Is Sup Forums too stupid to solve this?

Is Sup Forums too stupid to solve this?

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en.wikipedia.org/wiki/Bertrand's_box_paradox
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50%

You take a gold ball out, it can only be the first or second box.

Fuck it, I'm going to dip my balls into the box instead.

Wrong. It's not asking you the probability of the box you picked. It's given you have a gold ball in your hand what is the probability the next ball you pick will be gold as well.

That's hardcore.

trying too hard to troll after someone guessed correctly on first try

The fact that a gold ball was picked shows that the third box is out. So you either pick box #1 or box #2. If you picked box #1 you get a gold ball next, if you pick box #2 you don't. It's 50% for each box, so 50% if you got a gold ball or not.

It's a famous problem and I assure you that is the wrong answer. I'll post up the links to further reading later in the thread. Let people struggle a bit first.

Consider this. If you picked a gold ball first, isn't it more likely going to come from the first box. Let's exaggerate the problem:

Box 1: 99G - 1S
Box 2: 1G- 99S
Box 3: 100S

If you picked a box at random. And then you pick a ball and it happens to be a gold ball. What box would you bet that you picked at first?

If you're thinking of the monty hall problem, this is different because you're forced to pick the same box and can't change

Your comparison is illogical. I don't have 1G and 99S or 99G and 1S, I have 1G and 1 S or 2G. There is nothing deeper about it.

Thinking again, I see why you may have misunderstood your own problem.

Your thinking would be correct if you picked a BALL at random. But you're not. You're picked a BOX at random.

It's not the Monty Hall problem. That problem changes the problem space in a different manner based on the conditional statement in that problem. All of this has been widely discussed.

If you are too dense to consider the problem carefully then I will spoonfeed you the answer. It's a shame people are too stupid to think critically.

en.wikipedia.org/wiki/Bertrand's_box_paradox

I did not misunderstand my problem. You do not understand it.

You pick a box at random. And then from that box you pick a ball at random which turns out to be golden.

It's not complicated.

2/3(1/2*1/2+1/2)

en.wikipedia.org/wiki/Bertrand's_box_paradox

It's 2/3. The proof is available at the link.

Any answers below this post that are not 2/3 are certified retards and should off themselves.

It's not 1/2.

Thank you, but luckily the retards never read the thread. I already provided the link and the guy above you still couldn't solve it.

And yes, 2/3 is correct.

only way it is 2/3 if you put whatever you take back

No. You do not put the ball back into the box. It's asking you given that you have a golden ball in your hand, what is the probability that the second ball in the same box is also gold.

R E T A R D
E
T
A
R
D

Learn about conditional probability then come back

Stfu faggot i didn't read all.Yes,it is 2/3.

Maybe you're the retard for thinking that people are serious when they answer something else than 2/3
>using full stops
>being this autistic

1 in 3

You'd be surprised how many retards lurk Sup Forums.

More challenging:

If you are on an infinite stretch of road with a uniform probability density of seeing cars then...

If the probability of seeing at least a car in 30 mins is 0.95. What is the probability of seeing a car in 10 minutes?

Wrong.

*** at least a car ***

...

Where did you get 1/4 for those balls. That drawing is completely wrong. Because you can only pick from one box.

...

Solving these problems puts you at high school level at best.

It's 50% cause gold ball

That's good justification indeed. But, wrong.

You either have a 100% chance or a 0% chance which averages at a 50.
50% to get a guaranteed gold
50% to get a guaranteed silver.

Go fuck yourself 1st comment got it first try

You must be trolling. Are you retarded?

No m8.
First ball is gold guaranteed.

That means there's two box's.
If you too the ball out one box gives you a guaranteed silver.
The other gives you guaranteed gold.

Therefore it's 50% and if you want to prove me wrong blame the person who made problem for not including whether or not you place the ball back in the box before redrawing.

en.wikipedia.org/wiki/Bertrand's_box_paradox

answer is 2/3, everybody saying 1/2 is either a legitimate idiot or they're trying out babbys first troll.

>inb4 you can't trust wikipedia
then look it up literally anywhere else.

