Above 100 IQ faggots, show yourselves, are there any left?

above 100 IQ faggots, show yourselves, are there any left?

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Other urls found in this thread:

whyevolutionistrue.wordpress.com/2018/02/20/the-answer-is-2-3/
en.wikipedia.org/wiki/Bertrand's_box_paradox
twitter.com/SFWRedditImages

2 out of 5? How can this be any difficult?

minimum iq detected

out of 5?
5?
WHERE THE FUCK DO YOU GET A FIVE!

50% I guess ...

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nigger detected

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50 % chance. by pulling a gold, you totally disqualify the right box, leaving you with the center and left boxes. if you took a gold ball out from each, then the remaining possibilities would be 1 silver (if it is the middle box) and one gold (if it is the left box). so 50%

Hi bois. I have IQ 479. Ask me what you want to know.

the picture already did.
i find it hard to believe that your IQ is that high when you cant even see the obvious.

2/3...

wrong

but if you pulled a gold, you are twice as likely to have grabbed from the first box, so not equal probability between the two boxes

This is correct.

We know it was the gold/gold or gold/silver. It cannot be silver/silver.

It’s the same chance between the two boxes.

none because you didn't work for any of these things

is the ball replaced or removed from the box?

>It’s the same chance between the two boxes.
just look at this fucking nigger right here.

makes no difference since the question is about the OTHER ball in the box.

you can have some chocolate tho

assume removed

Close, but there are two ways to draw a ball from the first box. You could have picked either ball 1 or ball 2, and both will result in drawing another golden ball from the same box. So that leaves a total of three options, two of which will give you a golden ball.

So the answer is 2/3.

Checkmate libtards

there cannot be more than one correct answer to any situation ever so the answer is 70%
also if you get it wrong god hates you

brb need to go yell into my toilet

Of course.

The confusion comes from what sounds like an ambiguous wording of the experiment.

The way it’s worded, the experiment doesn’t start until only two boxes have already been identified, one with a gold ball and one with a silver ball.

Now, if you worded the experiment differently:
>what is the probability of first drawing a gold ball and then another gold ball from the same box?
you'd have a different probability.

The confusion stems from people mistaking the actual experiment for my greentext above

1/3, sum of probability of a.picking the first box and getting a the second one gold all the time 1/3*1 and b. picking the second one but not getting the second one gold 1/3*0 and c.no gold 0

thats a different experiment with a different answer because picking a gold ball at the start is a 50/50 so you would end up with a 1/3 as the final answer which is half of 2/3 which is the correct answer.

>The confusion stems from confusing*
the actual experiment with* my greentext above

That was weirdly worded, sorry.

There are 5 balls left in the boxes, moron. In five tries you will get 2 yellows, so that's the chance to get it.

P(B|A)=P(AnB)/P(A)
P(second ball gold|first gold)=P(Both Gold)/P(First gold)= (1/3)/(1/2)=2/3

any other answer is autism

no you retard. The probability is greater than 1/2 because you must account for the greater probability that you grabbed from the first box given that you pulled a gold ball.

Except this isn't Schrödinger's cat.
Theoretical science rarely translates well to the real world.

put you picked a gold ball, so it cant be the box with 2 silver balls.
are you REALLY this stupid or did you not even read the question?

50/50

1 2 3
GG G G
GG G G
SG S G
SG G S
SS S S
SS S S

OP never said the first box, which had the first ball withdrawn, was out of game. So, it has to count the three boxes in the chance. Then, it's 2 out of 5 tries, being 5 the number of balls left.

the correct answer is still 2/3.
your lack of ability to calculate probabilities doesnt translate well to the real world.

This. It's 2/3.

This is just a variation on the classic Monty Hall problem.

No.

>What is the probability that the next ball from the same box will also be gold?
That and only that is the question we need to answer.

Nobody asks what the probability of picking the first gold ball was. That’s completely irrelevant.

2/3 bc monty hall problem

Are you retarded? Schrödinger has nothing to do with this. It's simple probability. There are three possible ways by which you arrived at drawing a golden ball.

You picked ball 1 from box 1.
You picked ball 2 from box 1.
You picked ball 1 from box 2.

In the first two cases, you will pick another golden ball. In the third case, you will pick a silver ball.

2//3.

How is this difficult to understand.

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the answer is 2/3, but monty hall problem is different.

Mama always said stupid is as stupid does so I sepuku'd her kuz that's just sum dumb shit
420

This is the intelligent answer
Its either 1/2 or 2/3 depending on the actual question

That has nothing to do with the intelligence. It's just math - probability calculation.

seppuku is a suicide, your mama raised an idiot.

66.66666666666666666

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see
you picked a gold ball, you are agreeing with an idiot.

suppose you have two boxes. each box has a thousand balls. box one has 999 gold balls. box 2 has 1 gold ball. you pull a gold ball. how likely do you think it is you'll pull another gold ball?

hope that clarifies things.

IMBECILEEEEEES THIS IS A FAT TAIL DISTRIBUTION YOU GUYS DON'T KNOW ANYTHING ABOUT RISK TAKING!

