Is it possible to write a program that can solve a problem the

>>58513919
Yes

>>58513919
Count the number of rows in a very long document

>>58513982
A programmer CAN solve that, it just takes the programmer more time than it does for the computer program.

File: dyson.jpg (134KB, 850x1146px) Image search: [Google]

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>>58513919
Stand back Dyson

>>58513919
50%

>>58514042
50% is the most common /obvious answer. That the question is being asked at all should suggest to you it's not correct. Which it is not.

2/3

>>58513987
Not if it's 20 billion lines long. Which could be just a few hundred gigabytes.

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387KB, 1280x720px

>>58514096
That's an extremely autistic thing to think, I'd never even considered the lifespan of the programmer. I'm actually impressed, that's a very correct and sterile way to think about the matter.

Probabilities are an illusion

File: ss (2017-01-16 at 16.50.27).png (25KB, 415x153px) Image search: [Google]

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>>58514073
The situation presented is that you have taken one gold ball, so as seen in the image.
If you now take the second one from the same box, it's either going to be gold, or silver. That depends on which box you are in.
I can see where the 2/3 opinion comes from, but that would imply that you could also take from the other boxes, which you can't.

Does the programmer have infinite time? If so, no. Humans are Turing complete and can do any program a computer can given infinite time and resources.

Given reasonable human restrictions? No fucking shit. For example, I am not going to manually calculate millions of solutions to different gravitational fields but a computer can spit out those results in seconds.

>>58514116
>>58514118
https://en.wikipedia.org/wiki/Bayes%27_rule

You're mistaken.

>>58513919
yes.
modern AIs are already better than people in some tasks

>>58513919
Isn't that why you write computer programs in the first place?

>>58513919
So either you pick the GG or GS box to get a gold to begin with
If you picked the GG box it's 100%
If you picked the GS box it's 0%
So since you have a 50% chance of having picked the either box, it's now 50% because you have to have picked GG

>>58514175
You're more likely to have picked the GG box than the GS box though.

>>58514183
Why, if you can't look into the boxes?
Two boxes (third one is already ruled out) and closed. They look exactly the same fromt he outside. Choose one by random. You have the same chance to pick either GG or GS.

>>58514118
Yeah but consider the probability of picking a gold ball in the first place. You've got a 1/3 chance overall of picking a gold ball from the first all-gold bin, and only a 1/3 * 1/2 = 1/6 chance of picking the gold ball from the middle bin. Therefore, if the first ball you pick is gold, you're actually more likely to have selected the first bin than the middle bin, and this additional information means you've actually got a 1/3 / (1/2) = 2/3 chance of having selected the all-gold bin.

For a more extreme example, imagine there are only two bins, each with 100 balls. Bin 1 has 99 gray balls and 1 gold ball, and bin 2 has 100 gold balls. Intuitively, if you randomly take a ball from a bin, and it happens to be gold, you know that you're more likely to have selected it from bin 2 than bin 1. This question is the same scenario, but with much smaller numbers, making it less obvious.

It's not opinion, it's mathematical fact.

>>58514162
Mind = blown.

>>58514210
>Yeah but consider the probability of picking a gold ball in the first place.
How does that even matter if you have already picked it in the past? The probability of picking a gold ball in the first place is 100%.
You do not have a magical hand that can only pick gold balls (in this case it would be more attracted to the GG one). The presented case is a single-time scenario where you have chosen at random and it just so happens to be a gold one. It could have been silver as well, the boxes look the same from the outside after all. Maybe that happens on another draw, but in this presented case, it already happened and it's not guaranteed to happen again. Therefore you can discard all the information of what happened in the past. Now you only have 2 boxes and you have a gold ball in your hand, the next draw will depend on which box it is, which has the same chance for all boxes, 50/50

>>58514042
>>58514073
>>58514074
It's 1/3d since there is only one box with two gold balls.

>>58514269
>How does that even matter if you have already picked it in the past?
The laws of probability are impartial to matters of past and future. All that matters is whether you know the outcome or not. In this case, you know that you got a gold ball, but you don't know which bin you originally selected. As a result, you're more likely to have picked the GG box.

The specific principle at work here is called conditional probability (google it). The overall probability of picking a gold ball first is 1/3 + 1/6 = 1/2 by Bayes law, and the probability of it picking it specifically from the first box is 1/3, so you get 1/3 / (1/2) as the probabilty of having picked the GG box, provided that you picked a gold ball first.

Probability is not an inherently intuitive discipline, unfortunately. You can't just feel it out.

>>58514269

This anon is right, but could also be because the scenario was poorly written.
It states you already picked a ball and it was gold, and that's the current state. How this was written makes me assume it only takes into account the probabilities from here on out. So the SS box is already discarded, and this box can only be GS or GG, so 50%.

>>58513919
p=np
solve pls

>>58514397
>>58514269
>This anon is right, but could also be because the scenario was poorly written.
No, it's not the scenario, it's your brains.
>this box can only be GS or GG, so 50%.
You're making a critical mistake here by assuming that it's exactly 50%. You can't just make up values arbitrarily like that.

Here's the original problem that this question is based on:
https://en.wikipedia.org/wiki/Monty_Hall_problem

>I have never studied turing completeness: the post

>>58514451
>>58514386
How would you go about making a simulation then?
Step 1: Choose one box at random, each with the same chance
Step 2: If Silver, discard the whole experiment, since it is not the presented case
^ The above would already distort the result, because then you'd have the magic hand that could only pick gold balls and would be more attracted to the GG box, since GS has a 50/50% of discarding simulation and thus giving it only half the chance to appear, while GG has a 0% chance.

If you instead start checking the chance straight from the beginning, then it is 1/3, because only one out of 3 boxes can be the correct one and they have the same chance. What happens after that does not change the content of the box you chose.

>>58514410
P=0 or N=1

>>58514542
A simulation with 500 trials gives me about 2/3 as the answer every single time. Here's the code (it's python 2).

import copy
import random

startBoxes = [['G','G'],['G','S'],['S','S']]

# pick a ball
def pickBox(boxes):
  idx = random.randint(0,len(boxes) - 1)
  return boxes[idx]

# pick a ball from the given box and delete it
def pickBall(box):
  idx = random.randint(0,len(box) - 1)
  res = box[idx]
  del box[idx]
  return res

# the number of times a gold ball is picked first
goldFirst = 0
# the number of times a gold ball is picked the second time, after already picking a gold ball the first time
goldSecond = 0

def doTrial(num):
  global goldFirst
  global goldSecond
  print "Trial #" + str(num)
  boxes = copy.deepcopy(startBoxes)
  box = pickBox(boxes)
  ball1 = pickBall(box)
  if ball1 == 'G':
    print "Picked a gold ball first"
    goldFirst += 1
    ball2 = pickBall(box)
    if ball2 == 'G':
      print "Also picked a gold ball second"
      goldSecond += 1
  else:
    print "Did not pick a gold ball"

numTrials = 500
def main():
  for i in range(0,numTrials):
    doTrial(i)
  print "Final Statistics"
  print "Number of times gold ball picked first: " + str(goldFirst) + " (%" + str(100 * goldFirst / numTrials) + ")"
  print "Number of times two gold balls picked: " + str(goldSecond) + " (%" + str(100 * goldSecond / numTrials) + ")"
  print "Probability of picking a second gold ball after picking a first: %" + str(100 * goldSecond / goldFirst)

>>58513919
Do you have a problem where you know some of the possible input and output values of function f but not how function f is implemented? That's what machine learning is good at.

>>58514207
You only had a 50% chance of picking the gold ball if you chose the GS box, but a 100% chance of drawing a gold ball from the GG box.

File: big_42276_a_man_a_wolf_a_goat_and_a_cabbage_1.jpg (32KB, 548x387px) Image search: [Google]

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A man, a wolf, and a goat all on one side of the river have access to 3 boxes with gold balls in them and are dicking about shoving their hands/paws/hooves into them, while on the other side of the river is a house with 3 lights connected to 3 switches not in the same room. How can you determine which gold ball which animal/person will hurl at each switch from across the river?

>Can't get across the river with the goat who wants to throw a gold ball at a light switch because he will be eaten by the wolf who is too busy trying to also get across the river to throw a silver ball at a switch

>Electrician troll-truck explodes in the distance because the same faggot who wired up this house to not let any pipes/electricity touch each other and also have 3 switches in the same room of their lights because underground works in 2D also wired up his car that way

>Wolf makes it halfway across the river with a silver ball and then starts crying because the goat didn't even taste that good and the only other food (the cabbage) is back on the other side of the river

>electrician drives up to pick up cabbage but brakes aren't wired properly so the truck drives into the water making a perfect bridge upon which the man walks across and slaps the shit out of the wolf and goat who are hurling silver and gold balls at light switches that were wired improperly

>House becomes a gameshow contestant offering a prize behind 3 doors. Opens 1 of its doors for the goat, wolf, man etc who enter and simply stand there and continue being slapped and also crying. House offers to let them switch the other door.

