>implying 4chan isn't a bunch of retarded larps You should be able to solve this. Hint: The answer isn't 2/3.

File: 484 KB, 745x583, sci_is_mega_dumb.png [View same] [iqdb] [saucenao] [google]

Anonymous Sat Apr 13 02:08:43 2019 No.10553293 [Reply] [Original]

>implying 4chan isn't a bunch of retarded larps
You should be able to solve this. Hint: The answer isn't 2/3.

>>	Anonymous Sat Apr 13 02:25:59 2019 No.10553325 File: 3 KB, 125x118, bertrandBox_diagram.jpg [View same] [iqdb] [saucenao] [google] 2/3

>>	Anonymous Sat Apr 13 02:27:44 2019 No.10553331 >>10553325 What did you use to make that graphic?

>>	Anonymous Sat Apr 13 02:36:31 2019 No.10553356 >>10553325 "Wrong." The silver only box is discarded.

>>	Anonymous Sat Apr 13 02:48:46 2019 No.10553375 1/2

>>	Anonymous Sat Apr 13 02:55:12 2019 No.10553387 >>10553293 2/3. Naive intuition tells you that after you see a golden ball you have 1/2 chance it’s 1st box and 1/2 chance it’s second box, but that’s incorrect. Golden ball gives you information that there is 2/3 chance you picked 1st box.

>>	Anonymous Sat Apr 13 03:55:31 2019 No.10553509 File: 76 KB, 757x685, animebrainrot.png [View same] [iqdb] [saucenao] [google]

>>	Anonymous Sat Apr 13 04:01:17 2019 No.10553521 The question is asking a isolated event. Who keep yelling 2/3 are absolutely stupid.

>>	Anonymous Sat Apr 13 04:04:26 2019 No.10553529 >>10553293 33,3% i give up

>>	Anonymous Sat Apr 13 04:21:39 2019 No.10553556 >>10553293 Fifty fifty. And i never finished elementary school. You guys are such dummies.

>>	Anonymous Sat Apr 13 04:37:40 2019 No.10553572 >>10553325 >>10553387 >>10553509 Brain dead retards It doesn‘t matter if the second box only has one gold, because, according to the question, when i draw it “it‘s a gold ball“ That contrasicts the random parts of the question

Anonymous Sat Apr 13 04:58:50 2019 No.10553597

>>10553293
For all retards who miss Probability Theory 101:
Three hypothesizes:
H1 - we draw from 1st box;
H2 - we draw from 2nd box;
H3 - we draw from 3rd box.
P(H1)=P(H2)=P(H3)=1/3 - probability that we draw from box i.
Only H1 gives positive result. P(G | Hi) - probability to draw golden ball of we draw from box i:
P(G | H1) = 1;
P(G | H2) = 1/2;
P(G | H3) = 0.
Now, you retards who try to solve this intuitively, we could use Bayes’ Theorem:
P(H1 | G) = P(H1)P(G | H1) / (P(H1)P(G | H1)+P(H2)P(G | H2)+P(H3)P(G | H3)) = 1/3*1 / (1/3*1+1/3*1/2+1/3*0) = 2/3. Dumbasses.

>>	Anonymous Sat Apr 13 05:05:15 2019 No.10553603 It's 2/3, OP is either retarded, or trolling, either way it's not worth your time.

>>	Anonymous Sat Apr 13 05:11:55 2019 No.10553611 >>10553597 *corrections P(Hi) = 1/3 - unconditional probability that we draw from box i, before we look at the first ball. P(H1 \| G) - conditional probability that we draw from box 1 if the 1st ball is golden.

>>	Anonymous Sat Apr 13 05:38:19 2019 No.10553641 File: 670 KB, 1982x1662, 1555115482757.png [View same] [iqdb] [saucenao] [google] Is there a pseudoscience worse than probability?

>>	Anonymous Sat Apr 13 05:40:55 2019 No.10553644 >>10553641 What kind of brainlet made this?

>>	Anonymous Sat Apr 13 05:44:20 2019 No.10553645 >>10553356 what on earth do you think the bottom left box means

>>	Anonymous Sat Apr 13 05:45:23 2019 No.10553647 >>10553572 american idiot

>>	Anonymous Sat Apr 13 05:47:27 2019 No.10553649 >>10553641 >leaves you with 50/50 It does not. If you picked a golden ball, there is a higher chance that you picked from box 1.

>>	Anonymous Sat Apr 13 06:08:04 2019 No.10553672 File: 226 KB, 800x1100, Balls.png [View same] [iqdb] [saucenao] [google] >>10553641

>>	Anonymous Sat Apr 13 06:13:57 2019 No.10553684 My book has this problem Question b) is "Explain why 1/2 is wrong". Lmao

>>	Anonymous Sat Apr 13 06:34:53 2019 No.10553724 >>10553672 this

>>	Anonymous Sat Apr 13 06:38:05 2019 No.10553728 >>10553684 >3 possible starts >2 possible happy endings yeah it's a fucking mystery for the ages

>>	Anonymous Sat Apr 13 06:39:11 2019 No.10553729 >>10553672 moving goalposts, the post

Anonymous Sat Apr 13 06:45:25 2019 No.10553744

>>10553572
What is funny about this thread, is how always the people who say 2/3 are very polite and try to explain their reasoning as best they can.
On the other hand, people who say 1/2 are "Go fuck your self you stupid braindead nigger brainlet". Or some variation of that statement.
Guess which one of the two is right.

