Previous thread: >>67213689

50%. You've eliminated the double-silver box by drawing a gold. Worry about the boxes not the balls.

Reminder that both 50% and 66% are correct answers due to the shitty wording of the question which leaves it open to interpretation.

It's not a 50% chance that you reached into the silver-gold box though. If you randomly reached into a box and pull out a gold ball, it's more likely that all the balls in the box are gold.

No it's not. It would only be 50% if there were some verbiage stating that the balls are ordered in some way and that you pulled out the FIRST ball. If it's just like the picture, and the text stated that you pulled the leftmost ball out of the box and found it to be gold, the answer would be 50%. Since it says you picked the ball from the box "at random" (i.e., random is in the description twice), it's two thirds chance.

50/50

___If you picked a gold on your first draw___, then the probability that you chose from box #1 is 2/3, since it contains two golds out of a total of three.

You do not have a choice when selecting the next coin, therefore the chance you have to pull a second gold is the same as the selection chance of box #1, 2/3.

We've established that a lot of smart people think it's 50%. There's more than one way to solve math problems, user.

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progressive mathematics

LAUGH AT THIS 50% BRAINLET

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this kind of reminds me of the monty hall problem

Math: sends shit to other fucking planets
Common Core: produced brainlets who think the answer to this question is 50%

If focus on balls as options of six statistics would be 2/3. If focus on grouping in boxes statistics would be 1/2.

If you modeled this ball picking scenario over 3000 attempts at exactly average results you would get 1000 of each result. 1500 would result in a gold ball picked first. 1000 of those would result in a second gold, with 500 resulting in a silver.

2/3 wins.

just let it die already

You guys can try running my code in your browser, it's cool Also, it's proof that the result is 2/3

The wording actually makes it clear that the chance is 50%. We have two results that need to be satisfied by the wording of the problem. We have to pick only one box and the box must contain at least one gold ball.

>next ball you take from the same box

Its 50/50

Double silver box is eliminated

This leaves double gold and single gold one silver boxes.

That means you drew from one of those boxes.
The remaining balls that can be drawn(same box remember) are one gold or one silver
Its a 50/50 chance

en.wikipedia.org/wiki/Boy_or_Girl_paradox

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retardlettes cant read

It says
>next ball you take from the SAME box will also be gold

Same box
We remove box silver/silver from the equation.
Remaining boxes are
Gold/Gold
Gold/Silver

Given it states you already pulled a gold ball its a 50/50 chance

American education

I'm a fucking brainlet and need to know the answer to that water column picture. Would the pressure be higher? How do you calculate that?

I used to be in the same boat as you buddy, but after running my own code and reading , I've completely jumped ship.

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Hydrostatic pressure is a function of depth. So they exert the same amount of pressure.

en.wikipedia.org/wiki/Hydrostatics#Hydrostatic_pressure

Water pressure is only dependent on depth, not container shape.

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left is more pressure

more water = more weight = more pressure exerted

50%lets, when will they learn.

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>>next ball you take from the same box
>Its 50/50
>Double silver box is eliminated
This is your intuition playing tricks on you. You need to think abstractly, and realise that, the same way that the double silver box is eliminated, the middle box is eliminated half the time, leaving you in the double gold box more often than the mixed box.

Yeah, take a selfie.

PEMDAS
20/5(2*2)=20/5*4=20/20=1
common core
????????????????

But that sounds ridiculous. Consider a window with an ocean behind it. Also consider a window with a thin, 2mm thick surface of water behind it. The second window obviously has way less pressure behind it.

Its 50%

If you had 10 gold balls in one and 1 gold ball with 9 other silver balls its still 50% chance

The reason is that the text gives a 100% probability into the equation by stating you 100% drew a gold ball

More balls does not mean higher chance with a 100% certainly stated

What are the odds of pic related? You drew a gold ball out out of each

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The rock extends beyond the water column so pressure exerted is not uniform. It depends on where on the rock you're talking about

>More balls does not mean higher chance

That's literally what it means, champ. Remember, each individual ball has an equal chance to be drawn, but the grouping of those choices is not uniform.

