Are you truly White? Only non-Pajeets can solve this puzzle

Are you truly White? Only non-Pajeets can solve this puzzle

Other urls found in this thread:

en.wikipedia.org/wiki/Bertrand's_box_paradox
repl.it/HRWH/0
twitter.com/SFWRedditGifs

2/3

agreed, i'm sure we'll be told we're wrong though

Can you Pajeet?

50%

50/50.

50% it's either gold or silver

ez

This is correct.

>mfw brainlets say 50%

Proofs?

>be pajeet
>cheat on questions
>get all answers right
What was that OP?

not that person but, the pic says that the first ball is gold meaning its either one of the left 2 boxes meaning the balls left are 1 gold and 1 silver, meaning 50% i guess?

you picked one of three gold balls. in two possible situations, the other ball is gold, and in one possible situation, the other ball is silver. 2/3

Please use the appropiate meme.

5/6

You drew a gold ball. Therefore, you either picked box 1 or 2.

It is possible to draw another gold ball in only one of those two boxes.

Thats 1 box / 2 boxes, or 50%

The chance the other ball will also be gold? 40/60
The chance the box is the one containing two gold balls? 50/50

ur dumb son

LOL

I'm sorry you're having so much trouble with pre-algebra, champ.

>mfw the majority of this thread unironically thinks it's 50% and not 2/3

You pick boxes, not balls. You have two boxes to choose from.

Could you explain your reasoning?

>ur

okay.

As a certified non-Pajeet, the answer is 50%. I'm guessing the bait for Pajeet here is that they probably will misread that you're taking the next ball from the same box you picked originally.

40% because there are two gold balls and 3 silver balls left. 2/5 = 40%. It would only be 50% if there were an unlimited number of balls left to choose from, but technically that's also correct in a semantic sort of way.

If you take a golden ball first, this means that you

a) took the ball out of the G+S box
OR
b) took the ball out the G+G box

since you have to take the next ball out of the same box you took the first one, and that box was randomly chosen, meaning that the choice of the box does not not affect the overall probability, you now have 2 golden balls out of 3 total balls which means a 2/3 probability.

I already did here but this user explains it well too

this
2/3 would be the answer is you could change the (possible gold) box you picked from. 1/2 if you can't change the box after the first pick

Except there aren't. You have already removed one gold ball. Of the two boxes in which this is possible, the only two balls you could remove are 1 gold and 1 silver. I think people might be skiming over the part where you have already picked a box.

I just assumed they were Mac users

I'm convinced people just troll with these threads.

Right, but if you got a gold ball, you're more likely to have picked from the box with two golds than the one with one gold, so it's higher than 50%.

no, there are 2 boxes with gold balls in them, and 1 which would fullfill the conditions, and they're picked at random, i e 50%

It says in the OP that you're picking the second ball from the same box you started with

en.wikipedia.org/wiki/Bertrand's_box_paradox

t. pajeet

>since you have to take the next ball out of the same box you took the first one, and that box was randomly chosen, meaning that the choice of the box does not not affect the overall probability, you now have 2 golden balls out of 3 total balls which means a 2/3 probability.

Are you retarded? The image asks at that point in time would the ball be. Not what the second ball will be when you first started.

I know.

If I picked the first box, then the second ball will be gold. (100%)

If I picked the second box, the the second ball will be silver. (0%)

The choice between those two box is obviously 50%.

Here is why it is 50%

If i first take out a gold ball then the box i got is either the first one or the second one.

NOT THE THIRD ONE(so it can fuck right off)

So the other ball will be either:
From box 1 gold or box 2 silver

So yeah 50%

50%.
The first ball being gold means it's not the last box.
So it's either the first box and you get another gold, or it's the second box and you get a silver.
One or the other, it's 50/50.

It can be 50%, Raj

you have 2 golden balls out of 3, after the first pick.

you can't determine if you took the ball of box 1 or 2, you just know it was one of them because they are the only ones with golden balls.

you have to calculate the total probability

these are the possible outcomes, there are 3
box 1, ball1gold -> box1, ball2gold
box 1, ball2gold -> box1, ball1gold
box 2, ball3gold -> box2, ball4gray

2/3 of the outcomes result in a gold ball

who's Raj?

> this dumb
kys user

you forgot a case, there are 2 gold balls in box 1 meaning that it has to be counted twice

see

Sorry Pajeet, you all look similar

you're retarded.
it asks for the probability that the next ball from the same box will be gold, not which specific gold ball it is.
Box 1 ball 2 and Box 1 ball 1 is the same thing.