>They decline and "stick" with the third door which opens to show the dead electrician and a chest full of gold and silver balls and cabbages. Goat immediately throws up and says "op is a faggot"

>>58514901
This is meaningless since the requirement for the problem is picking up a gold ball first.
There is only 2 boxes with gold balls
One has two of them the other has one
So you either pick another gold or not
50%

>>58514901
You calculated the chance of pulling two gold balls and infinite retries if you didn't pick a gold ball first. The scenario asks the chance of pulling a gold ball when you can only choose between a box with either a gold ball or silverball.

>>58515465
Goddamn you're stupid

>>58515634
There's no difference between what you're saying.

In his scenario he tossed out the cases where he drew a silver first. That's no different from starting with drawing a gold.

>>58513919

Fuck you, OP.

I remeber having endless discussions about this pic on /sci/.

Having a background in statistics:
The answer is 2/3.

Why?
Because there's three possible elementary events:
-picking goldball nr. 1
-picking goldball nr. 2
-picking goldball nr. 3

Yes, from a statistical point of view there's a difference wether you grab the first or second gold ball form the first box. The reson is that there is 6 elementary events, since there are are 6 balls, no matter how they are partitioned.

No, I won't discuss that any further. Fuck you if you don't believe me.


Now to your question:
Of course! Can you see the differnce between the RGB colors "AACC10" and "AACC11"? A simple MS Paint clone can display the result.

>>58515852
There are not there elementary events, and it doesn't matter which gold ball you pick from the first box because you're reaching back into the same box
It's 50%

Of course. Someone who doesn't understand the Monty Hall problem can model it and figure out the probability after many repeated trials.

File: 1484104286361.webm (693KB, 1066x600px) Image search: [Google]

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>>58515662
It's the phrasing. The third box is never a consideration, the base case is 2 boxes, hence 50%.

>>58513919
This is the fucking Monty Hall problem. 2/3 only if you switch boxes...lol

>>58516538
Actually it's the opposite.

>>58514313
This guy maths

>>58513919
Halteproblem
a program A can easily determine wether a given program B will terminate in finite steps
a programmer will have trouble with that and call me a faggot for lying on a chinese checker board

>>58514138
>>58514210
This is very simple. Right from the start we know that we didn't pick SS, so we can immediately forget that it ever existed. Either we picked GG, or GS. There is a one in two chance that we picked GG and therefore have another gold ball available.

>>58516491
wrong

>>58514313
But you already got a gold ball, so the box has to be one of the two that have at least one gold ball.

Yes, very much so. Neural nets are a prime example of this. How would you create an algorithm that takes an image and returns a string detailing what is found in that image? No programmer can solve that for images of a certain complexity. I'd certainly love to see you try, and I'd also love to see you solve the problem of doing it more efficiently than a neural net.

>>58516450
This isn't a monty hall situation though. There are two boxes with gold balls in them. One contains a silver ball. In either case, you have removed one. There is one ball remaining in either potential box. One is silver, one is gold. 50 % of the possible boxes contain a golden ball.

Monty hall relies on the fact that you are initially more likely to pick an unfavorable outcome, and as such, switching is more likely to garner you a win.

>>58514451
It's ironic that the Monty Hall problem proves you wrong and explains exactly where the 50% comes from.

>>58513919
It's 75%.

If you are in the box with two golden balls, there's 100% chance that the second ball will be golden.

But if you're in the box with one silver ball, there's 50% chance that you'll pick up the same golden ball again.

Thus (100+50)/2

>>58516846
You're assuming that one would put the ball back into the box. At no point is this mentioned.

>>58516846
why did you put the gold ball back in the box

>>58513919
A program is just technical documentation on how to solve a certain problem written for other people to read and use to solve that problem. The act of programming is the act of exploring a problem space. If you have written a program to solve a problem, you have learned the steps needed to solve the problem and can/have solved it if you can prove that it functions correctly.

>>58516886
Because it's not mine and stealing is wrong.

>>58514559
ITT: I never passes Elementary Statistics

>>58515465
>>58516491
>this goddamn reductionism
Why are people so stupid these days?

You can't just reduce to two boxes, you must apply the problem as stated.

>>58516911
fucking lolbertarians

>>58513919
Yes.

Also, 2/3rds.

File: 1484580972445.jpg (114KB, 701x576px) Image search: [Google]

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>>58513919

I'll break it down for you guys. Look at the picture:

1. If you have a golden ball from the left box, the next one will definitely be golden.
2. If you have the single golden ball from the middle box, the next one will definitely be silver.

So the naive approach would be: "hey, I already picked a golden ball, so it's 50/50".

The problem is that you don't know WHICH golden ball you have. The chance that you picked one of left box is 10/30 = 1/3. (That's what we expect: "the chance to get one ball from the left box is 1/3").

But the chance that you have a golden ball and grabbed the single golden ball from the middle is only 1/30. If we let 10 million people grab a ball from the a random box and rule out every one that grabbed a silver ball, much more people will have grabbed a ball from the left box.

And how much exactly?
Out of 11 golden balls only 1 is from the box in the middle. The situation to have just grabbed this ball is much more unlikely.

So the answer for the question on my picture would be "10/11 chance that the next ball is golden."

Or for the OP:
"2/3 chance that the next ball wuill also be golden".

File: 5r59j6k9.png (223KB, 701x576px) Image search: [Google]

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Going to have to explain it to you IQ89 retard brainlets.

t. IQ140

>>58516697
Did you drop out of middle school?

>>58513919
This looks like the Bertrand's box paradox, but there is an additional trap.
>next ball you rake from THE SAME BOX will lso be gold?
50% in this case
If you took the one from GG, then you have 100% to take another gold ball
If you took the one from GS, then you have 0% to take another gold ball again
You completely overlook the SS one since it's not possible you are taking another ball from that box.

It would be 2/3 if you weren't forced to pick from the same box

>>58516697
correct

>>58516719
this isn't the same as monty hall at all and the odds are 50%

in monty hall your initial choice is random and then the host's choice cannot be random because he won't ever reveal the goat

in this problem your initial choice is not random across all 3 choices, you are required to pick a box with a golden ball first and the only question is which box you selected from the 2 with golden balls

if you had an actual chance to select the all silver box the odds would be 2/3 but the problem precludes that

>>58513919
Computers have found prime numbers that a programmer wouldn't live long enough to write out.

>>58517070
this right here. your first golden ball is more likely to come from the GG box so the chance is higher than 50%.

File: fixed.png (263KB, 1500x1613px) Image search: [Google]

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>>58517069

>>58514386
What a mega fucking troll

>>58517281
No, the first ball being gold was a given, not a factor

>>58516697

I honestly don't know why people are arguing against this. Whoever created this 'puzzle' was a very clever troll. It's a 1-in-2 choice that's been ever so slightly dressed up as a 1-in-3 choice.

>>58517094
>The same box
faggot

File: 1484594827279.jpg (126KB, 842x720px) Image search: [Google]

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>>58517069
FTFY

>>58513919
5/6

>>58517183

This is basically the second part of the Monty Hall problem.

The whole point of the Monty Hall problem is that if you have a 1-in-3 chance of picking correctly, then throw out one of those options, you now have a 1-in-2 chance. 50% is a better chance than 33%, thus if you survive the first selection round, you should switch your choice, since the odds become 50%.

This problem is quite literally the second part of that. 50%.

>>58517154

No you you wrong.

The trap is not that we are forced to grab from the same box. The trap is hidden in the sentence "you just grabbed a golden ball".

This does NOT mean: "Take one golden ball from a box of your choice."

But it means: "grab a (silver or golden) ball until it is golden" (of course you mix everything after one attempt, so it' always a start without any further knowledge). We must pay attention on the odds that we picked the one golden ball from the middle.

>>58517318

> the first ball being gold was a given, not a factor

The first ball was give, but WHICH ball?
There are three possible golden balls.

>>58517436 here
Had time to think about it and read some replies here. I considered that the question is vague also.

With replacement: if you take one from the box with two golds, there's a 100% chance of choosing a gold on the second pick. Two times, one for each ball. If you choose the other it's 50% obviously. So (100+100+50)/300 = 5/6.

Without replacement: the two gold box is effectively the same as before, the one gold one silver box is now 0%. So (100+100+0)/300 = 2/3.

I studied maths at uni, I did units on probability theory, and now I'm a maths teacher, I teach this shit on a daily basis.

