose a random ball that you don't have when choosing a box.

>As a simple conditional probability problem the way you described it? Sure.
OK, so let's play. I have the boxes here. Which ball do you choose?
>>
>>12936187
Congratulations, you get the most retarded answer award.
>>
>>12936665
>How are you supposed to pick from the same box?
It's given to you by the question, you didn't intentionally pick from the same box or not pick from it.

>>12936635
>Yes. With equal probability.
You reach out and choose a ball. What else do you think is happening?

>It is a precondition, since the balls are in boxes and you don't know which ball is in which box. No assumption needed.
Correct, you don't know where any balls or boxes are. You're assuming that you can pick a box from an unknown location, but that you can't pick a ball from an unknown location. That distinction isn't given anywhere in the question.

>It's the same thing. Otherwise you would be able to explain the procedure for choosing the balls with equal probability. But you can't.
What do you mean by “the procedure”? What “procedure” are you imagining? There’s no process. You choose a ball. Your choice is independent of color or box.
>>
>>12936248
>Meaning you can't carefully pick the boxes to make sure each ball is picked equally.
Good thing you’re not picking the boxes.

>Yes, so again, how is it that in your little world each ball is picked equally?
How does not being able to see something in either of two scenarios allow you to assume information that isn’t given in the question?

> I can't read your mind or know who you are.
>Neither can I.
I recognize your movie stills and your numbering of them, so you’ve successfully identified your last few posts, but it’s not connected to any incorrect 4/9 answer you may have given somewhere else in the thread.

>But your box denialism is recognizable.
>box denialism
>>
>>12936249
>there are two possible initial states: You took a blue ball from blue-blue with probability 2/3, or you took the blue from blue-red with probability 1/3.
I agree with your split, I split the states into equal thirds to make it clearer.

>By necesity, you must first
No
>(randomly)
(yes)
>choose a box to choose a ball from it.
Correct, per above.

>Each box has an equal probability of 1/3.
No.

>The single blue box has a probability of 1 for you to take a blue. blue-red has a probability of 1/2. red-red has a probability of 0 to take a blue.
If you pick a box first, yes. You’re answering the wrong question.
>>
>>12936364
>1/3
Wrong
>>
>>12936103
This is not me >>12936176

>For Q1
>There are only two boxes that even have blue balls. If the first ball you picked is blue - it must have been from one of those two. Of those two boxes, whichever one it is one will have a blue ball (box A) the other one (box B in the middle) would only have a red ball remaining.
This is true if your second ball pick isn't conditional on your first ball pick.

If the question started you from here
>There are two boxes, the left box contains 1 blue ball, and the right box contains 1 red ball.

The question starts you from a different point, which is why you have the right answer to the wrong question. Just like the posters who are answering Q2 incorrectly as 4/9.

> So 1/2.
The right answer to the wrong question.
>>
>>12936644
>1. 2/15
>2. 2/5
Wrong.
>>
>>12936646
It is, as I explained in the previous thread.
>>
>>12936665
The same way that it's a conditional given that the first ball you pick is blue. You just do.

(More precisely, you consider only the outcomes where you both pick a blue ball for your first choice, AND in which you pick the second ball from the same box as the first.)

For example, here are all the following sequences for what balls you can pick. The blue blue box is "box 1", the blue red box is "box 2", and the red red box is "box 3". In parenthesis is the total, non conditional probability for that outcome.

box 1 blue, box 1 blue (1/9)
box 1 blue, box 2 blue (1/18)
box 1 blue, box 2 red (1/18)
box 1 blue, box 3 red (1/9)
box 2 blue, box 1 blue (1/18)
box 2 blue, box 2 red (1/18)
box 2 blue, box 3 red (1/18)
box 2 red, box 1 blue (1/18)
box 2 red, box 2 blue (1/18)
box 2 red, box 3 red (1/18)
box 3 red, box 1 blue (1/9)
box 3 red, box 2 blue (1/18)
box 3 red, box 2 red (1/18)
box 3 red, box 3 red (1/9)

As you can see, this all adds up to a total probability of 1.

The condition of "given your first ball picked is blue" narrows these down to the state space as follows. The conditional probabilities have all been divided by 1/2, their total probability, to form the new conditional probabilities.

box 1 blue, box 1 blue (2/9)
box 1 blue, box 2 blue (1/9)
box 1 blue, box 2 red (1/9)
box 1 blue, box 3 red (2/9)
box 2 blue, box 1 blue (1/9)
box 2 blue, box 2 red (1/9)
box 2 blue, box 3 red (1/9)

The second condition of "[given] you pick from the same box", which is on-top of the condition of your first ball being blue, the state space is once more restricted down to the following. This time the probabilities are divided by 1/3, their total probability, to normalize the new conditional probabilities.

box 1 blue, box 1 blue (2/3)
box 2 blue, box 2 red (1/3)

As you can see, the only remaining outcome where you choose blue twice is when you picked from box 1, for a probability of 2/3.
>>
>>12936704
>You reach out and choose a ball. What else do you think is happening?
How do you do that without first picking a box? The balls are inside the boxes (except in problem 3)

In order to randomly pick a ball, you must first randomly pick a box, since you cannot even select a ball without first opening the box it is contained within.
>>
>>12936750
>How do you do that without first picking a box?
By choosing a ball. What properties are you imagining the boxes have, that prevent you from reaching out and choosing any ball from any box?

>In order to randomly pick a ball, you must first randomly pick a box, since you cannot even select a ball without first opening the box it is contained within.
Opening?
>>
>>12936704
>You reach out and choose a ball.
How do you do that with equal probability? I can reach out and choose a box and then a ball within that box. I can't choose a ball randomly, because I don't even know where the balls are. You already know this, you're just shitposting.

>You're assuming that you can pick a box from an unknown location
I can, just by feeling around for a box. I can't feel around for a ball without choosing a box because they're all inside boxes. Nice try at equivocation.

>What do you mean by “the procedure”?
A series of actions conducted in a certain order.

>What “procedure” are you imagining?
None. There is no possible procedure for what you're describing.

