How smart are you Sup Forums?

How smart are you Sup Forums?

Don't OVERTHINK this

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en.wikipedia.org/wiki/Bertrand's_box_paradox
youtube.com/watch?v=yuOzZ7dnPNU
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75%

66.67%

50%

Since we know it's not the double silver can't we just ignore those for probability and say its 50/50 on which box of gold you got?

these guys get it

50%, 1:2, 1/2

50%

Since it's out of the same box we can take the silver only box out of the equation. Then since you're picking out of the same box as the first gold, it'll be either gold or silver. Basically after the first gold it's "did I pick the gold only box or the one gold box?"

WRONG

correct

0.67

>explain

2/3 cuz if you picked a gold ball from one it would have to be the box with 1 gold and 1 silver or 2 golds

>3 gold balls
>1 silver ball

Op youre retarded

But you already picked one so 2 gold 1 silver

Which means it's 66.7% chance idiot...

50% Eithet you picka gold ball or you dont

Here's your (You).

exactly. the chance is higher than 50%

>2 boxes
>1 box has other gold
>1 box has silver

50%

the rest are trolls

this is so dumb. you took 1 out. which means you're down to 2 boxes possible out of the 3. you have 2 selections left 1 silver or 1 gold and that's it. there's no chance of getting the 2xsilver box ever in the universe. it is exactly 50%

I hate threads like this. I'm not smart enough to know if Sup Forums is just trolling me with false answers or if the REAL answers are just stupid.

If I stuck my hand in a box and got a gold then I'd rule out the double silvers. Then I figure the other ball can only be gold or silver so that's 50-50. If that's not the real answer then you professional mathfags are making shit way too complicated.

1 box has 2 golds, if you draw either you win.

2/3.

There are three 3 gold ball, thus 3 chances to draw gold.

Now look at the boxes. 2 out of 3 of those gold balls are in a box with another gold ball. So 2 out of 3 times, you'll draw a second gold ball.

But it said draw from the same box and you already have gold. So you'll draw a gold or silver

WRONG
kys

This is an open problem, smartass. You didn't specify whether the first ball goes back to the box or not before the next draw.

No, you're just dumb as shit. There's 3 gold balls and 1 silver ball. You take one gold ball out. That leaves 2 gold balls and 1 silver ball. What is the chance to get a gold ball out of 3 balls when 2 are gold??

I think it's pretty obvious that it doesn't.

You have to pick from the same box you already picked from, you don't know whether the other ball will be gold or silver so it's 50/50.

You aren't taking the other 2 balls. You ignored them when you chose the box you did. If you know you aren't getting dual silvers, and you just got a gold then you literally have a 50-50 chance of the last ball in your box being gold or silver.

Shouldn't it be 33.3%

Then it's 50%.

t. Math major

That's the right way to solve the problem, everyone else is talking shit. If all three balls were together it would be different but since you're picking out of the same box it's 50/50.

Jesus Christ you guys are retarded it's a variation of the classic montys problem the answer is 1/3.

If you pick a box at random the chance it will have two gold.balls is 1/3 the fact that the first ball you pick is gold doesn't change the probability of what box you picked. Is this bait or wtf?

>what burger education looks like

No, look at it again.

There's a box with two gold balls. If you pick the first one, the other ball in the box is gold. If you pick the second one, the other ball in the box is also gold. You draw a gold both times, and both times the other ball is gold.

Now look at the box with one gold and one silver. If you draw the silver, it doesn't count because you had to draw a gold first. But if you draw the gold, the next ball is silver.

So there are 3 cases where you draw a gold, first. And in 2 of them, the second ball you draw is gold.

The best education?

It would be 1/3rd if it didn't say a gold was already drawn

Yeah I can see this line of thinking - if you picked a gold, higher chance of being from the 2 gold box.

the question is: after you took out a golden ball what is the probability of the second one being golden, you eithere have a box with two golden balls (100% chance of taking another golden ball) or a box with a silver and a golden ball (0% chance of taking out a second golden ball) so the overall probability is 50%

also remember gambler's fallacy, past events don't influence future ones, the probability of you taking one box or the other doesn't matter

It's 66%

This has nothing to do with gambler's fallacy at all

>There's 3 gold balls and 1 silver ball.
in 2 different boxes, the balls arent just in one big box you dumb cunt.

Are you illiterate? You're picking from the same box, not from all three

Its still 50/50, no matter how hard you try to bullshit people into thinking otherwise. Have a (You).

The gold was drawn after the box was picked. The chance of picking a box with two golds is 1/3 why is everyone struggling with this.

Except you're counting boxes, not balls.

You have 2 chances in the box with two golds, and only 1 in the box with a gold and a silver.

Refer to

Exactly and only one box out of the three would enable you to pick one gold ball after another

Its 50/50, The rest are people who did not properly read the problem

Because if it was the two silver box then the question wouldn't be asked

Why would the average person need to know the answer to this? In what way does probability help me in real life?

>The chance of picking a box with two golds is 1/3 why is everyone struggling with this.
No one's struggling with that, because that's NOT WHAT THE QUESTION IS ASKING YOU.

Within the bounds of the question, you have either picked the box with two gold balls, or the box with one gold ball and one silver ball. The box with two silver balls does not enter into it at any point. You do not switch which box you've picked at any point. You only have TWO outcomes, thus the chance is 50/50.

don't be fucking retards. this is not the monty hall problem and even the MH isn't 2/3 it's just some insignificant percent over 50.

I'm gonna be the guy who spoils it. This is a real problem

en.wikipedia.org/wiki/Bertrand's_box_paradox

The real answer is 2/3, and the whole explanation is here

Its 66% saw this last night

The box was picked at random. You could.of chosen the silver ball but this is asking about a specific case where you draw a gold one.

Within the bounds of the question the box was picked at random.

50%

the first selection eliminated the 2 silver box from the possible boxes you just picked out of. This leaves the 2 gold and the 1 gold 1 silver box. Which means on your second draw you have a 50/50 chance to take another gold ball.

66.6%

It just shows whether you lack critical thinking.

If you still think it's 1/2 after reading thread, you probably have a low IQ

The question specifically tells you you've picked a box with a gold ball.

It is not asking you what the probability of picking another gold ball it from the onset of the choosing, it's asking you the probability of picking another gold ball after you've already picked one. At that point, the probabilities have been altered.

Wrong. The 2g and 1g box do not have the same probability of being the box you picked from. The 2g box has twice as much chance of being the box you picked from because 100% of the balls within it are gold balls.

at least 100

Another case of counting boxes, not balls. But you're not drawing boxes. You're drawing balls.

There are 3 gold balls, thus 3 out of 6 chances of drawing a gold the first time. So whatever you draw second, it must be X out of 3 chances.

And look at the boxes. If you draw either gold from the box with two golds, the other ball is gold. That's 2/3.

If you draw the gold from the box with a gold and a silver, the other ball is silver. That's 1/3.

Exactly, the question only exists because you already got a gold AND are drawing from the same box

>But you already picked one so 2 gold 1 silver
you're only dealing with 1 box. 2 balls. you can't have 3 possibilities when there were only 2 to begin with and you've subtracted 1.

>1/3

You're an idiot. So is the guy you're arguing with. It's 2/3.

nah brah that's just not how it works.

I would bet even money with anyone 100000 times that whey they pick a gold ball first, the next one will be gold.

The important thing that people are not realizing is that you are twice as likely to pick a gold ball AT RANDOM from the box with two gold balls in it. That weights the chance that the next one is gold to be higher than 50%.

I don't know how else it can be explained. People are using math and using logic and using simple explanations and people still pretend they don't get it.

No, you're drawing boxes. You have either drawn the GS box, or you've drawn the GG box.

Since you picked a gold ball AT RANDOM, there is a better chance that you picked a ball in the 2-gold box. Since aprox. 2 out of 3 times you're gonna pull a gold ball it's going to be from that box. 66.67

The question isn't asking which box you've picked out of three. You'd be right if that's what was being asked.

It's asking if you'll draw a gold ball again. You have two outcomes: gold or silver.

You have two chances of drawing a gold from the gold/gold box, and only 1 of drawing the gold from the gold/silver box.

lmfao are you telling me everyone in a 3rd year math course understands everything taught in class? Why does being a math major mean that your answer is right?

The fact that you are calling yourself a "math major" means you haven't graduated yet or don't have a job that uses math so you obviously forgot statistics. QED.

No, you're drawing balls. You're assuming that the problem ends once you pick a box, but it doesn't. You can pick the 2nd box, but the problem still isn't over, because after that point you still have a 50% chance of not drawing a gold ball at all. On the other hand, in the other box, once you've picked it it is guaranteed that you will pick a gold ball. Therefore, the probability that your gold ball came from either box is unevenly distributed in favor of the first box.

why is the first pick random? it's part of the scenario. it's the starting point. we started with already knowing what the first ball was. your text suggests we could have had a different first ball which is a different scenario entirely.

you know you picked a box that had gold in it. 2/3 of the gold pieces are on the 2-gold box. So there's a better chance that's the one you picked up

Because that's the fundamental basics of probability. Of course the first pick matters. You still have to factor in the chance that you hadn't gotten a gold ball from the middle box at all.

You have two chances of drawing a gold from the gold/gold box,
No, you've only got one. If you've drawn the first gold ball, you can only draw the second. You can't draw the first one again.
Remember: the question only askes which colour you'll draw NEXT, past the point of the first ball draw. Nothing else before that drawing matters.