>>	Anonymous Sat Apr 13 06:53:25 2019 No.10553757 >>10553293 What's the probability OP is a baiting brainlet trying to get free hw answers. In which case he's out of luck as everyone who didn't answer 1/3 in this thread is evidently an imbecile

>>	Anonymous Sat Apr 13 07:32:32 2019 No.10553813 >>10553729 >moving goalposts except by your logic the answer shouldn't change

>>	Anonymous Sat Apr 13 12:53:41 2019 No.10554591 >>10553649 >It does not. If you picked a golden ball, there is a higher chance that you picked from box 1. except the box is already chosen.

Anonymous Sat Apr 13 12:56:26 2019 No.10554602

>>10553813
>except by your logic the answer shouldn't change
specify the way you select a ball from an infinity of balls. there is no default uniform probability distribution for that case.
if the sole golden ball has a higher than 0 probability of being chosen, then the overall result stays 1/2.
it does not change, it is still 1/2.

>>	Anonymous Sat Apr 13 12:57:02 2019 No.10554604 >>10554591 And? I buy a random national lottery. The news comes on and says that the winning ticket is in my state. Is the chance that my ticket is the winning ticket the same as when I bought it, or higher?

>>	Anonymous Sat Apr 13 15:30:53 2019 No.10554991 >>10554602 >if the sole golden ball has a higher than 0 probability of being chosen, why should this matter? it's given that you selected the ball already.

>>	Anonymous Sat Apr 13 15:46:25 2019 No.10555046 >>10553744 1/2 Ever heard about conditional probability?

>>	Anonymous Sat Apr 13 15:50:05 2019 No.10555057 >>10553325 For this you have to consider the last 2 cases, which never really happen

>>	Anonymous Sat Apr 13 15:51:35 2019 No.10555064 >>10553293 >It’s another “Monte Hall” thread

>>	Anonymous Sat Apr 13 15:51:38 2019 No.10555065 >>10553684 There are more gold balls than silver. The silver silver box is eliminated.

Anonymous Sat Apr 13 16:14:35 2019 No.10555123

>>10555057
you do realize the picture isn't an abstract mathematical tree?
just put the 3 boxes on the floor, and set the balls on front of them for each branch
op's prompt says "It's a gold ball"
the diagram clearly shows there are 3 ways that beginning could happen.
only 2 of those ways lead to picking another gold.
so, 2/3

>>	Anonymous Sat Apr 13 16:38:23 2019 No.10555172 >>10554591 Take it to extremes Say there's a box with 1000 gold balls, and a second box with 999 silver balls and 1 gold ball. You drew a gold ball, which box are you likely to be in?

>>	Anonymous Sat Apr 13 22:36:26 2019 No.10555961 >>10555172 It's equally likely to be from either since the problem starts with it already being selected, obviously

>>	Anonymous Sat Apr 13 22:53:55 2019 No.10556010 >>10553293 When you pick a gold ball, it came from one of two boxes. One of them has another gold ball, the other doesn't. 1/20 + 1/21 = 1/2. /thread

Anonymous Sun Apr 14 01:47:16 2019 No.10556325

prompt: pick a rand box, pick a rand ball

ok, say we do this 1500 times

500 times you pick from SS
500 times you pick from GS
500 times you pick from GG

SS: 0 favourable
GS: 250 favourable
GG: 500 favourable
--------------------------------------------
250+500= 750 favourable (prompt: "It's a gold ball")

500 of those 750 times you have locked into the GG box.
500/750 = 2/3

>>	Anonymous Sun Apr 14 01:57:42 2019 No.10556339 >>10553293 It’s 1/2, brainlets.

Anonymous Sun Apr 14 02:12:57 2019 No.10556362

okay everybody, here's what you faggots need to do, number the balls. g1, g2, g3, s1, s2, s3.
You've picked a gold ball but you don't know which gold ball it is all you know is now there are 2 less silver balls and you're left with g1, g2, and s1 as the remaining results for whats in the box. do the math. 2/3rds

>>	Anonymous Sun Apr 14 02:17:46 2019 No.10556365 >>10556362 samefag here. It's the exact same as the monty hall problem but in a different context

>>	Anonymous Sun Apr 14 02:21:45 2019 No.10556374 File: 15 KB, 334x697, math niggers.png [View same] [iqdb] [saucenao] [google]

>>	Anonymous Sun Apr 14 02:28:43 2019 No.10556385 >>10553293 Is the answer not 1/2? I'm a complete idiot who fails in college and I can do this.

>>	Anonymous Sun Apr 14 02:51:28 2019 No.10556423 >>10556365 >monty hall Yeah I was thinking the same thing, but forgot the name of it. Thanks.

>>	Anonymous Sun Apr 14 03:11:01 2019 No.10556457 >>10556385 1/2 is wrong. People who say 1/2 are either trolling or brainlets. I'm leaning moslty towards trolling,

>>	Anonymous Sun Apr 14 03:20:50 2019 No.10556473 >>10556457 I guess I’m a brainlet then. Once a gold ball is selected the only option left is to pick up a gold or silver ball. What am I missing?

