Is anyone here RETARDED enough to believe the probability is 50/50?

Are you retarded? The question specifically asks about the probability of the NEXT set of balls, literally no other case can be made about it. Its 50% you made up a whole scenario that has nothing to do with the question

The next set of balls depends on the ball you currently picked, obviously.

Sad that you would make this thread to try and feel better about yourself by laughing at people who can't solve a simple probability question that can easily be confusing. Petty and small minded.

Y r they"re 3 balls in 1 bocks?

No, I just saw this on /pol/ and how everyone who picked 50/50 had a USA flag and then I remembered the times I was wrong about the Monty Hall problem.

The fact that you are even making any kind of assumptions about me and my psychology based on Veeky Forums posts is funny to me as well, because I am now laughing at you too.

so it's 50/50, you either picked the wrong box or you didn't.

if you picked a gold ball out of the box, you have chosen one of the two boxes with gold balls in it. there are two boxes with gold balls in them. So you have a 50/50 chance of your box being the correct box assuming you picked the box randomly and you can't switch boxes.

Remaining might be G/G/S, but that third gold ball is irrelevant since you would have picked it if given that you got the other box, which in turn will make the first box irrelevant and vice versa.
>math fags always trying to overcomplicate shit

>so it's 50/50, you either picked the wrong box or you didn't.
you wouldn't say that if I didn't tell you about the golden ball being first. You only consider the SS box to be at 0% once you are told that your box has at least 1 golden ball in it.

You now know there is a golden ball in the box you picked, and the problem now lets you know that there are (including your ball) 4 balls that can be drawn.

1 is out because it's your own ball
2 balls are gold
1 ball is silver

2 gold, 1 silver

2/3

Again, the ball you first picked is UNKNOWN. You know it's gold, but if they were all labeled A, B, C, you wouldn't know which one you got, only that it is gold.

Youre trying really hard with some low quality bait. Please return to

But that is the starting point, nothing matters before it, this IS the new case scenario that has been provided at 100% for he question to be asked

Kek. But you are literally told so and are asked on that specific ground

shit tier bait sage

Consider that picking the correct box is a success condition. the third box with two silver balls in it is irrelevant because you haven't picked that box. you now have two boxes, one of which you have picked. the balls in the box are irrelevant, the only thing that matters is that one box is a failure condition and the other box is a success condition. Assuming you picked the box at random, you have a 50/50 chance of getting the correct box.

...

It's funny because it's true pal. No assumptions were necessary.

3 boxes, 1:GG 2:GS 3:SS. You pick G. Only boxes 1 and 2 contain G. Your odds of having G pair in box 1 is 100%, your odds of having G pair in box 2 are 0%. Average the odds (100+0)/2=50
The odds of G pair being in a G box is 50%

christians: 1 atheists: 0

Listen I'm going to let you in on a secret. This thread gets made all the time. Everyone knows the answer. 50/50 replies are just having a laugh watching you earnestly try to explain it again and again.

Yes, I remember making the same argument for the Monty Hall problem, except that whatever comes 'before' DOES matter. There is no "new case", because it is still the same problem. You already have one condition. The way you put it makes it sound like we are picking between 2 boxes where one has silver and the other has gold- then it would obviously be 50/50, but that is not the problem.

That's exactly the reason why it goes from "anything is possible" at the beginning to "some outcomes are more probable than others", such as the chance of the outcome of having 2 silver balls becoming 0%

The balls in the box do matter, because they are what the win condition is- 2 golden balls. I know it's hard to understand, especially since I've been on the other side of the argument, but I hope you can come around to it. Maybe Wikipedia will help.

Can you tell me which of the golden balls it was? Left, Middle, or Right? No, you can't. It is an unknown golden ball.

>more assumptions

You are not calculating probability. Only possibility under the assumption that all outcomes have equal probability, which they do not.

