This should be easy.

# This should be easy

either it is or it isnt, so its 50 50

69

it cannot be the box with 2 silvers as a gold has already been drawn.

therefore there is either a 1 gold or 1 silver left depending which box was selected.

50% probability you draw another goldie.

42

2/3. The boxes are there to confuse your intuition. Imagine a sack of 6 balls, 3 silver and 3 gold. Each individual ball is considered paired with another ball, so that two golds, two silvers and one gold and a silver are paired. I reach into the sack and pull out a gold ball. What is the likelihood this gold ball is paired with another gold ball?

As there are three gold balls, two paired with another gold ball and one paired with silver, the answer is 2/3.

In general, one has to be very careful with "common sense" probability problems that feature a posteriori events. They can be very tricky.

Thank you!

The question is ambiguous, because it depends on exactly what you mean by the probability of something happening. The way you get the 2/3 answer is by asking, if there are a million rooms with this set of boxes in them, someone in each room picks a ball, and you ignore all the rooms where someone picked a silver ball, what share of rooms had someone pick from the box with two gold balls? But since the question actually asks something much vauger than that, it's ambiguous and can be interpreted as needed for the shitpost you're trying to make. And you can shitpost endlessly by insisting that your interpretation of the question is what it's really asking.

2/3

No.

They were wrong, don't thank them.

its

(1+1/2+0)/3 = 0,5 = 50%

of course it can be interpreted in different ways, EVERYTHING can, but its not the questions fault that our intuition is to interpret it in the wrong way.

You are not the one who gets to decide what this question means, its the guy asking it that decides that.

It could maybe be a little more clear, but not without giving a part of the answer away.

No, it isnt vague.

The wording is HYPER SPECIFIC.

This is the reason why people get confused, if you didnt just Google the answer, you would know this.

50% chance since one box has 2 silvers.

How?

You put your hand in a box and took out a gold, what are the odds you will get a gold from the same box?

There are only 4 possible balls at this point, the one you took, which is gold, 2 gold, and 1 silver.

The box with 2 silvers is not relevant at all.

2/3 chance.

the box with 2 silvers is irrelevant.

when u take out one gold, there are 2 gold and 1 silver left. therefore the chance the next is gold again is 2/3

No.

There are 2 golds capable of being picked up, to 1 silver.

You took a gold, but there were 3 golds.

There is only 1 silver left because of the rules.

3 total balls, 2 are gold, and 1 is silver, what are the odds of getting gold?

This is a likelyhood of 2/3

Lmao.

This one is just a meme

i feel like you got the right answer the wrong way.

in your scenario its more like all the balls are in 1 box, but they arent, you are not choosing 1 out of 3 balls after picking a gold, what you pick after it is already decided before you do the picking.

its conditional probability and you forgot the condition part.

now better

(1+0+0)/3 = 0,(3) = 33%

No, that is not my scenario, and this assumption is retarded.

I clearly reply to people who are confused, they will not understand a mathematical explanation, and having been in this thread before, I know they will not understand an equivocation.

This is a way to describe to someone who does not understand "HOW IT COULD BE?"

If you simplify it, it makes it easier to understand, no?Furthermore, it is WORDED to produce that effect; I'm absolutely correct, conditional probability ignores the condition, that's how you make a SINGULAR probability, and not multiple for each condition.

Do you know what you're talking about, or did you just wanna start something?

Is a 50/50....there are only two possible outcomes either you pick up a gold ball or a silver ball the number of balls is irrelevant because there are only two and they're separated I to boxes. If you were pulling from a bag excluding the two silver balls you'd be right it'd be a 2/3 but your forgetting the nature of the boxes themselves

There are not 2 boxes.

There are 3.

Conditional probability removes the 3rd box, and conjoined the final 2 into a singular.

You just answered the question: "What are the odds of putting your hand in the ONLY gold box, given the removal of the ONLY silver box?"

The question's subject matter IS the balls, so the balls are ALL that matter, not the containers.

It's a more complicated variation on the Boy or Girl Paradox which is admitted to be ambiguous.

If we posed OPs question as being presented to a million people discarding all the results where the first ball drawn isn't gold (that is removing instances of people selecting the Silver-Silver box), 2/3 of the time the 2nd ball drawn will also be Gold.