2/3

If you pull a gold ball, there is a 2/3 chance that it is pulled from the box which contains 2 gold balls. There is a 1/3 chance that the ball came from the box with 1 gold ball.

who's Pajeet?

wrong, there is a higher chance that the initial condition that "it's a gold ball" happens with box1 since there are 2 inside the box

50% brainlets who don't understand how it's 2/3 even though someone already posted the wikipedia link explaining why it is do you seriously not understand that by picking a gold ball at random there is a HIGHER chance that you're in box 1 than there is that you're in box 2?

see
>en.wikipedia.org/wiki/Bertrand's_box_paradox

It is a given that you did not pick the SS box.

Therefore you picked either GG or GS.

There are 3 possible gold balls you could have taken initially, so there are 3 cases:

G1 from GG: Your next ball will be G
G2 from GG: Your next ball will be G
G from GS: Your next ball will NOT be G

Therefore in 2 of 3 cases you will pick another gold ball, probability is 2/3.

It's not asking for the initial condition.
The condition is set, you have already drawn a gold.
It's like flipping a coin multiple times, yes the chance of getting a specific sequence goes down but each flip is still 50/50.

I fucking hate these threads because it reminds me of how stupid the world is.

If Sup Forums fucks up with it, imagine what brainlets are on reddit

Joseph Bertrand is a fucking retard holy shit.

u

that has nothing to do with it

2/3 is not the probability of the precondition, it's the postcondition, which is what the question asks. The chance of getting a gold ball in the precondition is 50%, but that doesnt answer the question

1/3, only one box has two gold balls

I seriously did not know how retarded the average poster on this board is, holy shit. How many different ways does it need to be explained to you that it's 2/3? It's really not that complicated. I understand thinking it's 50% at first but doubling down when being shown you're clearly wrong triggers me

you're projecting, user

No I'm pretty sure I know when I'm being retarded.

thanks

it's not wrong though, only an idiot assumes it's 2/3 because some retard wrote a wikipedia article about it.
The first outcome in a sequence does not suggest the next will be the same in any way, there are two possible boxes.

i said you were projecting, not being retarded. Now you're being retarded

Ran a simulation to confirm 2/3.

50% first ball was silver.
16% first ball was gold, second ball was silver.
33% both balls were gold.

(this 33% of total trials is 67% of the eligible cases.)

>you picked one of three gold balls
No, you picked one of two boxes. The fact that one of the boxes has two possible gold balls in it is completely irrelevant when determining the answer to the question. The answer is dependent on which box was chosen. Since we know the box chosen contains a gold ball, it must be one of two boxes. If it was box a, the next ball will be gold also. If it was box b, the next ball will be white. 50/50

Yes but you're more likely to have chosen the box with two gold, because if you chose the box with one half the time you would have chosen the silver ball.

Possibilities:

Picked gold 1, the next is gold 2
Picked gold 2, the next is gold 1
Picked gold 3, the next is silver 1

Get it faggots?

yes and that makes sense except you've already picked the gold ball so there's zero chance of the first ball being silver.

the first two possibilities are the same thing, it doesn't ask which specific ball.
So listing it like this just misrepresents the question because it makes it look twice as likely.
the probability of it being either box is the same.

You bought some lottery tickets.

You like whiteboards, so you picked the numbers 2 3 5 7 11 15

Your neighbor is a normie who likes Lost, so his numbers were 4 8 15 16 23 42

It's the middle of the drawing. They have drawn the numbers 2 5 15 so far.

Do you and your neighbor have an equal probability of winning the big prize?

>some retard wrote a wikipedia article about it
You mean an accomplished mathematician conceived of it as babby's first probability puzzle and demonstrated how it was 2/3

No I mean if I was wrong about something like this and got called out I'd probably realize earlier that I was just retarded. You post very nice pictures though, friend.

the precondition isn't part of the calculation, it's merely separating the relevant data from the irrelevant.
why does the third box exist, given the final question? no reason other than misdirection, since it doesn't satisfy the precondition. this leaves two boxes with one ball in each. why? because the first ball is already taken. it's confirmed gold.

Yes but that still prunes off that part of the probability space. Just like if Monty Hall shows you a goat, that prunes off the probability space where that door had the car behind it.

math isn't an exact science, people can make shit up to suggest anything they want

picked box a, next is gold
picked box b, next is white

get it faggot?

...

There are four balls in the first two boxes. You choose one of the three gold balls. What is the probability that the other ball in the same box as the one you chose is silver?

The question doesn't say you choose a ball, it says you choose a box. If you choose box a, it is gauranteed the next ball will be gold. If you choose box b, it is gauranteed the next ball will be white.

