ITT we determine who should get to reproduce to gift the world with intelligent offspring

ITT we determine who should get to reproduce to gift the world with intelligent offspring

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%50 or 1/2? Organalizaon

50%
I studied maths at uni so I guess I have an unfair advantage, I'm socially retarded.

50% since one box is already ruled out

50% as only two of the boxes were eligible to have gold balls. You know it's not the one that had silver.

congrats, you're better than most of the people in the last one i made on Jow Forums. have a girl.

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but I answered wrong on purpose as bait >:(

You fuck. I know what you're doing. This problem is very similar to the flipping two coins and one lands heads problem but the answer is 50% which is the incorrect answer for the coins one. So people that think the answer to the coins one is 50% will use this to prove that they're right because they think it's the same problem.

bitch i dont care im already running like a nigger from the police to sell that gold ball

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well it's dicky snippy time for you then, no brainlets allowed in the new world, sorry user :(

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ITT: brainlets don't understand the concept of conditional probability

Yeah, a brain. It's 2/3 because two of the three gold balls have another gold ball in the same box. It's that simple.

Get a refund on that degree

He's right
This shows the difference between simplified highschool probability classes and actual university probability theorems.
Its more probable that you chose the first box since it has the most gold balls, and if that's the case then you'll always get a gold ball on the next try.
Oh boy

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I don't understand? Please someone explain this to me, I'm horrible at maths.

I thought that the fact that I got a golden ball means that it can't be box 3 so I either picked from 1 or 2 in which case there can only be either 1 gold or 1 silver ball remaining meaning my chances are 1/2? I don't understand how it can be 2/3, sorry for being stupid.

physically i can feel each of the boxes weight im just not allowed to open them. this is the simple solution, feeling the weight.

unless mr thrickster puts hevier pieves of silver. than that of gold. which case i would make silver pointed arrows and kill the le happy merchant for tricking me and take his gold by force.

Label the balls as gold 123 and silver 123. Picking Gold 1 and Gold 2 are different events even though they're in the same box.

Reframe the question: if you have picked a gold ball, what is the probability that it came from the box with 2 gold balls rather than the one with only 1 gold ball?

When you pull out a gold ball, it's twice as likely to have come from the first box because there were two gold balls in it. If this isn't intuitive to you think about it like this: Imagine there are three boxes,
Box 1: 100 gold, 0 silver
Box 2: 1 gold, 99 silver
Box 3: 0 gold, 100 silver
You pick a box at random and pick a ball blindly and it happens to be gold. Isn't it much more likely to have come from box 1? Therefore, isn't it much more likely that the other balls in the same box as your ball are gold?

This. The ball you pick can either be g1, g2, g3, and two of those results give you another gold ball after. 2/3rds.

fuck off with ur intuitive bullshit bitch. Next you're gonna explain the monty hall by saying "doesn't it just feeeeeel like you got a goat??

What are some other brainlet filter questions?

Probability of getting a gold ball in of itself: 2 gold balls, 5 balls total. 2/5 odds.

Probability of the other ball being gold: only one box had two gold balls. There are three boxes. 1/3 odds.

Probability of taking one gold ball the first time: 6 balls, 3 gold of which are gold. 1/2 odds.

Probability of picking gold the first time and the other ball being gold: Product of two previous probabilities - 1/3*1/2= 1/6 odds.

i have one.

>What are some other brainlet filter questions?

anyone who asks this is a guaranteed brainlet.

Pick any positive integer n. If n is even, divide it by 2. If n is odd, multiply it by 3 and add 1. Prove that repeating this process will always eventually get you down to 1.

If you can't solve this you're literally retarded

bro what the fuck are you even saying
you're not stipulating a problem
frame your shit correctly

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Imagine not understanding what conditional probability is

I have discovered a truly marvelous proof of this, but this post to narrow to contain it.

Thanks a lot, I get it now.

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If you can't even comprehend the question you're an even bigger brainlet than someone who gives the wrong answer

fuck of pierre

>ignoring givens to over complicate the problem
first gold ball has already happened. Now, looking at the boxes, all we know is that the ball can be gold in the leftmost box or silver in the middle box. That's it. That's the whole problem lmao. Not your fault I have a massive, almost uncomfortably large brain though desu.

>pick n
>if n is even divide by 2
>if n is odd multiply by 3 and add 1
>prove it equals 1
guess that explains why you're here

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look up the monty hall problem to feel like a brainlet again.

2/3, the first gold ball can be any of the three.