If we look at one individual instance, rather than across a range of repeated cases, it's 1/2 that the next ball will be Gold. They've either selected a box with another gold or one with a silver.

Normally I try to assume my opponent is earnest - it's better practice - but you're so clearly a troll it's not even funny. 0/5, better luck next time.

It's a 50/50 there are only two possible outcomes you pull out a gold ball or a silver ball...which one you pull out is determined by the box you selected from so it is in fact all about the boxes

sure it might help people get the right answer easier, but it doesnt help them understand conditional probability.

if we painted a silver ball in the double silver box to gold, people would arrive at the wrong answer if they followed your logic.

second ball drawn is from the same box which the first ball was removed

Except that the boxes are actually important you mongoloid.

The silver only box doesn't matter. You already pulled out a gold box so that one isn't involved at all. It's not a matter of 6 balls, it's about 4 balls now.

You already picked a single gold ball so your only options are either the gold ball from the gold only box or the silver ball from the gold and silver box. It's 50/50.

or the other gold ball from the gold only box.

2/3

There are legit two boxes you fuck. You already pulled out a gold ball. The third box isn't involved at all.

Fucking based answer

can a wrong answer be based?

Once you've pulled out a single gold ball your only choices are the second gold ball from the first box or the silver ball from the second box.

1/2

having two options that are randomly chosen from isn't 2/3

or the first godl ball from the first box

2/3

you are making the assumption that the ball that gets drawn is then placed back into the box from which is was drawn. this is not stated, but it is better to assume that it was removed and not placed back in

Reword it:

>You pick a box at random. What is the probability that it is the box with 2 gold balls.

A: 1/3

the first gold ball is already the one you pulled out. Once you've pulled out a ball from the first box, you don't magically still have two choices. There's only one ball left.

1/2

no im making the CORRECT assumption that you dont know which gold ball you drew, it could be either one and they are both separate events with the same probability, aka 1/3, meaning your chance to choose that box is 2/3.

and you dont know which one it is.

But since the question is posed in terms of something that's already happened, the straight Bayesian formulation isn't obvious either. Since it says I've already drawn a gold ball, rather than talking about future contingencies, the Bayesian argument is akin to asking, "I flipped a coin and it came up heads. What's the probability it actually came up tails?"

So perhaps the better point is that even "educated" intuition isn't often very reliable.

2/3

6 possible outcomes: SS, SS, SG, GS, GG,GG

only outcomes with a gold first are relevant leaving: GS, GG, GG

2/3 of these get a gold on the second ball

its conditional probability, get with the program.

It doesn't matter if you know. If I pull out the first gold ball, the second gold ball is still left. If I pull out the second gold ball, the first one is still left. This question is about gold balls, not which specific gold balls are pulled out.

Your logic is just fucked lol

The only relevant outcomes are GG and GS since you already pulled out a gold one. 1/2.

the box with 2 silver balls becomes pointless and can be forgotten so its just a question of which box of the other 2 did you choose, and if you got a gold ball already then you have a 50/50 chance to either get a gold ball or a silver ball based on the presumption that you dont know which box you chose.

so its 50/50 you either get it or you dont

i think i've done all i can, now its up to you.

the name of the problem is Betrand's Box paradox, its used in teaching of conditional probability and the answer is 2/3, go look it up if you dont believe me or forever stay ignorant, your choice.

Doesn't matter which one is drawn. You can only either draw a gold and a silver. Being able to draw either of the gold balls from the first box doesn't give you a higher chance.

No, it has some of the window dressing of conditional probability, which triggers the intuitions of people who've taken a probability or stats class and think that makes them clever.

except it does, see im not backing up, i KNOW the answer is 2/3.

you are wrong.

it says the gold ball is "taken" from the box which means it is not replaced.

you are drawing from the "same box" so there is only 1 silver ball or 1 gold ball left to depending on which box was originally selected.

can't you read?

>undergraduate level logic problem

>smart

okey

This

again people thinking they are smart for taking a class is not in any way relevant to this problem, which btw is used as a teaching tool for conditional probability, its a famous problem, not just something i came up with to troll this site.

