Could someone explain to me how the second choice isn't the correct one?

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# Could someone explain to me how the second choice isn't the correct one?

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Could someone explain to me how the second choice isn't the correct one?

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Let us say n is 80.

There are now 100-80 =20 green balls

You take a red ball out. There are now 79 red balls and 20 green balls. You still have a roughly 80% chance of getting a red ball. The trick here is that we don't know what n is equal to.

Thank you.

You can't apply normal rules to random variables, that's why you have to perform transformations.

This is also somewhat of an ambiguous question, not due to math, but due to how we are supposed to interpret natural language. As this anon correctly points out, there are choices of n for which the probability of retrieving a green ball on the second pull is lower than your probability of retrieving a red ball. If the second answer is interpreted in terms of the probability P(pulling green on the second pull), then the second answer is false, as illustrated by this example.

However, the second answer is ambiguous and could also mean "more likely to be green (than it was before)". In this case, it actually would be true that

P(pulling green on the second pull | pulling red in the first pull) > P(pulling green green on the first pull).

In particular, if the balls are not being returned to the urn, then a bayesian agent should update their belief each round, so that if a ball of color X is pulled in round n, then the posterior probability of retrieving a X ball in the next round should be lower and their probability of retrieving a not-X ball should be higher.

>roughly

He said 100 balls, not infinity

79/99 is roughly 80%.

No shit, doesn't make the answer C though.

Okay, then why did you take issue with the word "roughly" if you agree with it? I'm not the anon who first said it, by the way, I think the answer is D because there's not enough information to make a meaningful conclusion.

BEGIN

set {

total: 100

probability red: n [0,total],

probability green: total-n [total, 0],

quantity pulled red: x

};

if (pulled red = 1) {

x++

}

when (x > 0) {

n-x >= 0,

};

test {

total --,

pulled red [0,1]

}

TEST

>pulled: red

>total: 100 -> 99

>x = 1

>n-x >=0; -> n >= 1

>[0, 100] -> [1, 100]-1 -> [0, 99]

>[100, 0] -> [99, 0]

>[99, 0] = [0, 99]

END

Result: same probability

Think of it like this. The % chance a red ball would come from a particular fill of an urn corresponds to the number of red balls in that fill.

So an urn with 1 red ball would have an 1% chance of producing a red ball, but an urn with 100 red balls would have a 100% chance of doing so. So the expected number of red balls in the urn after drawing 1 red ball would be the sum of x*(x/100) for x=1 through x=100 divided by 100 minus 1, or 49.5.

There are 99 balls in the urn. 49.5/99=50%

So it's still a 50:50 afterward.

Or put simply, the increase in likelihood of it being a green ball next due to the reduction in red balls is exactly offset by the increase in likelihood of it being a red ball due to the fill favoring red on first pull.

You are far more likely (3:1) to be in a high red urn if the ball is red.

>if you flip a coin and it lands tails that coin is likely not have a heads on it

Simplify it to a single ball.

If you pulled 99 balls out and they were all red, what are the odds that the last ball is either color, given that the 99 red balls case and the 100 red balls case were equally likely initially?

You legitimately cannot get away from equally likely by pulling at random.

Now do that for all 101 cases and average the result. Wow it's the 2nd one.

Coin flips are mechanical and not random.

Not bothering with rest of thread. You are wrong.

>Now do that for all 101 cases and average the result. Wow it's the 2nd one.

Show your work, so I can point out your mistake.

>Not bothering with rest of thread.

Probably should have because the right answer is explained in detail.

>Now do that for all 101 cases and average the result.

It is literally 100 times likelier that you got a red ball at random from the all-red urn than from the one-red urn. It is still 50 times likelier that it was 50-50 than one in a hundred. The fewer red balls there were in the urn to begin with, the less likely the scenario. All urns were equally likely at first, but not after your first pick.

You eliminated the 100 green ball possibility and 1 red ball from all 100 possibilities containing red balls. The expected value did not change.

[eqn]P(text{Next ball is green}| text{First ball is red}) = frac{P(text{Next ball is green} cap text{First ball is red})}{P(text{First ball is red})}[/eqn]

[math]P(text{First ball is red}) = frac{1}{2} [/math] by symmetry.

For the other probability you use the law of total probability.