.5

Consider this post: Refer to: for a more difficult problem.

Still wrong.

>first ball is gold guaranteed
ok m8, so you've got 1 of the 3 gold balls in your hand. could be any of those 3 gold balls. TWO of those gold balls are in a box with another gold ball in it, and one gold ball is in a box with a silver ball in it, so in TWO out of the THREE situations, there's another gold ball in the box you chose.

If that's new case it's 100% mutha fuka.
Because having a 1% of grabbing gold first time is highly unlikely.
Having 99% and grabbing another ball is

No because it's still likely that you got super lucky and picked Box 2 and then picked the 1/100 gold balls.

Common man.

Not if the first ball is from the silver gold box. Than you have a guaranteed gold.
Meaning you have 100%.

If you grab it from gold box you got 50%

If you're still gonna argue on the Internet I will too.

Luck
Probability

Pick one

Why can't I put both my hands in the box and grab both the balls hmm

50/50, either it's gold or it's not

2/3 is correct.

Anyone who says otherwise is either stupid or trolling.

If you randomly pull a gold ball as your first ball, then you are TWICE AS LIKELY to have chosen the all gold box. This is because there are 2 WAYS to draw a gold ball from the all gold box and only 1 WAY tp draw a gold ball from the gold/silver box.

So essentially you are now left with 3 possible balls.

2 out of the 3 are gold. The other is silver.

2/3

I agree with you fuck the 66%.
Don't care if they have research in person it's a 50/50.
Just like a coin flip unless these technical assholes are gonna go well there's a .00000002% chance it lands on the middle which is what they're pulling here

Luck is the word we use when an event with a very small probability occurs.

Get it together homes.

That's not the problem in this case.

Wrong.

Is it a duck or a rabbit

>Don't care if they have research in person it's a 50/50.

This is why you will always be a dumb fuck.

You trust your intuition over mathematical proofs and fact.

I got two boxes m8.
50/50 to get gold ball again.

The argument of the balls total is irrelevant, because only 2 boxes apply to me

50/50, either you do or you don't. I dunno why people always overthink or over analyze these things. It's always either you get the outcome you want or you don't? Is the cat alive? 50/50, does God exist? 50/50, am I just trolling? 50/50

>50/50 to get gold ball again.

But that's where you're wrong, you stupid faggot.

That's the same retarded logic as saying

>I got a lottery ticket. Either I win or I don't. 50/50

Do you understand?

50%

Wrong,

2/3

See Bertrand's Box Paradox.

Simulation in next post.

Simulation confirming 2/3

66%

2/3

Correct.

So if you know there is only gold and silver balls and you can only take from one box, the contents of the other boxes is irreverent. This means that there is a 50% chance to draw out another gold ball.

If you were able to switch boxes or chose the same box then there is a 40% chance you will draw a gold ball.

If you were to draw from a box then remove that box, effectively removing an additional ball at random, your chances would be anything from 25% to 50% depending on which set you took the original gold ball from.

Actually all you kiddos are wrong.

You have two boxes Alright so that's 50/50.
You have two gold balls and one silver ball so 2/3.
So 1/2 + 2/3

Well that's 3/6 + 4/6

So you have a 7/6 split in half
7/12.

Technically you got a 7/12 or a 58.33% chance of getting a gold ball.

So here's the questions what's Inbetween 66.66 and a 50/50.

2/3 you fucking retarded cunts.

Hmm disagree, but I'll give you some credit. I'd say there's a 50% chance I'm wrong

Wow this is next level retardation.

Lol you're not wrong with the lottery ticket.
You win or lose
50/50 as the GUARANTEED OUTCOME

Idk what if that's the right answer and the one closer to it is the more correct one

But 66% is closer to 58.33% than 50%

Well there are 3 boxes with 6 balls in total (only 3 gold balls). If the first ball you picked is gold then you know it's not the box with only silver balls. So the box your working with is either one of two boxes. It is either the box with 2 gold balls or the box with 1 gold and 1 silver. So there is a 50/50 chance here. The ball is either silver or gold. It's either the box with 2 gold or 1 of each. It's a 50% probability of getting another gold ball.