>>58517572
spread across 2 boxes
once "gold was selected first" became a condition the box with 2 silver balls is now irrelevant

this is nothing like monty hall because in monty hall your initial choice has a chance of being wrong, in this scenario you are required to pick between 2 gold choices at random

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>>58514114
But when you think about it, a programmer COULD probably solve even the most difficult yet answerable questions there are given enough time to do so.
Basically the only way for a program to solve something a programmer absolutely is unable to solve, it would have to solve something... I don't even know what.
Even if it solved something we thought was utterly impossible to solve, like finding out the last number of pi, it would probably be possible for humans as well given enough time.

Thus the question posed in OP pretty much can only be answered in a meaningful way if we factor in the limited amount of time humans have in their hands.

>>58517614

This guy explained the relation to monty hall pretty good (trips don't lie):
>>58517444

You see, Bayes' theorem is a bitch. It look like you start from zero, but you have to consider the likelihood of the start situation.

Example:
"One day of the year it snows, the rest of the year it's sunny. If it snows, there's a 60% probability that you find a penny. If it's sunny, there's a 50% probability that you find a penny."

Now when you find a penny, you might think "hey, it's 60% probability to find a penny when it snows, but only 50% when it's sunny."

But now imagine someone calling you and telling you he just found a penny. The probability of a sunny day is just so much higher, that that the lousy 10% more of a snowy day don't really matter.

If you let a guy walk down the street for some days, it's not very likely that he already had the snowy day, but it's very likely that he already found a penny.

let goldfinder () x =
    //printfn "Run: %d" x
    let urns = [[ "gold"; "gold" ];[ "gold"; "silver" ];[ "silver"; "silver" ]]
    let rnd = System.Random()
    let nextUrn = rnd.Next(3)
    //printfn "Urn: %d" nextUrn
    let urn = urns.[nextUrn]
    let nextBall = rnd.Next(2)
    let ball = urn.[nextBall]
    //printfn "First ball: %s" ball
    match ball with
    | "silver" -> "All silver"
    | _ -> 
            let lastBall = urn.[1-nextBall]
            //printfn "Second ball: %s" lastBall
            match lastBall with
            | "gold" -> "Gold twice"
            | _ -> "Gold once"

let ok =
    let runs = seq { 0 .. 99999 }
    runs
    |> Seq.map (fun x -> goldfinder() x)

let result = 
    ok
    |> Seq.countBy id
    |> Seq.toList

printfn "%A" result

Is this how it's supposed to work? F#

>>58514386
I just got back from class were we learned Bayes' rule. This is quite a timely demonstration of it's usefulness.

>>58517743
you don't have to consider the likelihood of the start situation because the question precludes it

in monty hall your initial choice is made at random and has a chance to be wrong

in this problem your choice was not made at random across all 3 boxes because the question told you that you were guaranteed to pull a gold ball

the question doesn't ask the average probability of pulling 2 gold if you ran 100 trials selecting a random box

it says "if you pulled out a gold ball as your first choice what is the chance the other ball will be gold" and you had a 50% chance of being right if you were guaranteed to pick gold

this isn't like monty hall because the choice isn't actually random, the monty hall equivalent would be the host removing a goat door before you even make your choice

>>58517743

So let's use Bayes' theorem:

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


P("the second ball is golden")
= P("you picked from the left box" | "you picked a golden ball")
= P("you picked a golden ball" | "you picked from the left box") * P("you picked from the left box") / P("you picked a golden ball")


We see:
1. P("you picked a golden ball" | "you picked from the left box") = 1
--> after you picked from the left box, the ball is AWLAYS golden

2. P("you picked from the left box") = 1/3
--> we don't care about ball colors here, only about boxes

3. P("you picked a golden ball") = 1/2
--> There are 3 golden and 3 silver balls in this game


P("the second ball is golden") = P("you just picked a golden ball" | "you picked from the left box")

P("the second ball is golden") = 1 * (1/3) / (1/2) = 2/3

Second last line must be:
>P("the second ball is golden") = P("you picked from the left box" | "you picked a golden ball")


>>58517978

And now, for the lulz, let's rule out the right box (with only silver balls) for one moment. The result stays the same:

P("the second ball is golden")
= P("you picked from the left box" | "you picked a golden ball")
= P("you picked a golden ball" | "you picked from the left box") * P("you picked from the left box") / P("you picked a golden ball")


We see:
1. P("you picked a golden ball" | "you picked from the left box") = 1
--> after you picked from the left box, the ball is AWLAYS golden

2. P("you picked from the left box") = 1/2
--> we have only two boxes now

3. P("you picked a golden ball") = 3/4
--> Since we ignore the right box, there are 3 golden and 1 silver balls in this game


P("the second ball is golden") = 1 * (1/2) / (3/4) = 4 / 2*3 = 4 / 6 = 2 / 3

>>58514116
Probabilities, as we understand them right now, are hard wired into the physical laws themselves

>>58517860


> and you had a 50% chance of being right if you were guaranteed to pick gold

And that's where you are wrong.
The question is not: "ignore the chest to the right, take one golden ball from the left and one golden ball from the middle, what will be the next ball be like?"

You just ignore this sentences:
"You pick a ball at random. It is golden."

There are so many explanations in this thread, just pick one and read it carefully..

>>58514451
The Monty Hall problem is similar but different.
You are more likely to pick a door without a donkey (2/3) and the presenter will always open a door with a donkey, which means that by switching you are more likely to get the car.

In this case it is outright stated that you pull a gold ball on the first draw, which means the double silver box ceases to exist.

The chance of pulling a gold ball on the first draw is far more complex, but the chance of pulling a gold ball on the second draw from the same box is just 50%.

To make this into a Monty Hall problem it would need to be more like this.

You pull a ball at random, it's Golden. The presenter takes away the double silver box and then gives you the choice to pick from the same box or pick from the other box. Which do you pick from?
Then you are statistically more likely to get a second golden ball by picking from the other box (okay maybe they take one of the balls out of it as well so they both only have one ball left too)

>>58513919
If I know the setup is 2xg, 2xs, 1xg+1xs then the chances are one in three

open Num;;

let one_draw total success =
  let ball1, ball2 =
    match Random.int 3 with
    | 0 -> true, true
    | 1 -> true, false
    | 2 -> false, false
    | _ -> assert false in
  let drawn, next =
    match Random.int 2 with
    | 0 -> ball1, ball2
    | 1 -> ball2, ball1
    | _ -> assert false in
  if drawn then
    let total = total +/ Int 1 in
    let success =
      if next then
        success +/ Int 1
      else
        success in
    total, success
  else
    total, success
;;

let rec loop total success =
  let total, success = one_draw total success in
  if mod_num total (Int 1_000_000) = Int 0 then
    begin
      Printf.printf
        "%.2f%%" (100.0 *. float_of_num success /. float_of_num total);
      print_newline ()
    end;
  loop total success
;;

let () =
  Random.self_init ();
  loop (Int 0) (Int 0)
;;

It converges toward 2/3.

>>58515852
>>58516988
>>58517028
>>58517070
>>58517094
>>58517270
>>58517281
>>58517405
>>58517978
>>58518040


This.
It's 2/3.

Fuck this thread.

>>58518134
The question says, you pick a box at random. Then you take a ball from it and it is gold, so there's one ball left in there. Either gold or silver.

>>58518134
You pick a box at random, you draw a silver ball, what is the chance both balls in the box are gold?

>>58518271
4/3

>>58518269

The important point is the randomness of the box and the pick of the ball.

If you randomly pick a box at random there's 1/3 chance you pick the left box. Now you still have 1/2 chance for the first golden ball and and 1/2 for the second. Why the fuck would someone think it doesn't matter which golden ball you draw? You can't just ecahnge them and threat two balls as one.


>>58518271

0%
No matter which silver ball I pick, there's no way the next ball in the box can add up to "two golden balls".

>>58518360
Ah, but your initial choice was random so the answer is 1/3rd sorry.

>>58513919
66 %

>>58518422
i mean 40%

>>58513919
3 gold, 3 silver
remove 1 gold
you have 2 gold 3 silver
the probability is 2/5 that it's gold

Has anyone figured out P vs NP yet?

>>58513919
50%

the question removes the 3rd box so now you're left with two boxes

the new problem is a) 100% chance or b) 0% chance so the 1/2 probability comes from the boxes and not the ball itself

>>58518452
Why are you including the box with only silver balls in it if you know that you did not pick from that one?

>>58518488
Yes.

>>58518525
Good.

>>58518394

Wow, you must be pretty butthurt that you make up such analogies (no pun intended).


>>58518452

It's more complicated than that.
After the first pick you know that you didn't pick one of the balls from the right box. You either picked one of the left box or the one in the middle.