[eqn]P(text{Next ball is green} cap text{First ball is red}) = sum_{k=0}^{100} P(text{Next ball is green} cap text{First ball is red} | text{There are k red balls in total}) P(text{There are k red balls in total}) = sum_{k=0}^{100} frac{k}{100} frac{100 - k}{99} frac{1}{101} = frac{1}{6}[/eqn]

So [math]P(text{Next ball is green}| text{First ball is red}) = frac{1}{3} [/math]. Logically answer 1 must be correct.

This is correct but for fun I also decided to simulate it.

As says, the answer is [math]1/3[/math], but there's a nicer explanation than just evaluating the sum.

Start instead with 101 white balls and line them up in a row. Randomly choose three balls and put a small mark on each. Paint the first ball you chose black, all balls to the left of it red, and all balls to the right of it green.

The chance that the second and third balls you've chosen are now different colors is [math]1/3[/math], as this only happens if the black ball was inbetween the two.

Of course, this also shows that the exact number of balls does not matter.

Really nice way of solving the problem. Finally a good post on this board, thanks anon.

Where's the frikin upvote?

I agree with the anime girl.

the smartest posts always come with anime girls attached, we need to study this to figure out what's going on

>P(First ball is red) = 1/2 by symmetry.

whhhhaaaaaatttttt!?

First off, this is not a Bayesian problem. The selection of a ball does not change the population distribution, and no new information is added to the problem.

Second, the probability of selecting a red ball is totally dependent upon the population. Since the population can be anything from zero red balls to 100 red balls, there is no way to determine the probability that the first ball will be red, and simply saying that it is 1/2 because there are only red balls and green balls is pants on head stupid.

The only answer is that the question cannot be answered. A sample of one tells you nothing.

there's a reason questions like this are posted. it's because there's lots of mathematically valid aproaches, and that causes fights.

if it's going to make you angry don't participate, they want you angry.

So not only is it Bertrand's Box extrapolated to 100 balls, the odds work out exactly the same. Does that mean Bertrand's Box is generalisable?

>Does that mean Bertrand's Box is generalisable?

Yes. What's the likelihood of you drawing from a box that has n red balls and x-n green balls? It's 1/x. What's the likelihood of you drawing a red ball? It's n/x. What's the likelihood of you drawing another red ball? It's (n-1)/x. Therefore, overall probability is n(n-1)/x^2= (n^2-n)/x^2

Summing up yields n(n+1)(2n-2)/(x^2*6). When n = 100, this yields 100*101*198/60000 -> 33.33, divide by a 100 since we want to find the average and we get 0.333 = 1/3.

>When n = 100

It's not, though, is it? X = 100. N is a random number between 0 and 100. And the odds of getting red after red are 2/3.

>It's not, though, is it? X = 100. N is a random number between 0 and 100. And the odds of getting red after red are 2/3.

I just noticed that I skipped over some points that.

The probability for a single urn would look like that: elihood of you drawing a red ball? It's n/x. What's the likelihood of you drawing another red ball? n(n-1)/x^2 with x the number of balls in the turn. Since we're interested in what the probability is without knowing which urn we've got, we have to sum up over all the different urns (of which there are a 100 since we don't specifically care about any intrinsic order). Doing so results in n(n+1)(2n-2)/(x^2*6) where n = x = 100 because it's a sum formula. Calculating the number yields in 33.33. Divide by that a hundred to get the average probability per urn.

We want to divide by a hundred because we're interested in the average, that is for each urn. Now 1/3 = 33.33 % is the total probability. If we were interested in calculating what's the likelihood of drawing a red ball and then a green ball, we would get n(x-n)/x^2 = (xn-n^2)/x^2 = n(n+1)/2x - n(n+1)(2n+1)/6x^2. Summing it up results in 10100/200 - 10100*201/60000, roughly equal to 50.5-33.3 = 16.6. Divide it by a hundred to get the average probability of 16.666.

Compare the two numbers, realize that drawing a red ball after having drawn a red ball is twice as likely as it is to draw a green ball.

I agree that red after red is twice as likely as green after red but shouldn't the odds add up to 1?

Ah, wait, I believe you're calculating the odds of drawing a red ball twice in a row, instead of the odds that you draw a red ball after already having drawn a red ball. It makes sense then that 1/2 scenarios don't involve you drawing a red ball first.