Its time for you to put your big boy pants on and stop having these emotional responses to fucking math problems.

where did i use the word smart?

>ignorant

he is incorrect though because the sample is being taken after the first selection.

in any case the probability of selecting the combination GG from a sample of

GG SS GS is only 33% which is 1/3 not 2/3

you being ignorant of something doesnt even imply my smartness, do i need to teach you english too?

incorrect the ball is "taken" from the same box, not selected and replaced

this is not betrand's as the paradox states the second pick does not have to be from the same box, while OPs image does

wrong, you choose a gold ball not a box, 2 of the 3 gold balls are in the same box, meaning 2 of the 3 possible results end with you picking another gold ball from the same box.

and where are you getting this?

the only Betrand's Box paradox that i know of forces you to choose the same box twice.

but it does? provision of a derogatory adjective is a direct implication that you do not possess such a quality.

Wrong you do choose the box and select from it both times.

is says so in the rules, read it again and come back later

It's 50%.

3 boxes.

gg (gold/gold)

gs (gold/silver)

ss (silver/silver)

ss is excluded since we drew a gold ball. There is 1 event remaining - drawing a new ball. If original box was gg, then gold remains. If the original box was gs, then silver remains. It cannot be ss, since we drew gold.

Hence 50% probability. Number of balls remaining etc is irrelevant since we do not have more than 1 event remaining. Even if one box contained 10 billion gold balls and the other only contained 1 silver, the probability for the next event would still be 50%.

you are right now (before reading this) ignorant of the fact that im holding scissors, does that mean you are stupid?

no it does not.

but even if it did, does you being stupid mean that i am smart?

no it does not, get a fucking grip.

this is shit quality bait or you cannot even read the problem.

if it is the latter then you are obviously not capable of solving this and either way you are a waste of time in this thread

youre all fucking retards

I know most people claiming the answer is 50% are trolling, but just in case anyone truly believes it, here's a simulation that runs the experiment 1,000,000 times, with the percent chance that the second ball to be drawn will be gold in blue text printed to the right.

aha you are pivoting your argument, very humorous. I suppose we must now plunge into the mixture of contextual word usage and word definitions. Both of your uses of "ignorant" are in no way comparable. The first usage was in direct reference to the comprehension of a basic piece of logical theory, a topic covered in academic coursework. You also coupled that usage with the words "forever stay", which imply that the reader has missed out on something important, in this case, an academic topic. Those who miss out on academic topics (classes) are generally regarded as not smart. You then switch to an example involving the intimate knowledge of the presence of an object within your grasp, which is not an academic topic. Obviously an user having no knowledge of another user's basement attire is not a metric for "smartness", but the grasp and knowledge of a basic academic concept is.

1 glorious bastard also did this experiment by hand, 100 times. the results were similar.

>python

>he doesn't like to play with kiddie code

so because i have never studied lets say biochemistry, i am stupid because there is a concept that i have never heard of and therefor am ignorant of?

bottom line, words dont have meanings, they have usages and you are not the one who decides how i use words. you are free to have your opinion on how i should use words, but i dont care.

>non-CLE

sorry but your program is not taking into account variable change.

the question is not what is the probability of choosing combinations GG or GS, as the gold ball is already chosen before we begin.

now we are left with either a G or S possibility depending which box was chosen at random and there is a 50% chance of it being one of those.

you must read the question clearly when solving probability and take into account variable change in cases such as this

have you even been reading the thread dumbass? everyone of you is making the same mistake, the first gold ball in the gold/gold box is already gone, taken out, theres only one left in that box, so in that box you would pick gold and with the gold/silver box you would pick silver. its 50/50 which box you get and its 50/50 which color you draw

this program does not take into account variable change.

the first ball is chosen before we begin calculating the probability so you do not include it in the code as you do not include the SS box.

therefore the only 2 choices are 1 S ball and 1 G ball

You are ignoring context once again, likely in attempt to go to bed "correct". You then pivot a third time to a cop-out rebuttal, suggesting that nothing said in any language has ever had any meaning. sad day for user, your undergrad is showing cutie

Imma be honest here, English isnt even my first language and i have never visited an english speaking country.

