Does Yas Forums suck at basic probability theory?

When you took out a green ball you discard the other box, because you know there is no green ball in the third box, so the probability to take out another green ball relies within the two first boxes: one with two green balls and the other with green and red balls.

You already took one ball, so no matter which one of the two first boxes you took out the green ball: it only has one ball left.

There are two colors, and one ball. 50%

but what are the chances that OP will get blue balls?

2/3

40%

>but what are the chances that OP will get blue balls?
Lifetime probability = 0%

Wrong.

420%

Before we get 200+ replies in, I'm just gonna post this explanation I made last time this thread was made. Just replace "silver" and "gold" with "red" and "green," respectively.

Attached: Explanation.png (2850x1482, 153.72K)

2/3

so basically what you do is you look it up

66.66666666666666666666666666666666666...%

This

50%

There are three possible green balls you could have picked. One of them is in a box with a red ball, so that one is a fail. The other two are in a box with a green ball, so those two are wins. Win fail ratio of 66.6%

>does Yas Forums suck
Yas Forums sucks a lot of things

The confusion comes from whether you start the probabilistic analysis from the viewpoint of already having guaranteed a trial with a green ball or whether there is the "discarded trial" in which you pick the box with only red balls.

It's either 50% or 66% depending on how you interpret the problem posed.

> English is imprecise
> Math is precise

Oh shit I was right! I mean its the same as the 3 doors question I guess.

FOR FUCKS SAKE U MORONS

IF U PULL A GREEN BALL, IT MEANS THAT THE BOX U HAVE IN FRONT OF YOU IS EITHER THE DOUBLE GREEN OR ITS THE 50/50 BOX , IT CANNOT BE THE DOUBLE RED.

SO EITHER YOU WILL PULL A RED (50/50 BOX) OR YOU WILL PULL A GREEN (DOUBLE GREEN).

ERGO, YOU HAVE A 2/3 CHANCE OF PULLING ANOTHER GREEN.

Well, it's explicitly conditional, meaning the experiment doesn't work if the ball drawn is red, meaning it's already been decided that the first ball drawn must be green, meaning the probability of drawing a second green ball from the same box is 2/3.

If you take away the condition that the first ball drawn must be green, then the answer is 1/3 because then it only matters which box you chose, of the 3.

There is no way for the probability to be 50%, under any circumstance, unless you explicitly state that Box 3 (Red, Red) WASN'T chosen.

Actually, reading over this again, there's no way the probability would be 50%, even then.

It simply isn't possible.

the answer is either 1/2 or 2/3, both answers are true depending on the exeact setup of the experiment, but your explanation is wrong.

If you’re talking Monty hall, nah it’s not the same

and that's what's so messed up about this famous problem.

The probability is no longer about what chances there were for you to arrive at this situation. We already have arrived at this situation. The math about the all red box is no longer relevant.

The newest math is that you've either got the double green or the split green red box. We are not time traveling back to whether you might have picked a red ball out of the split box. You didn't.

You have one of two boxes now. You didn't do any of the other possible things. The one green proves that you either have the GG or the RG box. That IS 50/50 starting at the point in the timeline described.

What's the setup that yields 1/2?

1/3?

It's already been decided that the Box 3 wasn't chosen, because the condition is that the first ball you took out IS green, so it's impossible to took out two reds if the first ball you took out is green.

Read

this user just typed it out:

1/5 of course

>The one green proves that you either have the GG or the RG box. That IS 50/50 starting at the point in the timeline described.
What exactly is 50/50?

*1/6
typo

Even if you remove the box with red balls, the probability stays the same, 66%

the probability is 150%