Surely you don’t expect to make it if you can’t even accurately determine your odds, right?

# Can Yas Forums into probability?

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2/3rds

1/3

1/2

you already narrowed down the options to 2 when the ball you removed turned out green

WRONG

EXTRA WRONG

>There is 10000000000 boxes.

>one of them have 2 green ball, one only 1 green ball, and the rest of them only red balls.

>You pick a green ball. What is the probability of the other ball being green ?

Instinctively here we can see that the number of only red only boxes do not impact the probability of the other ball being green.

The probability is simply :

Number of all green boxes / (Number of at least one green ball boxes) = 1/2.

No he's right. Only brainlet faggots execute these types of problems on sequence. At the very moment of calculation given what is absolutely known, the chance is 1/2. Kys

50%

t. A brainlet

not science or math

This is not Monty Hall faggot. TV presenter do not open the doors at random. He specifically avoid to open the car.

Entirely 50%. You see, upon picking a green ball you have ruled out the red red box. Knowing this, you can deduce there are only 2 possible boxes that you’ve picked from. One these boxes has another green ball and the other does not.

Don’t fret if you don’t understand this. OP picked a tricky question. Lesser mortals would struggle.

Nope. The answer is 2/3rds.

It is more likely that the green ball has come from box 1 than box 2.

Wrong. 3 possible outcomes:

1. You have picked green ball #1 from box 1, next ball is green

2. You have picked green ball #2 from box 1, next ball is green

3. You have picked the green ball from box 2, next ball is red

2/3rds of the possible outcomes lead to a second green ball being picked.

>Knowing this, you can deduce there are only 2 possible boxes that you’ve picked from.

Correct, and the probability that it was box 1 is 2/3rds.

Nigger

This problem assumes all green balls are the same, so there is no reason to separate the balls in box one. We have

1. You picked a green ball in box one, next ball is green.

2. You picked a green ball in box two, next ball is red.

Oh yes indeed. Was too focused on box selection, forgot about choice axiom.

All green balls are the same, but not all boxes are the same.

Think of it this way:

Two boxes, each with 1 million balls.

Box 1- 1,000,000 green balls

Box 2- 1 green ball, 999,999 red balls

You randomly choose a box and pull out a ball. It’s green.

Which box do you think it most likely came from?

Incorrect.

You have already picked a box

You have already selected a ball

That ball is green

One ball remains inside the box

That ball within the box is either red or green.

50%

Both boxes are isolated instances.

if you select 1 box and pull out a green ball, It is guaranteed that the next ball will be either green or red.

there is no question about it. the 3 box situation is a red herring.

you reduced the equation to a [1,0] and [0/0] and pulled out a 0.

therefore you either only have the choice of a 1 or a 0 to choose from, a 50% chance.

Is this Yas Forums humor??

Absolutely diminutive intelligence. 110 is my guess.

>so there is no reason to separate the balls in box one

Indeed.

But think of it this way :

You have 6 easter eggs. Inside of them is a piece of paper with a number written in it.

There is 3 green easter eggs, 3 red easter egg.

2 green easter egg have written "Box 1" written in it.

1 green easter egg and 1 red easter egg have written "Box 2"

2 red easter egg have written "Box 3"

You pick a random green easter egg.

What is the probability "Box1" is written inside ?

You picked a green ball. 2 out of 3 possible green balls have another green ball as their neighbor.

Read here: It becomes easier to comprehend if you expand the scale of the numbers

You will never convince me that it isn't 50%.

If choice 1 was a green ball, the double red ball option CEASES to exist in the equation. You either have the double green or 1 green 1 red.

It is more probably that the green ball in your hand game from box 1.

observed probability vs actual probability

>You either have the double green or 1 green 1 red.

Yes, and there is a 2/3rds chance that you have the double green box.

The green ball in your hand is one of 3 possible green balls. 2 of those possible green balls have another green ball as their neighbor.

But it's not a 1/3rd chance. I have to options in the next step. Green or red. That is 50-50.

The only way of getting two greens in a row is to pick box one, which would be a 1/3 chance

But assuming the first ball was green, you would have to have either box one or box two, which means the probability of the other ball in the box being green is 1/2

This is also a helpful way to think about it

>I have two options on my D100. 1 or 2+. That is 50-50 chance.

No, it’s a 2/3rds chance. Only one green ball that could’ve been picked has a red neighbor; 2 of the green balls that could’ve been picked have a green neighbor.

no because you dont pick a green ball. you pick a box you retard.

reread the question

You are adding an element that isn't stated in OP's problem. The balls aren't different (numbered in your case) all we know is that they are green. Thus, we have no basis to separate them.

The number of balls in the box doesn't matter at all. The next pull will still have to be green or red.

DOESNT MATTER.

we are picking boxes not balls.

>But assuming the first ball was green, you would have to have either box one or box two, which means the probability of the other ball in the box being green is 1/2

Your mistake here is thinking that Box 1 or Box 2 is equally likely to have produced the green ball in your hand. They aren’t.

wrong

You're combining stages in the problem. A better analogy is rolling a 1 out of a d100 and then doing a coin toss afterwards. a coin toss is 50-50

PROVE ME WRONG

My balls aren't numbered. They have just different boxes written inside. There is no way to specifically pick one.

Yes, and the fact that you have a green ball in your hand means there is a 2/3rds chance that you have selected box 1.

Step 1

>pick a ball that is green from the crate

Step 2

>Flip a coin that lands either green or red

That is this problem in a nutshell. A coin flip cannot be 1/3 in chance. There is a 50-50 chance of it being green or red

You have two boxes.

Box 1 has one million apples

Box 2 has one apple and 999,999 bananas

You randomly choose a box, stick your hand in, and pull out an apple.

Which box did it probably come from? Is it a 50/50 coin flip because there are two boxes?

doesnt matter, we only are asked what the chance is after we have picked a green ball. and this ball can only have come from 1 or 2

its doesnt mater because all instances of a red ball are ignored. there is no difference between box 1 and 2 there is a 100% chance we get a green ball because thats the part where we start calculating. we ignore all instances where we pulled a red ball from box 2

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hurr durr muh 1/2

kys brainlet, pic related

Already saying I've picked a green ball creates an axiom that eliminates the 3rd box and even selecting a box from existence.

yes because you ignore all the times you picked a banana. in fact you just showed it is a 50% chance

yeah. Both boxes have apples that can be picked out from them. It equally could have been either one.

Ok I've reconsidered

There's no reason to separate them into boxes at all, it's better to think of it as putting all the balls into a black bag and pulling two random ones.

In this this case, your odds of pulling a green ball the first time is 1/2, and the odds of pulling one a second time is 2/5,

so there is a 20% probability (unless i'm a retard and fucked the math)

16, I didn't look at the image but the answer is always 16

There we go. Fellow big brain chad. The green ball is a given and thus eliminates the other probability which isn't relevant.

Yas Forums I am disappoint, basically you’re all fucked

This is a 131 year old problem and the answer is 2/3rds.