Meta: Jow Forums is now full of dumb teenagers with huge egos

Nah, I will not, this problem is too hard and dumb.
Have this random anime girl instead.

Attached: 002250.jpg (572x800, 75K)

you're taking the probability after he's already taken one golden ball out of the box not before
so its 50%

Possible outcomes:
- First gold ball from first box -> gold ball picked next.
- Second gold ball from first box -> gold ball picked next.
- Gold ball from second box -> silver ball picked next.
Since the next ball is gold in the 2 of the possible 3 outcomes, the answer is 2 + 3 = 600%.

import random

boxes = [["s","s"],["g","s"],["g","g"]]
passresults = 0
failresults = 0

for i in range(10000):
random.shuffle(boxes)
container = boxes[0]
random.shuffle(container)
if container[0] == "g":
if container[1] == "g":
passresults = passresults + 1
if container[1] == "s":
failresults = failresults + 1
print(str(passresults) + ", " + str(failresults))

Attached: file.png (1376x316, 45K)

Solved

it is 50%, the situation provided you first got a gold ball, meaning the bal filled entirely with silver was out of the situation entirely. therefore there were only two boxes you could’ve plucked from, the one with both gold, or the one with one gold. meaning, it’s a 50% chance you got the right box

The Monty Hall problem states you have a better chance if you switch doors, though.

it's not 50%. no amount of rationalization fixes this

too bad you didn’t account for them always getting a gold ball first

Go back and read the problem again. Specifically the last sentence.

fine, explain your reasoning then. you might want to read the question again, though, particularly in what it’s asking

one sec gonna try this out as a fun code exercise since I'm rusty

This is literally a statistics problem

see You don't "reason" your way through it.

There are three possibilities here:
You got the left ball in the two gold box in which case the next ball will be gold.
You got the right ball in the two gold box, in which case the next ball will be gold.
You got the gold ball in the one gold box, in which case the next ball will be silver.

All of these are equally likely therefore the answer is 2/3

>left and right matter

I accounted for them "putting their hand in and taking a ball from that box at random" and only considered the cases where "it's a gold ball".

guy who was arguing it was 50% here, totally neglected to account for the fact that it would be more likely i’d have gotten the all gold box than the half-nad-half box due to the fact it was a gold coin i picked originally

The people that say 50% are the same retards that say you have a 50% chance in the Monty Hall problem.

Attached: Monty Hall.png (1056x706, 57K)

no, that problem is a lot more straightforward. the odds of me picking the correct door in the beginning is 1/3rd, but once the announcer reveals the goat, i’ll have a better chance of switching since it’s more probable that i picked the wrong door than it was that i picked the right one originally

There are 6 balls, half of them are gold: 6 / 2 = 3.
2 of the balls are needed to be picked, so 2/3.
You can't refute it.

Okay, I spent some time dicking around with this.

import random

def draw(boxes):
# randomly select a box and a ball from that box
box = random.randint(0,2)
ball = random.randint(0,1)
color = boxes[box][ball]

if color is not "gold":
return draw(boxes) # redraw
else:
boxes[box].remove(color) # draw the gold ball
return boxes[box][0] # return the remaining ball in the selected box

def draw_n(n, boxes, probability_of="gold", i=0):
results = {"silver": 0, "gold": 0}
while i < n:
ball = draw(boxes)
results[ball] += 1
boxes = [["gold", "gold"], ["gold", "silver"], ["silver", "silver"]]
# how do I avoid having to assign boxes again in Python?
i += 1
result = results[probability_of]/(results["gold"]+results["silver"])
return result

boxes = [["gold", "gold"], ["gold", "silver"], ["silver", "silver"]]

n = 10000
result = draw_n(n, boxes, "gold")
print(result)

Anyone know how to fix the redundant assignment in draw_n()? Without it, I get an index out of bounds error. I think this is related to list mutability in Python, but I can't seem to fix it by creating a second list.

SyntaxError: unexpected EOF while parsing

missed a paren

Attached: IQ.jpg (968x968, 230K)

Half the balls are gold so the answer is 50%.

Refer to and python boxes.py
0.6688

how's it feel being wrong, NIGGER?

There is an error in that program.

which one?

a, a, a, c

Sauce for the image?

Both. You pick a box. Your first ball is gold. Now pick your second ball. It skews the box choices by picking first balls until it finds a gold one.

>Your first ball is gold
>It skews the box choices by picking first balls until it finds a gold one

Attached: patrick.png (471x354, 231K)

I know the right answer, but I'm not sure my reasoning is correct: you've taken a gold ball, that eliminates one of the boxes. Out of the remaining two boxes, there are three balls left, two of which can only be gold.
It seems way too simple, what's the catch?

>meta: Jow Forums is now full of dumb teenagers with huge egos.
Jow Forums is Jow Forumsprogrammerhumor tier now

Some people think it's about the number of golden balls (3) and not the number of boxes (2)

the secret is that since you are told that you drew a gold ball already, you can infer that it was more likely that you withdrew from the first box and impossible that you withdrew from the third box. Thus, you have already taken from box 1 or 2, but you were TWICE as likely to have taken from box 1

If you're in box 1 (gold, gold) you've taken a gold and there's one gold left, so you get a gold. If you're in box 2 (gold, silver) you've taken the gold and you take the silver now. 50%

50%, you'd have to be retarded to think its anything else.