>There’s no process.
I'm glad you finally admit there is no process for choosing balls independently from boxes.
>>
>>12936765
This retard just avoids the question and tries to make it about choosing any ball instead of choosing any ball with equal probability. He is arguing in bad faith. Ignore him, he already lost.
>>
>>12933011
1/3
4/9
2/5
Kiss me anons
>>
>>12933011
My answer is 2/3
It isn't 1/2
>>
>>12936795
Meant to say 2/3 for the first question
My apologies
>>
>>12936765
>By choosing a ball. What properties are you imagining the boxes have, that prevent you from reaching out and choosing any ball from any box?
That they are physical boxes, and it doesn't become any less likely to choose any one box regardless of how many balls are in it. If you were to choose equally from between the remaining balls, you'd have to choose the box with the one ball only 1/5 of the time, and the other two boxes 2/5 of the time each, which is utterly nonsensical since you have no information to differentiate the boxes or know what's inside them.
>>
File: 4.jpg (15 KB, 322x298)
15 KB JPG
>>12936713
>[...] assume information that isn’t given in the question?
Some things are implicit and make themselves known by simple deduction.
As I said earlier, one must be rigorous in their quest to epiphany.

>box denialism
Yes, you the box denialist.

Of course, I don't confuse 2/5 truthers for the same person.

The box denialist is not a person, but a people. An archetype.

Antiboxxers channel the essence of a very autistic hatred of packaging.

But in their crowd, a particular individual stands out.

Antibox-Man, Messiah of Thy Retardation, The Great Revealer of Ambiguity

His Majesty The Liberator of Balls shows the peasants the alienating nature of containment devices with his Post-Amazon dialectic.

I, however, choose not to believe. I like boxes. I guess you could call me a storage enthusiast.

Unfortunately we can't keep this going forever, and I think I'm all out of se7en box pics.
>>
Anyway, this has been great, I had fun.
Many anons fell for this and I love it.
Somebody should make a thread with a similar problem, like the coin flip or the croaking frogs problem.
>>
>>12936768
> How do you do that with equal probability? I can reach out and choose a box and then a ball within that box. I can't choose a ball randomly, because I don't even know where the balls are. You already know this, you're just shitposting.
Are you imagining that you can't reach out and put your hand into a box? You're imagining some sort of lid or other magic wall that isn't given by the question.

>I can, just by feeling around for a box. I can't feel around for a ball without choosing a box because they're all inside boxes. Nice try at equivocation.
Feeling? This is a simple conditional probability problem. You're not supposed to hallucinate yourself into a movie where you're "feeling" around for undescribed boxes.

>None. There is no possible procedure for what you're describing.
I’m not “describing” anything. I’m choosing a ball from a box, per Q2.
On the other hand, you’ve admitted that there's no possible procedure for what you're describing that doesn't involve "feeling" around for boxes.

I'm sorry you can't read.
>there is no process for choosing balls independently from boxes.
In a conditional probability problem, you use the information that you're given. There's no default "process" for how or what you choose. Q2 says that you choose a ball from a box, after having removed 1 blue ball from the initial 6-ball state.
>>
>>12936781
>This retard just avoids the question and tries to make it about choosing any ball instead of choosing any ball with equal probability.
I’m not trying to do anything, I’m answering the question.

>Ignore him

>>12936804
>That they are physical boxes
No, it’s a simple conditional probability problem.

>and it doesn't become any less likely to choose any one box regardless of how many balls are in it.
You’re not choosing a box, you’re choosing a ball.

>If you were to choose equally from between the remaining balls
What remaining balls? What balls do you imagine disappearing and why?

>you'd have to choose the box with the one ball only 1/5 of the time, and the other two boxes 2/5 of the time each, which is utterly nonsensical since you have no information to differentiate the boxes or know what's inside them.
>you’d have to choose the box
Good thing you’re choosing a ball, not a box.
>>
>>12936850
>one must be rigorous in their quest to epiphany.
>you the box denialist.
>2/5 truthers
>The box denialist is not a person, but a people.
>An archetype.
>Antiboxxers
>channel the essence
>of a very autistic hatred of packaging.
>But in their crowd, a particular individual stands out.
>Antibox-Man
>Messiah of Thy Retardation
>The Great Revealer of Ambiguity
>His Majesty The Liberator of Balls
>shows the peasants
>the alienating nature
>of containment devices with his Post-Amazon dialectic.
>I, however, choose not to believe.
Seriously. If you take anything from your participation in this thread, please let it be your fucking meds.

>I like boxes.
Yes, I like boxes too.

>I guess you could call me a storage enthusiast.
>Unfortunately we can't keep this going forever, and I think I'm all out of se7en box pics.
If you’re out of meds, try asking whoever usually gives you your meds to give you more of them.
>>
>>12933011
50% All probabilities are 50%. Something either is or isn't. Lotto drawing, you either win or not... 50%
>>
>>12933011
2/3
4/9
2/5
Anyone who dissagrees is retarded.
>>
>>12937138
>2/3
Yes

>4/9
No, 2/5

2/5
Yes

>Anyone who dissagrees is retarded.
I disagree, and it has nothing to do with your spelling. Your answer to Q2 is the right answer to the wrong question.
>>
1/2
1/2
1/2
it either happens or it doesn't
>>
>>12937138
>2/3
Wrong. You've been proven wrong multiple times this thread stop coping.

>4/9
Actually correct congratulations.

>2/5
Also actually correct. Look at you you're only 1/3 retarded.
>>
>>12937127
>meds
>meds
>meds
NEVER!

I love how I am so obviously fucking with you and you still bite.
You've demonstrated your autism in many different ways, very informative.

>>12937136
thanks anon, I'm buying a ticket right now

>>12937138
>Anyone who dissagrees is retarded.
lmao you just wait
>>12937159 there it is
>>
This is the real autism test.
>>12937116
>>
>>12937159
>Has nothing to do with my spelling
Ah fuck off Its almost 3 am where I live and Im phoneposting to.
I can see how one might answer 2/5 since it would be a dubble probability for picking a red ball out of the batch with 2 red balls meaning you would get 2/5 either way, but that would depend on how you interpret the question.
>>
>>12937177
>I love how I am so obviously fucking with you
>I'm only pretending to be retarded
Good for you, now take your meds.

>and you still bite.
>No I'm trolling you, I'm not the one being trolled
Seriously, take them.

>You've demonstrated your autism in many different ways, very informative.
Inform yourself where your meds are, and then take them.
>>
>>12937193
4/9
>>
>>12933011
blue and red now?
at which point did you fags change the colors?
>>
>>12937200
No, it's a solid 8/9 rounded up to 8/8, m8.
>>
>>12937205
we want to be more inclusive for balls of color
>>
>>12937206
Yes the ball is always blue, you just have to hallucinate it anon. simple conditional probability
>>
>>12937244
Doesn’t follow.
>>
>>12936850
>The box denialist is not a person, but a people. An archetype.
>Antiboxxers channel the essence of a very autistic hatred of packaging.
>But in their crowd, a particular individual stands out.
>Antibox-Man, Messiah of Thy Retardation, The Great Revealer of Ambiguity
>His Majesty The Liberator of Balls shows the peasants the alienating nature of containment devices with his Post-Amazon dialectic.
>I, however, choose not to believe. I like boxes. I guess you could call me a storage enthusiast.