66.6 recurring you tards

Holy shit you really are that fucking stupid

Retards. If you picked a gold ball that means you're in the middle or left box, the right box has no gold so it can only be middle or left. Given as you can only draw a gold or silver at that point, its a 50% chance. The right box never mattered it existed to draw attention to IQlets like you, especially the stupid fuck, you absolute arrogant cockbag calling othet people tards while falling for the most obvious fucking shit ever. You deserve to have your dick straight up obliterated with an orbital cannon, your genes are as useless as a perforated condom

If you drew a gold ball at first, you are twice as likely to be in box with 2 gold balls than the box with only 1 gold ball. It's a very simple experiment to do, and if you perform it, you will see that the right answer is 2/3.

I'm not going to bother to give a serious reply to bait, so just have this (You) instead.

is this bait or just dunning-kruger

Thats not how it works whatsoever. If you drew a gold ball, there are only 2 more options, drawing another gold, or a silver. You're looking at some initial percentage bullshit and thats not the case. Once a gold is drawn, only 2 options are left, a silver, or a gold, so its a 50/50 chance.
See above. OP will confirm im right, anyone with a brain will confirm im right. There would only be 2 options left. 2 options cant make a 2/3 chance.

>Once a gold is drawn, only 3 options are left, a silver, gold #2 from box 1, or gold #1 from box 1, so its a 2/3 chance.
FTFY

50/50, it states that one box contains no golden balls.

1/3 so 33%. Whoever said 50 is a fucktard.

The probability of whatever happens is 1, otherwise it wouldnt' happen

explain your reasoning process

I can almost grasp this but at the same time, not.
Ifni draw gold all I know for sure is I didn't pick the all-silver box. The the question is what are the chances of picked one of two possible boxes. That's gotta be 50%.
Like after I already drew gold, the remaining question doesn't care what the probability was that i drew the first one--whether I took it from a box of all gold or not.
All that remains at the core is, there's three boxes: you've confirmed that you didn't pick box 3. What is the remaining probablility you picked box 1 rather than box 2... And that is a coin toss, aka 50/50

Holy braincell. This is the most retarded shit I have ever read

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It's 2/3 you retards.
If you got a gold ball it's more likely you got it from the two gold ball box than the one gold ball box.

Just shake the box to make sure you know beforehand

Im And like I said, after you already drew the gold ball it doesn't matter which box from which you were more likely to do so. It happened,and it rules out one of the three boxes. That probability doesn't affect the coin toss of which of the two gold-containing boxes you drew from

I just typed this up, you can try it yourself if you have an IDE.
Imagine the situation is similar but in each box are 10 balls instead of 3, in one box you have 10 gold balls, in another you have 10 gray balls, and in the third you have 1 gold ball and 9 grey balls.
If you have gotten a gold ball by random chance, which box do you think it comes from?
Look up the goat problem.

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*monty hall problem, not goat problem.

Clear bait IMO

I did it in matlab.

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Can you explain why it's not 50%?
Since there's a gold ball, it has to be either first or second box, there are only 2 outcomes either i pick up gold ball again or a silver one since box third is out of the question by logic.

6 possibilities

Picked the box with two gold balls, picked its first ball
Picked the box with two gold balls, picked its second ball
Picked the box with a gold and silver ball, picked its first ball (gold)
Picked the box with a gold and silver ball, picked its second ball (silver)
Picked the box with two silver balls, picked its first ball
Picked the box with two silver balls, picked its second ball

It's not the latter three as they all involve picking a silver ball first. Of the three remaining possibilities, two lead to picking a gold ball the second time around. The probability is two in three.

This is a good explanation.

>there are only 2 outcomes either i pick up gold ball again or a silver one
Just because there are two outcomes doesn't mean the chance of each outcome is 50%. When you buy a lottery ticket, you either win, or you don't win, but that doesn't mean you have a 50% chance of winning.

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There being two outcomes doesn't mean that each is equally probable.
All you know is that you got a gold ball, you don't know which box you got it from.
It's more likely that you got the gold ball in your hand from the box with two gold balls (two times more likely, to be exact)
Since the question says GIVEN you drew a gold ball you work backwards to determine which box it likely comes from.
I think most people when they answer 50% have this mindset:
>choose random box, which box did you most likely pick if it was one of the gold boxes?
When they should have this mindset:
>given that you drew a gold ball, which box is it more likely to have come from?

Imagine responding to a bait
Imagine responding to a bait seriously
Imagine that "running a simulation" on a 3rd grader math problem will stop the trolls

Just sage

>it can be one of two things
>thus the probability of those two things are an even split
yikes

I'm gonna reply to your post so your sage was useless :)