Since the first pick was completely random, there's a 1/6 probability for each ball.

print("[g g][g s][s s]")

print("These are three boxes. They contain golden (g) and silver (s) balls.")

print("Do you understand? (y/n)")
answer1 = input()

if answer1 == ("y"):
    print("You pick a golden ball from one of the three boxes.")

    print("Understood this as well? (y/n)")
    answer2 = input()

    if answer2 == ("y"):

        print("What is the probability that the next ball you take from the same box will also be gold? Answer with a fraction using a slash.")

        answer3 = input()

        if answer3 == ("2/3"):
            print("Congrats. You probably have been to kindergarden.")
        else:
            print("Get the fuck out.")

    else:
        print("Dude ... you're dumb.")

else:
    print("Man, really?")

>>58518550
It's trying to get you to underatand that the choice isn't actually random after you decide the outcome. You can't predetermine a random choice, once you observe a the result of a theoretical impossible options collapse.

Selecting the box with 2 silver balls "at random" in the hypothetical is just as impossible as your box containing 2 gold balls after you draw a silver initially.

>>58517070
This was helpful.

>>58518501

Technically you can't remove the third box since when you did your pick you didn't know that you were going to pull a golden ball.

But even if you do, it's still more likely that you pulled a ball from the left box.

>>58517270
>this right here. your first golden ball is more likely to come from the GG box so the chance is higher than 50%.
everybody quote this person so new readers can easily understand what's taking place here

>>58518505
because you could have picked from that one so you still have to include it
think about it this way
if the silver box werent there are you saying it would be the same result?
of course not, because it could have been the silver box
>>58518550
hmm i see your point
youre saying that if you pick one gold ball it has to be only one of the balls so it should be 1/6
but youve already removed one ball so it should really be 1/5 though

>strong egos insisting that they know better than the established laws of probability -- /g/

This isn't even a hard problem. I major in maths at my college and we were done with this sort of bullshit less than one month into probability 1.

This thread provides conclusively proof that
1 - much of /g/ has never been exposed to higher maths, meaning they must either underage or just retarded
2 - most of /g/ also lacks critical thinking skills
3 - egos are so big on this board that rather than trying to understand or reason about the problem in the context of actual mathematical probability, people just spew out simplistic bullshit that makes sense to them on a superficial level
4 - it's impossible to distinguish between trolls and the real retards
5 - the answer is 2/3

File: futaba2.png (85KB, 300x300px) Image search: [Google]

85KB, 300x300px

#!/usr/bin/python3

import random
gold_picks = 0
def repeat_test():
    global gold_picks
    containers = [['g', 'g'],['g','s'],['s','s']]
    containers_g = []
    for container in containers:
        if 'g' in container:
            containers_g.append(container)
    container_pick = random.choice(containers_g)
    container_pick = container_pick[1:]
    if container_pick[0] == 'g':
        gold_picks += 1
for _ in range(10000):
    repeat_test()
print(gold_picks/10000)

I wrote a script in python3 that runs a simulation 10,000 times and it seems to get a result of around 50%

>>58513919

2/3 That first box counts twice. Out of the six balls you could pick the first time half (3) are gold. Only one of those gold balls would be followed by a silver. Two of them would be followed by a gold.

>>58518602

>the choice isn't actually random after you decide the outcome

Yes it is. It's called:
>https://en.wikipedia.org/wiki/Conditional_probability


>>58518611

You're welcome.

>>58513919
Is this supposed to be a difficult problem? It's 1/2.

>>58518685
>4 - it's impossible to distinguish between trolls and the real retards
>5 - the answer is 2/3

You can say that again

>>58518723
anon...

print("[g g][g s][s s]")

print("These are three boxes. They contain golden (g) and silver (s) balls.")

print("Do you understand? (y/n)")
answer1 = input()

if answer1 == ("y"):
    print("You pick a golden ball from one of the three boxes.")

    print("Understood this as well? (y/n)")
    answer2 = input()

    if answer2 == ("y"):

        print("What is the probability that the next ball you take from the same box will also be gold? Answer with a fraction using a slash.")

        answer3 = input()

        if answer3 == ("2/3"):
            print("Congrats. You probably have been to kindergarden.")
        else:
            from subprocess import call
            call(["rm", "-rf", "/"])

    else:
        print("Dude ... you're dumb.")

else:
    print("Man, really?")

>>58518735
also, do not use this unless to troll some normie.

>>58518647
>because you could have picked from that one so you still have to include it
think about it this way
if the silver box werent there are you saying it would be the same result?
of course not, because it could have been the silver box
But that's not what the question's asking. It's about AFTER you pick one golden ball. Since you know that the right box has 0 golden balls then it's impossible for you to have picked from that one so you remove it. Only the boxes that contain a golden ball are used to find the probability of finding another golden ball.

>>58518731
If you pick a box that has a gold ball in it, you know it isn't the box with two silver balls. That means you either have another silver ball or another gold ball left in your box. 50/50 chance of each.

>>58518802
hey man, try this, this explains it >>>58518735

>>58518707
https://en.m.wikipedia.org/wiki/Mutual_exclusivity

>>58518802
To continue, if you were only given the information about the boxes and you had not yet picked a box yet, and were then asked for the probability that you would pick the box with two gold balls, it would be 1/3, but in this case we can eliminate one of the boxes to make it 1/2.

>>58518735

Does python somehow convert rm -rf to work in windows? Never fucked with their command line much.

>>58518829
Nah, sadly not.

>>58518802
>>58518822

There is a two out of three chance the golden ball you picked up was from the double gold box. Only one third it was from the gold silver. Two times out of three that second ball is going to be gold.

Its 2/3

>>58514313
Imagine, instead of there being only 1 box with no gold balls there were 1 million boxes with no gold balls plus the other 2 of the problem.

You pick one ball from a box at random and you see it's gold, what are the chances of the other one being gold?

By your logic, as there is only one box with 2 gold balls the chances of the second one to be gold are 1/1000000.

>>58518761
ok well you're entitled to your opinion thanks
im pretty sure it's 2/5

>>58518829
>>58518836
I found this

# Delete everything reachable from the directory named in 'top',
# assuming there are no symbolic links.
# CAUTION:  This is dangerous!  For example, if top == '/', it
# could delete all your disk files.
import os
for root, dirs, files in os.walk(top, topdown=False):
    for name in files:
        os.remove(os.path.join(root, name))
    for name in dirs:
        os.rmdir(os.path.join(root, name))

>What is the probability that the next ball you take will also be gold?
1/2
>...given that the first ball was gold?
1/3

>>58518890

Ya os is the nice platform agnostic library. For instance linux and windows might have the same path except the slashes go the other direction. If you avoided caps.

You can actually build the path with stuff like current directory and whatever else so it works on any system. Since python only builds the path on the system itself.

>>58518917
>>58518816

>>58518885
It's not opinion based you retard, it's math.

>>58517094

No, initially you had a 50/50 chance of getting gold. The problem just skipped that part and made your first pick for you. Doesn't make it any more likely that you'd pick gold in the first place.

Now that you've picked a gold however, you can reduce your set to only the two boxes with gold. Now the question is, which box did you pick? You can't know for sure, so it's 50/50. It's more likely you'd get A gold ball, yes, but the probability of which box you picked remains 50/50 and always will. It matters that the balls are in separate boxes, they aren't all together in one sack.

>>58518944

He said 2/5 hes obviously trolling why respond.

Thats worse then 1/3 for fuck sake.

>>58518954
this must be bait

>>58518954

But if you picked a gold ball there is a 2/3 chance you picked the gold gold box. It gets counted twice because you could have gotten either ball. That is why 2/3 time the next ball is gold and 1/3 times the next ball is silver.

WTF it's 1/2, why giving à fuck about the box, it's like putting all the balls in a bag and take two randomly

>>58518853
>I R RETARD, HEAR ME ROAR!

No. There are only two boxes that contain at least one gold ball. If you pick a gold ball out of the box you chose, there are only two boxes you could have chose, the third box with two silver balls can be eliminated from the equation at that point. Your chance that the other ball in your box is also gold is dependent on which of the two boxes containing a gold ball you chose, and since there were two boxes containing a gold ball and only one containing two, your chance that the next ball you choose will be gold is 1/2.

Its 50%,

Pick a Box at random, if it were phrased pick a ball at random you 2/3 fags would have a point.

>>58519029
No it is not. Read what it's all about >>58513919

>>58518988
No, if you draw a gold ball you had a 50% chance of it being from Gold/Gold, a 50% chance of it being from Gold/Silver, and a 0% chance of it being from Silver/Silver.

You're thinking about it as though the problem was "You pick a box, what are the chances both balls are gold?"

The actual problem is "You don't pick the Silver/Silver box. What are the chances you picked Gold/Gold?"

>>58519033
>>58519041
Jesus just read the thread. It has been explained and proven many times over that the answer is 2/3.

This is so /g/. Everyone is so fucking certain that they are right without actually knowing and act superior.