>>
>>12937170
>2/3
>Wrong. You've been proven wrong multiple times this thread stop coping.
See >>12936718

>4/9
>Actually correct congratulations.
See >>12935127

>2/5
>Also actually correct.
>Look at you you're only 1/3 retarded.
>>
>>12937252
It does, just stop taking your meds for 2-3 days. The truth is right there. Just watch out for helicopters around your house, or robot spy birds.
Simple deduction really.
2/5
2 +2 /5 + 4
4/9
You have to be blind to not see it.
>>
File: Untitled.png (8 KB, 1690x521)
8 KB PNG
>>12937263
Pick a ball anon.

Careful! Don't pick a box, this wasn't mentioned in the question. Don't start hallucinating things now!

Just pick a ball
>>
>>12937305
Incredibly based.

Ballfags blown the fuck out time and time again. Boxboys are king's of probability. Your simple smooth marble sized brains are unable to comprehend basic concepts. You argue the probability of a false construct.

>By ignoring the boxes you ignore the word of God himself, you ignore nature, you ignore the realm of possibility.
>>
>>12937301
>2 numerated helicopters <-> 4 denominated robot spy birds.
Holy fuck, this solves everything. Thank you
>>
Mfw I have unequivocall proof the answer is 1/2.
>>
>>12937305
You first
>>
>>12937327
The probability of a ball being picked depends on the number of balls in the same box.
Therefore, all balls don't have the same probability of being picked.
This probability is directly related to the box in which the ball is contained.
>>
>>12937112
>Are you imagining that you can't reach out and put your hand into a box?
That would be choosing a box. Good job. Which box you put your hands in determined which ball you can pull out. If you don't know which box has the lone ball, there is no way to choose the lone ball with equal probability to the other balls.

>You're imagining some sort of lid or other magic wall that isn't given by the question.
Where did I say anything like that? I'm sorry you can't read.

>This is a simple conditional probability problem.
Then simply describe to me how you can choose a ball idleness from the boxes.

>I’m not “describing” anything.
Yes, that's my point. I'm sorry you can't read.

>I’m choosing a ball from a box, per Q2.
You're not, what you claim to be doing is impossible.

>On the other hand, you’ve admitted that there's no possible procedure for what you're describing that doesn't involve "feeling" around for boxes.
Where did I say there's no other possible procedure? I'm sorry you can't read. The only impossible procedure is your own.

>In a conditional probability problem, you use the information that you're given.
Right, the information given shows your interpretation is impossible.

>There's no default "process" for how or what you choose.
Where did I say anything about a default process? I'm sorry you can't read.

>Q2 says that you choose a ball from a box, after having removed 1 blue ball from the initial 6-ball state.
Yes, therefore the answer is 4/9.
>>
>>12937339
Which box? I thought you were trying out a new visual argument. What box do you see here?
>>
>>12937347
Sorry, I’m phone posting now. Hold that thought and I’ll get back to you in the morning.
>>
File: blu.png (7 KB, 103x84)
7 KB PNG
>>12937327
Alright...

Wow! A blue ball! Not that this is surprising, because I had 1/2 chance of getting one on the first round.
But I wonder we will get another one next time. I'll just keep the ball, so this box will only have one ball now.

Now let's shuffle the boxes and reiterate.
Your turn again anon, pick a ball. Not a box! A ball.
I know you can't see them but I want you to consider giving them equal opportunities.
Pick a ball anon
>>
>>12937115
>I’m not trying to do anything, I’m answering the question.
Then explain how this
>What properties are you imagining the boxes have, that prevent you from reaching out and choosing any ball from any box?
responds to the fact that you can't choose balls with equal probability.
>>
>>12937371
You cannot pick a ball without first picking a box as the box contains the ball you brainlet scum
>>
>>12937372
>the fact that you can't choose balls with equal probability.
That’s your mistake, not mine.

>>12937371
Gibberish.
>>
>>12937388
Don't interrupt me, I'm just demonstrating the absurdity of antiboxman's argument
>>
>>12937371
>>12937388
Lol.
>>
>>12937397
You’ve demonstrated that you need to take your meds. I approve.
>>
>>12937397
Their arguments are just normally so ridiculous I couldn't even tell your post was satire.

Sorry boxbro.
>>
>>12933011
1. 2/3. Box 1 is eliminated. Think of the final two boxes as one straight line of 4 balls, the first 3 of which are blue. We know that one blue was selected, so the last ball is out. So then there's a 1/3rd chance of selecting any of the blue balls, since there's two blue balls in one box it must mean the probability that the first ball comes from that box is 2/3rds.

2. 1/3 chance the first ball was from the 2nd box, as previously established. If it's from this box, there's a 1/3rd chance of the next ball being picked being blue. 2/3 chance it was from the 1st box. If it was from this box, the chance of the next ball being blue is 1/3 from the first box+ (1/3)*.5 from the 2nd box, so exactly 1/2. Multiply the ultimate probabilities by the initial probabilities and add, so (1/3)*(1/3)+(2/3)*(1/2). Answer is 4/9.
3. It's 2/5, if you can't figure this one out yourself you're retarded.
>>
>>12937411
Run, anon, leave this thread before it's too late. RUN!
>>
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9 KB PNG
>>12937403
Pick a second ball anon.
Let's see if it will turn blue as well.

I shuffled the boxes for you, so you don't know which box contains the lone ball.
Just pick a ball now
>>
>>12937411
>If it's from this box, there's a 1/3rd chance of the next ball being picked being blue.
No, see >>12935127
>>
>>12937411
1 problem faggot.

You forgot about the boxes.
>>
>>12937419
Why are you hallucinating three black rectangles you can’t see? Pic related is the only thing you can see
>>
>>12937436
Nigga I don't know what the fuck any of that means I'm a finance major
>>
>>12937394
>That’s your mistake, not mine.
How is it my mistake? You're the one that claimed it can be done but have failed countless times to show how.
>>
>>12937451
The boxes are in front of you anon. One a bit to the left, one a bit to the right and one right in front of you.

