Anonymous Internet

0 becvause he put the hand in the silver ball

tfw you summon gold balls from nowhere

>0 becvause he put the hand in the silver ball

Attached: brain.jpg (800x450, 58K)

>1/3 I guess.
You don't guess math problems, you solve them

2/3

This is just monty hall problem phased differently
It's 2/3

brainlet

2/3 I guess.
There are 3 balls you could have selected at first:
The first gold ball in the first box, the second ball in the first box or the first ball in the second box.

The chance of you picking either one in the first box are thus 66%. And if you picked one of those, the chance of you picking another Gold ball is 100%.
That leaves 33% that you got the ball the second box, which has a 0% chance to get a gold ball.
Thus, 66.3333... percent to get a gold ball.

>2/3 I guess.
You don't guess math problems, you solve them

It isn't.

you are picking from the same box though, not another random box.

if you pick a gold ball you can discard the other box as you know it's 2xgrey

50/50

Depends. Who are you hiding from? ISP, websites and your government then yes, use anything listed on privacytools.io
Using a VPN means trusting the provider and also trusting their ISP. Tor is good if you want a "free VPN" and the browser itself is excellent for removing fingerprinting.

>66.3333
God damn it I keep getting it wrong.

Well if you picked a gold ball, then you are either taking from Box 1 (gold and gold) or Box 2 (gold and silver). Eliminate Box 3 from the choices.

That means you will either pick a gold ball or a silver ball when you choose another ball. 50% chance of getting a gold ball.

Alright, so what does it depend on exactly? I mean, let's say I'm looking for the alternative, what would I be looking at? Thanks for the tip about privacytools.io

Is tor compromised or is it just a CIAnigger psyop to keep the goyims complacent?

No.
You got a gold ball.
Which one is it?
It could be one of three. Either the first in the first box, the second in the first box or the first in the second box.

At that point, the game is over.
If you picked either one from the first box, which has a 66% chance, you'll draw a gold ball.
If you picked the one from the second box, you'll draw the silver ball.

Thus, you have a 66.666...% chance to win if the first ball you drew is a gold ball.

>pick a box at random
>yeah actually it isnt random you picked a box with a gold ball in it

Attached: hmmm.png (367x403, 5K)

it's 50%

it is guaranteed that you picked the [G G] or the [G S] box, it is guaranteed that the first ball you pull out is gold so just remove one gold ball from both boxes and the boxes now look like this: [G] and [S], so the chance of the next ball being gold is 1/2

intuitive answer but intuition can go astray when god wants so
you pulled a golden ball, so it's either box 1 or 2 (left to right). but it's twice more likely that it's box 1 because it contains 2 gold balls. so it's 2:1 for box 1 and the answer is 2/3.

Yes.
you got a gold ball, yes there is three left. but only TWO boxes. you have 2 choices to make, it's 50/50

same question goes if you picked a grey ball, you'd just be asked what the chances are the next one is grey.

everything up to picking the 2nd ball is a guarantee, it's a setup to the problem, so it doesn't affect the 2nd pick, which is what you study
see

Yes, if the chance was 50% that you picked from either the first or the second box.
But it wasn't. There were two balls in the frst box and only one in the second box.
Thus, the chance that you picked from the first box is twice as high as the chance that the initial ball you picked.
And if you picked one from the first box, you win. If you picked the one from the second box, you lose.

It would have been 50-50 if you could only draw the first ball from either box.
But you can pick any of them, the one on the left or the one on the right.
The left box has two golden ball, while the right box only has one.
If the golden one you picked from the left box, which is twice as likely since there were two of them, you win, since the remaining ball is also golden.

I cannot possibly explain this any better for you. At this point, you're either braindead or Americans.

For people who can't into probability:

You pull a gold ball. This means you either have a box with two gold balls or you have a box with one gold ball. You obviously do not have the box with two silver balls. It is out of the equation. So

Scenario 1: You have the box with two gold balls. The next ball you pull is gold.
Scenario 2: You have the box with one gold and one silver ball. The next ball you pull is silver.

Out of the two possible scenarios, only one results in a gold ball being picked. It is a 1 in 2 chance. 50%.