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If you go into the water, cup your hand and push it against the water, you're pushing against the ocean. Where do you think it's going to be harder to push the water, at the top near the surface, or at the bottom?
There are seawalls that are basically water-tight, metal fences. It's why the hoover dam is so narrow at the top, and so wide at the bottom. The pressure goes up the further you go down. It doesn't matter the shape or volume of the container, depth is what determines pressure.

Prove me wrong.

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The problem is poorly stated. You can't "pick randomly" and guarantee picking a gold ball at random.

If the problem is "A box is picked at random.
Given that one of the balls in the box is gold, what is the probability that the box is the gold/gold box?", then the answer is 50%. If the problem is "You pick a ball at random. Given that the ball is gold, what is the probability that the box is the gold/gold box?, then the answer is 66%."

The problem as stated is ambiguous. It's like the missing dollar riddle where you try to con someone by setting up a confusingly worded problem.

Say we have two setups. One like in OP, with 2 balls per box, one like in picrel, with 10 balls per box. One box is all gold, one all silver, one has a gold and the rest is silver.
We set up a game. You pick which setup (2 or 10 balls/box) to play in. You win the game if after going through the process in the OP, you end up getting a second gold ball. You win $200 if you win the 2-ball game, $199 if you win the 10-ball game. Which game do you choose?

Attached: balls.png (1920x1080, 30K)

not 2mm deep, 2mm thick, resting against the pane.

20/5(2*2) = 4(4) = 16

>Given that one of the balls in the box is gold, what is the probability that the box is the gold/gold box?", then the answer is 50%. If the problem is "You pick a ball at random. Given that the ball is gold, what is the probability that the box is the gold/gold box?, then the answer is 66%."
It's 2/3 in either case, as it's twice as likely that you picked a gold ball from the gold/gold box.

Let's label the balls
Gold ball 1 is in box 1
Gold ball 2 is in box 1
Gold ball 3 is in box 2
Silver ball 1 is in box 2
Other silver balls don't matter since if you drew a gold ball already, you know you have box 1 or 2.

There's an equal chance you have any of the gold balls because you drew at random.
If you got gold ball 1, you will draw gold ball 2.
If you got gold ball 2, you will draw gold ball 1.
If you got gold ball 3, you will draw silver ball 1.

Ergo, 2 out of 3 you get another gold ball.

Problem with your solution: You pick a box not a ball.

2/3?

Boxed balls general when?

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No, you still pick a ball. The point is that the probability of gold is dependent on which box you pick in your first selection.

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3 events involve picking a gold ball
2 are in the first box where the other ball is gold
1 is in the second box where the other ball is silver
so, the answer is 3 + 2 + 1 = 6%

2/3

This is somewhat similar to the Monty Hall problem in that the odds change upon additional information.

Pick a box at random:
Box A: 1/3
Box B: 1/3
Box C: 1/3

Pick a ball from that box at random:
Ball A1 (gold): 1/6
Ball A2 (gold): 1/6
Ball B1 (gold): 1/6
Ball B2 (silver): 1/6
Ball C1 (silver): 1/6
Ball C2 (silver): 1/6

Now we state that the first ball was gold, so we eliminate all cases where there was a silver ball. Since all cases were equal likelihood, they still are:
A1: 1/3
A2: 1/3
B1: 1/3

If you picked A1, you will pick A2, which is gold.
If you picked A2, you will pick A1, which is gold.
If you picked B1, you will pick B2, which is silver.
Therefore, again, 2/3 chance your other ball is gold.

>Choose container
>"it's a gold ball"
>impossible to have picked 2silver container
>impossible to have picked silver ball out of 1gold container
>picked 1gold container, get silver fail
>picked 2gold container, get gold pass
50/50

Wrong, look up the monty hall problem it's similar.

When you pull a ball out in your hand and it is gold then you know that there are only 3 balls left that are unknown to you. 1 is silver, 2 are gold. 2/3

2/3

en.wikipedia.org/wiki/Bertrand's_box_paradox

But there are 2 ways you could have got a gold ball out of container 1.