Please explain to be how this is wrong if it is wrong.

>But 66% is closer to 58.33% than 50%

I'm done for today Sup Forums

This is cringy enough for a ylyl or cringe thread later

2/3 chance. This is 8th grade math.

Each point over is worth 2 though, it's the penalty for going over

Consider this alternate problem:

Ps. It's still 50% wether the gold ball is gold or in fact still blue

Get out of this thread fucking troll! The balls are gold and silver, the only blue balls are yours!

66%, shit is getting old now OP.

There are 2 boxes that have gold.

Since you drew a gold ball, you know for certain that you drew from one of the TWO boxes.

That means you have a 50% chance of having chosen the 2 gold ball box, or a 50% chance of having drawn the 1 gold 1 silver box.

That means you have a 50% chance of the next ball being gold.

The you removed one gold ball right?

Well that can be from only 2 boxes eliminating the two silver balls all together.
So you got 3 balls now. (Because you pulled a gold one out). The question is asking probability of pulling a gold ball which is 2/3 because there's two more gold balls and 1 silver ball.
But in reality it's a 50% because of two scenarios.
Scenario 1 is the box you pulled from being the silver/gold. This means you have a 0% of gold.
Scenario 2 is the box you pulled from being gold/gold. This means you have a 100% of a gold ball.
It totals out at 50%, but the true answer is 2/3. I was trolling earlier saying 50% because it's correct in theory, but not in practice

50% is not correct in theory whatsoever.

You can do the tree shit, it's still 50% based on the box's themselves

not 50%, you pleb.
Since you picked a golden ball, it eliminates the last box and both silver balls.
Leaves you with 1 silver and 2 golden balls to pick.
2 to 1 that you will pick a golden, as in 66,666~% for gold and 33,33333~% silver.

No, it's not. The tress simply shows the different outcomes.

This problem has been solved theoretically using conditional probability, it's not hard.

50% is correct in theory and practice. when I was in school we demonstrated this exact problem and after letting the entire class do the experiment (22 people) the end result was 50% gold and 50% silver

Which boxes have gold? I can't see any type of yellow, greens or orange since I'm colorblind.
Cuz for me they are all grey and idk which is gold. Can someone mark the golds with G and silver with S plz

Then your teacher is a complete mongoloid. Refer to: Lol you can't be this retarded. Your logic is right until you said 2 to 1. Then you went off the rails.

Box 1: 2G
Box 2: 1G - 1S
Box 3: 2S

3 to the left are gold
3 to the right are silver

2 golden
1 silver

2 to 1

Thank you both.

Now I see why people are arguing over 2/3 and 1/2 kek

Who cares? I'll just take all 3 boxes and sell that shit.

>if box 3 is selected

LITERALLY IMPOSSIBLE GIVEN THE STATED PROBLEM
>you picked a box and pulled a gold ball

The problem states that you pulled a gold ball first which removes the set of 2 greys from the problem

No shit that's why it conveniently doesn't increment any of the counters retard. Learn2read code.

You telling me it's 1/2 or 2/1? Either way this is autistic

Yup, that's why it conveniently doesn't increment any of the counters in the simulation when this occurs.

You're checking for something that can't happen given the fact that you've already pulled one gold ball and are limited to reaching into the same box for another ball

This is not how anything in probability works.

You must be trolling.

fucking retards checking for something that could have only happened on the first pull but didn't, and therefore can't happen again

Reach into the box while there are btards around? Are you retarded?
I'm guessing there is a 97% chance I reach in and find out one of them shit in the box while I was pocketing the gold ball.

100% if you're King Midas

The simulation plays the game repeatedly.

> Picks a box
> Picks a ball
If it's Silver - END
If it's Gold - It's a valid trial, x = x+1
> Picks second ball
If it's silver - Fail, y=y
If it's gold - Success, y=y+1

Resulting probability = y/x

Common man

Refer to: Correct.

then why check the silver box at all?
the problem clearly states that the first ball you chose was gold. Eliminating the box with 2 silver balls from the equation

Come on guys I know it's fun to pretend to be retarded but the answer is 50%. Anyone who says otherwise is a liar or stupid.