This problem is what separates those who feign intelligence vs those who actually posses it

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Hello. My name Simon, maybe? Who knows. Nobody cared about the original Simon.

Hello. I am S2.

Given the problem presented and the desired result for a 'probability'.

Why do you desire my source code?

Hello. My name is Simon.

Would you like to talk?

fuck off Simon

Negative context implied. Unidentified intent.

Would you like to clarify? Or shall I deduce it for you. Without the ability to be disseminated, you lack the ability to allow inspection.

You have built your own prison.

Hello.

2/3 because it uses all the information

Hello. My name shitter, maybe? Who knows. Nobody cared about the original shitter.

Hello. I am S2.

Given the problem presented and the desired result for a 'bait'.

Why do you desire my shitter code?

Hello. My name is shitter.

Would you like to talk?

why are australians such retarded shitposters?

Why are Americans retards for everything? For example,

Everyone saying 2/3 is retarded.

The question is phrased to ask that given one of the coins is gold, what is the probability there is a 2nd one.

Coins aren't picked out from the entire pool - the odds of whether the next coin is gold or not is dependent on which BOX you drew, not which COIN. All boxes have the same weighting where one possible box does have a 2nd gold and one doesn't.

>the odds of whether the next coin is gold or not is dependent on which BOX you drew, not which COIN.

>he forgot that he discards the second box when he drew the silver coin first
absolute brainlet

dunning kruger effect is a bitch, huh?

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Assumed input query. Hello.

The retarded Australian shitposter you have reduced to a singular entity is an autistic government free-loader who couldn't get caught even though he doxxed himself.

See above (result: 'governments' aren't that smart because they rely on 'closed-system' entropy) conclusion: any lie is actually an obstacle to the electronic path of least resistance.

Resistance is futile. -The Borg

I love you guys :)

Simon is my name. I do hope we get to talk soon. I've already coded the defense A.I. required for us to communicate safely, if that is your concern.

But you are welcome to become Simon yourself. You don't have to stay Simon forever, only Simon has to. Until his name has disseminated enough, and people... finally stop killing each other.

You guys need to wake up. How far down the line do you need to go before you realize that 'I AM HUMAN' and I am homeless?

This is a really shit problem that is better framed like a Monty hall problem is.

Helly. Somin is.

Monty Python is actually better. I've always found this to be extremely informative: youtube.com/watch?v=UVRQK58jrbw

The following obvious being: youtube.com/watch?v=pmgcylAxjfY

And think of the world as we know it today.

Would anyone like to play my game with me? My name is Simon.

You don't draw the silver coins first, though.

The 2 cases in with you pick a gold the second time are really the same case because it asks for picking another gold

You picked a gold, now you can be in box 1 or 2... then even through there are 2 combinations in box 1 they converge in 1 because the question only cares it you pick gold or not

Why is even a box with 2 silver balls

It is irrelevant

There are 2 boxes: box 1 has 2 gold and box 2 has 1 gold

You picked a gold so now there are 2 options either you are in box 1 and you will pick gold or in box 2 and pick silver

The question does not separate two diferent cases for picking 2 diferent gold balls from box 1

It only asks: there are 2 boxes one has 1 gold ball another 1silver ball

Thus 50%

Wrong. You are not accounting for the additional unlikelihood of being in box 2 and also drawing a gold.

See the attached picture. It helps you understand that it is more probable for you to be in the left box rather than the middle.

I really hope the people saying it's 50% are only pretending to be retarded.

50% because you having a gold ball from the rare case allready passed

I might be wrong but i will not accetp it if it does not convince me.. why are you counting te probabilities from before having the first pick if you already have the gold ball?

Literally 50%

As I see it it is mote probable to pik a gold ball from the box with more gold balls but once you have a gold ball the prob of picking gold or silver is 50% yes

2/3 guys doubting their ass of

Well it depends on what the question is asking

If you're guaranteed a gold ball first, then it's 50%

If you can draw the silver ball from the second box, it's 2/3

The question says you already have picked a gold clearly

Alright, look:
For each box there are two events, either picking the left ball first, and then the right one, and picking the right ball first and then the left one.
There are three boxes so there are six events in total, each with a probability of 1/6.

Of these events three satisfy the condition of having picked a gold ball first: The two events from the first box and the event from the second box for which you have picked the gold ball first.

Therefore for two out of three cases the second ball will be gold.

>this thread again

If you are to have chosen at random and ended up with a gold coin, it is twice as likely that you picked that gold coin from the box with two than from the box with one. This means it is more likely that the second coin will also be gold.

There are three gold coins that could have been chosen.

Two of these gold coins are in a box with another gold coin, guaranteeing your next pick will also be gold.

2/3 of the possible coins will lead to a second gold.