You just need to reach out and grab a ball. Unless you also need to be handheld physically?
>>
>>12937449
They mentioned the boxes, that's why they're 100% correct.
>>
>>12937455
>finance major
Impressive, are you minoring in Elliott waves or are you minoring in φ
>>
>>12937476
I haven't chosen my minor yet, I kinda think technical analysis is mostly bullshit so I'm leaning towards the latter. What do you recommend?
>>
>>12937467
>You just need to reach out and grab a ball.
Correct. That’s all you need to do.

>>12937464
>You're the one that claimed it can be done
When did I claim you can’t choose balls with equal probability?
>>
>>12937483
> What do you recommend?
For random finance work? Network and use your network to network into other networks, and then retire whenever you get tired of that
>>
>>12937497
>>12937483
>>12937476
This is a thread about balls and boxes not fagance minors.
Stop derailing from the fact Q1 is 1/3
>>
>>12937502
>Q1 is 1/3
Prove it.
>>
21 KB JPG
>>12937487
>Correct. That’s all you need to do.
Thanks for reassuring me, I'll give it a go now...

Oh anon! Did you put meds in the box? What are you implying here? Not cool.

Now let me pick a ball from a box which I will not pick.
I know the balls are not pooled but I want to be inclusive and give them all a fair chance.

Oh! It's a red ball. Looks like we did this for nothing. I mean I had a 5/9 chance of getting a red ball after all.
Better luck next time I guess.
>>
>>12937515
Sorry, I’m phone posting and your comment is too desultory and exclamatory to read and reply to. Hold your thought, I’ll get back to you tomorrow morning.
>>
>>12937487
>When did I claim you can’t choose balls with equal probability?
When did I claim you claimed that? I'm sorry you can't read.
>>
>>12937525
I've been phoneposting this whole time.

You cannot beat the argument antiboxfag
>>
>>12937436
See >>12935363
>>
The real autism test is whether or not you get into an hours long argument about balls on a Filipino competitive hopscotch forum
>>
>>12937537
> I've been phoneposting this whole time.
Good for you.
> You cannot beat the argument antiboxfag
I haven’t read your argument and I’m not against boxes, or against you sucking another man’s dick if that’s what you want to do, but I’m sure you’ve explained it all very well in your tl;dr argument that I’ll probably read and reply tomorrow.
>>
>>12937577
Don't forget to reply to >>12936684
>>
>>12937535
> When did I claim you claimed that?
In your post I replied to. Take your meds.
>>
>>12937585
>In your post I replied to.
No, that post says you claimed it CAN be done. I'm sorry you can't read.

>>the fact that you can't choose balls with equal probability.
>That’s your mistake, not mine.
>>How is it my mistake? You're the one that claimed it can be done but have failed countless times to show how.
>When did I claim you can’t choose balls with equal probability?
>>
>>12937583
Remind me tomorrow. It’s too much effort to even source garbage like this >>12937569 from the thread,
let alone your longposting efforts
>>
>>12937615
Quote my post. I’m not going to source shit on my phone for you.
>>
>>12937622
How do you choose a ball such that each has an equal probability of being chosen when there is a box, unknown to you, which has a lone ball?
>>
>>12937628
I just did quote your post. Just follow the replies if you don't believe me. I'm sorry you can't read.
>>
>>12933011
Q1: 2/3 (assuming balls are picked at random, which is not explicitly stated)
Q2: 4/9
Q3: 2/5
Q4: What's the blindfold for? An awkward attempt to disambiguate the OG bait pic that still wasn't entirely successful.
Some of you need to remember that outcomes that look the same are still different outcomes.
>>
>>12937628
How do you choose a ball such that each has an equal probability of being chosen when there is a box, unknown to you, which has a lone ball?
>>
>>12937636
You said this, not me
>>12937372
> the fact that you can't choose balls with equal probability.
>>
You're all fools, holy shit
>>
>>12937664
Yes, and then you implied that I claimed that you claimed you can't choose balls with equal probability, when I did the opposite. I'm sorry you can't read.
>>
>>12937633
Per Q2, you don’t choose a box, you choose a ball. It doesn’t matter what you do or don’t know about the contents of any random box, because you’re not choosing a box.
>>
>>12937680
>Per Q2, you don’t choose a box, you choose a ball.
How do you choose a ball such that each has an equal probability of being chosen when there is a box, unknown to you, which has a lone ball?
>>
>>12937678
> the fact that you can't choose balls with equal probability.
This isn’t my fact or any fact.
>>
>>12937690
See >>12937680
>>
>>12937692
>This isn’t my fact or any fact.
Then it should be easy for you to explain how you can choose balls with equal probability. But you've spent two threads failing to answer this question. So, yes it is a fact, as we both know.
>>
>>12937697
How do you choose a ball such that each has an equal probability of being chosen when there is a box, unknown to you, which has a lone ball?
>>
>>12937699
> Then it should be easy for you to explain how you can choose balls with equal probability.
Yes, you choose a ball. Any other schizo shit is all on you
>>
>>12937707
>Yes, you choose a ball.
I'm asking you how, and you're just saying that you do. If you can't answer the question you admit that you can't choose balls with equal probability.

How do you choose a ball such that each has an equal probability of being chosen when there is a box, unknown to you, which has a lone ball?
>>
>>12937704
See >>12937697
>>
>>12937710
How do you choose a ball such that each has an equal probability of being chosen when there is a box, unknown to you, which has a lone ball?
>>
>>12937709
What do you mean, how?
>>
>>12937676
Offers no solution?

100% a storage enthusiast.
>>
>>12937712
See>>12937710
>>
>>12937714
In what way or manner, by what means, do you choose a ball with equal probability.

For example, you said that you would play this game:

Let's play a game where you give me \$2 and I let you choose a ball from either of two boxes, one with 1 green ball and the other with 1000 yellow balls. If you choose a yellow ball I'll give you \$3. If you choose a green ball you get nothing. Would you play this game?

You were asked which ball you would choose, and you failed to answer. This is because you know there is no actual way to do what you're claiming can be done. You know it and I know it.
>>
>>12937719
So you admit that you can't choose balls with equal probability.

