Can Jow Forums find the answer?

1/3, you dont know that is "middle box"

57.5%

en.wikipedia.org/wiki/Bertrand's_box_paradox

62.5%

>mfw brainlets post 50% before reading the thread and realizing they don't know basic statistics

Attached: 1535481011618.jpg (805x604, 132K)

#include
#include
#include
#include
#include
#include

std::mt19937_64& get_random_gen_ref() {
thread_local static std::random_device rd;
thread_local static size_t thread_id = std::hash{}(std::this_thread::get_id());
thread_local static long long time = std::chrono::high_resolution_clock::now().time_since_epoch().count();
thread_local static std::seed_seq seeds{
static_cast(time >> 32), static_cast(time),
static_cast(thread_id >> 32), static_cast(thread_id),
rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), rd(), //partially fill mt's state with random data
};
thread_local static std::mt19937_64 mt{seeds};
return mt;
}

int main()
{
enum Ball {silver = 0, gold = 1};
using Box = const std::vector;

Box boxes[3] {
{silver, gold, gold, gold},
{silver, silver, gold, gold, gold},
{silver, silver, silver, silver}
};
auto random_range = [](int a, int b){ return std::uniform_int_distribution{a,b}(get_random_gen_ref()); };

int possible_outcomes=0;
int event=0;
for(int i = 0; i

55%

5/9

Not this shit again.
>A GOLD BALL IS NOW INSERTED INTO THE MIDDLE BOX.
Thanks Satan.

Assuming the middle box is random you can have three different configurations (4+1, 2+2, 0+4), (3+1, 3+2, 0+4), (3+1, 2+2, 1+4) You have an equal chance for all three.

With the first configuration you have an equal chance to choose any box HOWEVER since the box you choose HAS to have a gold ball in it, it cannot be the last box. This leaves you with two situations where you either pick the box with 4 gold balls and 1 silver ball which means you are guaranteed to have 1 silver ball left or you pick the box with 2 of each which means you are guaranteed to have 2 silver balls left. Now we have to figure out how likely you are to pick a gold ball from between these two. 4 gold balls give you 1 silver ball and 2 gold balls give you 2 silver balls so you have a 2/3 chance to get one silver ball..

With the second configuration you have an equal chance to choose any box HOWEVER since the box you choose HAS to have a gold ball in it, it again cannot be the last box. As above you cannot be the last box and the same rules apply, 3 gold balls and 1 silver ball or 3 gold balls and two silver balls. 3 gold 1 silver ,3 gold 2 silver. 1/2 chance to get 1 silver ball

The third configuration is different as the last box can have a gold ball in so we factor in all three. 3 gold 1 silver, 2 gold 2 silver, 1 gold 4 silver 1/2 chance to get 1 silver ball.

Adding these all up gives us (2/3+1/2+1/2)/3 which is 5/9 or 0.55555555555555555555.

If the picture is to be taken literally then it is 1/2 as you take the second configuration as the whole problem.

nice highschool math problem, no one is going to solve your homework

It's always 1/2, you either pick it or not.

Attached: 1516233975379.jpg (657x527, 32K)

>every day you either die or not
>therefore you have a 50% chance of dying every day

Yes. Life is a fragile thing.

based and redpilled

55.55%

There's a 75% chance of picking a gold ball from box 1, and a 60% from box 2. That's a ratio of 5:4, add those together to get 9 and divide into 100 for 11.11%. Multiply by 5 and 4 and you've got 55.55% / 44.44%.

2/3

0%
It can't "only contain 1 silver ball" since all boxes have at least 4 balls. So any of the boxes would have 2 (or 3) other balls in addition to the silver one and in addition to the gold ball that was taken out.

There are 2 boxes. One has 1 silver ball and the other has 2 silver balls. What is the probability that you pick the box with 1 silver ball? 50%.

5/9

There is no answer to be found, only a correct explanation to invent.

based and retard pilled

Only consider the 2 boxes with golden balls. The single silver ball probability is 1/2 at the box level. In-box golden ball probability is 3/4 and 3/5.

1st box golden ball probability is 1/2 * 3/4 = 3/8. 2nd box golden ball probability is 1/2 * 3/5 = 3/10. Golden ball probability ratio of box 1 relative to box 2 is (3/8) / (3/10) = 1.25 : 1.

Therefore, total probability of picking the first box when observing only golden ball pickups is 1.25 / 2.25 = 55.55%.

The key to understanding is 1. Write down the obvious probability chains like 1/2 * 3/4. And 2. Interpret the question as calculating the ratio of "correct" qualified observations relative to all qualified observations, which in this case are golden ball pickups from first box / golden ball pickups from any box.

.333 * .25 =.083
.333 * .4 =.133
.333 * 1 =.333

SUM = .549
This is high school tier statistics

Wow scrolling through these other answers I didn't realize Jow Forums was this retarded.
I will never overestimate you again.

