Sophistry

Prev thread archived, you know the deal ... it's 50pc

Attached: its50pc.png (1024x768, 49.53K)

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Dumb leafs and swedes fell for the turk sophistry ... you begin from having the green ball already eliminating the (R,R) combo so the 50/50 choice remains

Thanks mohammed, i was hoping for a new thread. Here's a simulation of 1000 ball draws.
The first ball was green 513 times. Out of these, the second ball was also green 355 times.
That makes the odds 355/513 = 69.2%
BRAINLETS

Attached: sim.png (945x559, 104.02K)

wikipedia.org/wiki/Bertrand's_box_paradox
Shut it down, 2/3rds won. There's nothing left to debate. user ran a simulation in the last thread and got 69%. It's over

why is it only the nords and leafs who cant see past the sophistry ... I fully understand bayes theorem but you have to account for 'logic' aspect of the question ... u fell at the first hurdle so the entire program run by jama's girlfriend's autist fin husband is wrong

suck my dick retard

there is no hope for leafs, enjoying having nordic cock crammed into your shitter ... both the worlds most insufferable regions in a midwit 'intellectual' circlejerk

50%

The third box (RR) is irrelevant.
The correct answer is 50%

If you pull a green ball your box is one of two, either 100% green or 50% green. 100 x0.5. + 50x 0.5 = 75% chance that you pull out a green ball

You're in the first 2 boxes, so 75% chance to see green first (total space), which causes the GG box to be weighted more.

({GG} / {GG, GR})
/
({GG,GG,GR} / {GG,GG,GR,RG})
=
(1/2) / (3/4) = 2/3

Decent reduction excluding the RR box though and a proof that it's irrelevant.

Close. It's either 100% green for the first green ball, 100% green for the second green ball (both in the first box) and 0% for the last green ball. Add those probabilities up and you get 200%, divided by three balls is an average chance of 66% or two thirds.

No, if you pull a green ball then there is only one ball left in the box. If the original box was (GG) then the remaining ball will be G. If the original box was (GR), then the remaining ball will be R. Since the (RR) box is irrelevant, there are only two possibilities left. One is G and one is R, therefore, 50% chance for G.

If the first ball is green the there are only 2 boxes out of the 3 which have green balls then it's a 50/50 chance.

No, because it's more likely that you pulled G from GG in the first place rather than GR

anybody saying 50% is trolling or a brainlet

That's not how probabilities work

...

>Aryan toothpaste answer
>Brainlet shitskin answer
Thanks Yas Forums

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Holy shit Zoomers don't know 7th grade math.

>reads thread.
What the hell are you idiots doing?
It's already certain that the first ball you pick will be G, which means it cannot be the RR box which is then eliminated for being irrelevant, which means we now have TWO boxes, GR and GG but we do not know which one the box we picked is, there is a 50% chance it's the GR and 50% chance it's the GG.

It's the odds of drawing a green first if you're in a valid box, which the problem states you are.
He only did the first half of the answer that the 50% brainlets ignore. Then he ignored the 50% logic that applies after.

[G out of GR] out of [G out of GGGR]

(1/2)/(3/4) = 2/3

Thank you I was really stuck on this but you cleared it up for me with the equation

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All you are wrong.

You have a 40% chance of picking a green ball at random when 2/5 of the choices remaining are green (assuming the first green ball was removed otherwise it would t be included)

It's B and the hint is a red herring

It's 66.67%.
You fools need to learn probability and statistics.

psst it's not B

1 in 3, right?

You should run your simulation a few million times to converge on the correct answer.

More people got the right answer than I expected.

Now let's do Monty Hall!

You pull out a green ball. So the red box is gone by default. What you have left is a box with one green ball and a box with one red ball/ 50/50. Right?

Yeah this is an attempted leftypol raid thread. 1/3 is correct, they want pol to appear stupid.

Correct. It's 50/50

If it's not B then there's two points and this problem is a sham.

With enough ingenuity you can solve any problem

Attached: kurisu solved.png (900x2924, 204.75K)

How can one man be this based?

somewhere inside, he can't make it in there so it'll take really long

This

2/3

This is baby conditional probability

Run it longer