The odds should add up to 1/2 because you have a 50 / 50 chance of drawing a red ball first. Similarly, if you draw a green ball, you have a 66.66 percent chance of drawing a green ball the next time.

[math]P( text{First ball red}) = frac{100}{n}[/math] literally one of the first things outlined in the problem.

everything else is correct, but you end up with 100/6n which is greater than 1, the nonsense is because of

typo, otherway round obviously

But N is a random number between 0 and 100, so it will average to 50/100

in that case P(first ball red) is really a conditional probability (also a random variable) and 1/2 is the expectation of it

101x100 rectangle, pixels are half green half red. One pixel is picked from one row, and is red. Initial probability is 1/2 for either color, but sum has to be done to predict likelihood of next color being the other color. 1/101 is the green line initial probability. You cut a slice of red pixels taking the one, giving you k/100. The remaining options being green are (100-k)/99, since option all green (1/101) is excluded and you've taken one column out. So the chance of the next pixel in the randomly selected row being green ends up being the product of the three, since it could be any row the chances are sum of all chances for each given row, 1/6

1/3* im rarted

Linguistic tricks.

"more likely" he says. More? More than what?

israelite.

I interpreted as "more likey [than the previous pull]" to be red/green.

I believe the intended interpretation is "more likely to be red [than green]" "more likely to be green [than red]" "equally likely [to be green or red]"

It is a stupid linguistic trick. The kind that mathematicians make because they just have to jerk off to themselves in the mirror.

Obviously, if there's an urn, the urn will contain a clear pre-determined number of red and green balls. Just because you don't know how many red or green balls are in there, doesn't mean that the number just so happens to be indeterminate. Except if you're a literal toddler and don't get object permanence.

If you choose a red ball, the likelihood of choosing a green ball next time will either increase or remain the same (that is zero, if there are no green balls in the urn).

The only way the likelihood of choosing a green ball could not increase (other than the urn with no green balls) is by choosing over the sets of all possible urns, then pretending that your urn is simultaneously every single other urn that could potentially exist. Only then can you take a red ball and not increase the likelihood of getting a green ball because your mathematical schizo urn model assumes that the probabilistic median of drawing a green ball is the same for an urn with 99 balls as it is for one with 100 balls.

The part that you miss is that urns with the majority of green balls are not likely to be part of the problem, as you draw a green ball first from them. While red majority urns are likely to stay.

Reminds me of the Monty Hall problem. But your description doesn't refute my point. Drawing a red ball will increase the likelihood of drawing a green ball in any given urn because the urn has a pre-determined number of balls. On the other hand, assuming that any urn has the chance of having that many red and green balls based on binomial distribution, choosing a red ball first will neither increase nor decrease the odds for drawing a green ball next time. But in reality, it makes no sense to choose over the set of all possible distributions because you only got one urn which you would have chosen in advance.

>choosing a red ball first will neither increase nor decrease the odds for drawing a green ball next time

This has already been demonstrated to not be the case in this thread. As the chance will decrease.

>This has already been demonstrated to not be the case in this thread. As the chance will decrease.

Let's set n = 3, then we have the following combinations of red = r, and green = g balls: rrr, rgr, rrg, rgg, grr, ggr, grg, ggg. All these combinations are equally likely to happen.

Having chosen red will delimit our options to gr, rg, gg and rr. The likelihood of drawing another red ball is then 1/2.

However, inspired from

:

The likelihood of the first pull being below or equal to n is n/100, the likelihood of the second pull being above n is (100-n)/100 = 1-n/100. The likelihood of both happening is x(1-x) where x = n/100 with any number between zero and hundred. We assume the two events to be independent. Integrating x(1-x) from 0 to 1 yields 1/2-1/3 = 1/6. The likelihood of getting a red ball after already having gotten a red ball is equal to x*x, subsequently we will get 1/3 upon integration. Comparing the two numbers gives a ratio 2:1, in other words you're twice as likely to draw a red ball than you were to draw a green ball.

Curious, isn't it.

So fricking what? The urn is fixed. There are x red balls and y green balls on the first pull, with probability x/(x+y) to pull a red. On the second pull there's an (x-1)/(x-1+y) probability to pull red, which is lower than the previous probability. I swear to fricking God mathematicians are legitimate schizophrenics

>You take a random ball out of the urn

>RANDOM ball

>It's red.