>there are people on Jow Forums that don't even know Bayes' theorem
I guess I shouldn't be surprised.

50/50
You either picked the gold ball out of the left or middle, and picking either of the two is equally likely when picking the box you're going to choose from at random. The question tells you that you've picked a gold ball so the chance of getting a gold ball in either doesn't matter.

Why does it say "A GOLD BALL IS NOW INSERTED INTO THE MIDDLE BOX" instead of "A gold ball is inserted into one of the boxes at random"?

Do your own homework, OP.

I prefer doing JS for these threads because then anyone can just open their JS console in their browser and put this in.

(function(){
const SILVER = 0
const GOLD = 1
const ITERATIONS = 1000000
let boxes = [[SILVER,SILVER,GOLD,GOLD], [SILVER,SILVER,GOLD,GOLD,GOLD], [SILVER,SILVER,SILVER,SILVER]]
let times_chosen = boxes.map(_=>0)

for(let i = 0; i < ITERATIONS;i++){

let selected_box = Math.floor(Math.random()*boxes.length)
let selected_ball = Math.floor(Math.random()*boxes[selected_box].length)
if(boxes[selected_box][selected_ball] !== GOLD){
i--
continue
}
times_chosen[selected_box] += 1

}
let sum = times_chosen.reduce((x,y)=>x+y)
let ratios = times_chosen.map(x=>x/sum)

console.log("box pick ratios:",ratios)
})()

Oh I fucked up the boxes variable. You guys can fix it.

you didn't read the question.

hint: they told you that you picked a gold ball. one of the boxes has no gold balls.
there are really only two boxes in this problem

read the question
there are 3 boxes
you pick a box
you pick a ball

it's 5/9 or 55.56%

Besides the fact that I fucked up the first box by making it 2 gold 2 silver, instead of 3 gold 1 silver, what's wrong with it?

what the fuck does this have to do with Jow Forums or technology?
take your fucking cancer to /b/, faggot.

>bayes nets and probability are unrelated to technology

Attached: its-ok-youre-just-retard-17680126.png (500x476, 107K)

The question wasn't "what is the probability that the next ball you pull will be silver" the question is "what is the probability that your box contains exactly 1 silver ball?"

3 boxes. One contains zero gold balls, therefor we do not have this box since we previously pulled a gold ball. That means we have 1 of 2 boxes. One box contains 2 silver balls, the other contains 1 silver ball. Therefor the probability that our box contains 1 silver ball is exactly 50%.

This isn't an exercise on critical thinking or probability. It's literally kindergarten level reading comprehension. You failed.

Attached: 1505759465846.jpg (1241x914, 114K)

I can't believe there are people on here that failed 6th grade math and insist that they are correct.

you can multiply, but apparently you can't read.

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

It is 1/2. But Jow Forums is too dumb to read and thinks the question is harder/smarter than it really is.

> thinks probabilities and technology is the same topic
what a fucking spastic.

Let A = Event of selecting left box
Let B = Event of selecting gold ball

P(A) = 1/3
P(B) = (1/3)(3/4) + (1/3)(3/5) + 0 = 0.45
P(B|A) = 3/4

P(A|B) = P(B|A) P(A) / P(B) = (1/3)(3/4) / 0.45 = 0.25 / 0.45 = 5/9 = 0.5555...

If I'm wrong, it's because I got P(B) wrong. I think this is the right answer though.

Read the fucking question again dipshit. You are answering a question that was never asked.

How about you read the question, you dipshit?
Hint: the question doesn't say that you randomly select between box 1 and box 2.

The question asks what is the probability of selecting the box that contains only 1 silver ball. Only one box contains only 1 silver ball, and that is the box containing 1 silver ball and 3 gold balls (the left box).

Solution to original Bertrand's Box problem is 2/3.
Solution to current variation is 5/9.

Both are best solved using Bayes Theorem.

>what is the probability that the same box contains only one silver ball

Read it. Your box contained a gold ball which means it could only be one of two boxes. One box has only one silver ball.

not 5/9
not 55.56%
it's 1/2, 50%

what does your ilk find so funny about pretending to be retarded?

>you either win the lottery or you don't
>it's 50/50

Just because there are only two possible boxes does not mean that the boxes have equal probability. Indeed, the knowledge that we drew a gold ball gives us greater information about which box was more likely to have been selected.This is a conditional probability problem. What is the probability that we selected the box with only 1 silver ball, given that we drew a gold ball from it? The answer is 5/9.

If I hand you two lottery tickets and tell you to scratch one, what is the probability of you winning the lottery?

Oh, BTW, one of the tickets is a winner.