>literally proven wrong by code

Code that's doing the wrong thing. You pick a box. There's 2 you can pick to get a gold first ball. You have a gold ball. Now pick a second ball. In the first box you get another gold. In the second you get a silver.

>if you have a gold ball then that means you are either picking from the mixed box or the all-gold box. It is impossible to pick a gold ball from an all-silver box, so therefore, assuming you are picking the next ball from the same box, there is exactly a 50% chance the next one will be gold. If you are picking from a different box though, then its 66% chance.

This anyone disagreeing is a steamgroup raider

wrong

40%
50% if you wanna be a smartass

Post correct answers, user.

40%? That's a new one. How did you get 40%, user?

read it again user:
>What is the probability that the next ball you take from THE SAME BOX will also be gold?

6 balls in total, take 1 gold. you are left with 5. 40% are gold 60% arent.

to turds

O

but its not asking about global probability, localized.
Read the last sentence again, it clearly says "what is the probablility from THE SAME BOX", which means it can only be box 1 or 2 left. Thus 50%

hmm should i blame my bad english or my retarded brain. thanks user

If english is your main language, the brain is the problem.

Actually a, a, a, b

the first one is c, retard.

even seasoned engish veterans fail to read that last part (thus all the confusion in thread).

There aren't two connecting lines close to the circle.

After you've made the initial selection of 1 gold ball, there is only a 50% chance of getting another one from the same box. The box with non-gold balls is out of the running, so to speak, at that point.

The next time you pull a ball out it will either be gold or it won't be gold, hence you have a 50% chance.

Bertrand was a fuckhead and utterly wrong.

2/3 niggers
50% brainlets should go back

there are 3 scenarios where you pull a gold ball because there are 3 gold balls. in scenario 1 you pulled from the first box with 2 gold balls in it, in scenario 2 you pulled the from the first box and got the 2nd gold ball. or you pulled from box 2 and got the only gold ball in there.

so from each of those situations there are 2 outcomes with lead to an additional gold ball and 1 that leads to getting a silver ball.

so yeah 50%

There're only two scenarios, one where you choose box one, one where you choose box two.

there are 2 balls in the first box,so you are saying the balls are not distinct? because if they arent distinct then the 2nd box has two balls that are 50% gold and 50% silver so that cant be correct. Therefore the balls are distinct and there are 2 scenarios with box 1, ball 1 and ball 2.

He choose a box, and got a gold ball, both balls in box one are the same choose.

IT'S FUCKING 2/3

AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

Attached: 1518550254821.png (400x416, 41K)

...

this

Anyone who thinks it's not 50% is a retard.

what i remember from the last thread
#include
#include
#include

int main()
{
srand(time(NULL));
int b[3][2]={{0,0},{0,1},{1,1}},i=0,k[2]={0,0},j,l;
for (;i332878,667122,66.712200%

This. There are only two possible scenarios, one where you take the first gold ball from the one with two gold balls or one where you take the gold ball from the one with the silver ball. Therefore you have a 50% chance of it being the right box a second time. The Box with two silver balls might as well not exist.

Attached: Boomerang_silver_age.jpg (1101x1642, 543K)

Does it still look like 50% in this case?

Attached: balls.png (1125x240, 722)

they cant be the same ball because the balls in the other boxes are separate and distinct and in a real scenario they would be distinct as well

yes, since the question is asking about the NEXT ball in the SAME box. Every other ball might as well not exist.

can any of these 50% retards just code the problem and see for themselves the solution? pretty please

So in your "real" scenario it would be 66%, my bad we are talking about OP's scenario.

>>it is 50%, the situation provided you first got a gold ball, meaning the bal filled entirely with silver was out of the situation entirely.
Wrong, the fact you got a gold ball to begin with means you are more likely to have chosen the box with 2 rather 1. The 2 silver balls are relevant for this distinction, it makes the question much more intuitive.

fpbp
(using threading) spbp

Everyone else are redditors who treat wikipedia as blind gospel

>the first two paragraphs of this image
>you pick box at random
hmm

Attached: 1495930591295.gif (256x256, 490K)

This. Retards think their dumb little program they whipped up will make up for their lack of logic and critical thinking when they're no better than random fotm circlejerks about irrelevant shit nobody else cares about, like this image.

We can end this now.

Attached: DSC_0296~01.jpg (2446x2268, 879K)

unreadable

you pull a gold ball
which one are you more likely to have pulled from, bud?

So, what you're saying is you got it wrong, and you can't code, and you don't accept the general consensus—and the output of every single program here—that it's 2/3?

Okay man, whatever you say.
Imagine being so delusional you think being a code monkey makes you smart lol

>Imagine being so delusional you think being a code monkey makes you smart lol
nobody said that you insecure retard

You only have to imply for others to get what you mean you brainlet, if you actually used critical thinking for once you'd know
You'd also have the side effect to realize it was 50% too

there was no indication in the problem statement that the problem should not be based in reality so its pretty simple to assume that the problem is based in reality, and so 2/3rds

the first.

you inferred something incorrectly
also I assume you're just b8ing at this point
or actually stupid
either way, "hurr I was only pretending to be retarded"

so then you can see that it's not 50/50, as evidenced by literally ALL of the programs ITT demonstrating the concept

anyone who uses the hurr retarded analogy as an argument is an instant dismissal of all opinions in my book
I'm glad I'm able to shrug off everything you said with that one sentence