Attached: 6RpTZgI.gif (320x180, 3.63M)

You're wrong because you have 100% chance to pick the gold ball. The game only begins once you have that gold ball in your hands so the probability the next ball is gold is 50%.
This problem is fundamentally different from the monty hall problem

A) You don't have a choice to make
B) There are 6 possible worlds before you pick the first ball out. 3 where you pick a gold ball, 3 where you pick a silver ball.
After you picked out the ball, you know you're in one of 3 possible worlds:
>World where you pick first of the two gold balls in box one
ck a silver ball.
After you picked out the ball, you know you're in one of 3 possible worlds:
>World where you pick first of the two gold balls in box one
>World where you pick the second gold ball in box one
>World where you pick the gold ball in box two

There's only one world out of all 3 where you don't pick a second gold ball.

>ctrl-f bayes
>0 results
BAKA

>condescending tone
>gets it wrong anyway
Jow Forums in a nutshell

whatBall=[0,0] //initialized ball colour counter
drawBall = function(){
pickBox=Math.floor(Math.random()*3) //draws a random box out of 3
boxes=[[1,1],[1,0],[0,0]] // initializes the bozes
randBool=Math.random()>=0.5 //true / false 50%
pickBall=randBool?boxes[pickBox].shift():boxes[pickBox].pop() //takes one of the balls
if(pickBall==0){
//not gold, do nothing
}else{
nextBall=boxes[pickBox].pop() //ball was gold, pick another ball from the same box
whatBall[nextBall]++ //increment
}
}
for(i=0;i 49.7394741570342

Please post your explanation, Doc.

in my head this is how it goes:
you pick a random box and take a ball. if it's a silver ball you put it back, scramble everything then choose at random again. in this method, the boxes have the same hope for being chosen, but box 2 has one silver ball, so half of its hope is gone. on the other hand, box 1 has 2 gold balls so all its hope stays. it is just natural that this way, box 1 will be elected more often.
ofc god is wiser in this matter and i might be mistaken.

t. total brainlets over complicating things
you do have a choice to make, and just one. that choice is which of the two gold ball containing boxes you pick. it is only guaranteed that the box you pick contains a gold ball, it is up to you to pick it, that being the only random part of the problem

Yes, I know that. But you picked a Gold ball, didn't you?
Which. One. Is it?
There are two in the first box that you could have picked.
EIther one of them causes you to draw the other golden one, and thus win.
The chance that you picked one of those was 2/3.

It would ONLY be 50% if you could only draw the leftmost ball from either box.
Then it was 50% that you drew the one on the left of the first box and 50% that you drew the one on the left of the second box.
Thus, in that situation, you would have a 50% chance to draw the next golden one.

But that's not what happened.
You didn't just have those two golden balls on the left for choosing.
You had three.
And the two in the left box would cause you to win.
You can't draw the silver one on your first turn, so the only one from the second box you could have drawn was the left one.
And since there were three golden balls to choose from - the two from the 1st box, which guarantee that you'll draw the other golden one, and of which you have one in two of three cases, or the one from the second box, which guarantee that you'd lose and which you'd only have in one of three cases.

Based God poster.

>poster explains the correct answer
>you, a faggot, claims he is condescending and wrong
nu/g/ in nutshell

Attached: 4134131.png (966x665, 156K)

Attached: aivo_aukko.jpg (1218x1015, 212K)

>that being the only random part of the problem
This. You're asked about the outcome of the final step in a specific scenario. You're not asked about the outcome of getting into that scenario.

it's literally wrong
the answer is 2/3
you are stupid

>tfw i failed markov chains and probability and will have to retake them
how do i into probability, bros?

Attached: 1520678470534.jpg (1437x1079, 199K)

are you drunk mate? technically speaking, you literally countered your own argument.
>It would ONLY be 50% if you could only draw the leftmost ball from either box.
i'll follow you on this idiotic thought
technically, YOU CAN ONLY DRAW THE LEFTMOST BALL, you faggot. in the [G S] box, the golden ball is on the left, in the [G G] box both balls are golden, so technically if you pick the rightmost ball, you could just pick the leftmost instead and it would be LITERALLY THE SAME FUCKING THING, you twat. god damn, i understand thinking it's 2/3 out of shitty intuition but are you actually retarded to the point that you don't see that the only choice in the entire problem is picking a box with a golden ball out of 2 possible boxes

50% obviously. How is this even a riddle?

>is tor compromised
It's not. But that doesn't mean websites aren't compromised. JavaScript can be genuinely malicious, not just for your privacy but also security while CSS can now be used to track users.

>what does it depend on exactly
It depends on who you're hiding from.

Just visualise all paths you can take.

Thanks for the explanation, I understand it now. This big EU brain is starting to pay off.