Are the mods literally dead?

This is shopped btw.

The common core way is right, brainlets.
And that's not even "common core," common core is a state standard, not a teaching method.
Also, that's how PEMDAS is supposed to work.
It goes:
Parentheses (or brackets or whatever the fuck)
Exponents
Multiply/Divide (done left to right)
Add/Subtract (done left to right)
First you take the parenthesis
2*2=4
which leaves you with 20/5(4)
You do 20/5 first, since you do whichever comes first.
You are then left with 4(4) which gives you 16.
The first way isn't right, and it was obviously shopped.

2 ball game, because it pays more and the probability of wining is the same.

You have a cute calligraphy.
Be my friend.

No, since you eliminate a box after you reach in and not before. You're equally likely to get a silver or gold ball.

I'd play the 10 ball game. In 10 out of 11 plays I would be guaranteed a second gold ball.

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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))

Same anime posting retard gaslighting dumbnons into thinking it isnt 50:50. Post your reasoning or gtfo

3305, 1639

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>There's more than one way to solve math problems, user.
Just run a Monte Carlo, jesus.

>there are people that still won't believe it's 2/3 after seeing this and a dozen incredibly simple explanations

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Let's give 's monte carlo analysis a go and see whether you're right. I've added a couple extra 0's to the range line to ensure we aren't just seeing a fluke.
334249, 166274
Total is 334249+166274 = 500523 (about half of the total of 1 million ball picks I did, since it discards when the first ball is silver)
334249/500523 = 0.6678
Better luck next time, user. It's 2/3.

the laughing anime qt shitposting is probably more effective at convincing people that it's not 50/50 in the end

God, If you think the answer is 50% , you don't deserve more than being mocked.

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No 50% fags will reply to this.

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Why are you doing multiplication before division?
In PEMDAS, multiplication and division have the same priority and you solve them left to right.
So it would be,
20/5(2*2)
20/5*4
4*4
16

The only reason you would get 1 is if you follow the rule that implied multiplication is done before signed multiplication and division, which is an old way but hasn't been taught in school for decades.
I actually think that rule makes more sense and would rather put a sign into my equation if I want multiply something by the sum of enclosed brackets.

More like no 50% fags can code.

What's there to code? already wrote it. It's very simple and legible code too, so if any 50%fags can spot an error in it that might cause it to incorrectly report 66%, they're welcome to point it out.

#include
#include

int main(void) {
srand(time(NULL));
int box[3][2]={{0,0},{1,0},{1,1}};
int pass=0; int fail=0;
int i=0; for (;i

Exactly. 50% fags couldn't code a test to find out if they were right or wrong, and can't understand such code.

>if (box[k][0]==1) {
Top lol. Always picking 0 first when you could pick either 0 or 1.

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You'll never pick the second ball in box 1 first, and that one is gold. You're omitting a case.

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#include
#include

int main(void) {
srand(time(NULL));
int box[3][2]={{0,0},{1,0},{1,1}};
int pass=0; int fail=0;
int i=0; for (;i6663,3337

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> if (box[k][0]==1) {
You have to shuffle the balls in the box.

Let's tidy that up a little and make it clearer what it's doing.
#include
#include

int main(void) {
srand(time(NULL));
int boxes[3][2]={{0, 0}, {0, 1}, {1, 1}};
int pass=0; int fail=0;
for (int i=0; i

>int ball = rand()%2;
could just be int ball = rand();
the %2 is redundant

rand() returns an int between 0 and RAND_MAX, which is 4 billion or 65535 depending whether your compiler sucks. I use ball directly to index the array, so I do need it to be either 0 or 1.

For otherball, I just used !ball because it's 0 or 1 and I wanted the other one.

shit you got me there. i'm too used to java's random.

Remove the int otherball line and just use !ball in the second if statement.

too many useless tabs. so much wasted space.
#include
#include

int main(void) {
srand(time(NULL));
int b[3][2]={{0,0}, {0,1}, {1,1}};
int p=0; int f=0;
for (int i=0; i

>[bo][ba]
What do you think this is, a fucking bubble tea?