I have never studied probability, statistic, debating and i dont even know what an undergrad is.

But im still not wrong.

I also check all of those boxes but with more ink

well my whole point was that education etc are all irrelevant, its as if you attempted an argument from authority, but with no authority, never seen that before.

Are you the same guy who tried to argue with me over the two-coins logic puzzle a couple nights ago? You sound like him.

Either way, let me put it this way:

>half (1/2) the time you pick the [Silver, Gold] box, the first ball you draw will be silver

>every time (2/2) you pick the [Gold, Gold] box, the first ball you draw will be gold

>you therefore have twice the opportunity to initially pick a gold ball from the [Gold, Gold] box as you do from the [Silver, Gold] box

>if you pick the silver ball from the [Silver, Gold] box, the trial doesn't count, because the condition is that the first ball drawn is gold

>therefore, when the first ball you pick is gold (66% chance coming from the [Gold, Gold] box because of the above), there is an equal chance that the next ball you draw will be gold

Does that make more sense? x:

>(You)

This is just the difference between "given that" and "independent trial".

Given the knowledge of the first ball being gold, it's 2/3, as the fact that the ball was gold means you are twice as likely to be in GG than GS in the first place. So you are disproportionately likely (twice as likely) to get gold again. This makes sense, because GG has twice as many gold balls as GS. So, given that you have a gold ball in your hand, your twice as likely to be in GG. 2:1 = 2/3 : 1/3.

If you are NOT told the information about the first coin and you are only presented with two single-ball boxes, told one has G and one has S, THEN it's 50% BASED ON THE INFORMATION YOU HAVE.

the initial conditions. You chose a random box and in that random box you picked a random ball which happened to be gold. This changes the setup as you were slightly more likely to get the box with two gold balls. So it has to be slightly more than 50%.

the question is not asking the probability of choosing 2 gold balls or the box with 2 gold balls.

1. the box is "already" chosen at random.

2. you have "already" taken a gold ball out of the box.

3. from the same box you are about to take another ball.

this is where the probability calculation begins: what is the probability the ball is gold?

50%

you are over complicating it and not reading the question correctly.

>coin toss

nah wasn't me my isp was blocked so i have not been on for a while.

also i have no insulted anyone on this thread so that wasn't me either

This is another iteration of Bertrand's Box paradox. Google it.

Forgot to say, though you THINK it's 50%, if whoever is presenting you the boxes only does so AFTER getting a gold themselves, you'll still see a 2/3 gold rate. Probability doesn't care about what you know or think you know, it happens anyway.

>the question is not asking the probability of choosing... the box with 2 gold balls

Except it kinda is? With a condition, of course, that it only matters after you've picked a gold ball from the box you chose.

But, because you're twice as likely to pick a gold ball from the box with 2 gold balls, the trials will count twice as often as if you happened to pick the gold ball from the box with a silver and a gold.

"What is the probability that the next ball you draw from the same box will be gold?" is the same as asking "What is the probability that you picked the box with 2 gold balls," since if the first ball you drew was silver, it wouldn't count.

I don't understand where you're seeing things differently than me here.

It's really not 50%.

It's only 50% if you RANDOMLY choose boxes again and disregard half your current information (keeping only the knowledge that you're not in SS), rather than sticking with the original box.

Needless to say, if you're going to get to keep the coin, you should stick with the first box if you got a gold the first time. Likewise, if you picked silver, you should switch.

It's very similar to the Monty Hall problem.

i hope you don't calculate like this in poker, you cannot just play the hand 100 times over and hope to average it out lol

I actually love Poker, and do quite well.

That, of course, is a completely different discussion to be had than what we're talking about here.

Poker's got a whole plethora of external factors to take into account, because it includes an element of choice, giving way to the potential for human error, which may skew otherwise statistically sound results in a vacuum.

If you haven't already, you should pick up playing XCOM.

16.5%

The balls are misleading.

Imagine you had two jars of paint, one gold, the other silver. Your boyfriend drank out of one for lunch, but you don't know which. You do know you drank out of the gold one. That night you are fucking each others assholes, what are the chances you pull your dick out and it's covered in gold paint?

2/3 is correct.