>> 50%
but wait, there are 3 boxes and i don't understand maths.
>>you're an idiot, one of the boxes contains 2 silver balls its discarded
but wait! if you discount the 3rd box then theirs 3 gold and 1 silver ball potentially, that makes a 75% chance of getting another gold
>>kinda, if we put the first ball back then redraw a ball you'd be right but that's not the question.
BUT WAIT!! if im holding the ball in my hand and disregard that ball its still 2 gold vs 1 silver, making it a 66.66*% chance that the next ball will be gold.
>>imagine you're holding the gold ball in your hand and the box is in front of you, that box either had 2 gold balls or 1 gold 1 silver, you're holding 1 gold ball in your hand and infront of you is either the 1 gold ball or the 1 silver ball.

>So there's a better chance that's the one you picked up
Again, WITHIN THE CONFINES OF THE QUESTION, the s/s box DOES NOT MATTER. ODDS HAVE CHANGED.

>past events don't influence future ones

it does when the balls don't change and you have 2x the chance to pick a gold ball from the one that started off with two. The fact that the question states this has already happened doesn't change the way math works. Think about it as disregarding all of the times you picked silver; it's just not stated in the problem but it happened because it said "AT RANDOM".

If the problem said "you pick a ball at random from a random box, you get silver x times in a row, and put them back and re-randomize the boxes and balls, until you finally pick a gold one, what's the chance that the next one is gold?

Since the problem states the selections are AT RANDOM it implies that you have a higher chance to be picking from the double gold.

You grabbed a gold piece, and 66% of the gold pieces are form the 2g box

it's 2/3 guys.
if you still think it's 1/2 then you're wrong and don't get mad just learn why it is 2/3 and everything will be fine :--DDDD

Not the guy you're arguing with, but no it isn't a different scenario you faggot.

OPs question is basically asking "What's the probability of you getting a box with two of the same color?" Which is easier to think about, and is still 2/3rds.

no i don't because we started with the 3rd box phased out already. we started knowing we had 1 of 2. why are we pulling 3 back into the mix?

At this point, you're either a complete idiot, or a troll.

The three boxes:

gold 1 / gold 2
gold 3 / silver 1
silver 2 / silver 3

If you draw gold 1 or gold 2, your next draw is gold.

If you draw gold 3, your next draw is silver.

There are literally no other options where you can draw gold first. So you can draw gold first, in 3 different ways.

And in 2 of them, the next ball will be gold.

This.

>WITHIN THE CONFINES OF THE QUESTION

the question also stated specifically that you selected the gold ball by picking a random ball from a random box. That has a higher chance of happening in the g/g box. This has to be taken into account BECAUSE the problem says the initial selection was random. It DOES NOT state that you picked either g/g or g/s, it DOES say that it was RANDOM.

You still had a 1/3 chance of picking the s/s box, even though the problem implicitly disregards all of the trials where this happened, or the times when you picked the silver from the g/s box.

Guys, if you already got a golden ball out that rules out the possibility of box 3 being the one you selected. With this in mind its easy to see why the odds are 50%: if you got box one then you have 100% chance of getting another gold ball, if you got box 2 then the chance is 0%, so your averaged chance of getting another gold ball is 50%

yes it is
no it isn't
faggot
hey look you and i have equally informative posts!

Math major? And fucking up a simple conditional probability problem? What country may I ask?

We're not you retard. How did you even get to that conclusion from what I said? It's pretty clear you don't actually understand what I'm trying to tell you. Both boxes have an equal chance of being selected, but you don't know which box you've selected. All you know is that your first pick was a gold ball. Now, of all the scenarios in which you have already picked a gold ball the first time, how many of those times will you also pick a gold ball the second time? Because the two gold box had a guarantee of yielding a gold ball the first time, and the one gold box only had a 50% chance, you will pick a gold ball again 66% of the time.

>Case 1:
Box 1, Ball 1 = Gold
Next ball is gold
>Case 2:
Box 1, Ball 2 = Gold
Next ball is gold
>Case 3:
Box 2, Ball 1 = Gold
Next ball is silver
>Case 4:
Box 2, Ball 2 = Silver
Criteria for premise not met, start over
>Case 5:
Box 3, Ball 1 = Silver
Criteria for premise not met, start over
>Case 6:
Box 3, Ball 2 = Silver
Criteria for premise not met, start over

2 Gold results
1 Silver result

2/3

2/3

Pick first gold ball, other one is gold
Pick second gold ball, other one is gold
Pick third gold ball, other one is not gold

It states you have to pick another ball from the SAME box you picked the first one

>ITT Limmy
youtube.com/watch?v=yuOzZ7dnPNU

>why is the first pick random?
because it is explicitly stated in English words in the problem.

>your text suggests we could have had a different first ball

Yes, we could have, that is what random means. Those trials are disregarded and that is why the problem confuses people. They still affect the correct answer.

You're so fucking dumb it hurts. Once you select a gold ball you can only choose between ONE silver or ONE gold ball. 50/50 you fucking retard