>>	Anonymous Sun Apr 14 03:31:37 2019 No.10556487 >>10556473 There are three ways to hold a gold ball, 2 from the first box and one from the second. Hence, correct answer is 2/3. If you name the gold balls A, B and C it becomes much clearer.

Anonymous Sun Apr 14 03:35:21 2019 No.10556490

>>10556487

so for example,

You are holding A, you flip to B
You are holding B, you flip to A
You are holding C, you flip to silver.

These are the possible cases, 2 of them flips to a gold ball.

If this in still unclear to you imagine a case where the first box has infinite gold balls, and the question is similar. Clearly the answer is not 1/2 because you are much more likely to be holding a gold ball from the first box, rather than from the second.

Same logic applies to the 2/3 case.

>>	Anonymous Sun Apr 14 03:43:41 2019 No.10556507 >>10556473 Intuitively you think chances that you pick from 1 or 2 are equal but they aren’t. There is higher chance that you draw from box 1 and before you know what color the second ball is it’s all about chances.

>>	Anonymous Sun Apr 14 04:29:22 2019 No.10556565 >>10556473 >What am I missing 3 ways to pick 1st gold ball

>>	Anonymous Sun Apr 14 04:33:30 2019 No.10556572 >>10556490 >imagine a case where the first box has infinite gold balls no effect, once you've locked into the GG box, that's it - doesn't matter how many balls after two, since all you're ever going to pick from it is max two balls

>>	Anonymous Sun Apr 14 04:34:55 2019 No.10556576 >>10556362 >You've picked a gold ball you have picked a box. says so in the text of the problem. you pick a box and managed to pick a golden ball from it: 1/2. you pick a ball directly: 2/3. but this latter is a different problem.

>>	Anonymous Sun Apr 14 04:39:56 2019 No.10556587 >>10556457 you are trolling, of course.

>>	Anonymous Sun Apr 14 04:41:15 2019 No.10556590 >>10556587 gr8 b8 m8

>>	Anonymous Sun Apr 14 04:42:28 2019 No.10556591 >>10553293 This is a classic discrete maths problem it's called the pigeon-fucker equation.

>>	Anonymous Sun Apr 14 04:43:56 2019 No.10556593 >>10556487 >There are three ways to hold a gold ball, 2 from the first box and one from the second. irrelevant, because you don't start by picking a ball, you start by picking a box.

Anonymous Sun Apr 14 04:45:23 2019 No.10556594

>>10556365
>samefag here. It's the exact same as the monty hall problem but in a different context
absolutely different. in the monty hall problem you are given extra information about the other boxes/doors after you picked one. here all you know you picked at random a box which contained at least one golden ball, no new information is forthcoming.

>>	Anonymous Sun Apr 14 04:45:52 2019 No.10556596 >>10553293 I AM a retarded larp and I am one of the few who answers correctly. Who's side are you on?

>>	Anonymous Sun Apr 14 04:46:23 2019 No.10556599 the problem as stated can be rewritten as follows: you pick randomly a box which contains at least one golden ball... complete the description.

>>	Anonymous Sun Apr 14 04:47:25 2019 No.10556602 >>10556596 >math is a sport retard

>>	Anonymous Sun Apr 14 04:51:00 2019 No.10556607 >>10556602 I already said that. Whats your answer dummy?

>>	Anonymous Sun Apr 14 04:57:35 2019 No.10556614 >>10554991 yes, but it gives an out for the brainlet to say that picking the sole golden ball has zero probability wharrgarbl etc.

>>	Anonymous Sun Apr 14 04:57:38 2019 No.10556615 >>10556607 [math] \dfrac{P(\text{left})}{P(\text{left})+P(\text{middle})}=\frac{1}{1+\frac{1}{2}} [/math]

Anonymous Sun Apr 14 05:00:18 2019 No.10556620

>>10554604
>And? I buy a random national lottery. The news comes on and says that the winning ticket is in my state. Is the chance that my ticket is the winning ticket the same as when I bought it, or higher?

I don't know how this national lottery thing (I presume in the USA) works. when you say you buy a random national lottery, you mean you leave it to chance which state's national lottery you buy? or you assume you buy a national lottery ticket for the state you live in?

>>	Anonymous Sun Apr 14 05:01:01 2019 No.10556621 2/5?

Anonymous Sun Apr 14 05:01:47 2019 No.10556622

>>10555172
>You drew a gold ball, which box are you likely to be in?
you have either box with equal likelihood. you picked one of the two boxes from which a golden ball could be extracted. the timelines in which you picked a silver ball from the box that contains the sole golden ball is lost to the problem space.

>>	Anonymous Sun Apr 14 05:02:09 2019 No.10556623 >>10556615 I can't tell you if you're right unless you speak English. Did you forget that Im retarded?