The fact that you must take a random ball out first, and that it needs to be red makes you discard the majority of the urns with more green balls. Let's say you start with an urn with ten green balls, there is a 10% chance that you won't reach the "it's red" part. But if there is only ten red balls, there is a 90% chance that you won't reach it.

Irrelevant. Maybe you have an urn with 10 greens or 70 greens. It doesn't fricking matter. All that changes are what x and y are. The probabilities are still defined as x/(x+y) and (x-1)/(x-1+y), the latter of which is always smaller than the former unless y = 0. At best you're justified in saying there's a 1% chance there are no green balls and a 99% chance there are green balls.

You were told that about a fair coin. This is not a fair coin.

Not -necessarily- a fair coin, if I understand what you're getting at. As in, there are 101 coins before you, with various weightings, with one always coming up heads, one 99% heads, one 98% heads etc. all the way to 0%. You grab one at random, flip it, find that it comes up heads, and are asked how likely it is that it'll come up heads again.

There is precisely one fair coin among the 101, of course. But more likely than not you've got one that's weighted towards heads to some degree.

Never claimed it was fair.

THE URN IS FIXED.

Therefore there are x red balls and y green balls, and THIS IS NOT CHANGING.

You pull a red ball. The probability is x/(x+y).

Then the probability to pull another red is (x-1)/(x-1+y)

YOU DO NOT NEED TO KNOW WHAT X AND Y ARE

The second probability is manifestly lower than the first unless y = 0 in which case they're the same.

I think on the contrary this lack of the ability to consider what must have come before for this to happen or things to be like this is what make people narcissists and similar, because they don't realize how blatant their lying often is, so they lie all the time, even when normal people understand that THAT is definitely not what happened.

Even if there is 1 red ball in the urn and 99 green balls, there is a chance you could draw the red ball on the first attempt. It's an unanswerable question, there could be any number of red balls and drawing one on your first try doesn't guarantee anything.

Than not, idiot. But from your next post it's entirely clear you don't understand the first thing about probability and you cope by shouting "israelite".

Than before obviously. What kind of stupid nonsense question would it be otherwise?

If that was the case then OP's answer would be correct.

https://en.wikipedia.org/wiki/Rule_of_succession

This is the case of s=1, n=1 approximated at 100 intervals.

Its a linguistic trick of the sort MEGAPRECISE Bourbaki masturbators pride themselves on while using sloppy language and undefined references.

More likely than what? Equally likely with reference to what? Without an affirmed reference point the question isn't a dignified question.

Yea but what if it was the last red ball? What if theres only one green ball? there is no answer here without knowing atleast on variable, the one thing you know is that there's atleast ONE red ball, which doesnt change much now does it? The urn could be mostly green or red

The second option is objectively right and anyone who says otherwise is a cringe reddit pseud midwit who didn't understand the question or intentionally misinterprets it.

Initial probability of a green ball:

(100-n)/100 = 1 - n/100

Probability of green after pulling a red ball:

(99-n)/99 = 1 - n/99

Apply simple algebra

1 - n/99 > 1 - n/100

>b-b-but le distribution of n

Doesn't matter, you illiterate cretins. The question asks for the probabilities given a fixed n. At no point n is rerolled.

Since you already picked red, it's more likely that n is a closer to 100 than 0.

That wasn't the question though.

It may not have been the question but it is indeed part of the answer

Absolutely not.

This, basically the only time it doesn't change is in the case that n is zero or 100.

But given an N, which is static and unchanging, this is the truth.

Even with n being a random distribution the probability space is evenly split into n being nonzero and less than 100 and n being zero or 100 and all cases fall into this. This is a very easy calculation to make without any simulation.

Your odds are literally 98% that the likelihood of pulling a green ball goes up and 2% that there is no change.

Wrong. There are still (100 - n) green balls. Your probability of pulling green on the second pull is (100-n)/99. Your probability of pulling red is (n-1)/99.

Even your "simple algebra" is wrong. Simplifying 1 - n/99 > 1 - n/100 gives us n/100 > n/99 and this is false for any n.

I hate to break it to everyone here, but I asked my cs friend Dr. Monte Carlo (you may have heard of him) and the answer is indeed green

https://termbin.com/d1p5

>[math]texttt{for i in range(1, 100): # fill with random balls}[/math]

>[math]qquadtexttt{self.balls.append(}[/math]

>[math]qquad qquadtexttt{random.choice(["r", "g"])}[/math]

>[math]qquadtexttt{)}[/math]

Absolute moron.