It's similar case in poker where you see, say, a 7 on the board (but not in your hand). The chance that someone is holding another 7 is slightly higher than a naive expectation, because they're more likely to have not folded if they have a pair, 3s, 4s or full house using that 7.

Using what you see on the table, you can make statistical inferences based on the information beyond simple enumeration of outcomes (in which case you'd conclude an opponent is UNLIKELY to hold a 7, because there's one on the table).

Poker is MUCH more complex, and the "boxes" are strategizing against you, so it's very very hard to make inferences on the fly.

The fact that people trip up on this simple problem and disregard half the given information by mistake says a lot about why people lose their shirts playing poker.

That IS 50%, because he either drank it or didn't, and is independent of your drinking.

BUT, if two people drank the gold paint, will there be enough to properly coat the anal walls?

Well this is information you need to know. If that has an effect, but you didn't know it, your calculated probability will differ from the actual probability.

If we're talking a little Warhammer jar, probably not. If we're talking a 5 litre bucket, probably yes.

can one of you geniuses explain how this is any different than what you are saying?

Yeah but there's a 2/3 chance the box you picked from is the one with two gold balls.

a woman is pregnant with twins.

the woman gives birth to a boy.

are you saying you think that the probability of the next twin being a boy is 2/3?

Actually, it pretty much is. 1/3 of twins are born monozygotic (identical), while 2/3 are dizygotic. So, given that one of the twins is a boy and we don't know if the twins identical or not, there's 1/3 chance the other is identical twin boy, 1/3 the other's a non-identical twin boy and 1/3 it's a non-identical girl. So, 2/3 the other's a boy.

there is not a 1/3 chance they are identical.

it is 4/1000

and boy girl Dizygotic twins DO occur 50% of the time.

4/1000 BIRTHS are identical twins. Compared to global average of twins at all, we get that 1/3 of TWINS are identical.

Also yes, half of dizygotic twins are different sex. I used that in my calculation, so no idea what you're on about.

No. but let's consider three indistinguishable pregnant women: one with 2 boys, one with two girls and one with one of each. If the first one to give birth gives birth to a son, there's a 2/3 chance her second child is also a boy.

The difference is simple: if you don't know which child is older and you ask "is your older child a boy", and ONLY if she answers "yes", then ask her the gender of the other one, then it's 2/3 to be a boy.

This is because you're twice as likely to get a "yes" from Woman BB than Woman BG in response to the first question.

In your example, you don't do the first question, so you have no bias towards which one you ask. Then it's 50:50.

There's a simple way to visualize this problem I think. You can substitute gold balls for silver balls in the question, which means the odds of the second ball being silver of you picked a silver are the same as the odds of the second ball being gold if you picked a gold ball.

The question can be therefore simplified to: if you pick a ball at random, what are the odds of the other ball in the box being the same color?

you guys would argue over the color of sight to prove your mathematical "paradoxes" and "theorems" and "subjective opinions".

learn critical thinking and to measure and assess.

user is right, the evidence is here:

this is probably the simplest way for these mathematical magicians to see how their paradox theorems do not work lol

The genetics behind this means it's not a good analogy for the OP post

welcome to the real world son

I don't see your point, the simplification shows that the odds are 2/3. Either you picked from one of the two boxes with identical balls, or you picked from the one with different balls.

You have to first stop samefagging, then realise that we're not even arguing about the same thing. I'm the identical twins dude, and I'm not presenting my argument as relevant to OP's problem. It's merely a funny coincidence in my opinion.

you speak with the disdain of one who completely flunked their high school math papers. Just because you don't understand it, doesn't make it not real.

Probabilities are not meant to be an accurate representation of reality you mong. It's just a tool to make decisions based on limited data.

Yep

1 in 2, since there are 2 boxes with gold balls and you are taking the 2nd ball out of the same box, 1 box Will have a Silver the other box a gold ball. So 1 in 2 chance

66% basically prisoners paradox

balls : 6 - 1 = 5

gold balls : 3 - 1 = 2

===> 2/5

Wrong, you can forget about the Silver box. You are dealing with a box with at least one gold ball. Now you have a 50% chance the next ball will also be gold

Yeah but there's a 2/3 chance the ball you picked came from the first box.