I can rephrase the question:

There are boxes numbered 1-3
1: G, G
2: G, S
3: S,S

You picked a box, and you were told it had a golden ball in it. What are the odds that the box you picked is Box 1? The gold ball you were told about might be any of the G's, but box 1 (with 2 G's) is more likely.

make me

>but box 1 (with 2 G's) is more likely.

Why is it more likely?

You reach into box GG or box GS. Either way, you get a gold coin.

If you reached into box GG then you have a 100% chance to pick another gold coin. If you reached into box GS then you have 0% chance to pick another gold coin. You can't know which box you picked. You only had two options of boxes to pick. 50% chance.

People that say 2/3 assume there's a chance to pick a silver coins first, and if that happens then the coin is put back in the box, the boxes are shuffled, and the problem resets.

You can't pick a silver ball first. You can only choose between 2 boxes, not between 3 gold coins. You have no way of knowing which box you drew from, but you have a 50% chance of having picked the GG box.

> The boxes are chosen at random, if there is a golden ball in it is removed

you're not getting the two thirds GG to 1 third GS box rate, so it's 50/50

I assumed a few of these guys would be trolling like
>8603834
but I consider the amount of mental gymnastics required to write extensive posts where you attempt to give a solution you know to be wrong to a probability problem to be beyond the capacity of a normal well adjusted human.

I'm sure there might be a few guys who earnestly don't know

anyway thanks for telling me

The experiment starts when you already have eliminated the third box. So it is 50:50

>You can't know which box you picked
since you picked one with a gold coin, it's more likely to be box 1

I think the spin this brainino is trying to put on it is that everything is made up and the boxes dont matter. Im your host, Drew Carey.

>Can you tell me which of the golden balls it was? Left, Middle, or Right? No, you can't. It is an unknown golden ball.

>Left, Middle, Right

It is either the GB with another GB as a couple OR ot is a GB with a SB as a couple.
Theare are no Left, Middle, giraffes, dark energy and pagan magic involved.

50/50

ok fine, at least 2/3 of you are trolling

i leave this thread in shame

Why?

You pick the box, then you pick a coin, and that coin is gold.

You have an equal probability of choosing either box.

Getting a gold coin is a certainty.

Therefore it's as likely you choose the mixed box as the pure gold box.

The framework of the false paradox guarantees the first coin drawn is gold.

You aren't twice as likely to pick the pure gold box, and the first time you pick the mixed box you're certain to draw the gold.

And obviously you can't draw from the pure silver box because then you defy the framework of the scenario.

Picked ball is could be any one [G]
Box 1. Box 2. Box 3.
([G][G]) ([G]S) (SS)

What are the odds that within each (box) one picked [G] shares the (box) with another [G]?

100% for box 1, 0% for box 2, box 3 is not applicable as it contains no [G].

What is the average of 100% and 0%?

50%

>at least 2/3 of you are trolling
got me

this

It was made to troll the fags that fucked up on monty hall and are now damaged forever.

Not op and cant explain in mathematic terms but imagine having a sea of gold balls and a sea of silver balls with 1 gold ball n them. You shove your hand and pick a gold ball. Its more likely that you picked it from the sea of gold balls than he sea of silvers with 1 gold

...

amazing

You are picking a gold ball. It is irrelevant what the other contents of the set are as long as you can select 1 gold ball. The original selection isnt probable, it is static. There was 1 gold ball selected. There couldve been 69 billion rainbow collered balls and one gold ball in one set and 69 trillion black balls and 1 gold ball in the other, the odds of you selecting a gold ball if ypu must select a gold ball is always 100% given the set contains a gold ball.

Thank :▪)

It's more like your hand can only hold gold balls, so regardless of which sea you shove your hand blindly into, it's guaranteed to pull out a good ball.

You have an equal chance of choosing either sea.