What are you complaining about, ma'am?

What's the probability of there being 1 red ball in the box?

Nice catch ma'am, the problem states n red and 100-n green.

That makes it heavily red being the favored 2nd pull

https://termbin.com/a840

if n is between 1 and 99, n red and 100-n green is mathematically identical to n green and 100-n red.

it's mathematically the same as "pick two random numbers that sum to 100". order doesn't matter.

Green is definite winner here. Twitter man btfo yet again

https://termbin.com/npzk

The answers are intentionnaly vague.

It's basically Bertrand's bloody box again innit

In fact if I say you're given a box containing 2 coins, N of them gold, 2-N silver, where N is chosen uniformly at random in [0, 2], you've literally got Bertrand's Box.

Its 50/50

If you take out red. It must be that in this multiverse there likely more red balls anyway. The multiverse where there is equal amount of red and blue balls is very unlikely.

If you take a red ball out, it's 50/50. If you take a ball out that happens to be red, the next ball is more likely to be red.

I initially misread the question, but as written, more likely to be red is the correct answer. More likely to be green makes no damn sense unless you're interpreting "more likely to be green" as "more likely to be green in this state than in the previous state" instead of "more likely to be green than red". Technically that would be a valid reading and correct answer for that reading, and I certainly wouldn't fault someone for *intentionally* trying to argue that, but the more likely intended reading is the latter, which is wrong.

In all distributions except the case of 100% red, drawing a red ball makes it more likely to draw a green ball.

From this breakdown we know that both are equally likely, either you draw a red ball or you don't. Conversely, either you draw a green ball or you don't.

You misread the question. It's not "has it become more likely than before that you will draw a green ball"; it's obvious that you would, given that the proportion of green to red has shifted ever so slightly to red's detriment. The question is "are you more likely to draw a green ball or another red for your second draw?" which is counter-intuitively not the case.

To me this is just a simple question of asking what can you know about a distribution from a single sample.

You cannot know anything about the population from the description of its creation, because the description of its creation is a somewhat complex description of not being able to know the population, outside of the fact that it is made up of either red balls or green balls, and adds to 100, which tells you nothing about the probability density at all.

It simply boils down to the question of what can you know about a population from a single sample, which is nothing.

Technically, there is no answer, because it is not that you don't know; it is that it is unknowable from the sample so far.

it's double randomness

if your first draw is red, it's statistically more plausible that the urn contained more red balls in the first place

It is certainly possible that n is 1 and you pick the only red ball in the urn on your first try, but it's far more probable that there are more red ones in the urn and the next one will be red also

let's check case n:o 6

there is 15/21 that the first red came from case where n>3. from these cases its always more probable to draw red again

Let's go over it exhaustively. For every 105 red balls we get on our first draw, we expect:

30 red balls to yield another 30

25 red balls to yield 20

20 red balls to yield 12

15 red balls to yield 6

10 red balls to yield 2

5 red balls to yield nothing

So 70/105 balls will give us another red, or 2/3. Erry tiem.

Schizophrenia

Point out the flaw

See

Of course you don't know what's in the urn. That's why you assign a probability to the contents of the urn.

I bet you also think you're clever for pointing out that in the Monty Hall problem the car COULD be behind the door you picked initially because you don't know and it's not like it's being moved around.

You don't understand my post then. It clarifies the Answer is unquivaically that you're more likely to pick a green ball than before. Unlike the answer you say, which requires magically switching urns before pulls

Oh.

Well you're obviously wrong then and it's you who misunderstands. In fact I think it begins with you misunderstanding what is being asked of you in the first place. Then you also go on to misunderstand what everyone else is talking about.

Magic urns don't exist anon.

You're going to have to explain which part you think requires magic, because you are far from making sense.

By the way, when it's asking you if the next ball is "more/equally likely to be green/red", it means "as opposed to the other one", not "compared to the likelihood of drawing that same colour ball from the same urn before you took a red ball out". That's your first mistake. Maybe that'll help you understand my posts.

>"more/equally likely to be green/red", it means "as opposed to the other one",

No it doesn't.

Yes, it does. The way you interpret the question, the answer is trivial. The ratio of red to green obviously changed in green's favour. That's barely even a maths question. That's Sesame Street "do you remember the colour of the ball" level.

The more interesting question - the one they were clearly going for here - is whether or not you are more likely to get green after red, or another red, or what. The wording is, admittedly, somewhat ambiguous but your interpretation is the less interesting one, and the greater stretch, semantically speaking. It only makes sense when you restrict yourself to the first two options but "equally likely" is nonsensical if you read it your way.

>The more interesting question is the one that isn't reflective of reality and wasn't asked

Mathematicians are legitimately mentally ill.

I'm sorry you're incapable of parsing the sentence as intended. As for the answer not being reflective of reality, though, that is where you are simply wrong. That is not a matter of interpretation.

>One correct way to read problem

>Open to interpretation

You're brain damaged.

There are blatantly two ways to read the problem. Your refusal to even acknowledge one of the ways (arguably the more correct one) speaks to your own brain damage. That you are furthermore unable to recognise the correct answer to Bertrand's Box when it's in front of you and instead dismiss it as schizophrenia not rooted in reality shows that maybe you're not cut out for mathematics, any more than linguistics.

Funnily enough I guarantee you I've learned more math (masters) and linguistics (bachelor's) than you. This problem isn't Bertrand's box. My analysis works perfectly fine with Bertrand's box btw.

>Funnily enough I guarantee you I've learned more math (masters) and linguistics (bachelor's) than you.

Yeah? You can guarantee that you're the only person in the world to obtain either a master's in mathematics or a bachelor's in linguistics? And a fat load of good they did you, too.

The problem is exactly Bertrand's Box except instead of two balls per box there are 100. If you can answer Bertrand's Box correctly, then you wouldn't be ranting about "brain-damaged" "schizophrenics" with "magic urns" now because both the question and the answer are the same.

I've corrected people on their misunderstanding several times now. The problem seems to be that none of them is willing to acknowledge it. Seems one side is capable of seeing the problem from multiple angles whereas the other stubbornly clings to a flawed interpretation.

Anyway, that's not the answer to the question. The next pull is not -necessarily- more likely to be green, in the <1% of cases where we got the all-red urn. That is why this interpretation makes no sense for a multiple choice question. The answer would then be "it's A except sometimes when it's C". Also, the part about either colour being equally likely is plain wrong. It has been pretty conclusively shown that red is twice as likely as green.

>The problem is exactly Bertrand's Box except instead of two balls per box there are 100.

*with the number of boxes adjusted to accommodate for every possible combination, of course.

>My analysis works perfectly fine with Bertrand's box btw.

"Ah yes, I drew a gold ball, that means silver is now more likely" deranged

You pick a box. There are x gold coins and y silver coins.

You draw a gold coin with probability x/(x+y).

The second draw probability to be a gold is (x-1)/(x-1+y).

Either y = 0 in which case you're guaranteed to draw a second gold coin... Or y = 1 and x = 1 such that the probability to draw the silver coin is 100%, which is more than 50%.

Everything works as expected.

Ah yes, so your answer is basically "either you're more likely to draw silver or you aren't - because there isn't any silver". Which is still not an unequivocal "yes".

But I do furthermore believe that the point of Bertrand's Box is to evaluate the likelihood of being in either situation, is it not? You can say your method "applies" to Bertrand's Box but it's actually no help at all in solving it. And it's just as useful to the problem at hand, because it is essentially the same problem.

Well, what do I know. I only have an MA in linguistics.

Bertrand's box asks a different question. It wants to know a precise probability of picking a specific coin. That's not what the problem in op is. OP is merely asking if you picked gold, which ball is more likely to be picked next. The answer is silver. That's how it's phrased. Dismissing it and arbitrarily solving a different question is textbook schizophrenia

Anon, tell me this, if you're twice as likely to get a gold ball as a silver ball, which ball is more likely to be picked? The answer is not silver.

Knowing the precise probability of picking a specific ball also tells you which one is more likely to be picked next, obviously. Your solution fails to take into account the likelihood of selecting red in the first place, which is the key to both Bertrand's Box and this problem.