Thanks!
>>
>>12937727
See >>12937719
>>
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224 KB JPG
>>12937716
as a Boxxie myself, I'd like to state that we only claim those who provided demonstrative proofs of how they got to 4/9 on Q2.
We'd like to be more inclusive, but this warehouse is meant to be a safe space for box-loving individuals
>>
>>12937733
Which ball do you choose?
>>
>>12937727
I already answered this nonsense bet question way back in the thread. Rtft and if you still don’t understand conditional probability after reading the fucking thread then feel free to type out a long format question and queue it up into the queue of tl;dr rants I’ll probably reply to tomorrow.
>>
>>12937750
>I already answered this nonsense bet question way back in the thread.
No, you didn't. You never said which ball you want to choose. So you never actually played even though you claimed you would.
>>
>>12937732
Yes. Yes. You’re a genius.
>>
>>12933011
for question one do you put the ball back or take it?
>>
>>12937754
>>
>>12937766
I did. You failed to answer. All of this is moot anyway since by yet again failing to answer a simple question you admitted that you can't choose balls with equal probability.

>>
>>12937737
This is your last chance. After this, there is no turning back. You take the box pill – the story ends, you wake up in your bed and believe whatever you want to believe. You take the ball pill – you stay in Wonderland, and I show you how deep the rabbit hole goes. Remember, all I'm offering is the truth – nothing more.
>>
>>12937733
Not me but I cosign this post
>>
>>12937771
>>
>>12937785
You literally didn't respond to the post. Do you really think lying is going to help you. You already admitted your entire argument fails. Get over it.
>>
>>12937792
I certainly did reply to the \$2 \$3 retard. And if that was you, take your meds.
>>
>>12937773
Another Receptacle Defiant spreads the Holy Truth of Ball Loving.
They always try to convince others that only balls lead to salvation.
I will not be fooled by your crate-hating rhethoric.

No need for matrix references, only the weak believe they finally got out of plato's cave.
I see the box for what it is, a cuboid reservoir that contains objects.
>>
>>12937842
I do not hold hatred in my heart for those who have been misled.

Love the Ball
for it is the salvation of mankind

Obey it's words

for it will lead you into the light of the future

Heed it's wisdom

for it will protect you from evil

Whisper it's prayers with devotion,

for they will save your soul

Honour it's devotees,

for they speak with it's voice

Tremble before the ball,

for we all walk in it's immortal shadow.
>>
>>12937115
>No, it’s a simple conditional probability problem.
Correct, and the wording of the problem implicitly states that the balls are subsets of boxes, with the boxes having equal chances of being selected (they're randomly selected, and "random", sans any other qualifiers, is conventionally taken to mean a uniform distribution).

>You’re not choosing a box, you’re choosing a ball.
Choosing a ball is, by necessity, also choosing a box. In the action of choosing a ball, one must choose a box first.

>What remaining balls? What balls do you imagine disappearing and why?
The one ball you already took out of the boxes, as is stated in the problem.

>Good thing you’re choosing a ball, not a box.
Good thing you're a schizoid autist.
>>
>>12937803
No you didn't >>12936684
>>
>>12937571
You can never tell in this place whether you're dealing with someone who goes to obsessive lengths to pretend to be retarded, or a new and exciting form of retardation. Either one is psychologically interesting but trying to actually convince them they're wrong is probably a fool's errand.
>>
>>12933011
>Q1
B represents the event "the first ball you picked was blue"
BB represents the event "the next ball you pick from the same box is also blue"
P(BB|B) = P(B|BB)*P(BB) / P(B) # bayes rule
= (1)*(1/3) / (1/2) = 2/3
>Q2
BB represents the event "the box you drew the first blue ball from contained two blue balls"
BR represents the event "the box you drew the first blue ball from contained one blue and one red ball"
B2 represents the event "the next ball you pick from any one of the distinct boxes is also blue"
P(B2) = P(B2|BB)*P(BB) + P(B2|BR)*P(BR) # law of total probability
= (1/2)*(2/3) + (1/3)*(1/3) = 4/9
>Q3
2/5 # trivially
>>
>>12937571
The first guy to fail the test in the whole thread.
>>
>>12939947
Ballschizo is dead, boxbros we won!
>>
>>12939264
>but trying to actually convince them they're wrong is probably a fool's errand.
I think it's been a good logic exercise for me in a few of these threads trying to come up with inventive ways to explain the answer. But it can quickly turn into a waste of time, too. Gotta learn to bail when it becomes clear that the other person isn't just being dense.
>>
bump
>>
>>12942193
might seem like it but these antiboxxers are some stubborn mfs
especially mr. Lil Fuck-A-Box

just watch them come with the same enlightened arguments next time this post comes around
>>
I don't understand. Why must we distinguish between the two blue balls in the box? They are equivalent and interchangeable. Once you draw a blue ball, you only have two remaining possible boxes to draw from. How is the probability not 50:50 after that first key instance of already picking a blue ball? Why is the probability not reset after the key action of already picking a blue ball?
>>
>>12943313
It is.

The answer is 1/2 these fucking idiots are a bunch of ball loving bitches. They haven't embraced the 1/2. They haven't embraced 4 corners. They haven't taken box into their heart.
>>
>>12943313
>Once you draw a blue ball, you only have two remaining possible boxes to draw from. How is the probability not 50:50 after that first key instance of already picking a blue ball? Why is the probability not reset after the key action of already picking a blue ball?
If you have a box with 100 blue balls, and another box with 99 red balls and 1 blue ball, and you randomly pull a blue ball, do you think the odds are greater that you pulled one of the 100 blue balls from the first box, or the only blue ball from the second box?
>>
>>12943313
because drawing a blue ball is not equally likely from the two boxes, which means that it provides evidence for one box over the other
>>
>>12943447
However the determination means that you can only pick from two possible boxes. And since you can't pick a marble without picking a box anyway that's a 50/50 box choice. Also, like he said it's irrelevant the probability resets as soon as the blue ball is determined to be the first selected. Leaves two possible outcomes with equal likelihood. Blue or red. 1/2

2/3 fags are coping.
>>
>>12943313
you dont have to distinguish between them, simply acknowledge that picking one of the blue balls from the left is twice as likely as picking the blue ball from the mixed box
>>
>>12944136
It's not though. Since we know we pickup a blue ball the 3rd box is eliminated. Then it's a matter of selecting between the two boxes to select our ball. You must pick a box before a ball. 50/50 on what box you pick.