>>	Anonymous Sun Apr 14 05:02:27 2019 No.10556624 >>10556620 >I don't know how this national lottery thing (I presume in the USA) works stop assuming overcomplicated bs then

>>	Anonymous Sun Apr 14 05:03:31 2019 No.10556628 >>10556623 no, because I am not

>>	Anonymous Sun Apr 14 05:05:39 2019 No.10556630 >>10556628 Well. You disregard the third box so, you may be on to something.

>>	Anonymous Sun Apr 14 05:06:43 2019 No.10556632 >>10553293 4chan is fucking retarded the answe is 100%. If u take out gold from the 1 gold 1 silver there’s no gold left in that box to take. Only box u can take gold from and expect to take another is the first which would give 100% chance.

>>	Anonymous Sun Apr 14 05:06:48 2019 No.10556633 >>10556624 so, you just realized your analogy makes no sense, and you are now about to disappear in a puff of indignation?

>>	Anonymous Sun Apr 14 05:07:06 2019 No.10556635 >>10556630 retard

>>	Anonymous Sun Apr 14 05:08:23 2019 No.10556636 >>10556633 >can't answer the question not surprised

>>	Anonymous Sun Apr 14 05:10:27 2019 No.10556638 >>10556636 why should I know how lottery works in your neck of woods? "random national lottery". what the fuck does that mean? and yeah, I cannot figure out what goes on in your brain when you act like this. maybe nothing noteworthy.

>>	Anonymous Sun Apr 14 05:11:41 2019 No.10556642 >>10556638 >imagine being this retarded

>>	Anonymous Sun Apr 14 05:13:01 2019 No.10556643 >>10556374 It’s 50%. 100% chance in first box. 0% chance in second box. Average of 100% and 0% is 50%. Easy

>>	Anonymous Sun Apr 14 05:13:43 2019 No.10556644 >>10556642 you win. go have a fap.

>>	Anonymous Sun Apr 14 05:13:48 2019 No.10556645 >>10556643 >>10556325

>>	Anonymous Sun Apr 14 05:14:51 2019 No.10556646 >>10556385 >>10556457 It’s 50%. Once you pull a gold you know you’ve either got the first box or second box. The chance of another gold from first box is 100%z the chance from second box is 0%. Learn to average.

>>	Anonymous Sun Apr 14 05:16:44 2019 No.10556649 >>10553293 It’s strange how people tell correct answer - 2/3, but for completely wrong reason. No, “it’s 2 out 3 ways of doing something” is not correct. Google conditional and a posteriori probabilities, Bayes’ Theorem.

>>	Anonymous Sun Apr 14 05:16:54 2019 No.10556650 >>10556646 >imagine being this retarded

>>	Anonymous Sun Apr 14 05:17:02 2019 No.10556651 >>10556635 >>10556643 At least this guy knows how to use words. If i was smart enough, I could explain how dumb you are.

>>	Anonymous Sun Apr 14 05:18:36 2019 No.10556652 >>10556651 so did hitler >If i was smart enough guess not

>>	Anonymous Sun Apr 14 05:19:32 2019 No.10556653 >>10556645 Let’s say we do it one time. Pick gold Cold be from GG could be from GS If from GG, 100% chance of another gold. If from GS, 0% chance of another gold. Two choices. The average is 50%z

>>	Anonymous Sun Apr 14 05:20:35 2019 No.10556654 >>10556653 >50% no that's your IQ

>>	Anonymous Sun Apr 14 05:23:10 2019 No.10556658 >>10556654 >can’t refute. Resort to other means You’re a Low Energy algorithm

>>	Anonymous Sun Apr 14 05:24:16 2019 No.10556660 >>10556652 Oh!! The old Hitler trick. Defeated, I shall go with my tail between my legs.

>>	Anonymous Sun Apr 14 05:32:07 2019 No.10556671 File: 68 KB, 640x581, E89130DB-FD20-4A83-99C9-0BE81E0DB367.jpg [View same] [iqdb] [saucenao] [google] >>10556654

>>	Anonymous Sun Apr 14 05:50:32 2019 No.10556695 >>10556660 >has a tail no surprise

>>	Anonymous Sun Apr 14 05:56:05 2019 No.10556713 >>10556658 the refute is the 1500x case, sorry, I can't make it look like a crayon drawing or a cartoon, ask if mommy has time to read it for you

>>	Anonymous Sun Apr 14 05:57:08 2019 No.10556715 >>10556671 >seething

Anonymous Sun Apr 14 06:11:31 2019 No.10556742

>>10555123
>just put the 3 boxes on the floor, and set the balls on front of them for each branch
but that's wrong. you choose a box first. then it turns out you chose a box which had at least one golden ball.
I am sure you are a malevolent retard or a troll, mind you, so this is not an answer to you, but let's just drive this difference for the - possibly hypothetical - readers who care.

>>	Anonymous Sun Apr 14 06:11:56 2019 No.10556743 >>10556695 I said I would leave but, I came back because I really enjoyed this. Now I must sleep the night away while exposing myself to outside predators. I keep my right eye open but it can't see a thing, its just a deterrent.

>>	Anonymous Sun Apr 14 06:14:36 2019 No.10556745 >>10556742 its called laying out all the probabilities, retard

>>	Anonymous Sun Apr 14 06:21:07 2019 No.10556759 >>10556745 that should be t. retard. but then, being a retard, you obviously would not know how to sign a post.