Also, I just googled the answer - en.wikipedia.org/wiki/Bertrand's_box_paradox

The question in a formalized form: The probability that the second ball is Gold under the condition that the first ball is Gold:
P(second G | first G) = P(first G and secondG)/P(first G) = 1/2

>but are you actually retarded to the point that you don't see that the only choice in the entire problem is picking a box with a golden ball out of 2 possible boxes
HAVING A STARTING POINT WHERE IT'S MORE LIKELY THAT YOU ARE ALREADY IN A CERTAIN BOX YOU DUMBFUCK
autism rage mode engaged

It tests your reading comprehension, cognition and logical problem solving

>It depends on who you're hiding from.
can you give me a few hypothetical examples? I would reveal my intentions but I am very privacy-oriented

Since I seem unable to provide the special needs explanation you need, please just read wikipedia.

Listen up brainlets, it's from the SAME box when he picks again, which means that it's one of the first two boxes. Now that would mean that there's a gold ball left or a silver ball left depending on which box he selected, so it's 50%

Picking the rightmost golden ball isn't the same as picking the leftmost golden ball.

You're not picking a box. You're picking one of the three golden balls.

Here's another way to think about it: Forget the boxes. There are 3 golden balls, two have 'A' written on the and one has 'B' written on it. You destroy one ball. What's the probability the two left over have the same letter written on them?

2/3 sorry lads

ahaha you done goofed

Wait, that's not analogous at all. I'm retarded.
It's still 2/3 though.

you
had
a
higher
probability
of
picking
from
the
first
box

you are all stupid and are destined to be low-tier wagies

>how do i into probability
Probability is tough because we all have an innate "sense" of probability but it is skewed by our experiences and feelings. When you're given a problem you can't rely on your sense of probability and instead have to rely on maths and logic. Questions like the Monty Hall Problem have strange outcomes because our sense of probability fails us. We feel like we know the answer but we actually don't.

Timetravel to the start of 2018 at latest, goto Buenos Aires, enroll in FIUBA, take Grynberg's Probability course, achieve godhood if you're not a brainlet.
If you're a brainlet like me, you'll get a good enough grasp.

how did anyone manage to view this as hard?

50%
You either pick the gold ball or you dont.

kek, being this bad at everything

oh shit yeah, what a brainlet I am

The probability of picking a box with a gold ball at the first try is 1 because the pic says so. It's constant. Always true. The wording is important. It's not "assume you picked a gold ball first". In that case you would be right.

Attached: Screen Shot 2018-08-19 at 1.32.57 PM.png (994x362, 451K)

Yes, but the probability of picking from the first box with gold in it is higher than picking from the second box with gold in it.

Huh? What's wrong with it

What programming language is that? (Serious question. I don’t know shit about programming but would like to learn in the near future. This piece of code seems something I could learn.)

the problem in OP's picture is not the same as Bertrand's box paradox
en.wikipedia.org/wiki/Bertrand's_box_paradox
in OP's pic, it is stated that the ball being golden is a guarantee, as in
>It's a gold ball.
Bertrand's box paradox, on the other hand, says
>if that happens to be a gold coin
thus, in the case of Bertrand's box paradox, the correct answer is 2/3 because obviously most of the positive cases are going to be from the [G G] box
in OP's problem, the ball is guaranteed to be gold, so it's the setup to the real problem, and the only decision to be made, which is picking between [G G] and [G S]

Stating that it is a gold ball takes any probability out of it. It means that a gold ball was magicked into your hand. If the question had said you reset and repeat until you get a gold ball or assume the ball you drew is gold then you would be right.
So the start condition would be after you have drawn the gold ball and only one ball remains. Either a gold or a silver ball.

Imagine this situation, which is equivalent:
You have three golden balls. Two are labeled "You win!", one is labeled "You lose!"
You are blindfolded, the balls are shuffled, and you pick one.
What are the odds that you pick one that says "You win"?

Of course, 2/3.

This situation is basically identical. The second time you draw doesn't change the statistics at all - the game is decided the moment you drew the first ball.
You could only draw a golden ball, as you realized in your post.
But there were three golden balls it could have been.
Two of those are basically "I win" balls - if you picked the left one on the first box, you've already won, since you'll draw the right one on the left box.
If you picked the right one on the first box, you've already won, since you'll draw the left one on left box.
If you picked the left one in the right box, you've lost - since the ball you'll draw will be the right silver one.

Thus, there were two balls you could have picked that make you win, and one that makes you lose.
Since you choose one of them at random - they are the only valid choices after all, since you can't have drawn a silver one, as you've realized - you have
a chance of 1/3 to get the left one in the first box which makes you win,
a chance of 1/3 to get the right one in the first box which makes you win
a chance of 1/3 to get the left one in the second box which makes you lose.