1/2 chance the second ball is blue. You can simulate this.
>>
>>12933011
1) 50%
2) 33% ~
3) 40%
>>
>>12933011
you passed the autism test, gratz
>>
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>>12943913
>However the determination means that you can only pick from two possible boxes
the determination means that the box you picked was either the one with two blues or the one with mixed. However, because the box with two blues is more likely to produce a blue, it is therefore more likely that the box you picked was the two blue box. And since it is more likely that the box you picked was the two blue box, it is more likely than even odds that drawing again from that same box will produce another blue
>since you can't pick a marble without picking a box anyway that's a 50/50 box choice
only one box pick has been made, the original one between the three boxes. There is no 50/50 choice after that point
>the probability resets as soon as the blue ball is determined to be the first selected
why would the probability reset? The boxes didn't reset. You drew from a particular box and by doing that you learned some evidence about which box that one is. Then you drew from that particular box again

>>12944218
>Since we know we pickup a blue ball the 3rd box is eliminated. Then it's a matter of selecting between the two boxes to select our ball. You must pick a box before a ball. 50/50 on what box you pick
again, what you're missing is that when we pick up a blue ball, the chance that we picked the 3rd box goes to 0, but the chance that we picked the 1st and 2nd box do not each go to 0.5 . The chance we picked the 1st box goes to 2/3 and the chance we picked the 2nd box goes to 1/3. This is because drawing a blue ball is not equally likely between the 1st and 2nd boxes, so when it happens that is stronger evidence that you picked the 1st box than the 2nd
>You can simulate this
you can. Here is the simulation
>>
>>12944365
I knew ide get an autist to code it lmao.

Look I could keep this going forever but I feel like you've put in enough effort for me to throw in the towel.

I bow to the 2/3rd fag.
>>
>>12944380
The other tranny may bow but I can't code so your fuckin gypsey wizard writing means nothing to me. It's 1/2. Writing scribble to clearly produce the result YOU want is useless. You have a blue ball there's only to possibilities Ur in box one or box two. 2 possibilities, 2 outcomes, red, blue, Ur either in or the other that means it's 1/2
>>
>>12944365
Forgot to quote
>>12944386
>>
>>12944380
>>12944386
>you've put in enough effort for me to throw in the towel
don't flatter yourselves. I coded it and wrote out the math a long time ago and just copy paste into these threads. You guys are spending more effort pretending to be retarded
>>
>>12944386
>you either won the lottery or you don't, that means it's 1/2
>>
>>12944386
Do you agree the chance of picking each box before you choose is 1/3?

Do you agree that the chance of choosing a blue ball given you chose the all blue box is 1?

Do you agree that the chance of choosing a blue ball given you chose from the blue-red box is 1/2?

If you say yes to all of these then you admit the answer is 2/3:

P(box 1|blue) = P(box 1) P(blue|box 1) / P(blue) = (1/3)(1)/((1/3)(1)+(1/3)(1/2)) = 2/3
>>
>>12944463
If only two people bought tickets it does brainlet.
>>
>>12944479
>Do you agree the chance of picking each box before you choose is 1/3
No because the question specifically states the first ball selected is blue meaning the chance of picking a box is now 1/2

>Do you agree that the chance of choosing a blue ball given you chose the all blue box is 1?
Yes

Do you agree that the chance of choosing a blue ball given you chose from the blue-red box is 1/2?
No. It's 1. Since the question has already defined that we will pick a blue ball first.

If you say yes to all of these then you admit the answer is 2/3.
Good thing I didn't and it's not.

It's 1/2
>>
>>12944386
>your fuckin gypsey wizard writing means nothing to me.
Based retard.
>>
>>12944594
The lottery doesn't choose between ticket holders, it chooses between numbers, retard. If no one bought the winning ticket, no one wins.
>>
>>12944598
>No because the question specifically states the first ball selected is blue meaning the chance of picking a box is now 1/2
That has nothing to do with what I said. The chance before you choose is not the same as the chance after you are given information about what you chose.

>No. It's 1. Since the question has already defined that we will pick a blue ball first.
Again, this has nothing to do with what I said. I'm asking you for a probability under a different condition. Also something occurring and something having to occur are two different things. If I flip a coin, and get heads, that doesn't mean the probability of getting heads before I flipped was 1. It was 1/2.

Now please answer my questions as they are written, not with your added condition.
>>
>>12944617
Nobody gives a fuck about the lottery you autistic moron. This is a question of boxes and balls
>>
>>12944660
Except the question didn't ask for the probability of your first ball being blue. It told you it was blue and then asked you the probability of the second being blue.

The conditions are set up anew. Everything prior is irrelevant. if I flip a coin twice and the first coin Lands heads what is the probability the second coin is heads?

1/2 because the second coin doesn't give a fuck about the first or anything prior. It's a 50/50 toss.

The question has presented you with the facts. The knowledge we will draw a blue ball makes the 3rd box irrelevant from the question. The balls contained within the final two boxes are also irrelevant since we must first select a box. Knowing both contain blue and knowing we will draw a blue ball from either one we choose that means there's either a box with a red ball left and one with a blue ball left.

It's 1/2
>>
>>12944834
The question is whether you understand how to calculate probability. The answer is: you don't.
>>
>>12944857
>Except the question didn't ask for the probability of your first ball being blue.
Where did I say it did?

>Everything prior is irrelevant.
So Bayes Theorem is wrong? How interesting. Please tell me more about your revolutionary probability theory.

>1/2 because the second coin doesn't give a fuck about the first or anything prior.
Right, because they're independent. P(box 1|blue) is not independent from P(box 1), P(blue|box 1), P(blue). It is in fact completely determined by those probabilities.

>The knowledge we will draw a blue ball makes the 3rd box irrelevant from the question. The balls contained within the final two boxes are also irrelevant since we must first select a box.
Wrong.

>Knowing both contain blue and knowing we will draw a blue ball from either one we choose that means there's either a box with a red ball left and one with a blue ball left.
Correct, but that doesn't imply the answer is 1/2. The former is twice as likely as the latter.
>>
>>12944906
The question is whether or not you gargle balls like Tom Cruise gargles whitening toothpaste. The answer is: you do.
>>
>>12945006
Nothing you said refuted my argument. You are now twisting the words of the question to match your interpretation. The reality is you aren't more likely to select the box with 2 blues that's what we are arguing over. Nothing else.

You aren't more likely to select it because you are picking between two boxes. Not 4 balls 2 boxes. You know they both have already one blue and you know you'll receive it the box you pick is a 50/50 and since each box contains a different outcome for your second marble the answer is 1/2..

Bayes theorem is wholly dependant on the input. If you plug the wrong fucking data in you'll get the wrong result, I'm not arguing your probability logic I'm arguing your interpretation of the question.
>>
>>12945030
>Nothing you said refuted my argument.
LOL. I already refuted your argument with Bayes Theorem, you're just denying the basic concepts and definitions of probability theory.