You have 2 lottery tickets. 6 numbers have to match to win the lottery. One of them has 1 number scratched off, and it matches. One of them has 5 numbers scratched off, and all of them matches. You know that exactly 1 of the 2 tickets is a winner.

What is the probability that the first ticket is the winner?

0.55555

brainlets will never learn

not this fucking thread again

let positive = 0
let negative = 0

function sample(array) {
return array[Math.floor(Math.random() * array.length)]
}

for (let i = 0; i < 100000000; i++) {
const boxes = [
["gold", "gold", "gold", "silver"],
["gold", "gold", "silver", "silver"].concat("gold"),
["silver", "silver", "silver", "silver"],
]
const box = sample(boxes)
const ball = sample(box)
if (ball === "gold") {
const silverBallCount = box.filter(ball => ball === "silver").length
if (silverBallCount === 1) {
positive++
} else {
negative++
}
}
}

console.log(`Probability: ${(positive/(positive+negative)*100).toFixed(2)}`)

fuck your image quality

P(S | G) = 0.5
if there is a gold ball added to the middle box then
P(S|G) = 0.43

Actually it does. It says "you got a gold ball", which DOES limit it down to 1 and 2. What's interesting about this is a blue/white dress problem. There are in fact two ways of reading the question - your either reading a story for which picking a gold ball is the last relevant fact, or you read it as a statistical probability question from start to end.

The point is to not look into the box when you grab one brainlet. Have you never done probability?

>There are in fact two ways of reading the question - your either reading a story for which picking a gold ball is the last relevant fact, or you read it as a statistical probability question from start to end.
Both of them result in the same outcome. Which is 5/9 and not 1/2.

P(B1|G) = P(B1 & G)/P(G) = (3/13)/(6/13) = 1/2

The balls do not all have equal probability of being picked.

why not?

Read the question.

Since you already have a gold ball it's either left or middle box.
There are 3 gold balls in left and 3 in middle.
The number of silver balls doesn't matter, because you have already picked a gold ball.
So the chance is 3/6.

57.7?

my answer is consistent with my reading of the question

No it's not.

2/3?

TIL conditional probability doesn't exist.

import time
import os
import pyautogui
import subprocess

BOX_1 = ['gold', 'gold', 'gold', 'silver']
BOX_2 = ['gold', 'gold', 'gold', 'silver', 'silver']
BOX_3 = ['silver', 'silver', 'silver', 'silver']

# TODO finish code

oh shit okay.

P(B1|G) = P(B1 & G)/P(G) = (1/3 * 3/4)/(6/13) = 13/24

;(function() {
function sample(array) {
return array[Math.floor(Math.random() * array.length)]
}

let positive = 0
let negative = 0

for (let i = 0; i < 100000000; i++) {
const boxes = [
[ "gold", "gold", "gold", "silver"],
[ "gold", "gold", "silver", "silver"].concat("gold"),
/* image told us we didn't pick this box (removed) */
/*["silver", "silver", "silver", "silver"],*/
]
const box = sample(boxes)
/* image told us we picked a gold ball, no need to test for it (removed)
const ball = sample(box)
if (ball === "gold") {
*/
const silverBallCount = box.filter(ball => ball === "silver").length
if (silverBallCount === 1) {
positive++
} else {
negative++
}
/*
}
*/
}

console.log(`Probability: ${(positive/(positive+negative)*100).toFixed(2)}%`)
})()

Oh, look at that. The answer is 1/2 or 50%.

it is 1/2 anyone saying otherwise is a Swift programmer

*sips* ahh, now THAT was a theorem

do elaborate

That's not the question. I can't imagine how you survive in society without being able to read.

Since the question states i already got a gold ball its 0.5.

this

explain your case or go away

Literally just follow the steps in the question, which is exactly what the code did before you messed it up. You don't know that you picked a gold ball until you looked at it. You can't just magically decide you're going to ignore a box.

Picture effectively says you have one of two boxes, and it tells you how many silver ball are in each.

No. The picture says you pick a box AT RANDOM, which can include 1, 2 or 3. Only AFTER you look at the ball, do you see that it is gold. I'm amazed at your lack of reading skills.

my program with n = 1M gives

0.5543179815458922

>The picture says you pick a box AT RANDOM, which can include 1, 2 or 3. Only AFTER you look at the ball, do you see that it is gold

So fucking what? At that point you have 1 of two boxes and you know at that instance that there if a 50% chance it's the box with 1 ball and 50% chance it's the box with two balls. The bullshit before hand doesn't fucking matter any more.

No you don't. Read the fucking question. Look at . It performs EXACTLY the procedure described in the question, without taking shortcuts that are just plain WRONG.

There's a chance the ball you grabbed will be silver and the question would never have been asked. This chance is bigger when you pick the second box. So even if you pick boxes with equal chance, you don't pick boxes with equal chance when given that your ball is gold.