But look, I admit that the phrasing in this case is ambiguous, and that a case could be made for your interpretation, though I have my reasons for dismissing it. And furthermore I admit that, if you're intent on answering the trivial question you think it is, your maths check out; they just don't tell you anything interesting and give no definite answer. Your fundamental arrogance lies in, firstly, insisting that there is no ambiguity and that your interpretation is not merely more likely but unambiguously correct (it very plainly is not), and secondly, insisting that the correct mathematical answer to the alternative interpretation is invalid for reasons you've yet to clarify.

So, once again, tell me how the phrasing of the question explicitly precludes the interpretation "is red more likely than green", and then also why the solution to that question is invalid.

>Dismissing it and arbitrarily solving a different question is textbook schizophrenia

Now venturing into psychology and being wrong as well, huh? Next you'll tell me you have a PhD in psychology, I bet.

>if you're twice as likely to get a gold ball as a silver ball,

In what context? Certainly not the context as phrased in OP.

>In what context?

Well, certainly in Bertrand's Box, as I'm sure you'll agree, which is what we were talking about here. But also in the OP problem, which is analogous to it.

>If you're insistent on the bertrand box analogy, the silver -silver box doesn't exist and has been completely eliminated from the situation because you picked a gold.

This is the classic mistake to make with Bertrand's Box, too.

I've changed my mind. You are unambiguously wrong, and I've realised your mistake. You might be correct if all you know about the urn is that it has some amount of green and some amount of red. In that case, indeed, the most you can say is that green has now become a fraction more likely, unless there was no green to begin with. But because you know how the urn was prepared, you have more information, and this information renders it precisely analogous to Bertrand's Box. Because being given an urn that is randomly prepared as described is functionally identical to picking one random urn from 101 urns which together contain every possible combination of 100 red and green balls. And this means you can (and indeed, must) take into account the probability of having selected red in the first place - and thus, the infinitesmal gain in likelihood of picking green is offset by the overwhelming likelihood that the urn you have has more red in it to begin with. We do not know which of the 101 possible urns we have, but we do know the likelihood of each one. Unambiguously, then, we can state that the odds of drawing a red ball next are greater than the odds of drawing green.

Has anyone ever told you that you write like a literal flaming homosexual?

Penis.

Except if you read the op closely you'll see a parenthetical remark that explicitly tells you my interpretation is the correct one.

>You are given an urn

>[Rest is irrelevant, you have an urn with predefined red and green]

>You take a random ball out of the urn—its red—and discard it

>The next ball you pick (out of the 99 remaining) is...

If you're insistent on the bertrand box analogy, the silver -silver box doesn't exist and has been completely eliminated from the situation because you picked a gold. So the question becomes

>You discard the gold coin

>The next coin you pick (out of the remaining one coin) is...

>The next coin you pick (out of the remaining one coin) is...

... more likely to be gold, duh. Everyone knows this.

>OP is merely asking if you picked gold, which ball is more likely to be picked next.

Actually, when phrased like this I'd say it unambiguously is Bertrand's Box. Only a madman or a fool would insist on reading this as "which colour ball has, in absolute terms, now probably become more likely to be picked than it was before, unless it happens to be absent in this particular sample" instead of a question about their relative probability. How do you even write this and not realise your mistake?

You forgot the case where all of them are red. The random interval is [0,100], not [0,100).

I clearly see one all-read urn.

r g

r r

r rr

r rg

r gg

r rrr

r rrg

r rgg

r ggg

...

>correct one

Put all possible balls in one basket. Pull a red one , count the remain.

as with monty hall or the gold/silver coin question, the many verbose wrong takes itt show that the abstraction effort of probability seems to be some hard cut-off point in IQ.

The fact of drawing a red ball establishes that you're more likely - not *necessary* but *more likely* - to have an urn with more reds than greens.

There's also no linguistic trickery involved. It's perfectly well-stated, but in probabilistic language. The first answer is exactly correct, because no part of the problem involves asking what *is*, but what is more likely. And the answer would be the same for 3 balls or any other number, simply by the fact of "all balls green" already being excluded by premise.

you don't know n. It might be 100.

Unfortunately, we don't really pick n uniformly. We uniformly pick n UNTIL our first pick is red. We discard all choices of n if our first pick is green. n's distribution looks like this, courtesy of Dr. Carlo.

>ITT

>supposed great minds arguing while none them notice the opposition interpreted the question differently

The actual answer to the question is: "The next ball you pick is more likely to be green than in the last pull and the next ball is equally likely to be either colour".

Ah, yes. The staturbation thread.

In this problem, we have an urn containing 100 balls, where n balls are red, and the remaining (100 - n) balls are green. The value of n is chosen uniformly at random from the set {0, 1, 2, ..., 100}.

Initially, a red ball is drawn from the urn and discarded, leaving (n - 1) red balls and (100 - n) green balls in the urn.

To find the probability that the next ball drawn from the urn (out of the remaining 99 balls) is red, we need to consider the possible values of n and their corresponding probabilities.

Let's define the event A as "the next ball drawn from the urn is red."

We can calculate the probability of event A using the law of total probability:

P(A) = Σ P(A | n) × P(n)

where P(A | n) is the conditional probability of drawing a red ball given that there are n red balls in the urn, and P(n) is the probability of having n red balls in the urn initially.

Since n is chosen uniformly at random from {0, 1, 2, ..., 100}, we have P(n) = 1/101 for all n.

For each value of n, the conditional probability P(A | n) is given by:

P(A | n) = (n - 1) / 99, for n ≥ 1

P(A | n) = 0, for n = 0 (since there are no red balls left in the urn)

Substituting these values into the law of total probability, we get:

P(A) = Σ [(n - 1) / 99] × (1/101), for n = 1, 2, ..., 100

+ 0 × (1/101), for n = 0

Simplifying the summation, we obtain:

P(A) = (1/101) × [(0/99) + (1/99) + (2/99) + ... + (99/99)]

= (1/101) × (99 × 50 / 99)

= 50/101

Therefore, the probability that the next ball drawn from the urn (out of the remaining 99 balls) is red is 50/101 ≈ 0.495.

Literally the only not moronic post in this thread, amidst a sea of moronic code monkeys who can't even simulate the correct problem

It's wrong, tho? I'm not sure what it's calculating but it's not the answer to the question in the OP.

Isn't this just the odds of getting red on your first pick?

All right, no, I see what's happening here. This is just the odds of getting red from a randomised urn which has one red ball removed. Not from specifically the same urn that you already took a red ball from at random before.

>This is just the odds of getting red from a randomised urn which has one red ball removed

Almost. They forgot to exclude the 100 green balls case, since you can't remove 1 red ball from that. Changes the answer they were (mistakenly) trying to get to .5.

I believe that's because they're intentionally excluding it because they couldn't have gotten a red ball from it.

Either way it's wrong, of course.

Lol I popped in yesterday before bed after five beers and now I see I completely misread what you were saying.

>Changes the answer they were (mistakenly) trying to get to .5.

You're right, and this is also interesting to me because that's also what you get if you approach Bertrand's Box in the same way (i.e. treat it as an equivalent choice between two boxes containing either a gold or a silver ball). Seems the right approach and the wrong approach both scale exactly the same.

Actually, this makes perfect sense now that I think about it. If we once again consider the 101 urns scenario, then really what we're doing here is removing the all-green urn, taking a red ball from each of the 100 remaining urns, and then picking at random. So, in effect, we've removed 100 green balls and 100 red balls, and the system is in equilibrium. And if we do that again, we now remove one urn and 99 balls of either colour. And if we do that again and again and again, it's going to remain at 1/2, until we get to three urns with two balls each (Bertrand's original problem, where you also have 1/2 probability of picking red initially), and then finally two boxes with one each (the fool's solution). So in fact what this does is pare down any instance of Bertrand's Box with n balls to another instance with n-1 balls, where you are now making your first random pick. It's an endless chain of Bertrand's Boxes where your initial pick is always 1/2.

(1 + 1)/(1 + 0 + 2) = 66% red

It is, but whatever. It's impossible to truly know. It's like one of those gotchas Professor Layton's riddles when you can bullshit your way to any answer.

You are all making this too complicated. Out of all of the urns possible you have 10000 balls, 5000 red and 5000 green. By selecting red and not replacing it, you eliminate the urn with 100 green balls, meaning there are 99 more red balls (subtracting the one you took out) than green balls out of the total population of balls in all of the urns possible, making red more likely.

What urn they are in is irrelevant, since you cannot know what urn you have selected. The trick with Bayesian problems is to see past the irrelevant partitions of the population to the total population at the time of selection.

That works, since the question only asks which is more likely, but it still rather understates how much more likely red is.

well its more like to be green since you just took a red one out innit?

2nd answer is the correct one.