>You are now twisting the words of the question to match your interpretation.
How so?

>The reality is you aren't more likely to select the box with 2 blues that's what we are arguing over.
We are, and I proved it. Your response was to deny the entire basis of probability theory. If you are unwilling to learn, I suggest you replicate the problem for yourself. Count how many times you get a blue ball. Count how many times you get a blue ball from the all blue box. Divide the former by the latter and you will see it's 2/3.

>You aren't more likely to select it because you are picking between two boxes.
We picked between three boxes, then we learned we didn't pick one of them and that it's twice as likely we picked from one vs. another.

>the box you pick is a 50/50
Wrong.

>If you plug the wrong fucking data in you'll get the wrong result
Which data are wrong?

P(box 1) = 1/3

P(blue|box 1) = 1

P(blue) = 1/2
>>
>>12945030
>I'm not arguing your probability logic
Then how can you say prior probabilities are irrelevant? Why did you confuse prior probabilities with posterior probabilities? Why did you confuse the probability of an independent event with a conditional probability? It seems unlikely you have any clue what you're talking about.
>>
>>12943327
Why is everyone ignoring this? It's pretty obvious what's going on when the quantities are increased. You clearly have much, much better odds of drawing another blue ball in this scenario.
>>
>>12945729
Because they're retarded.
>>
>>12945729
>>12945734
>>12945232
You can't select a ball without selecting a box. Since the first ball is guaranteed blue it's literally irrelevant the ratio of blue to red. The box choice is 50/50 since both confirmed to have alreast one marble. You aren't selecting a ball you're selecting a box.

Brainlets can't comprehend it's all about the boxes
>>
>>12946257
>Since the first ball is guaranteed blue
What do you mean by guaranteed? Can you draw a red ball, and it will magically change to blue? Otherwise, you have a chance to draw a red or blue ball, and in this scenario, you draw a blue ball. The question even includes "if the first ball you picked was blue," which means it's not guaranteed, only that it's a possibility, and if that were to occur, what are the odds for the next draw.
>>
>>12946257
>You can't select a ball without selecting a box.
No one said anything to the contrary.

>Since the first ball is guaranteed blue
It wasn't "guaranteed" to be blue, it just happened to be blue.

>it's literally irrelevant the ratio of blue to red.
So Bayes Theorem is wrong? I look forward to your disproof. Until then, shut the fuck up.

>The box choice is 50/50 since both confirmed to have alreast one marble.
How does that make it 50/50?
>>
>>12933011
>>>/fa/cebook
>>
>>12946521
Bayes theorum reeeeee. Doesn't work if you ignore the question and provide the wrong inputs.

It's really simple peanut brain but I'll try make it simpler for you.

Pick box. Box U pick must have blue marble since question guarantees it. Mean 2 box U can pick.

Cancel out first blue marble in either box U pick. Now only 2 options left still based on the initial blind 2 box 50/50 pick.

2 outcomes 50/50 box pick tadaaa

It's 1/2.
>>
>>12946728
>must have blue marble since question guarantees it.
You keep using "guarantee," but that word is nowhere in the question. There's nothing guaranteed.

>Box U pick must have blue marble since question guarantees it.
No it must not. The question is only asking what would happen "if the first ball" from the box you select is blue.
>>
Alright halflets, your end has come.
There is no way you can force this to be 1/2 with your weird logic of forcing blue to be picked.
>>
>>12946728
>Doesn't work if you ignore the question and provide the wrong inputs.
Which inputs are wrong?

>guarantees
It wasn't "guaranteed" to be blue, it just happened to be blue.

>Now only 2 options left still based on the initial blind 2 box 50/50 pick.
Wrong, it's twice as likely you picked a blue ball from the all blue box.
>>
>>12933011
1: 1/2
2: 2/5
3: 2/5
easy
>>
>>12946895
shuffling does nothing, your still selecting from the same box. Its still 1/2
>>
>>12947597
>2/3
Correct.
>>
>>12947569
It being twice as likely is irrelevent, as you are asking the probability for the subsequent step.
>>
>>12947609
>as you are asking the probability for the subsequent step.
But you needed to get to that step from the previous step, and the previous step involved a random draw that weights the odds. Unless you design the question to totally eliminate the random draw, you can't eliminate the fact that it changes the odds for the subsequent step.
>>
>>12947621
Oh i see, you're more likely to get the blue ball from the box with 2 blue balls, so if you got the blue ball first time you're still more likely to get the blue ball second time.
I get it now
>>
>>12947590
>1: 1/2
>2: 2/5
Wrong.
>>
This is just Monty Hall
>>
>>12947935
Wrong.
>>
>>12947638
Faggot don't let them poison your mind

The prior is irrelevant we have blue ball confirmed first
>>
>>12933011
1. 1/2
2. 1/2
3. 2/5
I hate probability
>>
>>12933011
>Q1
1 if you picked the ball out of the left box, 0 if you picked it out of the middle box; median is 1/2
>Q2
1/2 if you picked out of the left box, 1/3 if you picked out of the middle box; median is 5/12
>Q3
Depends on the context
>>
>>12947935
Correct.
>>
>>12946728
Except you don't know which marble you picked first in the blue+blue box, so there are actually two outcomes there: blue 1 then blue 2, or blue 2 then blue 1, meaning 2 out of 3 total outcomes lead to a blue marble being picked second.
>>
>>12933011
Q1 - 1/2
Q2 - 1/3(1 + 1/2) = 3/6 = 1/2
Q3 = 5/2
>>
>>12948589
>dyslexia
Q3 - 2/5
>>
>>12948589
>Q1 - 1/2
Based and boxpilled. A man of intellect.
>Q2 - 1/3(1 + 1/2) = 3/6 = 1/2
Correct.
>Q3 = 2/5
Also correct.

They cannot dispute this don't let the shills dissuade you. They misrepresent the question to try prove themselves correct. Stay strong, stay based.

>1/2
>/
>2
>>
>>12933011
1. 2/3
2. 2/5
3. 2/5
>>
>>12948732
>2/3
Fake and gay. It's 1/2.

>2. 2/5
Fake and gay. It's 4/9

>3. 2/5
At least you got this one correct. Congratulations you're not completely retarded.
>>
Rest now. You were a good thread.
>>
>>12949376
>You were a good thread.
Wrong
>>
>>12948457
>1. 1/2
>2. 1/2
Wrong.