You asked why it was more likely to pick a gold ball, that means you need the probability before you pick the gold ball and its very unlikely that you are gonna pick the gold ball from the countless silvers

There is no such factor you made that up

>You have an equal probability of choosing either box.
No, you don't.
The part about "picking a box randomly" becomes irrelevant, old information as soon as you exclude the box with only silvers. It's extraneous information.
The problem explicitly states that you picked a gold ball at random, and 2/3 of the gold balls are in the box with 2 golds.

Try reading the original problem. You pick a box, then you grab a random ball, and the random ball is gold.

You can't operate on the framework that the random ball might not be gold. The random ball is certainly gold. You only picked between 2 containers that had any number of gold balls.

That is a plainly stated factor.

You don't calculate by balls, you calculate by boxes,
>What is the chance that the second ball that you pick is gold.
This is the question.
If the balls were not seperated by boxes, then "2/3" would make sense.
They are seperated. Go fuck yourself.

>to smart to stop responding in bait threads

I think you're confused. You pick a gold ball at random OUT OF the box you picked at random prior. You're not picking from 3 gold balls assigned to two different boxes, you're just taking a gold ball out of whichever box you picked.

Ok if we are to discuss solely on the problem then the question asked has as answer 50%

Isnt it?

Your brain is up your ass, btw.

But you do pick the gold ball. 100% of the time. Even if the probability of selecting the gold was 100% or 0.00000000000000000001%, the gold WAS picked. A set containing all gold or almost no gold are equal candidates of having a gold ball.

no, it's 2/3, see

That diagram assumes you didn't select a box before selecting a ball, or that there was a chance of selecting a silver ball.

Either way, the diagram is fallacious.

>no, it's 2/3, see

But the problem asks what the probbility of the NEXT ball being gold is, it doesnt take into account anythig else. The way the problem is worded its obviously 50%

>you're just taking a gold ball out of whichever box you picked.
Nope. You can't see inside the boxes. There's no way to guarantee that the ball you draw will be gold unless you're only choosing among the golds.
The silver ball inside the mixed box isn't an option, so you're not actually picking between the all-gold box and the mixed box as illustrated.

In the thread he made on /b/ he would just blandly reply "wrong" to everyone who made an argument that wasn't a verbatim copy of wikipedia.

One guy made a perfectly sound argument but he used a roundabout way and used 1/6 in his calculations somewhere. Because the small minded OP never saw that number on wikipedia he just claimed it was "obviously wrong and you don't even understand the problem" without trying to actually understand the approach that person took. Fuck him.

Assuming the golden ball was picked from the GS box, and the ball is not returned, then there are two possibilities

If you pick from GG you'll pick gold
If you pick from GS, you'll pick silver

1/2

Assuming that the golden ball was picked from the GG box, and the ball is not returned, then there are three possibilites

If you pick from GG, you'll pick gold
If you pick from GS, then
a) you'll pick silver
b) you'll pick gold

2/3 chance of picking gold

[math]\frac{1}{2} * \frac{2}{3} = \frac{1}{3}[/math]

The answer is [math]\frac{1}{3}[/math]

>anythig else
read again, it says 'same box'

You pick either the gold gold box or the gold silver box, because you pick a gold ball and there aren't any in the silver silver box.

We all understand that.

The same logic continues to apply.

Whether you pick the gold gold box or the gold silver box, you still pick a gold ball.

Since you already picked a box, it's just as likely your box has one silver left as one gold left. You were no more likely to favor the gold gold box because the constraints make either box as likely to give a gold ball the first time around.

>You were no more likely to favor the gold gold box because the constraints make either box as likely to give a gold ball the first time around.
Here's where you're getting confused: there's no "constraint" that magically changes probability. The problem is misleading you when it suggests your choice was truly random.
Your 50/50 choice was overruled, and a gold ball was taken from one box and placed in your hand. Anything else is impossible.

How is this wrong?

Didn't read the thread but going to have a go anyway.

If you pick a gold ball, chances are 2/3 it was in the left box and 1/3 it was in the middle box.

If it was the left box you're guarenteed to pick another gold. If it was in the middle box its guarenteed to be silver.