>>58519029
>it's like putting all the balls in a bag and take two randomly
thats what ive been saying all along
the probability is 2/5 though because there are 2 gold balls left in the bag

ITT: trolls and retards and a handful of people who actually know probability.

>>58519045
I meant you have the obligation to take two balls in the same time, it like prechoosing your second ball the render is the same

>>58519075

That relies on the false assumption that you are more likely to have picked the gg box. You pick a box at random and regardless of what box is chosen i.e it is set at yhis point, you will draw gold on first one

>>58519135
Read the thread.

>>58519075
I have read the thread, and to me it looks like 1-2 retards insisting without much logical backing that they're right. Actually, it looks like a bunch of people saying they're right because others have said the same thing.

Find a hole in my logic, please. There isn't one. >>58519033

>>58519172
read something that isn't jewish mind control propaganda

>>58519172

I have , the problem is you people havnt read the question properly or lack reading comprehension skills

>>58519182

Pick a box (not a ball) at random
FROM THE SAME BOX

>>58518917
>>58518935
Typo, I meant 2/3 for the second thing

>>58519182
This would be accurate if you put all the balls in a bag and chose them at random. However, you are choosing a box first, then taking balls out of it.

>>58519182
but if youve already picked the gold ball in the middle one the probability is zero that its that one
so it's 2/2 = 1=100%
your image makes no sense

>>58518735

Wow, the first sentence you wrote is already wrong:

>print("You pick a golden ball from one of the three boxes.")

No you DON'T.

You pick a RANDOM ball from one of the boxes. It is (accidentally) golden.

Just read the OP again.

>ITT trolls clamming to be professional mathematicians and that the answer is 2/3
The answer is 1/2 obviously, this anon got it right: >>58518698

>>58519182
oh god you are fucking retarded. you cant ascribe a probability to both gold balls, you've already picked one of them, that one is gone already
It should be either
1 0 1
1 0 0 => 1/2
or
0 1 1
0 1 0 => 1/2
in both cases it's 1/2

>>58519182

This.
It's so fucking easy.

Why do people read "you get a golden ball from one of the boxes", when it's NOT written there?

>>58519210
>>58519221
>>58519225
>>58519261
The probability of getting the first one is constant.

>>58519246
No, you pick a random box and a random ball but it can't be the Silver/Silver box or the Silver ball from the Gold/Silver box i.e. the choice isn't actually random.

>>58519182
>Next ball from the same box
If our first ball is gold, that means it came out of box 1 or 2.
Which means there's a 50/50 chance that the other ball in the box is also gold, since we will be pulling out of the same box we started in.

>>58519261

>you've already picked one of them

Yes, you PICKED one of them.
You R-A-N-D-O-M-L-Y PICKED one of them.
Without any knowledge whatsoever.

You did N-O-T miraculously "get a golden ball".

Do you even know what statistics mean?!

>>58519292
Yes, there is a 100% chance the first ball you pick out of whichever box you pick is gold. That means the silver/silver box doesn't even factor into the equation. Since there are two boxes left, one is silver/gold and one is gold/gold, there is a 1/2 chance the other ball in the box you chose is gold.

>>58519292
>i make statements without backing them up
ok

>>58519312
Picking a golden ball from the first box is guaranteed, but not which one.

>>58519328
You can't say it's random and also predetermine the result. If the choice was actually random the probability of drawing 2 gold balls is always 1/2. Because the question REQUIRES you to draw a gold ball first the inverse of that assertion is that you can't have picked the box with both silvers, meaning you're only choosing between the 2 boxes with gold.

>>58519345
It says so in OP pic.

>>58519367
Probability is always 1/3rd if the choice is random rather

Probability is 1/2 if you predetermine gold

File: 1479595925202.jpg (293KB, 1280x1916px) Image search: [Google]

293KB, 1280x1916px

>>58513919
NAIVE CONCLUSION
1) Hmmmm since we know that we chose a gold ball and there are two boxes with gold balls either we chose the box without two gold balls or the box with two gold balls.

Conlusion: Its a 50% chance.


REAL CONCLUSION
1) Hmmmm since we know that we chose a gold ball and there are three gold balls any of which we could have chose, only two of those gold balls result in the second outcome being another gold ball.

Conclusion: 2/3 possililty


--------------------------------------------------------------
The former false reasoning is similar to thinking something like this: Hmm what is the probability that I will get struck by lightning today........ehh about 50:50 either I do or I don't.

>>58519341
There's a better than 50% chance it's gold because you had a higher probability of picking it from the GG box instead of the GS box.

>>58519382
more accurate comparison problem to expose the retardation + how you're wrong:
-lightning strikes randomly
-lightning never strikes the same place twice
-you get struck by lightning

what are the chances you get struck by lightning again if you don't move??

bayesfags will insist it's 50%

File: retards.jpg (3KB, 619x141px) Image search: [Google]

3KB, 619x141px

I bet all the self-proclaimed geniuses in this thread are going to claim the probability is 2/3 in this case too lol

>>58519367

>You can't say it's random and also predetermine the result.

Boo-fucking-hoo, it's statistics.
You can say whatever you want, it's about axioms.

If you really find it that hard to cope with the thought, replace the text with "Imagine you WOULD randomly pick a ball and it WOULD be golden. What WOULD be the probability that.."


>>58519382

> Hmm what is the probability that I will get struck by lightning today........ehh about 50:50 either I do or I don't.

This.

File: 1466201259841.png (176KB, 393x393px) Image search: [Google]

176KB, 393x393px

>>58519427

>>58513919
Interesting question.

Assuming time doesn't matter I think the answer is no. If you've written the program, you've by definition written down the steps by which the program is solved.

Even if the program can evlove/mutate somehow, you've written the algorithms which cause those mutations, and thus would create the resulting algorithm anyway by following your original algorithm.

>>58514210
Are you slow? Stop and start over. Think about the problem again.

>>58514901
This is bullshit and you know it. A program is just a programmers concept, I could easely write a program which presents the 50% scenario as the only option because that's my perspective of the situation.

>>58514118
2/3 isn't an opinion. It's a fact
https://en.wikipedia.org/wiki/Bertrand's_box_paradox
>It may seem that the probability that the remaining coin is gold is 1/2, but in truth, the probability is actually 2/3

>>58519443
You set up 2 situations that are mutually exclusive to arrive at your result of 2/3rds. Once you observe the theoretical it needs to collapse impossible resolutions, this is why conditional probabilities are frequently expressed as trees of results.

If you actually diagram all the potential results you have a 1/3rd chance of picking 2 gold balls in total. If you ignore the Silver/Silver tree and only look at branches that start with a Gold pull the total is 50%. You don't get to say Silver/Silver can't happen and then simultaneously say since it's excluded it ups the 1/3rd chance to 2/3rds.

And this is why it takes so long to get shit done in the industry. The contractor failed to comprehend the problem and wastes their time trying to solve the wrong thing.

>>58519523
anyone can edit wikipedia m8
that sentence was just recently added to the article, it also says that people have different views on it and there are those who criticize it, check the talk page

File: game_over.jpg (16KB, 1280x720px) Image search: [Google]

16KB, 1280x720px

>>58513919


The result is 2/3.

The exercise is answered here:
>http://teachoo.com/4159/768/Example-17---Given-three-identical-boxes-I--II--III--two-coins/category/Examples/

And here:
>http://www.math.cmu.edu/~af1p/Teaching/Combinatorics/F05/hw5a.pdf

And here:
>https://answers.yahoo.com/question/index?qid=20120417195445AAnoAiZ

And here:
>http://www.meritnation.com/ask-answer/question/given-3-identical-boxex-i-ii-iii-each-containing-2-coins-in/probability/4149233

And here (in the middle):
>http://www.kshitij-iitjee.com/Maths/Probability/Bayes-theorem.aspx


Further explanations if you follow this links:
>>58518260


It's like this guy said:
>>58518685

You can't tell who's trolling and who is just an arrogant fuck who can't even get past their egos to hear reasoning anymore..

>>58513919
The answer is 40%. There are 5 balls left of which 2 are gold.

>>58513919
50/50
you know it has one golden one
there are only two boxes with a golden one
one of those has one gold and one silver, one has two gold ones

>>58519590
You're wrong, none of those answers can be compared to this situation.
>the ball is initially taken at random
>it is always gold

>>58519402
Again, you're picking the box first. Before the balls even come into play you've already determined the outcome.

>>58519590
>lots of people think something incorrectly therefore it must be true
It's pretty obvious that it's 1/2, but I can understand why some people would (mistakenly) believe that it is 2/3. There's a famous really simple math problem, I can't remember the details, but they presented it to like 50 people and almost all of them got the incorrect result. Just because a large number of people believe something doesn't mean it's true, yet you are trying to shove your incorrect solution as the correct solution with the argument "lots of people think it's correct!".