1/2, you can't pull from the whole pool once you pick a box. it would be 1/3 if you can switch between the two boxes for the second pick

I don't know anymore who's shitposting or who is just plain dumb at this point.

It says you took a ball. If you took the silver ball it doesn't count. So you took one of the gold balls. There are three gold balls. Two of them share the box with another gold ball.

Tried to make it as simple as possible

import random

iterations = 10000
twoGoldBalls = 0
oneGoldOneSivler = 0


balls=["Box1gold","Box1gold","Box2gold","Box2silver","Box3silver","Box3silver"]


while iterations != 0:
#A random gold ball is chosen. Either balls[0], balls[1] or balls[2]
firstBall = balls[random.randint(0,2)]

#If the gold ball came from the first box, the second ball will be gold
if (firstBall == "Box1gold"):
twoGoldBalls += 1
else:
oneGoldOneSivler +=1

iterations -= 1


print("Two gold balls happened " + str(twoGoldBalls) + " times")
print("One gold one silver happened " + str(oneGoldOneSivler) + " times")
print("The odds are of " + str(twoGoldBalls/(twoGoldBalls+oneGoldOneSivler)))

You can try it out here: repl.it/HRWH/0

If you think this is wrong, please explain why

Thank you, I use them as practice for painting

But if you pick the silver ball it doesn't count, so half of the times you take the middle box don't count.

No, you are dumb. Either you picked box a which results in the next ball being gold, or you picked box b, which results in the next ball being white. Those are the only two possible outcomes given the parameters of the question. This isn't rocket science.

You're misinterpreting the question

using System.IO;
using System;
using System.Collections.Generic;

class Program
{
static void Main()
{
List space = new List();
var ran = new Random();
for(int i = 0; i < 1000000; i++)
{
var b1 = new box() { ball1 = new ball() { gold = true}, ball2 = new ball() { gold = true}};
var b2 = new box() { ball1 = new ball() { gold = true}, ball2 = new ball() { gold = false}};
var b3 = new box() { ball1 = new ball() { gold = false}, ball2 = new ball() { gold = false}};
var p = new List();
p.Add(b1);
p.Add(b2);
p.Add(b3);
space.Add(p);
}

int count = 0;
int total = 0;
foreach(var bo in space)
{
var t = ran.Next(0,3);
var c = ran.Next(0,2);
var bal = bo[t];
if(bal[c].gold)
{
total++;
if(bal[c == 0 ? 1 : 0].gold)
count++;
}
}
Console.WriteLine(count + "/" + total + "=" +(double)count/total);
}

public class box
{
public ball this[int i] {
get
{
return i == 0 ? ball1 : ball2;
}}
public ball ball1;
public ball ball2;
}

public class ball
{
public bool gold;
}
}
well it says 66.6%

See

Yes but you're twice as likely to have picked a, because half the times you picked b don't count, just like none of the times you picked c count.

Or is the answer actually 40%?

Because you're lumping all the balls into one set, where you're deciding you're picking the balls and not the boxes.
You don't KNOW that the first gold ball was from Box1. Hence the entire fucking point of this problem.

66.(6)%

>half the times you pick the middle box don't count.
there is zero chance of that happening because you've already picked a gold
whether you observe it or not you have already started with a gold 100% of the time, the chance of the remaining ball being gold is up to which box you've chosen.

Look how I called them, I think my naming was pretty clear.

you idiot, *you* don't know it, the *program* knows it. It's how his program measures the fact that the next ball will be gold or not.

But I fucking knew someone would say that, so here's a more direct simulation.

import random

boxes = [['au', 'au'], ['au', 'ag'], ['ag', 'ag']]
for i in range(100000):
random.shuffle(boxes)
box = boxes[0]
random.shuffle(box)
if box[0] == 'au':
if box[1] == 'au':
print 'yes'
else:
print 'no'
else:
print 'skip' # first was silver

python goldball.py | sort | uniq -c
16819 no
49935 skip
33246 yes

No, you are.

Pick box 1. It's gold. The next is gold.
Pick box 2. It's gold. The next is silver.
Pick box 3. This never happens under the stipulations of the problem, it's useless information.

Two possibilities exist
11
and
10

Focusing on what ball you pick as though they were all contiguous isn't how this probability works.

Baby's first probability problem?

Not quite.

It's one of the famous probability paradoxes for a reason.

But there's a 2/3 chance you picked the first box.

the ball has been taken though, and it's gold. we know this already. every valid sample begins with pulling a gold ball. any sample pulling a silver one first doesn't fulfill the requirement to be involved in the question.
I think I understand where you're coming from. this might be a physical vs mathematical perspective issue, explaining the frustration.

this point either needs to be responded to or accepted as truth