Say you go to Vegas and spin a roulette wheel. Your chance of hitting red is slightly less than 50/50.

Now say you spin it 6 times, and it comes up red every time. On the seventh spin, is it more likely to come up red or black?

The answer is neither, the chance of either is the same as when you spun the wheel the first time.

Now, say you're on a TV Show and Monty Hall, the presenter, puts you in front of 3 doors. 2 of them have a sheep behind it, and the other's got a shiny new sports car. You pick one, and before you can open it Monty opens another door you didn't pick, revealing a sheep. He asks if you want to switch from your choice to the other unopened door.

And unless you're a smoothbrain, you should be choosing to switch doors. This is because information about Monty's door allows you to infer information and act on it. When you made your first pick at random, there was a 2/3 chance that you were getting a sheep. Now Monty's showed you where one sheep is, but that *doesn't* change the probability of what's behind your door.

What it does mean, is that if you're sure that your current door has a goat behind it, you should switch doors. And 2/3 of the time, you'd be right and win a shiny new sports car.

The OP question mashes these concepts up a little. You're trying to get the box with 2 gold balls in it. You draw one ball, and its gold, so you know it can't be the last box in front of you. That leaves boxes 1 and 2.

But, crucially, you don't get to pick which box your 2nd draw is from. You got to pick from the same box. This takes the "Monty Hall" aspect out of it. Your chances of picking either of the remaining boxes at the start was equal: both 1 in 3. Now that one box has been taken away, the chance of you having picked either of the remaining boxes *remains equal*.

Hence, it's 50/50 chances.

50%

Silver box eliminated, so you have a gold ball and a silver ball.

/sci/fags

>roulette

true

>monty hall

true

>box problem

try again. Or just look the solution up.

You fail to accurately assess the data you revived by taking out the first ball. You have taken out a gold ball, one of THREE gold balls and you don't know which. The chance is now equal for you to have taken any of the THREE gold balls. In TWO of the cases it is paired with another gold ball and in ONE of the cases is it paired with a silver ball. Thus the chance of a second gold ball from the same box is 2/3.

So you're also a smoothbrain.

The chance is 2/3, because it's more likely you selected the gold ball from the two-gold box.

Remember, you could have chosen the silver ball from the gold-silver box. Assuming you would have always chosen the gold ball first from that box is what is tripping you up.

If the boxes had two drawers, left and right, and you knew the GS box had the gold on the left, and you opened the left drawer, *then* it's 50:50, because that has given you no information about the right drawer.

Being unable to choose the drawer, but instead blindly drawing the balls is what fucked you up, because you have assumed getting a gold on the first try is the same likelihood for both boxes. It's not, GG is twice as likely as GS to yield a gold on the first try.

I looked the solution up. It's 2 in 3. I am a dumb.

Math is a language

Earth is flat

>wholly shit, i had to scroll this far before i finally understand the 2/3 probability.

its actually simple, you have the two boxes. in one we have ball (1) and (2), and in the other is ball (3) and (4). ball (1), (2) and (3) are gold, and ball (4) is silver.

now, since ball (1) and (2) look exactly the same you cant tell which one is pulled out, so both are considered when deciding which ball was picked up. so since the ball (3) is atached to ball (4) (the silver), it cant be an option for the second draw.

that leaves only (1), (2), and (4) as potential draws making the probability 2/3... i think

See for my admission of the same. Sorry guyz, I was a dumb.

>Math is a language

Yes.

>Earth is flat

No.

Smoothbrains everywhere

Admitting a mistake? Cred Forums first!

There is no magical deep philosophical debate on the meaning behind it.

Its a 66.6% chance to draw a gold ball.

