How intelligent is Jow Forums

...

>you pick a box at random
we know that we got a gold ball so box 3 is imposible. even though the first box has more gold balls we picked a BOX at random and not a BALL at random so unlike the 2/3 and 1/3 ratio like some people claim it is 1/2 and 1/2 as the boxes themselfs have equal propbability. this can be shown more clearly by saying the first box has 100 gold balls instead of 2. so now we have 100 gold balls in the first box, 1 in the second, and none in the third for a total of 101 gold balls. is it now a 100/101 and 1/101 chance or is it still arround 1/2 and 1/2?

Why the fuck do none of you retards use the obviously correct answer of 33%?
To pick a gold ball after a gold ball, you'd have picked a 2 gold balls box from the beginning, and the chance to do that is 33%.
It's the only way that two sequential picks from the same box net you two golden balls.

Because we are not stupid mongrels like

>it's another episode of computer cissies can't into basic math results

>e picked a BOX at random and not a BALL at random
both were picked at random

conditional probabilities and asymmetry are such massive fucking brainlet filters

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>The boxes have equal probability
Well, they have equal prior probabilities, but not equal posterior probabilities. Because we saw a gold ball after choosing the boxes, the probability that we picked the box with more gold balls is higher. If you don't know what the words "prior probability" and "posterior probability" mean, go read a book on statistics, because they are fundamental to understanding conditional probability, and this is a conditional probability problem.

Because picking the first gold ball is a given. The problem starts after you already have already picked a gold bold.
The word "random" is there to confuse brainlets such as yourself.

Except it's already stated that you have picked one of the two boxes that have a gold ball in them. The third box doesn't even factor in since it's given that you don't pick that one.

on the wording question: the important part is that the ball you pick is random, and only happens to be gold. this operation being random "enhances" the correlation between a gold ball being picked and the box with 2 gold balls being picked.
this is different, for example, from a scenario like "a third person looks into all 3 boxes, picks one, picks a gold ball from it and places it outside the box", to which the answer to the OP question would indeed be 50%

Let me help you guys out in a language that the people who don't understand speak. Someone circle a door.

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so what is the fiftypercenter-argument against
>drew a ball, it was gold
>this gold ball could be any of the three gold balls
>in two of the three (equally likely) cases the next ball is the other gold ball
>in the last of the three cases the next ball is the silver ball

I mean literally just think of the gold balls as red/green/blue and the silver balls as white
if you struggle with thinking of balls with the same colour as distinct

forgot to erase the postnumber I clicked to open quickreply

nice try FBI

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so would the chance of picking the box with the silver ball change if we put more gold balls in the first box?

>so would the chance of picking the box with the silver ball change if we put more gold balls in the first box?
the probability of picking the silver ball conditional on that you picked a gold ball would change

they probability of picking a silver ball in the absence of further information is constant.

No you fucking 66% brainlet, the question is two gold balls on purpose, and it wants to know the chance of picking a gold ball again, not gold ball 1 or gold ball 2, and certainly not a fucking blue/green whatever the fuck you wanted to use to obfuscate the problem.
66% brainlet btfo

I guess I will do it myself since no one will play a long.

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It's not the same problem. Gtfo with your obfuscation propaganda 66% shill.

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

Why??

I thought I was 50% smartlet but wiki says im a fucking brainlet... Why Jow Forums?

The trick is that the starting point is not having the three boxes, but having just two boxes:
One with a gold ball and one with a silver ball

Not the same problem. Gtfo 66% shill.

You now have the option to choose again. With the new information do you have a better chance of avoiding the border agent if you switch doors. Why or why not.

I am simplifying it, flat earther.

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OP assumes the chance the ball was drawn from the first box and the second box are equal. They're not.

en.wikipedia.org/wiki/Monty_Hall_problem

No you're not, your shilling 66% propaganda by confusion and obfuscation of the original problem under the friendly and helpful guise of "simplifying" it. Literal fake news gtfo.

I decided to type down some stuff without thinking and came out at 0.75
am I doing maths like a 50%'er now? c:
0.5*[0.5 + 0.5] + 0.5*[0.5 + 0] + 0.5*[0.0 + 0.0]
= 0.5*1.0 + 0.5*0.5
=0.75

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Box 1 has 100,000 gold balls.
Box 2 has 1 gold and 99,999 silver balls.
Box 3 has 100,000 silver balls.

You draw a gold ball. Are you braindead enough to say the chance to pick another one from the same box is 50%?

Fake news 66% propaganda.
The variables of other boxes are no longer in the original problem's equation. The box from which you took the gold ball is now isolated from all others and you only have a 50% chance at that ONE pick to take out another gold ball, just like you only have a 50% chance of tossing a heads of tails.
Btfo.

you are supposed to only change the 2 gold balls to a high number

Except that's not the question. Once again another 66% shill and it's obfuscation propaganda btfo.