>>	Anonymous Sun Apr 14 06:26:09 2019 No.10556761 >>10553293 1/2. duh.

>>	Anonymous Sun Apr 14 06:36:49 2019 No.10556779 >>10556646 it's not 50 / 50, once you pull a gold you have two ways to turn to a gold and one way to turn to silver.

>>	Anonymous Sun Apr 14 08:28:01 2019 No.10556958 >>10556759 Impressive, did you quote that from your PhD?

Anonymous Sun Apr 14 09:17:45 2019 No.10557051
File: 785 KB, 720x960, D1554496-88AB-4C96-A6A8-35A311D2C1D4.png [View same] [iqdb] [saucenao] [google]

You can google “Bertrand’s Box Paradox” the 2/3 answer is on wikipedia with full context and solution.

You can run the code that has been created to simulate the problem to get the 2/3 answer.

You can use the various pictures and comparisions in this thread to try and wrap your brainlet mind around the idea that you have a greater chance of your hand being in the left box if you pull out a gold.

Or you can spam these threads to the end of time because a problem that seems simple wasn’t immediately intuitive to you.

Anonymous Sun Apr 14 09:40:04 2019 No.10557090

>>10556649
Well it depends what exactly that means. If you enumerate the events properly then you can say that. There are three boxes that are initially equally likely to be picked and one of the boxes has two equally likely colors. Therefore give two events to each box: Two events for gold from the gold box, 1 event for gold from the half gold box. This gives us 2 ways to get gold from the gold box out of 3 ways to get gold.

Now change the number of balls in the mixed box and this still works. If the mixed box has x gold balls and y silver balls then give each box x+y events and the answer will be correct regardless of how many gold balls are in the gold box.

>>	Anonymous Sun Apr 14 10:01:40 2019 No.10557144 >>10553293 1/2. Decent bait

>>	Anonymous Sun Apr 14 10:13:08 2019 No.10557172 File: 419 KB, 745x612, gold3.png [View same] [iqdb] [saucenao] [google]

>>	Anonymous Sun Apr 14 10:19:34 2019 No.10557191 File: 438 KB, 745x583, gold4.png [View same] [iqdb] [saucenao] [google]

>>	Anonymous Sun Apr 14 10:32:14 2019 No.10557223 >>10557172 >>10557191 1/2

>>	Anonymous Sun Apr 14 10:45:18 2019 No.10557255 >you put your hand into a box what could possibly go wrong?

>>	Anonymous Sun Apr 14 10:56:38 2019 No.10557280 >>10557223 halfwit

Anonymous Sun Apr 14 11:00:53 2019 No.10557290

>>10557280
Each box is initially equally likely to be chosen. There is only one possible outcome from each box, 1 outcome leads to pulling iron from the half iron box and 1 outcome leads to pulling iron from the all iron box. Therefore the answer is 1/2.

>>	Anonymous Sun Apr 14 11:04:26 2019 No.10557301 >>10557290 >Each box is initially equally likely to be chosen wrong

>>	Anonymous Sun Apr 14 11:07:43 2019 No.10557303 >>10557301 How so? It even says they are chosen randomly.

>>	Anonymous Sun Apr 14 11:10:45 2019 No.10557307 >>10557303 3 gold balls one of which is in your hand 2/3 chance it was from the GG box >>10556325

>>	Anonymous Sun Apr 14 11:14:32 2019 No.10557319 >>10557307 That works for the original problem, not the problems I'm replying to.

>>	Anonymous Sun Apr 14 11:16:55 2019 No.10557325 File: 1.15 MB, 4030x4096, tfw2smartformemes[1].jpg [View same] [iqdb] [saucenao] [google] >>10553293 1/2, it either happens or it doesn't.

>>	Anonymous Sun Apr 14 11:17:01 2019 No.10557327 >>10557319 hurrrrrrr durrrrrrrrr ok, idgaf, carry on

>>	Anonymous Sun Apr 14 11:21:53 2019 No.10557334 >>10557327 Nice job reading, retard.

>>	Anonymous Sun Apr 14 11:35:02 2019 No.10557345 What if each box was divided into two compartments, and you're instructed to take the ball in the left compartment first?

>>	Anonymous Sun Apr 14 11:37:25 2019 No.10557349 >>10557345 Doesn't matter.

>>	Anonymous Sun Apr 14 11:46:51 2019 No.10557363 >>10557349 But then there's only one possible choice for the Gold-Gold box, not two. If you label the balls as GL and GR, you can only ever pull GL on the first pick.

Anonymous Sun Apr 14 11:51:14 2019 No.10557375

>>10557363
There are still two events for the mixed box and since the boxes are chosen randomly this means the gold box must have the same probabilistic weight. So 1 event in which we pick gold from the mixed box and 2 events where we choose gold from the gold box. 2/3.

>>	Anonymous Sun Apr 14 12:01:59 2019 No.10557404 >>10553572 How can you call the person who simulated this event occurring 100,000,000 times and got 2/3 a brain dead retard? You should yell at reality. He just showed you results you literal nigger

>>	Anonymous Sun Apr 14 12:04:16 2019 No.10557407 >>10553729 Isn’t this magnifying the core concept of the OP?

>>	Anonymous Sun Apr 14 12:05:56 2019 No.10557412 >>10557407 >no silver balls nope

>>	Anonymous Sun Apr 14 12:10:42 2019 No.10557425 >>10556642 I feel like this post was about yourself. Why won’t you give him the information he asked for??

Anonymous Sun Apr 14 12:11:52 2019 No.10557429

>>10557375
See, that's interesting. A lot of people (including already in this thread) look at the diagram in >>10553325 and think the two choices for the all gold box are because there's two gold balls, one choice for each ball being picked first.

But it's not, the two (identical) cases there are just to match the two (distinct) cases for the mixed box.

If the all-gold box had a million balls, the probability would still be 2/3, if the mixed box only had two balls in it.

>>	Anonymous Sun Apr 14 12:22:38 2019 No.10557457 >>10557425 it's too obvious

>>	Anonymous Sun Apr 14 12:26:48 2019 No.10557466 >>10557429 >If the all-gold box had a million balls then the branches coming out of the leftbox could be a million balls long, but it doesn't matter since you only take two balls

>>	Anonymous Sun Apr 14 13:23:06 2019 No.10557571 >>10556713 What you don’t realize is it’s the percentage AFTER already picking a golden ball. You model starts off PRIOR to picking a ball. That’s why your model is busted.

>>	Anonymous Sun Apr 14 13:24:12 2019 No.10557573 >>10556742 Agreed see >>10557571

>>	Anonymous Sun Apr 14 13:29:34 2019 No.10557588 >>10557571 >your model is busted. ok bud https://en.wikipedia.org/wiki/Bertrand's_box_paradox

>>	Anonymous Sun Apr 14 14:07:05 2019 No.10557699 File: 1.29 MB, 3264x2448, DB39D3C5-C1A9-4B92-A6FB-D8C6896B2C14.jpg [View same] [iqdb] [saucenao] [google] >>10557588

>>	Anonymous Sun Apr 14 14:18:47 2019 No.10557732 Can someone explain to me intuitively why it is 2/3? I'm a brainlet and thought it was 1/2 because according to the question you are either in box one or two and there is absolutely no way to tell

Anonymous Sun Apr 14 14:28:04 2019 No.10557749

>>10557732
>you are either in box one or two
And why aren’t you in box 3? Well a golden ball indicates that probability that you’re in box 3 vanished but why doest it have to be equal probabilities between box 1 and 2? Now imagine 1st box has 1000 golden balls and 2nd has 999 silver and 1 golden. You’re holding a golden ball, do you believe you so lucky you did draw 1 golden ball out of 1000 from box 2. No, it’s more probable that you’ve chosen box 1.

Anonymous Sun Apr 14 14:56:53 2019 No.10557841

>>10557732
Let's say you do this 300 times. 100 times you will choose from the silver box and get silver. 100 times you will choose from the mixed box. 50 of those times you will choose the silver ball. The other 50 you will choose a gold ball and a silver will be left. 100 times you will choose from the gold box and a gold ball will remain. So there will be 100 times a gold ball is left in the box out of the 150 times you chose a gold ball initially. Thus the answer is 2/3.

>>	Anonymous Sun Apr 14 15:18:38 2019 No.10557913 >>10557749 This made sense thanks

Anonymous Sun Apr 14 17:39:26 2019 No.10558312

>>10557749
>You’re holding a golden ball, do you believe you so lucky you did draw 1 golden ball out of 1000 from box 2. No, it’s more probable that you’ve chosen box 1.
The scenario only states that a gold ball was already taken out. So you've got 999/999 or 1/999 chance, but the random probability comes from picking the box first. The only information is that you picked a box at random and that a gold was taken out. Think of it this way:
The picture states that you pick a box randomly. So you have an equal probability of picking any box. It then tells you that you picked a gold out for certain. This doesn't mean that the chance of picking the gold out matters, it means that box that was randomly picked couldn't have been the one on the right. Your choice of box was equal between the 3, but since the right box is disqualified it's really only equal between the left and middle one at 50/50 rather than 33/33/33 . From that point the probability of getting a 2nd gold out from the left is 100% and 0% from the middle, giving you overall 50% chance of having chosen the left or middle box. The chance of picking a gold out of the picked box is still 100% as per the picture, picking the box is the part that matters.

>>	Anonymous Sun Apr 14 17:40:34 2019 No.10558316 >>10558312 *999/999 or 0/999

Anonymous Sun Apr 14 20:25:16 2019 No.10558693

>>10558312
That's not how conditional probability works. It's not enough to acknowledge that you are in one of two possible scenarios, the point is to figure out which of those two is more likely. The 100% scenario is twice as likely as the 0% scenario.

>>	Anonymous Sun Apr 14 21:20:14 2019 No.10558790 >>10558693 >the point is to figure out which of those two is more likely. Not according to the OP pic

>>	Anonymous Sun Apr 14 21:23:58 2019 No.10558799 >>10558790 why are you still debating? This is not an open question, just read the submitted articles and shut up until you can post something that's not retarded. You've been refuted countless times already.