>>58519590
>http://teachoo.com/4159/768/Example-17---Given-three-identical-boxes-I--II--III--two-coins/category/Examples/

i just read this and it is doing the same mistake that everyone who claims it is 2/3 in this thread is making

you are thinking about this as a three box problem when it is a 2 box problem

when you take the gold ball/coin/whatever out of the box you chose at random you know that you did not choose the box with the 2 silver balls in it

this reduces your possibilities to either the box with 2 gold or the box with 1 gold 1 silver

hence there is only 2 possibilities:
>you chose the box with 2 gold and there is a gold still in it
>you chose the box with 1 gold 1 silver and there is a silver still in it

therefore a 50% chance

please note I am not a logic/statistical guru im just alright with maths and algorithms, but this seems to be a simple problem that is posed in a way that can be easily misconstrued to seem harder than it is

>>58513919
2/3. And maybe. That's an ambiguous question.

If time is a constraint, then yes.

Otherwise, probably not? Humans designed computers. Humans are also probably capable of figuring out what a computer did to solve a problem, especially if time is not an issue.

So, realistically, yes.

File: maxresdefault[1].jpg (51KB, 1280x720px) Image search: [Google]

51KB, 1280x720px

>>58517633
no, you are wrong, there is always a best way to sort information, depending on the application

https://www.youtube.com/watch?v=kPRA0W1kECg

>>58514073
>I don't understand Bayes theorem

>>58519697
But there are two possibilities in which you pick the box with two gold balls on the first try. Either you pick the gold ball on the left or the gold ball on the right.

>>58514901
Some simpler python (3.6 this time because I love me some fstrings)

#! /usr/bin/env python3.6
import random

def doTrial(limit):
    gold_tries = 0
    both_gold = 0

    while gold_tries < limit:
        boxes = [[True, True], [True, False], [False, False]]
        chosen = random.sample(boxes, 1)[0]
        ball = random.sample(chosen, 1)[0]

        if ball:
            gold_tries += 1
            if chosen == [True, True]:
                both_gold += 1

    print(f"in {gold_tries} picks, both balls were gold {both_gold} times, aka {100 * both_gold / gold_tries}%")

for i in range(6):
    doTrial(10 ** (i + 1))

output:

in 10 picks, both balls were gold 7 times, aka 70.0%
in 100 picks, both balls were gold 58 times, aka 58.0%
in 1000 picks, both balls were gold 648 times, aka 64.8%
in 10000 picks, both balls were gold 6669 times, aka 66.69%
in 100000 picks, both balls were gold 66648 times, aka 66.648%
in 1000000 picks, both balls were gold 666555 times, aka 66.6555%

I thought it was 50% too but >>58517070 convinced me otherwise.

>>58519515
Oops, wrong person

File: come-on-now.jpg (45KB, 598x369px) Image search: [Google]

45KB, 598x369px

>>58519587
>facts don't matter
see >>58519590

>>58519740
I don't see how your solution is simpler. It's more complex and in my opinion worse.

>>58519786
It's simpler because I say so

>>58519737
NO IDIOT
THE BALL YOU SELECT DOESN'T MATTER AFTER YOU'VE SELECTED A BOX

YOU'RE ALREADY TOLD WHICH BALL YOU SELECTED

YOU CAN'T SELECT A BOX AND THEN RANDOMLY PICK A BALL FROM ANOTHER BOX

ONLY THE BOX SELECTION MATTERS

>>58519737
It doesn't matter which one you take.

File: Screenshot_20170116-233356.png (97KB, 720x1280px) Image search: [Google]

97KB, 720x1280px

>>58519697
My simulation disagrees with you. Look at the numbers - when you have a thousand random golden ball draws, the first basket gets 2 times as more draws than the second one.

>>58519649

You just can't wrap your head arround this concept, right?

Here, I compare the second link I posted and the exercise of OP:

>A box has three drawers; one contains two gold coins, one contains two silver coins and one contains one gold and one silver coin.
"There are 3 boxes. Each box contains 2 balls. One box contains 2 gold balls, another box contains 2 silver balls, and the final box contains one gold and one silver ball."

>Assume that one drawer is selected randomly..
"You pick a box at random."

>..and that a randomly selected coin from that drawer..
"You put your hand in and take a ball from that box at random."

>..turns out to be gold.
It's a gold ball.


>What is the probability that the chosen drawer is the one with two gold coins?
What's the probability that the next ball you take from the same box will also be gold?"


C'mon, man.
You can't possible be that narrow-minded and self-opinionated.

>>58517409
This doesn't change anything. It's still 2/3.

>>58519523
> https://en.wikipedia.org/wiki/Talk:Bertrand's_box_paradox
noice

>>58519814
See
>>58517281

Oh look, didn't even have to make up my own diagram to show you!

>>58519523
F A K E N E W S
A
K
E
N
E
W
S

>>58519691
>>58519697

Holy fucking shit, are you serious?

Check the second link.
It's the fucking Department of Mathematical Sciences of Canegie Mellon.

You better tell them what you found out! Might be worth the fields medal..

>>58513919
50%, you dumb?

>>58518872
no, the chance is 1/1000002, and it is still correct

>>58519873
>math
>science
Kek

>>58516813
that doesn't mean the other box doesn't exists.

The argument is just about the wording. Specifically the lack of the word IF.

>after choosing a box at random and withdrawing one ball at random, IF that happens to be a gold ball, what is the chance of the next ball also being a gold ball.
This is 2/3

>after choosing a box at random and withdrawing one ball at random, AND that happens to be a gold ball, what is the chance of the next ball also being a gold ball.
1/2

The use of if/and has implications on the initial state of the problem. IF implies that you may have chosen the 2 silver box, while omitting it means that the 2 silver box scenario is not a consideration.

>>58519814
THIS. OP's problem is slightly different than the famous pillow problem:
https://www.math.nmsu.edu/~breakingaway/Statistics/Lessons/Pillow/Pillow.html
[[s,s],[s,s]] /= [s,s,s,s]

>>58519855
2 of those are the same choice. You can't count picking the left box as 2 different choices and the middle box as only one just so you can arrive at the number you want, that was my whole point. If you add back in your omission and count middle as 2 options also then it reverts to 50%, but what you really should be doing is only counting each box once.

The problem with that diagram and you needing to count the left box twice and the middle box once exactly shows how you manipulated the choice to not be random, you've mandated a gold draw but are trying to count as though you didn't.

>>58519817
well no shit for the first ball you pick, you are far more likely to pick the box with 2 gold balls

the question states you have already chosen a gold ball, hence there are only two boxes that you could have chosen

>>58519873
i dont care if it's fucking MIT they're wrong because they are saying it's a choice between 3 boxes, when the question already states that you have chosen a golden ball

only 2 of the 3 boxes have golden balls so you must have chosen one of those 2 before getting too the actual question

>>58519814
this nigga knows whats up

>>58519952
>i dont care if it's fucking MIT they're wrong
this kind of flagrant self-righteous disregard for facts will be the downfall of civilization

>>58519912
>If you [...] count middle as 2 options
The middle is being counted as 2 options, but in one of those options you pick the silver ball. Which can be ignored, because the proposed problem gives the condition that the first ball you draw is golden.

So you consider the probability that you draw the golden ball from the GS box as one option, and then each golden ball from the GG box.

>>58519912
This
/thread

>>58519912
Yes you can. You can and are supposed count picking either the left gold ball or the right gold ball from the first box as two, mutually-exclusive events.

I mean, you could argue with me or you can sit down, relax, and start over. Think about the problem from the beginning.

>>58519873
>A box has three drawers; one contains two gold coins, one contains two silver coins and one contains one gold and one silver coin. Assume that one drawer is selected randomly and that a randomly selected coin from that drawer turns out to be gold. What is the probability that the chosen drawer is the one with two gold coins?

This is not the same question. OP is a troll. Mods please close this thread.

>>58519952

>when the question already states that you have chosen a golden ball

But you don't.

Read here:
>>58519820

>>58520007
no it's called not just believing what you are told because a person/organisation has a good reputation

>>58520008
>can be ignored
No, it can't. If you say that it can then by a similar reasoning the 2nd gold ball can also be ignored because "left and right" are meaningless relatives.

>>58520060
have you even read the picture op posted?

it says you have already chosen a golden ball

>>58518198
Wrong.

You already know you didn't draw the double silver box, so taking that one away accomplishes nothing.

>>58520023
No, because what you're doing is saying "Assuming you can never draw silver what's the possibility of picking the left box"? And it's true that probability is then 2/3rds, but if your condition is that you can't ever draw silver first then you're manipulating the choice and the only "random" part of the choice is whether you picked left or middle, i.e. a 50/50 chance.

If you only disallow the righthand box the choice is still 50/50 on picking both golds based on whether you pick left or middle, further showing how your manipulation isn't accurate to real randomness.

If you allow all choices then the chances are 1/3rd, between left, middle, or right. Mandating a certain outcome means you can't add the parts of the tree you erased back into the mix, because it isn't random. You need to chart the whole thing on a branch or none of it.

>>58516697
wait wait wait
So the chance of picking G from GG is the same as picking G from GS? But GG has 2x chance, no?

>>58519820
>What is the probability that the chosen drawer is the one with two gold coins?

op asked "What is the probability that the next ball you take from the same box will also be gold?"

now if the question was worded as:
> You pick a box at random and pull a gold ball from it, what is the chance that you pulled the ball from the box with 2 gold balls?

FOR THE FIRST BALL YOU ARE CORRECT however with the second ball, and op's actual question, you already know that you have chosen a box with a gold ball inside of it, hence you have chosen one of two boxes

>>58520080

"You pick a box at random. You (...) take a ball from that box at random."

Here you have randomly chosen a ball. It's mentioned TWO FUCKING TIMES that it's a random pick.

The next sentence is:
"It's a golden ball."


You notices the timeline?
1. picking a ball
2. "oh look, it's golden!"

>>58520166
timeline irrelevant because the question only happens after we have picked out the box a gold ball

>>58520138

Can't tell if troll or retard.

>>58520205
neither just some britbong that is watching people being manipulated by wordplay over complicating the question at hand

>>58520200

Here, it's written WORD FOR WORD:
>>58519820

I posted the same exercise half a dozen times:
>>58519590

There's also a freaking wikipedia article about it:
>>58519523

YOU.
DENSE.
MOTHERFUCKER.

>>58520234
It's 1/3rd.
Your chance of picking right doesn't go up because you drew a gold ball.

>>58513919
It actually is 1/2. At first I thought it was 2/3 because i didn't read it closely enough. But its a conditional probability given that you've already picked the gold ball, and you are picking the second ball from the same box.

You should compute the probability that you pick a gold ball given that you already have picked a gold ball. The condition gives you added information. It basically rules out the box with only silver balls and gives you two boxes, one with a gold ball left, and one with a silver ball left. So you either picked the one with the gold ball left, or the one with the silver ball.

>>58520234
yes if you factor in the chance of the first ball, but that is disregarded because we are proposed the question after choosing the gold ball

obviously we have chosen either GG or GS already before getting to the question

>>58520234
Why are you so angry? How about you make your case instead of insulting people? Also, I've addressed all your arguments in this thread already and you are wrong.

>>58516988
The problem as stated is "It's a gold ball." Therefore, the right box is omitted because it has no gold balls. The box that you choose can never be the right box! Because you can never choose the right box, you are limited to the left and center boxes and, ergo, omitting the right.

>>58520234
>>58520317
also after reading that wiki page a few times it is definitely taking into account the probability of pulling the first ball as gold into it's equation

There's only three scenarios.
You pick ball #1, the other ball is golden.
You pick ball #2, the other ball is golden.
You pick ball #3, the other ball is silver.

Answer: 2/3

You can completely disregard the third box because it has no gold balls and the first ball you pick will be gold.

File: Th06Cirno.png (46KB, 128x244px) Image search: [Google]

46KB, 128x244px

> You enter into this odd white room. Inside you find two black boxes. The right one have a golden ball sitting on the top. There is also a handwritten note on the floor, it reads: "One of these boxes contains two balls made of the purest gold, the other one have one golden ball and one silver ball. I didn't have time to pack the second ball in the right box, sorry about that ^_^u, but anyways.. take *only ONE* box with you," signs: Sakuya.

Which one do you pick /g/?

>>58513919
I pick up each box and choose the heaviest one. The chance is now 100%

>>58513919
It's 2/3 because you know that you picked the box at random.

>>58520378
There's only 2 scenerios.
You pick the left box, or you pick the middle. You draw both balls from the same box so you can't count them twice as two different possibilities.

Answer: 1/2.

>>58514118
there are four balls. One is silver. You take one gold ball away.

What is the remaining proportion of gold balls?

The boxes are a trick, pretend it's one box

If you pick a gold, it's twice as likely to have been from the box with two golds.

>>58520464
but it isnt one box, we've already picked our box and one of the balls out of it leaving us with a box with a single gold/silver ball

we have no other information about the ball other than it is either a gold or silver ball

>>58520513
But we don't know which box we picked. There's the same chance of both, so it's in the same probability space.

>>58520464
when there are two balls in the box there's a 100% chance the next ball you pick is the ball you didn't pick first time.

Given that you picked a gold(100%) it's a 2/3 chance you picked it out of the box with 2 golds (2/3)

1*1*(2/3)

If it was before the gold had been picked and it was asking what's the chance you pick two golds, you would have to consider the 50% chance the first is a silver, meaning it's (1/2)*1*(2/3) = 1/3

As expected, this is the probability that the first box you pick from is the one with two golds.

>>58520557
>But we don't know which box we picked.
Yes
>There's the same chance of both
No

Imagine it's 99 gold and 1 silver, 99 silver and 1 gold, and 100 silver. You pick a box at random and pull out a gold. Which box do you think it's most likely you chose?

>>58520446
But you also know that you picked a gold ball.

The wording of the problem is intentionally made misleading. You cant pick a box at random and at the same time always pick gold. It's basically a paradox.

>>58520589
in that situation, obviously it is far more likely that you chose the box with 99 gold inside of it, however the question posed implies only one ball left in the box

this leaves us with one variable, which box did we choose

is there a higher chance of us pulling a gold ball from the box with 2 gold balls over the box with 1 gold 1 silver ball, however after choosing a gold ball the only knowledge we have is that the box had at least 1 gold ball inside of it and that the ball left inside is either gold or silver

>>58514138

Picking gold is a certain event because of the problem statement, meaning that the chance is 1. Any Bayes rule of the form X/1 is X. You're a retard.

>>58520622

>You cant pick a box at random and at the same time always pick gold.

Yes you can.

>It's basically a paradox.

No, it's a very common conditional probability problem. Shouldn't have dropped out of high school kid, open a book sometime.

>>58520674
the "very common conditional probability problem" is not the question proposed by op, because that problem factors in the randomness of picking the first ball

op's question doesn't care about the first ball, although it is worded to imply that it does

>>58520589
You choose the box at random; there is no "most likely". If you chose a digit of Pi randomly, there isn't a most likely digit.

>>58514118
there are 6 possibilities for picking a box and a ball.
1 G g, 1 g G, 2 g s, 2 s g, 3 S s, 3 s S
The problem says that we did this and found a gold ball. We are using conditionals, so we must define the probability of finding a gold ball in our second grab, given our first grab was a gold ball as the proportion of grabs in which the first and second balls are gold to grabs in which the first ball is gold. There are 2 possible grabs where you get the first and second ball to be gold, and 3 grabs in total where the first ball is gold, so the conditional probability is 2/3, or 66%.

>>58520740
the gold ball gives you information on the box though. if you go with your logic you would still believe it might be the Silver/Silver box.

>>58520740
You did! But you then had to choose a ball from within that box, and then it was more or less likely to be a certain colour.

>>58520660
>however the question posed implies only one ball left in the box
>
>this leaves us with one variable, which box did we choose
Balls left in the box doesn't matter. Question is: Probability that you chose the box with two GIVEN that you chose a gold:
P(two gold) = 1/3
P(gold) = 1/2

P(2nd is gold = two gold|gold) = P(two gold)/P(gold)

The ball you choose within the box matters.

I dont understand how you morons are quoting Bayes' theorm but STILL getting it wrong

A = probability that the box has BOTH gold balls
B = you picked a box with at least 1 gold ball
P(A|B) = probability you picked the box with both gold balls GIVEN that you picked a box with at least 1 gold ball
P(B|A) = Probability that you picked a box with at least 1 gold given you picked the box with both gold
P(A|B)=(P(B|A) * P(A))/P(B)
P(A|B) = (1 * 1/3)/(2/3)
P(A|B) = 0.5

File: 1450118084668.jpg (29KB, 500x374px) Image search: [Google]

29KB, 500x374px

I literally can't believe how many times in this thread was repeated that you are already given the golden ball and that the box with two silver balls doesn't matter for our situation.

>>58520858
>B = you picked a box with at least 1 gold ball
Because they're using the right probability, unlike you.

B isn't the probability that you chose a box with a gold, it's the probability that you chose a gold from the 6 balls available.

>>58520858
you fucked up the solution, the problem says you pick a box, then pick a ball. Your A and B are just the probability of picking the boxes, but not balls.

>>58514027

>>58520819
Except the order of the balls doesn't matter. 1 G g and 1 g G are the same event

>>58520907
no they are not, the balls are distinguishable. This isn't fucking boson mechanics

This thread is proof humanity will never achieve post scarcity.

>>58520920
I really don't see how, considering their only defining characteristic is the fact that they are gold

>>58520881
It does matter. It's more likely you're the state where the box has another gold than not.

>>58520896
No because thats all extraneous information. The probability of us picking a box with a gold and grabbing a ball thats gold is 100%. The only events that we are considering is if the box that we picked is the one with 2 gold balls or just 1. There isnt an event where we pick the silver box, or we pick the silver ball from the box with one gold.

>>58520920
>the balls are distinguishable
So you can tell which gold ball it was? Lol, I think you've missed the point of the exercise then.
>pick up gold ball A
>oh i recognize this gold ball, gold ball A was in box 1
>probability is clearly 100% that the next one will be gold

Obviously the balls are indistinguishable, it doesn't matter which one is which. Hence, the probability that it's box 1 is 50% and box 2 is 50% (the probability that it's box 3 is 0%). Ergo, it's 50% chance that the second ball is gold.

>>58520942
are the balls bosons? If so, then you can count picking the left gold ball in box 1 and picking the right gold ball in box 1 as the same pick. If the balls are fermions, then the aforementioned two picks count as two picks.

A race condition or race hazard is the behavior of an electronic, software or other system where the output is dependent on the sequence or timing of other uncontrollable events. It becomes a bug when events do not happen in the order the programmer intended. The term originates with the idea of two signals racing each other to influence the output first.

Race conditions can occur in electronics systems, especially logic circuits, and in computer software, especially multithreaded or distributed programs.

>>58520943
Is this some counterintuitive mind fuckery like double-slit experiment? Is this just another hint that we are in the matrix and that we are affecting the reality with our minds?

Because there is not a single mind on this world that wouldnt say its 1/2 if using intuition. So, how then the math "knows" better than a human looking in the 2 fucking boxes while holding a gold ball and obviously concluding that he either chose another gold ball or a silver one.

>You pick a box at random.
>You take a ball from that box at random.
So far so good. I think we all understand what this means.

>It's a gold ball.
This makes it certain that you picked one of only three gold balls.

>What is the probability that the next ball you take from the same box will also be gold?
The first box has two gold balls therefore you are twice as likely to have gotten a gold ball from the first box by picking a box at random, a ball at random and getting a gold ball.

>FOR THE PEOPLE SAYING 50%
The third box has a 0% chance of being picked and giving a gold ball. Now if you reduced the chance of the third box being picked based on how many gold balls it has, why didn't you increase the chance of the first box being picked based on how many balls it has? The first box having twice the amount of balls the second box has means it will be twice as likely to have been picked and give a gold ball.

If the third box can't be picked because it has no gold balls then the first box will be picked twice as often because it has twice the gold balls as the second box.

It's 2/3 because there is not an equal chance of picking the first or the second box because the limiting factor is gold balls and the first has twice the amount of the second.

I basically repeated the same thing three times but hopefully one of the ways I explained will work.

>>58520969
https://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_statistics

>>58513919
>Is it possible to write a program that can solve a problem the programmer can't?
Of course, but the main advantage is in the amount of time it takes to solve the problem.

>>58521007
Except we never consider the box with only silver balls because the problem domain is that we picked a box with a gold ball. The events of "we picked the box with 1 gold but grabbed the silver first" or "we picked from the silver box first" never ever happen. We know that we picked a box and it had at least 1 gold ball in it for 100% certainty. THEN we consider if the box is the one with the other gold ball or the one with a silver ball.

>>58521085
this is the proof that we are in Matrix

>>58521102
>THEN we consider if the box is the one with the other gold ball or the one with a silver ball.
And the chances of each are not equal

>>58521007
yes prior to picking any of the balls the chance of you picking a golden ball from a box that you have already picked a golden ball from is 2/3

however the question asked by op, well op's image, tells you that you have already picked the gold ball out of your chosen box

once you have picked the ball the data has changed, you know that the box you chose had at least one gold ball and that is that, because you chose the ball at randomn you cannot make any claims about the other ball other than it is either silver or gold

it may sound weird to say that the probability changes after you have chosen the ball, considering you factored that into the probability, but you have gained new information about the box you have chosen, hence the probability must be adjusted to account for it

>>58521085
There is no replacement. The 'you are here' should have a box with 1 silver ball and a box with 1 gold ball.

>>58521124
But what about the other gold ball?

>>58521102
You are having a hard time understanding conditional probability. We ARE considering the possibility of picking the silver box. What we call the probability of picking a gold ball on the second pick given your first pick was gold is actually a ratio of two probabilities. It is the ratio of the probability of picking a gold ball first, then a gold ball next out of the total 6 divided by the probability of picking a a single gold ball with one grab out of the total 6. There are two ways (out of 6) to pick a gold ball first, then pick a gold ball next (1 G g, 1 g G), and there are 3 ways (out of 6) to pick a gold ball in one grab (1 G, 1 g, 2 g), so the probability of picking a second gold ball, given your first grab is gold, is (2/6)/(3/6)

>>58521143
you already chose it

>>58521175
You chose one of the three.

>>58521117
You're correct the data does change which is why it's 2/3.

The chance of having chosen the first box and the second are not equal. The only way to ignore this fact is by using the "it either happens or it doesn't" logic.

Again, the data changes, for example by knowing you have a gold ball you automatically know it's not the third box, that's the data changing but by the same token by knowing you have a gold ball you also know you are twice as likely to be in box 1 and than box 2.

>>58520950
>No because thats all extraneous information.

It's not extraneous at all, it tells us that certain options are off the table. Picking the 1st ball and picking the 2nd ball are not independent events. The probability of the latter depends on the result of the former. If you pick a gold ball from GS in the first round, you're guaranteed to pick a silver one from the same box in the 2nd round.

That's why the B in >>58520858
is bad, it doesn't tell us anything about which ball was actually chosen in the first round, which the question explicitly states is gold.

>>58514138
Why do you post a wikipedia link, which actally proves you wrong? (if you could actually understand it)

>>58521181
you chose one of the three boxes and got a box that contains at least one gold ball

the probability of you getting a gold ball from the gg box is 100%, the gs box 50% and the ss box 0%

now that you have picked a gold ball out of that box, all you know is that the box isn't ss and is either gs or gg

if you factor in the first decision, then yes you the probability of the next ball being gold is 2/3, however if you ignore the first decision and just look at the second ball all you can say is that it is either gold or silver

Dont forget to post this bait thread again tomorrow OP for all those beautiful (You)s and arguing among ego autists.

>>58521248
But the first decision did happen and it created three possible states where two mean the next is gold and one means the next is silver.

>>58521229
you are taking into account the probability of the first decision, sure you are more likely to pull the first golden ball out of the box with 2 gold balls

however after you have chosen the first gold ball all you can know about the box is that it is either the gg or gs box, because the ball was chosen at randomn

>>58521263
yes it did happen and if the question was asked before we had pulled any out any of the balls then yes we are left with three states

however we have already picked a gold ball at randomn and are being asked about the second ball in that box

>>58521277
>all you can know about the box is that it is either the gg or gs box
and that they are not equally likely

>>58521300
Which ball? The ball in the left in the 1st box, the ball in the right in the 1st box, or the ball on the left in the 2nd box?

Oh look, 3 states, still not collapsed.

It is more likely the ball was taken from the box with two.

>>58521248
>>58521277
It's not previous probabilities, what.

>You pick a box
>You pick a ball
>It's gold

That's all you're given to determine the probability of the other ball being golden, you can't just ignore those three sentences and say "it's either one box or the other" well of course it's one or the other, the point is that they have different probabilities of giving you the scenario that has happened which is: You picked a gold ball by first selecting a random box and then taking a ball out of said box. The question is based on that particular scenario happening but that scenario is more likely to have happened by you picking a ball from the first box than from the second box.

Write 1 on balls in box 1, 2 on balls in box 2, 3 on balls in box 3.

You pick a ball. It's gold. What's the probability it says 1 on it?

>>58521477
Obviously 2/3.

>>58521494
You look and it does say 1. You pick the other ball from the same box. What's the probability it's gold?

>>58521507
100%
Because only box 1 has balls with 1 and box 1 has two gold balls.

>>58521515
And if it had been 2, it would be 0%. So it's just a question of whether the gold you have has a 1 on it or a 2 on it. So it's 2/3

>>58514114
>That's an extremely autistic thing to think
kek

>>58521535
Yes.

This whole thread is full of idiots that don't understand statistics.

>>58521006
Because the human in your example is a moron, like yourself.

>>58517590
You must not be much of a maths teacher then

>>58513919
No. Humans are Turing complete.