So the answer is 2/3

fight me faggots

>If it was the left box you're guarenteed to pick another gold. If it was in the middle box its guarenteed to be silver.

so 50/50

P(X=0)=0

>If you pick a gold ball, chances are 2/3 it was in the left box and 1/3 it was in the middle box.

The people in this thread really suck at exlaining it, but it is indeed 2/3. Just read the fucking wikipedia article as said

This is why I love probability theory. All these counterintuitive results

OK brainlets, try this

0%

Wrong.

2/3

Wrong.

50 %. Gee, it's not that hard...

Wrong.

>"1" in the top left corner indicates that there has to be exactly one mine in the adjacent fields
>There's only two fields left
>The others do not contain the mine
>P=1/2=0.5=50 %
YOU DENSE MOTHERFUCKER

OK since this basic as fuck question seems way too hard for you brainlets, just give an argument for what the best move in this situation.

Wrong. The two fields do not have equal chance of containing a mine, due to the other information you have.

That would only be true for two or less overall mines.

It's 2/5 retards.

Wrong. Calculate it yourself and show your work.

[math]\frac{5}{14}[/math]

Wrong.

1/3

Sounds annoying to do.
10 mines over 76 squares, some of which are mutually exclusive.

Wrong.

It's really not that hard.

Well I guess Veeky Forums really is full of underage brainlets. I'll post the answer since you apparently think this is some kind of trick question.

Interesting.

I think the answer is again 2/3

DESU famalams I would've said 1/2 before I read en.wikipedia.org/wiki/Bertrand's_box_paradox, but they actually convinced me.

For those who aren't convinced yet:
What is being done here is we're putting the three boxes, and selecting a box randomly (each box has 1/3 a chance of being selected) and taking a ball from it randomly; if it's a silver ball you'd repeat the experience, but if it's a gold one you'd continue.
Now repeating the experience a bigger and bigger number of times, the following happens with an approximately equal number of times:
1-GG box selected and the ball picked is first gold.
2-GG box selected and the ball picked is second gold.
3-GS box selected and the ball picked is gold.
4-GS box selected and the ball picked is silver.
5-SS box selected and the ball picked is first silver.
6-SS box selected and the ball picked is second silver.

Since the picked ball is golden, 4,5 and 6 are eliminated, and 1,2,3 happen with equal probabiliy, so the GG box gets selected 2/3 of the time, meaning that the second ball taken is golden 2/3 of the time.

Nope, you incorrectly assumed that each of those cases is equally likely in a randomly generated board. You have to take into account the fact that there are more ways for 8 mines to be distributed in the rest of the board than there are for 7 mines to be distributed in the rest of the board. See But you did start out correctly.

But 1/3 of all boxes have X = 0.

Yep, true. But for the purposes of playing a game of minesweeper I'd happily accept the 2/3 approximation.

The question doesn't say you selected a box at random, you selected a ball at random

Your 1 and 2 are the same event

I'm talking about after you select a box and take a ball out of it.

Are we taking a random ball out of the box or are we taking the first (left) ball out of the box?

yeah but, thats bullshit.
it says in the problem that when you take the ball from the box, its gold, already supposed within the problem itself is that you either pulled the first box, and it doesnt matter what ball you took, or you took the seccond box, and the universe makes you take the gold ball, every time you pick up that box.

and so, since you will pick up only box one and two[in order to pick up a gold ball first] and you will pick up both of those boxes an equal number of times, because you pick them at random, you have a 1/2 chance of picking gold and 1/2 chance for silver.

>coming to a science board without understanding math problem mechanics.

Suppose you are on a game show, and the host brings out these three boxes. Each box has a left and right compartment holding one ball. The host has full awareness of the position of each ball.

The aim of the game is to pull out two balls of the same color.

You first pick a box and open a compartment at random.

The host then opens a compartment in another box revealing a ball of a different color.

He then gives you free reign to pick any remaining compartment, from your box, his, or the completely unopened box.

What is your best option?