>The box from which you took the gold ball is now isolated from all others and you only have a 50% chance at that ONE pick to take out another gold ball
yes but there are three equally likely but distinct ways you could end up in that situation
two of those situations will give you a gold ball
one of those will give you silver

I see that the brainlet filter is working like a charm though

you might have picked from one of two boxes. if you picked from the one with gold balls, drawing another would give you a gold ball.
if you picked from the one with 1 gold ball and the rest silver, your will draw a silver one.

literally 50%.

from random import randint

boxes = ((1, 1), (1, 0), (0, 0))

N_EXPERIMENTS = 1000000

n_success = 0

n_total = 0

for _ in range(N_EXPERIMENTS):
box = randint(0, 2)
ball = randint(0, 1)
if not boxes[box][ball]:
continue
n_total += 1
another_ball = 1 - ball
n_success += boxes[box][another_ball]

print(n_success / n_total)

> Outputs 0.6677187284284113

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Yes and you exist only in one universe with one of those distinct ways actually happening, therefore a 50% chance by the elimination of a double silver box purely because you drew a gold.
66% shills and their interconnected parallel universe obfuscation propaganda once again btfo.

Except it illustrates the point that because the first box has more gold balls, if you pick a gold it's more likely you picked it from the first box, and thus more likely you'll pick another one.

Not the same problem.
Yet again another 66% shill and it's obfuscation propaganda btfo.
The truth will prevail.

You are objectively stupid and you're grasping at staws. Just because you can't comprehend a concept doesn't mean it's incorrect, it just means you will never be good in stem.

What do all of these things have in common:

medicaldaily.com/nigerian-student-debunks-homosexuality-magnets-university-lagos-doctorate-uses-science-prove-gay

youtube.com/watch?v=8GyVx28R9-s

en.wikipedia.org/wiki/Dunning–Kruger_effect

(You)

Why is it not the same problem, user?

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That's not the problem.
>more obfuscation and irrelevant 66% shill propaganda
Swatting 66%ers like flies today.

Im so intelligent
that im not going to waste my time on your stupid question, and instead fuck my assisstant and go for a swim in my scyscraper sized money bin. So long fuckers, quack quack

>herefore a 50% chance by the elimination of a double silver box
wow, not only did you trigger the usual brainlet-trap in the problem of not considering the two possible gold ball draws from the first box to be distinct
but you also went ahead and fell for the silver ball box mattering too

you are either on the tallest dunning kruger mountain ever, or a troll with a lot of sparetime

GOING TO BED

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Because you didn't get 50%
Stop trying to gaslight me 66% shill. Accept your brainletism and try to learn why you're wrong.

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>Poe's law

Do the math on that one, oh you can't, 66% brainet BTFO

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ITT 50% user is retarded and after being repeatedly BTFO for the next hour he'll say he was just pretending to be retarded.

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>and after being repeatedly BTFO for the next hour he'll say he was just pretending to be retarded.
wishful thinking

>more 66%er demoralisation and obfuscation propaganda
Imagine doing it for free too.
Yet once again 66% shills btfo

really getting your troll in aren't you

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>exposure of truth is trolling
Nice try. Once again 66% shill and it's gaslighting propaganda btfo.

what brainlet 66.(6) percenters do not understand is that once the gold ball has been drawn, the 2S box is out of the question.
furthermore one of the remaining 2 boxes is left without a gold ball. and it's only a question of which one.
B1 - draw and you get G
B2 - draw and you get S
no other outcomes possible, 50% simple as

common core at its finest, OP.

State your education OP

That's like saying the chances for you to win the lottery is 50%, because you either win or lose

Nope, not comparable at all. Read the damn question.
Law graduate from Harvard.

probably should have taken high school statistics

My answer is actually 2/3, these are my first two posts in this thread, but the example to the lottery above is disingenuous.

Now imagine every ball has a number that you can't see
Box 1 has G1 and G2
Box 2 has G3 and S4
Box 3 has S5 and S6

You pick a box at random and somehow manage to get a G ball. There are 3 scenarios:
1. You picked G1, so the second ball is G2. You managed to pick box 1 (1/3)
2. You picked G2, so the second ball is G1. You managed to pick box 1 (1/3)
3. You picked G3, so the second ball is S4. You managed to pick box 2 (1/3)
In the end, the chances for you to be in front of Box 1 is 1/3+1/3 = 2/3, while the chances for you to be in front of Box 2 is just 1/3

The original post postulates that the first ball you draw is a gold one, i.e the third box is never picked and the probabilities of the first two are all that matter.

it's FROM THE SAME BOX you RETARDS
just how difficult is it to read the damn problem goddamnit 66% shills the double silver box was never an option

>shill
you know shill means you're trying to sell something right

and they're trying to sell us misinformation you goddamn npc

What would be the odds if there were an infinite amount of boxes with two silver balls?