Thus, your total chance to win is 2/3. In 2 of three cases you will have picked one of the two balls which are guaranteed to win.

This is the most retard friendly explanation I can come up with.
If you still don't get it, please consider a career in maths or IT so you can thoroughly fail and embarass yourself so badly that you'll see no other way out than to kill yourself.
And as you tie that rope around your neck, always remember user from that one golden ball thread that offered you a helping hand out of the retardation that made you the failure you are.

alright, so if we look at the pic in the op the probability would be 1/3? see pic for my train of thought.
>Monty Hall Problem
will have to look into it. i have to learn this shit or else what's the point of even studying eyy ayys?
don't they have a course online? my uni has some materials but they're not public, though.

Attached: probability.png (691x557, 20K)

1/3 chance.

Only in one in three boxes it is possible to take out two gold balls, and if you have 1/3 chance of picking any box, there is a 1/3 chance in taking the one box where this is possible.

This doesn't seem to be like the Monty Hall problem. Is the answer 50% or is that too obvious?

>Imagine this situation, which is equivalent:

It's not equivalent. OP's pic removes the first drawing from the equation, leaving it at 50%.

import random # note that random.randint(x, y) is inclusive on both ends

# Keep two integers, one which counts the tries of OP's scenario, and one which counts how many times it was the gold gold situation.
tries = 0
goldPickedSecond = 0

# You have an array with three "box" arrays [ [g, g], [g, s], [s, s] ]
g = 'gold'
s = 'silver'
boxes = [ [g, g], [g, s], [s, s] ]

def trial():
global tries, goldPickedSecond

# Pick a random item from the array, do not record anything.
chosenBox = boxes[random.randint(0, 2)]

# Pick a random item from the chosen box array that was chosen in the previous step, do not record anything.
firstBallIndex = random.randint(0, 1)
firstBall = chosenBox[firstBallIndex]

# If the item is "s", then discard everything and start over.
if (firstBall == s):
return

# If the item is "g", add 1 to the integer with the tries of OP's situation, then proceed to the next step.
tries += 1

# Pick the remaining item from the chosen box array, if it's "g", then add 1 to the gold gold situation integer.
secondBallIndex = not firstBallIndex
secondBall = chosenBox[secondBallIndex]
if (secondBall == g):
goldPickedSecond += 1

# Repeat until you have a decent amount of tries to get conclusions, do not reset the integers.
for i in range(0, 1000):
trial()

print('number of valid tries: ' + str(tries))
print('number of gold+gold situation (gold picked second): ' + str(goldPickedSecond))
print('percentage of gold+gold situation: ' + str(goldPickedSecond / tries * 100) +'%')

Attached: image.jpg (750x151, 51K)

Seems like very basic syntax used by various languages. Try copypasting it into the JavaScript console, just change the print() to alert().

>What's wrong with it
pickBall=randbool?boxes[pickBox].shift():boxes[pickBox].pop()

Not that user but this part looks terrible and I would change my code so to avoid that mess.

Looks like JavaScript since it has an anonymous function expression
drawBall = function()...

Wrong screenshot, sorry.

Attached: image.jpg (750x208, 62K)

the third box is out of the question because it contains no golden balls

Attached: Screen Shot 2018-08-19 at 1.43.53 PM.png (2422x424, 826K)

It's 2/3,because you have a 50% chance when picking the middle box to pull out a silver ball, nullifying that try.

IT. DOESN'T.
It says "You pick box [sic] at random. You put your hand in and take a ball from that box at random".
It is part of the equation.

Why can't it be a silver ball?

it's not poised as a guarantee - you pick the ball at random and then question is asked.

it's the same as monty python problem in a way

>and if you have 1/3 chance of picking any box
You've been given a box with a gold ball in it.

Guys, am I retarded? I think the answer is 50%, here's how I got there.

The problem states that you've already pulled out a gold ball. This means that you could not have selected the first box. (In diagram box is in eliminated)

You're left with two boxes. if you select the middle, you have 1 ball left and that is silver. If you pick the one on the right, you have 1 ball left and that's a golden.

You have two outcomes. 1 golden and one silver. This means that you have a 50% chance of getting a golden ball.

Guys please correct me if I'm wrong or there's a mistake in my logic. I would love to improve.

Attached: goldenball.png (581x321, 19K)

>Yes but what if it wasn't!
>t. 1/3 poster

jesus christ, what i'm reading.

See