>>	Anonymous Sun Apr 14 21:28:36 2019 No.10558812 >>10558799 "What is the probability that the next ball will also be gold?" Not "What is the probability that you will pick a gold ball following a gold?".

>>	Anonymous Sun Apr 14 21:33:33 2019 No.10558826 File: 54 KB, 1483x604, 2.png [View same] [iqdb] [saucenao] [google] >>10557699

>>	Anonymous Sun Apr 14 21:37:38 2019 No.10558837 >>10558812 stupid is as stupid does

>>	Anonymous Sun Apr 14 21:40:34 2019 No.10558842 >>10558837 Semantics is as semantics does, that's how it's written

>>	Anonymous Sun Apr 14 21:42:05 2019 No.10558846 >>10556457 nah they brainlets. First time I saw this I was like them. "It's 50% you stupid niggers REEEEEEEE." Then some guy explained it and I got it. Guess some people just never learn.

Anonymous Sun Apr 14 21:46:13 2019 No.10558853

>>10553293
There are a total of six possible outcomes for the order of balls picked before the conditioning.
Box 1 is the double gold
Box 2 is the gold and silver
Box 3 is the double silver

Box 1 Gold 1 - Box 1 Gold 2
Box 1 Gold 2 - Box 1 Gold 1
Box 2 Gold 1 - Box 2 Silver 2
Box 2 Silver 2 - Box 2 Gold 1
Box 3 Silver 1 - Box 3 Silver 2
Box 3 Silver 2 - Box 3 Silver 1

As the question is conditioned to be "the first ball picked is gold", formally; Given that the first ball picked from a random box is gold, what is the probability that the next ball from the same box is gold?

the state space is now:
Box 1 Gold 1 - Box 1 Gold 2
Box 1 Gold 2 - Box 1 Gold 1
Box 2 Gold 1 - Box 2 Silver 2

Out of these three possible outcomes, the first two result in the second ball being gold.

The probability of picking a gold ball on the second pull is 2/3.

>>	Anonymous Sun Apr 14 21:55:30 2019 No.10558871 >>10556779 I don't know. You must draw from the same box. There is no all silver box and there is no "other" box. It could be asked: "what os the probability that you picked box one?"

>>	Anonymous Sun Apr 14 21:57:18 2019 No.10558879 >>10558853 Not being able to see into the box as you pick is equivalent to making Gold 1 and Gold 2 equivalent though. Replace the individual balls with a variable representing the number of each type of ball, and the 3 scenarios become 2.

Anonymous Sun Apr 14 22:10:49 2019 No.10558911
File: 9 KB, 762x599, 8f9d977d8a[1].png [View same] [iqdb] [saucenao] [google]

>>10558879
Not at all; just because two balls are identical doesn't mean its the same outcome. Choosing a box first is actually a red herring meant to confuse you, and it can be replaced by a model describing sequences like I did.

Lets ignore the "pull a second ball" part of the question and slightly modify it so I can explain this concept.
You have two boxes. One box has 2 gold balls. The second box had 1 gold ball and one silver ball.
You randomly choose a box (this is equivalent to the boxes being opaque) and pull out a gold ball.
What is the probability that you pulled the ball from box 1?

Formally, this is "What is the probability that you chose box 1, given you pulled a gold ball?"
Event A is you choose box 1 (and thus event A' being you choosing box 2)
Event B is you choosing a gold ball (and thus event B' being you choosing a silver ball)

P(A|B) = P(A intersection B) / P(B) = P(A ^ B) / P(B)

As we can see in the shitty probability tree to the right, you'll notice that the probability of A intersection B (choosing a gold ball from box 1) is 0.5
The probability of choosing a gold ball, however, is the probability of choosing a gold ball from box 1 plus the probability of choosing a gold ball from box 2, 0.5 + 0.25, or 0.75.

P(A ^ B) / P(B) = 0.5 / 0.75 = 2/3

It's left to the reader to show that this example is actually identical to the OP's original problem.

>>	Anonymous Sun Apr 14 22:13:03 2019 No.10558918 >>10558911 Guess I'm just gonna have to disagree with your interpretations of the problem then, 'cause I fully understand what you're saying but I just disagree on the specifics of the what the question is asking.

>>	Anonymous Sun Apr 14 22:16:06 2019 No.10558925 >>10558918 Which is exactly why I restated the problem in the way I did; notice how there's no statement of "gold 1" or "gold 2" in that probability tree; only that there's a probability of 1 of pulling gold from that box.

Anonymous Sun Apr 14 22:17:14 2019 No.10558931

So the answer is clearly 2/3 based on a very straightforward tree diagram that many have already done. This doesn't interest me. I'm more curious about what specifically makes 50% such a compelling answer. This is the best argument I could come up with, feel free to pull it apart, but also think whether or not this is the thought process that is leading folks to the wrong answer:

Because the box selection is uniformly random, and every box obviously has a 'first draw', every ball has an equal likelihood of being the first draw. Therefore, the odds of a gold ball being the first drawn is 50%, P(G1)=0.5 . In the subset of cases where a gold ball is the first drawn. We're looking for P(G2|G1). We can partition this along our choice of box (which was equally likely) so we obtain P(G2|G1) = P(G2|B1|G1)P(B1|G1) + P(G2|B2|G1)P(B2|G1). The box selections are equally likely so they have probability 0.5, but g2 given b2 is impossible and g2 given b1 is guaranteed, so we find P(G2|G1)=0.5

Does this seem about right? The issue is assuming P(B1|G1) is the same as b2 given g1, but I suspect this is how the mistake seems most reasonable to make.