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same logic

FFS if pick a ball and don't look at it that ball is BOTH silver and gold because the probability of picking one color is the same as the quantum chance. So, if there's no observer, the ball is gold and silver at the same time (that assuming that the gold and silver have the exact same mass and shape). Do you even physics?

SIMILAR logic.
if it was monty hall you'd be changing the box you pick from.

>but monty hall problem is different.
yes, because there the game show host will ALWAYS open a door with a goat.

THAT affects the outcome.

here, we have no such effect. the probability of picking a gold ball from one of two boxes, only one of which contains a gold ball, is obviously .5

I'd say it's about 77% chance that the other ball is gonna be gold

except its not .5
what an idiot you are.

semantics

Then you have 999/2, so you are absolutely likely to get another gold one.

Lmao, like "above 100" is impressive. Even 100 is monkey tier.

here semantics are important.
if it was a math test you wouldnt go and argue with your teacher for not giving you the points after you misunderstood the question.

its not, but i like saying that anyone who gets this wrong is below monkey tier.

no, i would just think im smarter than him for not understanding me, while acknowledging he probably thinks similar

You just disproved yourself dum dum.

>here semantics are important
exactly.

the experiment is cleverly but misleadingly worded to sound like a monty hall problem, when it actually isn't.

I''ll rate you a 2/10, the trolling is way too obvious

If the teacher is a fucker who likes to mess with people by chosing confusing words I'd beat the shit out of him till he gives the points. It's not my fault he doesn't speak like a civilized human being.

posibility 1: we put our hand in the first box therefore the probability of the second chosen ball to be gold is 100%

posibility 2: we put our hand in tge midle, therefore the next ball would be silver => 0% gold

the diference is what box we randomly chose at the begining, for one it will be always gold and for the other always silver (second ball) so its 50%

2/3

nope, see: and

7 out of 11 chances to take gold ball.

Then it's 1/4, 25%, because you divide the possibility in two because there are two balls in each box then you divide again because there are two boxes.

first ball is always gold. read op pic

there is literally 0% reason someone would write that problem

Since we don't know what there is in each box and can never look inside, it automatically disqualifies the boxes' as something we can talk about. What's in each box is completely irrelevant to the actual meaning of gold, silver, ball, and their every day use

Most people are wrong with this.
Even smart ones.
When the question is: the ball you get is golden or silver, so how high is the probability that it is golden? ... then people say: 50%
That's not stupid.
In this case it is not correct, but maths are often not obvious.

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sure there is. confuse Sup Forumstards into thinking a straight 50-50 chance was a monty hall problem

we already picked one ball. why do they count it?.....

that's a really good way to piss a 8 year old off

Being a complete idiot with good memory I remember mechanically solving such problems in school using bayes theorem, I've completely forgotten how to do it

if god exists why would someone be retarded enough to make this problem

And you forgot to tell that the boxes are completely racists for being a problem created for white privileged man who wants to trouble the mind of the poc with math.

It's true that we already picked one, but there are three possible ways by which we could have picked it.

>Most people are wrong with this.
most people are wrong when they try to solve the monty hall problem.

but the question here is different.

in order to make THIS problem into monty hall, you'd have to ask,
>what are the chances than user draws two balls from the same box

Do you agree or disagree that changing the question changes things?

the answer is you can't play hearthstone in hell/10

1/2 faggot

The answer is 2/3 for those of you still arguing.
This is call the Bertrands Box.
Heres an answer sheet for those of you who couldn't get it right.
whyevolutionistrue.wordpress.com/2018/02/20/the-answer-is-2-3/

This guy didnt read.

en.wikipedia.org/wiki/Bertrand's_box_paradox

If one of the balls are gold, you have chosen one of the two containing at least 1 gold ball. Since it is equally likely that you chose one or the other, and you have only confirmed that it is not the one containing two silver balls, you must have a 50 % chance of having picked the one with 2 golden balls.

>being a complete idiot with good memory
>I've totally forgotten
Man, you better accept that then you are just a complete idiot.

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You fucking moron.
That IS the question, read the OP.

>>what are the chances that* user draws two gold* balls from the same box

these are the type of comments i really should proofread

No.
2/3.

At first I thought how the fuck can it be 2/3 but then I actually thought about it as opposed to surface skimming it and felt like a dumbass for not noticing it in the firstplace.

>it is on a blog
>it must be true and correct
Oh you.

It says from the same box retards. The box with the two silver balls is out of the equation.

you start with GGSGSS after the first pull you have GGS so the probability is 2 (gold balls) out of 3 (total balls).

Don't be fuckwits.

THAT DIDNT CHANGE THAT YOU DIDNT READ THE OP.

OP here.

This is know as the Betrand's Box paradox and the correct answer is 2/3.
If you dont believe or understand me, go to youtube and search for a video that explains it.

No it's not, lol. If you first take out a gold ball, you already eliminated the box with no gold balls. So the next ball will either be another gold ball, or a silver ball, since those are the only 2 options left. 1 possible next gold ball in 2 possible outcomes, 1/2.

>someone else posted the wiki article showing the same thing
Oh you, I chose the blog as the first result, no matter who you google though, it will be 2/3.
You're not very smart if that's your best defense against being clearly wrong.

see
you are an idiot.

This: