Can you properly explain why you think it's 50% or 2/3?

can you properly explain why you think it's 50% or 2/3?

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There's an equal amount of balls of the same color, it's a 50% probability.

No I can't because I'm retarded

1/2, as soon as you know the ball is green, you can remove on of the boxes (all red) from the equation.

2/3 you have to count both of the green balls in the box.

It's 50/50 because the box you pulled a ball out of either has a second green ball or it doesn't.

Jesus christ anons... If you don't know math, just don't answer. Some of you fucking retards who make a big long case for 50% know fucking nothing of statistics and you know it. It's 2/3 because statistics says so. If you don't know statistics, run the experiment yourself 100 times. The answer will not be 50%.

2/3

Each box has an equal chance of being chosen, but the problem set only includes the results when at least one ball is green.
Half of those results will be two greens in the box.
You are not choosing one of four balls. You're choosing one of two problem sets, and only one of those two can result in two greens being drawn.

This isn't the Monty Hall problem.

2/3

Explanation:

You are starting from a position of having already picked a green ball. There are three green balls which have equal chance at being the one you picked. Two of those green balls will have you picking a green ball on the second grab. One of those green balls will have you picking a red ball on the second grab.

One you’ve pulled a green ball, you know you’ve chosen from either the first or second box above. The third box, containing only red balls, becomes irrelevant.

You’ve thus chosen a green ball from the first two boxes. There are three green balls among them, and thus if you picked a green ball on your first draw, the probability that you chose from box #1 is 2/3. The probability that you chose it from box #2 is 1/3.

If you chose from box #1, the probability that you will then draw a second green ball is 1 (100%).

If you chose from box #2, the probability that you will then draw a second green ball is 0 (0%) for there are no green balls left.

Thus, the overall probability that if you got a green ball on the first pick then you will get another one if drawing from the same box is (2/3 X 1) + (1/3 X 0), or 2/3 (probability 67%). The answer is thus 2/3 (probability 67%).

If I ate a whole fucking watermelon you'd still be more of a nigger than me

LMAO, you have to be trolling

LMAO you have to be getting a handjob from a muslim

The key is to realize that the separation of balls into boxes is irrelevant to the probability of drawing green EXCEPT that drawing the green allows you to eliminate two red balls. There are only three balls remaining in the boxes you could have, and two of them are green.

fucking this. 2/3

This is the answer. Anyone who says 2/3 is either dumb or trolling.

If EITHER the green or red ball is alive, you should be very wary of putting your hand in the box. You should also be very careful of getting your balls entangled!

are you kidding me you mongaloid self-poster? did you even read

I still think the answer is 1:1 (i.e. half-and-half).

Maybe this is because I’m a “frequentist” — that is, I understand numerical claims of probability statistically, as relative frequencies.

Here’s a virtual test (in javaScript), which seems to confirm my view:

var rand, box, ratio;
var drewRedFirstSoStopped = 0;
var drewGreenFirstSoContinued = 0;
var greenCount = 0;
var RedCount = 0;
var boxes = [[“G”, “G”], [“G”, “R”], [“R”, “R”]];
while(drewGreenFirstSoContinued < 1000) {
rand = Math.floor(Math.random() * 3);
box = boxes[rand];
if (box[0] == "G") {
if (box[1] == "G") {
greenCount ++;
} else if (box[1] == "S") {
redCount ++;
}
drewGreenFirstSoContinued++;
}
else if (box[0] == "S") {
drewRedFirstSoStopped++;
}
}
ratio = greenCount / redCount;
alert(ratio);

2/3rds is right and I'm angry you idiots can't fucking read or do math.

"You pick a box at random"
"You pick out a ball"
"The ball is green"
"You cannot see into any of the boxes"

There's no fucking excluding red boxes you dipshits. if there's three choices, and two of those choices have a positive then it's 2 / 3rds.

50/50 fags are illiterate or have autism.

you're right but your wrong.
"You take out a random ball from that box without looking into it. It's a green ball"
Those two sentences exclude box 3.

This is simple enough to look at all the options:
gg > g > g
gg > g > g
gr > g > r
gr > r > g
rr > r > r
rr > r > r

There are only 3 possibilities where you can pull a green ball first. There are only two possibilities after that where the other ball is green. As such, the probability is 2/3.

How are you excluding it if you don't know what's in the box? The example says you chose from THREE boxes, and you can't see into them.

Maybe mentally challenged people who can't read shouldn't call others mental midgets

Shut up candyass

It is ratios not statistics. You are a dumbass.

He's right but he came to the wrong conclusion. The double red box is excluded because it's explicitly stated that the first ball picked from a box is green, there are only two boxes with green balls.

The problem starts literally when there is a box with a red ball in it and a box with a green ball in it. There isn't 1 fucking box with two green in it and you might reach into that one. You've already eliminated 1 ball and it was green. It's either going to be red or green next. There is no extra chance that it is fucking green.

It’s 50:50
The question asked about the probability of the NEXT ball, and not of the probability from millions of tries, which would then be 66%.

Yup. It convolutes the equation with unnecessary information. You know the potential color combinations. You know the first ball you drew. This eliminates the double red box as an option. You know that if you removed the green ball from a box, it's either box 1 or box 2, thus either red or green. While it's more likely to draw a green ball blindly on your first pick, that has no bearing on what you draw next. It's either red or green, with equal odds, so 50%.

If the rule was you could choose to switch boxes after you got the 1 green but the way the puzzle is defined, you can’t. You made one choice. You already know that it wasn’t box 3, so given that, you how is the situation you now face for your 2nd draw any different than just starting with 2 boxes in the first place, one with no green and one with one green?

>This is simple enough to look at all the options:
>gg > g > g
>gg > g > g
>gr > g > r
>gr > r > g
>rr > r > r
>rr > r > r
best so far

Probability to pick a box where u can have these green ball 2/3
Probability out of those 2/3 to get the green ball 3/4
Total probability of getting a green ball
2/3 * 3/4 = 6/12= 1/2 =0.5 = 50%

Pick red first? excluded. Pick green first? we're in business and we start looking at the next pick. The next pick will be green 50% of the time. Try it in real life with coins or something. Or run a program like that one fatass.

damn so of yall kinda retarded
its 50%

u already know u have one green, meaning there are only two boxes reaming, hence 50%

the last only red ball box is there to throw u off

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The answer is 50%.
There are only two colors two choose from. When you select a ball from a box the first time there were three boxes with three options: red-red, green-green, and green-red. When you selected the ball the first time, we are told you pulled a green ball, so we can eliminate the box with the red-red ball combination. That means the box you have is either green-green or red-green. As such there are still two colors left; red and green. Additionally there is now and equal chance (one ball remaining in the box) of pulling the either color. So you have a 1/2 chance.
1/2 = 50%.

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ITT: people who don't know Bayes' theorem

If you choose a ball at random from one of the boxes, and you didn't look at the ball you've chosen or if it's covered so that color is not a factor until the end of the experiment, at this point your chances of having two green balls after drawing from the same box becomes 33%.
If the ball you draw out is known to be green as said in the experiment, then you know that one box, 3, is no longer among the expected outcomes, leaving you with two out of three outcomes, hence the 2/3.
But, out of those two outcomes, only one will give you a second green ball, while the 2nd outcome will not. So 1 out of 2, 1/2, or 50% is the probability of getting a second green ball, assuming that the first is green.

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