If u pull a gold ball, it’s more likely that you got it from the double gold box than the 50/50 box, meaning there’s a higher chance of the second ball in that box being gold. Donno what the exact percents would be but it’s more than 50%, maybe 75%

40%, cuz you cannot see what is in the boxes right?, then if you grab a ball and is a gold one then you think, that could be the box of 2 golden ones or the one who has one gold and one silver but you cannot discard the double silver because how do you know which box contains those balls? then it's 40% of get another gold, why?.

as a i know nothing more than that i have a golden ball and there i have 3 boxes to grab a ball that 2 of those could be golden because i have one, 3 of those could be silver because there a 3 golden a 3 silver in total. finnaly you have 2/5 of getting a golden and 3/5 of getting a silver so it's a 40%

The answer is 2/3

There are 3 different possibilities for the first ball you took. Box-1 Ball-1, Box-1 Ball-2, or Box-2 Ball-1. Of those 3 different possibilities, 2 of them will result in the next ball grabbed being gold. 2/3.

You discard the double silver box because you already picked a gold ball, there are no gold balls in that box so you could not have chosen it.

absurd, you don't know which box it's that contains those 2 silvers you can't dicard a box

can you not read?

you ALREADY chose a gold ball, since we are now only talking about the box that i picked a GOLD ball from, we know its not the double silver box.

50% obviously.

>3 different possibilities. 2 of them with positive outcome

>obviously 50%

there are 6 possible outcomes since there are 6 balls.

three outcomes are immediately discarded because of the prerequisite that the first ball you draw is gold, because of this you are left with three options, first second and third ball, each of which is equally probable. only the first & second lead to another gold ball, so it's 66%

No, because let's suppose that you picked from the box of 2 gold balls, then you have 3 boxes to pick another ball. the same box, or the other 2 in total now are 5 balls, 3 silvers and 2 golden so it's a 2/5 or a 40%

Let's suppose that you picked the gold ball from the box who has a silver and a gold you don't know that in the same box it's a silver or a golden one so again you have 3 boxes to pick a ball then it's 2 golden and 3 silver. 40%

So nigger ebonics is your first language?

>No

so you admit you cant read, read the OP again, you always pick the second ball from the same box you picked the first one from.

you take next ball from same box

no im Finnish.

PERGELE

This is just a reworded Monty Hall

Spoiler: If you said 2/3 your retarded, it's logic not math

If you pick a gold ball out, then there's only two possible boxes it could have come from: The one with two gold balls, or the one with a gold ball and a silver ball.

Now you can only either pick one gold ball or one silver ball.

It's 50/50

math is the most logical thing we have, they ALWAYS agree with eachother.

your math AND logic are wrong, its 2/3

>3 possibilities

>you either pull out a gold ball or you don't

>3

Not sure if trolling or actually retarded. We know you select a gold ball first. That narrows it down to 1 of 3 different balls. 2 of those picks would result in the next ball also being gold. 1 of them would result in the next ball being silver. This isn't fucking rocket science dumbass.

There's two ways to look at the problem, one way leads to 50%, one way leads to 66%

The question is, do you pick a box and then automatically pull a golden ball (if there is one), or does the problem continue from drawing a random ball and then discarding all the options where you didn't draw a gold ball.

Both are weird, but depending on which way of looking at it you take, the answer changes

>There's two ways to look at the problem, one way leads to 50%, one way leads to 66%

One of those ways is objectively wrong and the other is objectively correct.

>Both are weird, but depending on which way of looking at it you take, the answer changes

Statistics is not a matter of perspective you dolt

50%

I come up with either 1:2 or 1:3 and I cannot give my reasoning because I have to go take a shit right this minute.

>changing the rules to fit your opinion

When you pull the gold ball from the box, there is only 1 ball left in the box, it's either gold or silver

>One is not a nickel

Is not an argument

>Assuming you would have always chosen the gold ball first from that box is what is tripping you up

/thread

>When you pull the gold ball from the box, there is only 1 ball left in the box, it's either gold or silver

yes, but it's twice as likely that you picked the box with 2 golds compared to the box with 1 gold and 1 silver. What is so hard for you to comprehend about this?

What's tripping you up is you're harping on the first choice. The first drawing of the ball was simply to reduce the number of boxes; WHICH gold ball you picked is irrelevant

>the other is a nickel

Not an argument

Hey retard: **en.wikipedia.org**

Why don't you take a day to study statistics on khan academy or something?