>>	Anonymous Sun Apr 14 22:24:57 2019 No.10558949 its 100% because you will expect it to be gold therefore it is.

Anonymous Sun Apr 14 22:25:38 2019 No.10558951

>>10558925
P(Gold | Box 1 selected) = 100%, P(Gold | Box 2 selected) = 0%, P(Gold) = 50% is my reasoning. My interpretation assumes that the probability of getting a gold out initially is set at 100% based on picking one of the boxes at random, at 50% each since that's how I interpret it. Doesn't make sense to me that it'd tell me that I picked a gold but that it also wants me to include the probability of not picking a gold first.
>>10558931
My reasoning is that I assume the question is telling me I've already picked a gold which would make the 50% chance of getting a silver first moot, and instead I just subtract a gold ball from each. In the circumstance that I'd picked a ball from one of them and I was guaranteed it to be gold, either has an equal chance since the chance of getting a silver is 0 (based on how I read it which must clearly be wrong).

Anonymous Sun Apr 14 22:28:55 2019 No.10558962

>>10558951
Don't worry anon, the error you're making is very subtle. While it's true that all the boxes are equally likely BEFORE the conditioning, once we've introduced the condition of having chosen a gold ball first, the probabilities of choosing box 1 and choosing box 2 are no long equal.

>>	Anonymous Sun Apr 14 22:39:58 2019 No.10558987 >>10558842 run Forrest run

>>	Anonymous Sun Apr 14 23:05:29 2019 No.10559046 >>10558911 Why would silver even be considered? Or even the fact that there is more than ONE out of two boxes? I know im dumber than /sci/ but i see an unquestionable half a chance.

Anonymous Sun Apr 14 23:09:12 2019 No.10559050

>>10559046
In the original question, because there's still a chance you might pull silver for the second ball.

In the formal definition of conditional probability, you still need to define the definition of pulling gold first to re-scale the probability space, and besides just being the sum of all outcomes where gold was pulled first, it's also equal to 1 minus the sum of all outcomes where silver was pulled first.

Anonymous Mon Apr 15 06:19:16 2019 No.10559728

>>10558790
Yes, according to OP pic, because if and only if you chose the all-gold box, then your next ball will also be gold, and so the probability of that is also the probability of getting another gold ball. The trick is that getting a gold ball actually gives you more information than you realise.

I'll use a different example to demonstrate. Suppose you have two six-sided dice. One of the dice is loaded and always comes up six. The other die is perfectly fair. You select a die at random and roll a six. Now, the odds of that are pretty good. BUT the odds of having rolled a six with the fair die are substantially lower than the odds of having rolled it with the loaded die. Now, you roll the same die a second time and get a six. And a third time, and get a six. At this point, it is very probable that you selected the loaded die rather than the fair one, and so the odds that your next roll will also be a six are very high. Even though nothing you did affected your initial choice 50/50 between the two dice, even though you cannot affect past rolls any more, and even though every roll is independent, you can still use the information gained from the rolls to draw a conclusion as to which die you are probably using and hence, what the next result is likely to be. After three sixes, you wouldn't say that it's still 50% likely that you selected the fair die.

How does this relate to the boxes? Well, you selected a box and randomly pulled out a gold ball. There are two boxes from which it could have come. BUT the odds of having taken it from the GS box are substantially lower than the odds of having taken it from the GG box. In fact, it is half as likely. The GG box is essentially our loaded die. And because you now know that it's more likely that we have selected the loaded "die", it is also more likely that the next result will be favourable.

>>	Anonymous Mon Apr 15 06:38:20 2019 No.10559755 >making huge posts trying to justify wrong result 1/2 Retard, it’s basic problem learned in the first course of probability theory

>>	Anonymous Mon Apr 15 06:41:11 2019 No.10559762 >>10559728 Good post, but don’t expect from 1/2 brainlets to understand

>>	Anonymous Mon Apr 15 06:48:43 2019 No.10559772 The majority of alien life in the universe look like human just because we exist. That is probability for you.

>>	Anonymous Mon Apr 15 06:52:40 2019 No.10559782 File: 84 KB, 1000x686, ignore_bait.jpg [View same] [iqdb] [saucenao] [google] >ITT: faggot OP baits faggots by pretending not to understand conditional probability, faggots who don't understand conditional probability argue with one another

Anonymous Mon Apr 15 06:58:36 2019 No.10559792

P(second ball is gold|first ball is gold) = P(second ball is gold AND first ball is gold) / P(first ball is gold)
P(second ball is gold AND first ball is gold) = 1/3, because it happens iff you choose two gold ball box
P(first ball is gold) = 1/2, as you need to pick one of the three gold balls out of 6 total, 3/6 = 1/2
Finally,
P(second ball is gold | first ball is gold) = (1/3)/(1/2)=2/3
Tldr: get off this board retards

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