The absolute state of canadian teachers

pulling green or red eliminates one of the boxes, it leaves only 2 possible outcomes after you have drawn one. either green or red. Regardless of the first ball drawn..

it's 2/3. I showed this to one of my students with ADHD last year who's struggling in math and even he thought people saying 1/2 are retarded.

Box 1 plus Box 2 has a 75% chance a green ball is pulled, BOX 1 has a 100% chance, Box 3 has a 0 % chance, if you average all 3 the answer is 58.3. We pulled a green ball in this situation. NOW ONTO PULL 2 FROM BOX 1 OR 2 we have 3 balls left, 1 is red 2 are green, there is now a 66% chance the next pull is green.

yeah, fascinating stuff, this last thread with the canadian
>I make questions like this for my classes
messing up his own logic several times was something; I literally could not predict the lengths these people go to defend either 50% chance or 33% chance
the man said
>G1 and G2 are the same result because the question is only seeking color, not orb + color.
as if we changed it from
>What is the probability that the other ball in that box is also green
to
>What is the probability that the other ball in the box is also a green ball
would actually affect the math

What conditions? I picked one of the three boxes, reach in and pull out a green ball.
It's a 2/3 probability that when I reach back into the same box, I get a green ball.

You niggers really got nothing better to do?

read the pic, very slowly.

The conditions are correct. The answer is 2/3rds (66.666_%).

You're not looking at the arrival to the start correctly.
There are three boxes. I pick one. Reach in and pull out a green ball.

That's the phrasing, and that's how you get to the start.

I'm likely to get the G/G combination 2/3 of the time. Why? Because the game never would have started if I pulled the G/R and pulled the red. So I'm twice as likely to start with the g/g box than the g/r box, so it's 2/3.

The end.

However, the fact that you drew a green ball in the first place means it’s more likely you picked box 1 than box 2, because box 1 has the highest chance of giving you a green ball.

>I'll demonstrate why I'm a retarded nigger.
You already have.

50% if you pick from one of two boxes after your first pick.
66% if you pick from one of three boxes after your first pick.

No. It clearly says you pick both balls from the same box.

Outcome 1. The 2nd ball is Green
Outcome 2. The 2nd ball is Red
Outcome 3. The 2nd ball is Red, also the Green ball you already took out is actually Red too :^).

Still 66%, but screw your rules, I'm opening all boxes and throwing the balls at you.

It would not be 66% if you randomly choose a new box after the first pick.

P(Second Green|First ball was green)
= P(Second Green and First Green)/P(First Green)
=(1/3)/(1/2)
=2/3

The numerator is 1/3 because you can only have the first and second balls being green in one of 3 boxes
The denominator is 1/2 because there's a 1/2 chance that your first ball is green

I thought autists were supposed to be smart

is this post meant to imply 50/50 odds?
if not, ignore
if yes,
imagine looking at only outcomes
i have a sack of 99 green and 1 red ball, so 100 balls
i pull out a ball
outcome 1: red
outcome 2: green
50/50 chance lol

Think about it this way. Forget about the wording of pulling out "the green ball first" What are the odds you pick the same colored ball twice in a row of either color.

No, it's a 50% chance, because you already pulled out a green ball dummy. there are three balls of each color. two boxes have the same color, the final box has one of each.
So, you will either be pulling a red or a green ball, the last box has two red balls, meaning, if you picked the box with red balls from the get-go, that first ball you pulled would've been red you retard. so, the 2nd ball will either be red or green, a fifty percent chance of either color. Same goes if you pull a red ball first. How do people not get this?

Good explanation. This makes sense I wrapped my head around this immediately after reading it.

The Monty Hall problem someone posted on here one time took me weeks to really understand.

See

ITT: retards, the probability of the mixed box is neither 1/3 nor 1/2, but 2/3.
Here is the proof:

Let X denote the event picking the purely green or mixed box (i.e. the event we draw a green ball).
Likewise let Y denote the even of picking the purely red or mixed box (i.e. the event we draw a red ball).

It is true that:
>P(X) (= the probability of event X occurring) equals P(Y) equals 2/3
>P(mixed box | X) (= the probability of having drawn the mixed box when observing event X, i.e. having drawn the green ball from the mixed box) equals P(mixed box | Y) = 1/2

Now by the law of of total probability we have
> P(mixed box) = P(mixed box | X) * P(X) + P(mixed box | Y) * P(Y) = 1/2 * 2/3 + 1/2 * 2/3 = 2/3

poltards are truly too stupid for math, sad!

Thx - that's the first (you) I've gotten that hasn't called me a nigger or retard in I can't remember when.

you pick up a green ball. that eliminates the box with two red balls.
You do not yet know if the box you picked from is green/green, or green/red
if the other box is green/green, your chance of picking another green ball is 100%. if the other box is the green/red, your chances are only 50% getting another green ball. your chances are 2/3rds.

No, you are wrong, because as soon as you pull the first ball, you change the conditions. the question asks what the odds of the next color you pull are, this means, you remove a ball from the set, doesn't matter if it's a red or green ball, once you remove a ball, the odds are then 50%.
If you pull a green ball, then the box you chose must have at least one green ball in it. if you pull a red ball, it must have at least one red ball in it, you can't pull a red ball from the box with two green balls, and you can't pull a green ball from the box with two red ones. The opening move determines the outcome and automatically eliminates one of the three boxes.

shit i meant if YOUR* box in both of those scenarios.

I feel like I missed out on something...

You need to learn how to read.

No I don't, I can read and comprehend just fine, no matter which color you draw, the odds of you drawing either red or green on the 2nd ball are 50/50.

Are you the retard that thinks mm/dd/yyyy is superior?

For people that don't get this at all, it the same thing as Shrodinger's cat.

Theres only two boxes with green. Red box is irrelevant. Its not so much of a probability problem as it is a word problem which implies you are capable of reading. Stop making us look retarded you mart sharting cunt.

I have chosen one of three boxes and pulled a green ball from it. We on the same page here? That's the starting condition, right? I'm never in possession of the red/red because I wouldn't be holding a green ball - we agree, right?

So how do I get to this starting point if I do it again and again and again? We're trying to determine the probability of the next action, which is the "what color is 2nd ball going to be when I reach into the box?"

Do you agree with that?

what lmao
this is literally en.wikipedia.org/wiki/Bertrand's_box_paradox