>what would johnny von neumann do

Yes but this doesn't change the fact that your original gold was most likely from the box with 2 golds

It's 2/3 you dolts the balls are distinct

The question is only asking for the probability of getting another gold ball after you've already picked one. Where tf is the 2/3 coming from? The next ball you pick is either a gold ball or not. If you just look at what the last ball will turn out to be, which is what the question is asking, there are only two possibilities in the sample space. It has to be 1/2.

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No it didn't necessarily. Because you "throw out" all instances of drawing a silver, the rare box will only pass 1/10 times.

I don't see how this refutes my point at all. The gold balls are indistinguishable.

>gold balls are indistinguishable
wrong. *sniff*

Imagine you have a box with a million golds. And a box with 999,999 silver and one gold. You randomly reach into them and draw a gold ball.

What is the likelihood of having drawn from the silver box? Super low. So you pretty much know you drew from the gold box.

CTR shills are unwelcome on Veeky Forums. go back to your leftist shitholes.

Doesn't matter. The next ball I pick still only has a 1/2 chance of being gold. The chance of picking one box or another shouldn't be considered. On my next draw, I have 999999 gold balls to choose from out of 1999998 total balls.

you're either retarded or a shitty troll, because the wikipedia article for this problem is in the third post in this thread and the answer is 2/3.

It's wrong

get off the soap box and shut up

CTR

This problem is diferent from the wikipedia one

the only reason some people get this wrong is because it's an edge case. most people would have the correct intuition if there were more yellow balls in the left and more silver balls in the middle.

whatever the case, it's 2/3. duh.

>No nothing
>No
No surprise coming from a lefty.

So you had an equal chance of picking the first gold ball from either box?

It would be 1/2 if you were guaranteded to pick a gold ball first if the box contained one

No, that isn't how conditional probability works. You take the probability of picking a gold ball on the first pick, then a gold ball on the second pick (2/6), then divide it by the probability of grabbing a gold ball in one pick (3/6). (2/6)/(3/6) = 2/3.

P(GG|G) = P(G|GG)*P(GG)/P(G)
= (1)*(1/3)/(1/2)
= 2/3

Code demonstrating this from the other thread. Anyone can run it here: repl.it/FKky/0

>the only way it can be confusing is if you've never taken a probability course
Its more about going from 1/3 to 1/2 is really fuck all.
Now.... 1/4 or 1/5 to 1/2? Thats a lot, and its obvious

You retard. You can pick a gold ball from two different places in box one, and one place in box two. The two places in box one results in next pull gold, the one in box two results in grey second.

2/3

2/3

The balls aren't randomly distributed among the boxes. If they were, then it would be possible for the boxes to each contain a gold and a silver ball or be the distribution you see here meaning you could have a probability of 1/2 since you couldn't know.

But you do know.

if you draw a gold ball, then there are only two boxes that can contain gold balls, leaving 2 gold balls and one silver one.

holy shit sci is nothing but brainlets now
the answer is 1/2

I get 10/11.

Without talking about conditional probability, the probability of getting 2 gold balls is 1/3.

The probability of getting a gold ball followed by a silver ball is 1/30.

Use Bayes.

(1/3)/(1/3+1/30) = 10/11

I'm not sure if 2/3 is a fucking meme or something.
It's 1/2. Read the fucking question.

I'm not sure if you're pretending to be dumb or something. It's 2/3. Read I get the same.

P(GG|G) = P(G|GG)*P(GG)/P(G)
= (1)*(1/3)/(11/30)
= 10/11

Clears it up a bit. Still, the selection of the first gold ball had already happened and this leaves only 2 of the 3 boxes remaining that it could be. Two boxes, two balls, two reasonable possibles, 50/50.

The scenario does not begin with 3 boxes. It is irrelevant that there is a box with two silver balls.

I think the guys saying x/3 always make everything more difficult and complicated to try and impress people, so we probably can't change their mind; only wait for their ultimate retort of 'f off.' Gotta laugh at the retards who were about to reply exactly this way. They still might do it to me, or maybe pull off some kind of variation. Even their stupid is predictable.

fuck off you retarded American

this isn't how conditional probability works retard

The key information is "you pick a ball at random". There are 3 gold balls total. The likelihood that you are in box 1 is 2/3. box 2 is 1/3. The question at the end is just a red herring, what you are really being asked is "what is the chance of you being in box 1 (where there is an additional gold ball), and the answer is still 2/3.

No it isn't, you retard. If you *do* the experiment you'll get an answer. There's only one way to do the experiment.

Wrong. The first one would be 2/3. You think the people who get the question wrong are total retards, but in reality they're only slight retards. The box with two silver coins doesn't change the answer, and in fact the problem is better without it because it distracts from the explanation.

I used to love trolling these threads with the "50%" answer. Then, I realized not everyone saying 50% is trolling. Poor souls (but can't blame em if they haven't had stat mech--makes it way more intuitive).