>>12948457
>1 if you picked the ball out of the left box, 0 if you picked it out of the middle box; median is 1/2
Median would imply those outcomes are equally likely. They aren't since the left box is all blue but the middle box is only half blue.
>>
>>12948489
>1/2 if you picked out of the left box,
1/3 if you picked out of the middle box; median is 5/12
Wrong. Same mistake as Q1.
>>
>>12948534
This only leads to the correct answer coincidentally. It's not the number of blue balls but the ratio of blue to red in each box which matters.
>>
>>12948589
Wrong. See >>12949655
>>
>>12937305
>Pick a ball anon.
okay, I will

I have a 2/3 chances where both balls will be the same color as each other, and 1 1/3 chance where they will be different colors from each other

Therefor there is a 2/3 chance that the second ball will be the same color as the first ball that was picked. Lets assume I have happened to grab a blue ball, because of the previous conditions, I have a 2/3 chance the second ball will also be blue.

Boxbois mogged out of oblivion. Imagine being so wrong about math that is theoretically and empirically correct.
>>
>>12950062
Fuck off ballfags. You can't select ball only a box.

Ballfags are the scum of 4chan truly the lowest of the low. We need another containment board called /retard/ for people like you
>>
>>12950075
Use Bayes Theorem to prove the answer is 1/2.
>>
>>12950062
>>12950075
I'm confused, I thought everyone realised the first question was simply Bertrand's Box, the point is that some schizo was suddenly treating questions 2 and 3 as if they were identical, as if the boxes suddenly weren't there. If you realise that the answer to Q2 is 4/9 for the precise reason that the balls are, in fact, in boxes and cannot be freely selected, you should arrive at 2/3 for Q1 for the exact same reason.
>>
1/2
1/2 or 2/3 im not sure
2/5
>>
>>12950062
>>Pick a ball anon.
>okay, I will
Oops I get your point

Ill have to pick a box first then hence;
I pick a box, there is a 2/3 chance that the balls of the elected box are common in color, and only a 1/3 chance they are different. Thus the subsequent ball I pick from the same box has a 2/3 chance to be the same color. Lets assume I have happened to grab a blue ball, because of the previous conditions, I have a 2/3 chance the second ball will also be blue.

Boxbois mogged out of oblivion. Imagine being so wrong about math that is theoretically and empirically correct.
>>
>>12950097
what? yes q2 is 4/9
>>
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171 KB PNG
>>12933011
1. 1/2 because there are only two boxes from which you could pick a blue ball. The blue ball taken out, one box contains now a red and the other a blue ball.
2. 2/5 because effectively the boxing doesn't matter. There are only 5 balls left, two of which are blue.
3. uncertain. the balls will roll all over the place. there is a reason you put them in a box baka.
>>
>>12950136
Because of the boxes, yes.

And so Q1 is 2/3, because of the boxes.
>>
>>12950075
>You can't select ball only a box
OK I choose the box first. 1/3rd of the time i picked the mixed box, after that only half the time I pick a blue ball (so in total the blue ball in the mixed box has a 1/6 chance of being picked. However if the probability is dependent of the first ball being blue, then that means 50% of the time I pick the mixed box, i then grab a red ball which does not satisfy the probability requirements, The result of halve the mixed box being picked isn't included in the conditional probability. So even if you pick a box first, the fact that question states that "the first box you picked was blue" means the box with two blue balls is more likely to be picked.

Boxbois mogged out of oblivion. Imagine being so wrong about math that is theoretically and empirically correct.
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>>12950149
Thats what im saying? im:
>>12950136
>>12950131
>>12950062
>>12950153
and new to the thread
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>>12950156
Then aren't you a "boxboi"?

I don't fucking get it any more, man
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>>12950167
no, i saying even if you go with boxboi's logic, you still get 2/3 4/9 etc.
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>>12950167
wait i though the logic of the box boys is that because you have to pick a box first, the chance of you picking the box with 2blues and 1 blue are the same, and because only 1 of the 2 outcomes results in 1/2 for the first question hence >>12950075 >>12946728
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>>12950171
So is "boxboi" the box denialist? I'm just trying to wrap my head around the terminology here.
>>12950177
I thought the box only became an issue for Q2. I think someone is deliberately trolling by taking the reasoning for Q2 and misapplying it to Q1.
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>>12950192
cool imma sleep this thread is dumb, as all threads about balls are.

2/3
4/9
2/5
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>>12950097
It's probably just some tard trolling.
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>>12950108
>1/2
>1/2 or 2/3 im not sure
Wrong
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>>12950192
>>12950262
Not gonna lie boys I knew it was 2/3 in the first thread but it's been a fun ride. I do seriously think Q2 is 4/9ths though
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>>12950140
>1. 1/2 because there are only two boxes from which you could pick a blue ball.
Since one box was only half blue, it's twice as likely you chose a blue ball from the all blue box. 2/3.

>2. 2/5 because effectively the boxing doesn't matter.
The boxes do matter, for the reason stated above and because a lone ball is twice as likely to be chosen as a ball in a box with two balls. 4/9

>3. uncertain.
2/5
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>>12950491
It is.
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>>12948807
I dont get how someone can get 4/9 on Q2 but 1/2 on Q1. Both answers literally use the same logic
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>>12951882
By both answers I meant 2/3 on Q1 and 4/9 on Q2
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>>12948807
>Congratulations you're not completely retarded.
Well I'm not the one intentionally trolling the thread, so I at least know I'm above you.
>>
bump
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>>12933011
Hi puzzle anon.
I think the answer to Q1 is 50%
So we picked a blue ball. That means there's two blue balls left. Either we took a blue ball from the box containing two blue balls, in which case we will grab the second blue ball from the box, or we first grabbed a blue ball from the box containing one blue ball and one red ball. 50-50 chance.
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>>12933011
Why does the first paragraph list the boxes in the order they aren't displayed?
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>>12933011
1/2, 1/3, 2/5.

How is it more complicated than this?
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>>12953600
> So we picked a blue ball
No, you picked a random ball, and discovered that it was blue, the probability is conditional of that outcome

Here is a more extreme example to more easily illustrate this
Imagine you had two boxes full of sand, red sand and blue sand.
The box of red sand has a single grain of blue sand mixed in
You randomly select a box and pick up a single grain of sand
You find that the grain in your hand is blue
It would be then reasonable to conclude with almost 100% certainty that you had picked that grain from the blue sand box
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>>12955282
does that trigger something inside you?
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>>12955305
>How is it more complicated than this?
Because your mind is simple.

It's twice as likely you chose a blue ball from the all blue box vs. the box that is only half blue.

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