Use your brain, retard

There is a box that has either a gold or silver ball, what are the chances it's gold? There's only 2 options, not some third color. You're not reaching into a box with the remaining balls from box 1 and 2. It's a single box with a single ball that is 1 of 2 colors

regardless of which box was first picked, it's 50/50 with the picked box because the other boxes contents were excluded upon picking the box, and the contents further clarified when picking the first ball as gold.

You aren't picking between two boxes in the second turn, so only one ball remains to be picked out of two possible colors in the only box picked. One ball, Gold or Gray, 50/50.

A bonus question. Can you determine the probability of correctly naming the remaining boxes contents prior to picking the last ball?

The odds of picking a box with a gold ball is 2/3. Once that box is picked the other 2 boxes don't apply.

The total number of gold balls, doesn't apply since you're limited to the box you already picked from. If you are given the choice of boxes to pick the 2nd ball from they do. If all the remaining balance are in play, they apply.

But you are stuck with exactly 1 box, 1 ball. The it's are identical to just being told there is a ball in a box, gold vs silver.

You are either a very dedicated troll or a complete moron. Either way, there is no point in continuing this conversation with you.

Thanks for admitting you're too stupid to understand

Back 2 Reddit, pseudointellectual

50%, you either picked the one with two gold balls or just one.

>THREAD

2 out of 3 implies there are either three possible colors or two possible turns, or vice versa.

The rules of playing also affect the probability.

If you alter the rules, you alter the probability. Like combinations versus permutations.

This is not valid Bertrand's paradox, because OP set the question wrong.

This is exactly why I cant stand probability theory. Its more than counterintuitive.

Intuition says p=.5 but it aint as it is a dumb fucking paradox with p = 2/3 or whatever.

Fuck this shit.

>OP

/thread

you dont /thread your own posts, are you guys new?

/thread

pic is awesome!

that would be better if the thread was about proving that these guys are newfags, but its not...

It's still 2/3.

There are 3 gold balls. Each of them have an equal probability of being selected during the first step. We'll call them gold ball A, gold ball B, and gold ball C.

Gold ball A: the next ball you select is gold

Gold ball B: the next ball you select is gold

Gold ball C: the next ball you select is silver

Wow, so hard. Why are right wingers so fucking stupid?

>

>It's still 2/3.

>There are 3 gold balls. Each of them have an equal probability of being selected during the first step.

Step one is irrelevant to what is being asked

both of the ways are objectively wrong

Only if by "step 1" you are referring to "you pick a box at random". Hence the fact that I ignored that part. It has zero bearing on problem. There are 3 gold balls. None of them has a greater chance of being selected than the other. This isn't a difficult concept to understand.

This is the setup at the end.

The answer is 2/3 Google it brainlets

First, let's name each ball -

Box 1: Gold-1, Gold-2

Box 2: Gold-3, Silver-1

Box 3: Silver-2, Silver-3

Now lets map out every single possible way that you could select 2 balls from a single box

Box 1: Gold-1, Gold-2

Box 1: Gold-2, Gold-1

Box 2: Gold-3, Silver-1

Box 2: Silver-1, Gold-3

Box 3: Silver-2, Silver-3

Box 3: Silver-3, Silver-2

We know the first ball selected is gold., so half of those are irrelevant. That leaves us with the following -

Box 1: Gold-1, Gold-2

Box 1: Gold-2, Gold-1

Box 2: Gold-3, Silver-1

2 out of 3 of those are a double-gold pick. Anyone who reaches any other conclusion is a brainlet who has never studied statistics.

But if you number the gold balls there is one more situation hence 2/3

There is no reason to name them.

OP is trolling. Trying to rewrite the original problem.

>180+ reposts

Are you fucking kidding me?

/thread

About three fity

1/3

Oh boy, it's another "user was only pretending the wrong answer was right and vice versa" episode.

another newfag trying to /thread his own post.

Please give us the "original" problem and explain how its worded different.

The real /thread

Wow, I think Bertrand would have loved to seen this conversation. Over 130 years since he first published the question, and it still generates such strong discussion today. Let’s not forget though that the reason this is an interesting problem is that it is a paradox. Coming to the conclusion of 50% is the “expected” answer, but what makes it interesting is that on further review the answer actually reveals itself to be 66% you fucking mongoloids! Please remember that people much smarter than you already solved this problem over a century ago.