How good is Zig Forums at basic probability theory?

how good is Zig Forums at basic probability theory?

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en.m.wikipedia.org/wiki/Bertrand's_box_paradox
twitter.com/AnonBabble

50%

50% and anyone who disagrees by saying there are two boxes and three undiscovered balls leading to 2/3rd is a contrarian shill.

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50%

88%
always

If one ball is green and one ball is red what is the chance that OP is a faggot?

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watch kikes not compherensing that it's 2/3, not 50%.

NO YOU FUCKING RETARDS!!! IF YOU GRABBED A GREEN BALL THERE’S A 2/3 CHANCE YOU’RE IN THE ONE WITH 2 GREEN BALLS SINCE YOU HAVE 2 CHANCES OF THAT COMPARED TO 1 CHANCE IN THE OTHER. YOU FUCKING NO BRAIN NIGGER MONKIES!!! AAAAAAAAAAAAARH!!!!!!

This

The next ball pulled will most likely be red.

And the entire premise is based on whether the question asker is trustworthy. How do we know there's only the two colors? It's just a trick the ball could be blue for all we know, the box could have a dead cat inside and till we open it the ball is both green and red, there and not there, ball and dead cat simultaneously

The question doesnt ask "whats the overall chance of pulling 2 green balls"
It asks, whats the chance that the next ball is green. After eliminating double red, you only have 2 boxes left. 1 will have greem, one wont. Its 50%. You overthought the question, and are probably a gay commie autist.

66.66%

So jews are gonna jew and that is the real lesson here. Thanks user.

33,3%

Flawless logic. I have a 50% chance of winning the lottery. I either win it or I don't.

If you grabbed a green ball, there’s a 2/3 chance you grabbed one from the one with 2 greens. Since 2/3 greens balls are in that box. So there’s a 2/3 chance the next one you grab will be green. It’s COMMON SENSE

I also got that. 1/3

actually its 66.6% sorry

Monty Hall problem
There are 3 green balls in total, 2/3 come from box 1, 1/3 come from box 2, 0/3 come from box 3. If I pick a green ball, then there is a 2/3 probability that it came from box 1, thus 2/3 chance the next ball will also be green.

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You eliminated the redbox and now there’s 2 boxes left but there’s a 2 out of 3 chance you’re in the one with 2 greens so it’s 66.6% not 50%

You're right it's the same as monty hall but it's not 33.3 it's 66.6

Think in terms of boxes not balls.

I didn’t overthink anything, you’re missing an entire component to this. And you say I’m overthinking it. No you’re under thinking it, your IQ is nigger tier

I'd assume 2/3 at first glance. Cuz you'd most likely get a green ball first from the one with the most green balls

One draw left, two boxes. One is guaranteed red and one is guaranteed green.

With the whole probability tree, it is 33.3%. The subtree, after having eliminated options with the green ball, is 50%.
/thread

First ball is green. Therefore, there is a 2/3 probability you picked from Box A, 1/3 probability you picked from Box B, and a 0 probability you picked from Box C.
The conditional probability the second ball in Box A will be green is 100%. Box B and C is 0%.
Therefore, 2/3 * 100% + 1/3 * 0% + 0 * 0 = 2/3

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its' 50% mr.double digit iq, you are in the first or second box, 2 choice so 50 percents.

The boxes don’t actually matter. If you grabbed a green ball, there’s a 2/3 chance you’re in box 1 (the box you need to be in to draw a second green ball), and a 1/3 chance you’re in box 2 (the box you will not draw another green ball). It makes perfect sense you’re not thinking about the one term in game theory that I forget what it calls

He is right. Don't believe me, use conditional probability:
P(RG|G) = P(G|RG) P(RG) / P(G) = 1/2 * 1/3 / (1/2) = 1/3.
Numbers don't lie.

50%. If you bring up up "muh monty hall," you're a brainlet redditor trying to sound educated.

You stupid nigger. If your first ball is green that does mean it is equally probable that it came from Box 1 or Box 2. Use Bayes theorem.

66%...

This is gay. Stop talking about balls.

This is a famous problem for anyone interested:
en.m.wikipedia.org/wiki/Bertrand's_box_paradox

Alright jackass. This isnt that tough of an experiment to replicate. Get some dimes and pennies and 3 boxes. Start making random pulls and record. If you get anything other than trending towards 50% after 100 runs, then tell me all about your superior logic. To make it faster, whichever coin you pull first can be "green" for that round

i hate redditors but it's 66.6 sorry

t. brainlet somali

based

What a thick French. Jesus

Law of Total Probability:

1/3 * (0) + 1/3 * (1) + 1/3 * (1) = 0 + 1/3 + 1/3 = 2/3

Yes, but you're pulling one ball out of the same box, and one box does not have another green ball. The answer is 50%. God damn britfag.

Correct.

Look up the Monty hall problem.

The set up of a choice can affect your chances later. It's counterintuitive but true.

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If I flip a coin 1000 times, and it came up heads each time, what are the chances the next flip will come up tails?

50%.

this:

If I pick a green ball out at random, it's most likely picked from box 1, so how can it be 50/50. Burger education is laughable.

Nope, 0%
It's obviously a two-headed coin.

There's a 50% chance that the remaining ball is green and a 50% chance that the remaining ball is red.
We picked a box at random and then revealed one of the balls at random. Because the ball was green, it became impossible for that box to be the box with two red balls, but it was still possible that the box contained two green balls, or one of each ball. We therefore are picking from two boxes. Of the two boxes, one still has a green ball and one still has a red ball.

1 / 2 = 0.5 = 50/100 = 50%

2/5?

It would come out to 66.6% after an infinite amount of trials. You’re an absolute retard. Once you have a green ball in your hand you don’t have to think about boxes. You have to think about it like a tree of probability. With the information you know there’s a 2/3 chance the next one will be green

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Chance that any given flip is heads = 50
Chance that flips continously land on same side in sequnce- statistacally improbable.
You're question doesnt really define what you want to know

Quick call the police because this nigger is using counterfeit money!

it would depend on what box you pick first. The question is flawed because it asks you to provide an answer prior to having correct data.
You have all the green and red balls there now, but asks you what the probability is after you remove 2 of the balls from the scenario and you cannot know which ones will be removed until you pick a box.

Wow, /pol average IQ is way lower than I expected

100%.
You're either a faggot or you're definitely a faggot.

sorry ahmed but you're the brainlet here

he got it

2/3 before you start picking at random
50% after first pick.

50%