Does anyone know what the answer is?

A bit under 100%, using Bayes' theorem.
Let A=Choosing a gold box
Let B=Choosing a gold speck
P(A|B)
=P(B|A)*P(A)/P(B)
=(1)*(1/3)/(slightly over 1/3)
=0.99...

If say there are a 100 gold specks in box 1 and 1 gold speck and 99 silver specks in box 2. then the odds of having picked from box two is 1/101, and the odds of having picked from box 1 100/101.

Pretty close to 100%.

near enough 1

Assume N specks per box, where N is "large".

There are 3N specks total. N specks are in the gold box. N+1 specks are gold.

If you pick a gold speck, then it was one of the N+1 gold specks. Since it's "given" that the speck must be gold, then N+1 is the total number of possible outcomes. (If you had picked a silver speck, then it's not the "given" outcome, thus the question is not even applicable, and the outcome can be ignored.)

Since N specks are in the gold box , then the probability that the rest of the box is gold is N/(N+1).

As N grows large, the probability approaches (but never exactly reaches) 1. For example: if N = 1000, then the probability is approximately 0.999.

- - - -

To quickly convince yourself that N/(N+1) is the correct probability, draw a diagram with N equal to a small number, such as N = 2:

Gold box: GG
Silver box #1: SG (contains one errant gold speck)
Silver box #2: SS

If you pick a G, then it's one of the three Gs, and two of those are in the gold box. Thus, the probability is 2/3, in agreement with N/(N+1) as the general answer.

Just use Bayes' Theorem. Too lazy to write it out or explain, but I know from the wording of the problem that Bayes is the way to go.

P(B|A)= P(A U B) / P(A) yes i had to look up the formula, it's been a while since stats
P(Box being gold, given that the drawn speck is gold) = P(Choosing the gold box and getting a gold speck) / P(Drawing a gold speck)

There's a 1/3 chance you draw the gold box at random, which of course has a 100% chance of giving a gold speck. The chance of drawing a gold speck at all is also 1/3, plus an infinitesimally small amount to account for the minuscule chance of drawing the gold speck from the second box. So the answer is (1/3)/(1/3+N), where N is a really really small number.

Thus, the probability is just under 100% that if you drew a gold speck, the rest of the box is gold.

Assuming there's 100 specks in each box, there are 101 gold specks.The odds of getting a gold speck that is not in the gold box is 1/101. Thus the odds of the box being gold is 100/101 or 99.0099%.

It's probably 66% chance

If I was presented with this problem in real life, I would say the probability is extremely high. The actual odds of you selecting a singular speck of gold in an otherwise silver filled box are so astronomically low that the opposite is nearly guaranteed to be true. I wouldn't bet on my box being full of silver.

Nobody knows the answer because this question does not give enough information. As far as I'm concerned a "speck" could be the size of the entire box and the image could be deceptive since no concrete measurements were ever given.

>idiots who would bet their life savings on a 50/50 chance

Wouldn't it just be 50% lol? I'm no mathematician, in fact I failed algebra 2 in high school. I'm just thinking that if you pull out a gold speck then it's either all gold or all silver after the fact. Two possibilities so 50% right? Why would that be wrong?

I like how that one speck of gold is still counts as 50% of that container filled with silver

Cause if you pick a spec at random from the mostly silver box chances of getting the gold one are very low, however in the all gold box chances of picking a gold speck at random are absolute.

I hope you're baiting and not genuinely denying an actual mathematical formula

Just imagine yourself doing what is described. Pick a speck from a box without looking at what box. Every time you get a gold speck, check what box it was from. Once you do it a couple of times it should be clear to you.

Because that isn't the question. The problem states that even before you pick the first grain, the box has been chosen at random, which means that the probability that you choose a golden grain is greater than 50%.

OP's problem is just a retarded and incomplete version of the Bertrand paradox (since we don't know how many grains are in a box, could be 2 or 1 million) en.wikipedia.org/wiki/Bertrand_paradox_(probability)

The reason most people think the answer is 50% because they consider that when they're given a box that it's a 50% chance which one they get which would be true if the wording of the problem was slightly different. But the real problem is that you keep a record of choosing at random from all the 3 boxes, and then also keeping a record of picking the 1st grain and then keeping record of every occasion when the 2nd grain is gold or silver. Then you'll see that 66% of the time when the first chosen grain is golden, the second grain will be golden too. This is very different from what most people think the problem is.

Well phrase the question differently. What is the probability that you got the one gold spec out of the silver box? Since the answer to that requires actual measurements, you can't answer the question accurately.
It is only 50/50 if both boxes are 100% gold.

then it wouldn't be 50/50 it would be 100%

Shit I linked the wrong article, it's this: en.wikipedia.org/wiki/Bertrand's_box_paradox

No it would be 50/50 between the two 100% gold boxes.
100% / 100%
50 / 50

The question asks what the probability is that the rest of the box is gold, not that you picked the box on the left.

Your explanation is horrible. This reminds me of the old adage "You never truly understand something until you can teach it to a child." Regardless, so everyone is all in on the .9999999999 whatever odds when it reality it's just 2/3 as it's asking about the box you pulling at first instead of the speck? I guess we're all idiots, huh?

Oh yeah you're right I'm retarded.

The answers to OP's problem is above 99% even if the particle count is low.

2/3 is the chance in the classical Bertrand's box paradox, where there are only 2 gold and silver coins per box.

The answer is almost 100%

origigigi

So Bertrand's Box Paradox is irrelevant to this? Why are you such a pseudo-intellectual? If you don't understand how something works don't copy-paste wikis to seem like you do.

>So Bertrand's Box Paradox is irrelevant to this?
It isn't you fucktard, OP's problem is just a version of it and an incomplete at that.

Depends on how many specks in each box.
If there are 10, then you have 11 specs of gold dust. 10 of these are in the gold dust box, 1 in the silver box.
So the probability is 10/11.
In general, N/(N+1), where N is the number of specks in each box.

1/2 because theirs two boxes that have the gold inside of them and one of them has more than that and the other one does'nt

You also think you have a 50% chance you win the lottery? Either you do or don't?

I think you replies to me by accident I wasn't talking about the lottery iwas talking about the gold in the boxes

No you idiot. You already picked a gold speck. There are only two possible options so it's a 50-50 chance. The question isn't what the probability of you picking a gold speck is. The question is is it the gold or silver one

lol this person is an amazing troll
Australian/10

The answer is 2/3 -

Attached: box-problem.png (989x533, 65K)

REASON SAYS: However small the chance of picking the solitary gold speck is, the event has already happened. The problem puts us at the point in time where we're holding a box from which we already drew a speck and it was gold. So looking forward, we either have a gold box or a silver box, the chances are 50% each.

INSTINCT SAYS: If I was asked if I want to trade boxes (I'm thinking about the three-door, one-car problem here) I would say no, because the chance I grabbed the one gold speck is amazingly low, so it's very likely that I'm holding the gold box.

>REASON SAYS: the event has already happened
But you don't have KOWLEDGE of which event happened, so you still have to weight them on the basis of which is more probable. So your reasoning was bad and your intuition was correct.

Another approach: What's the chance of flipping a coin 1000 times and getting 1000 heads? Pretty low. But now that I already have flipped the coin 999 times and already got 999 heads, what's the chance of the next flip being heads? It's 50%. My point is, it doesn't matter how low the odds were, they're already behind and thus irrelevant to the calculation.

it is impossible to say because the amount of specks in each box aren't known.

end of story fags. prove me wrong?

We'll settle for a "pretty high" or "pretty low" or "about half" answer.

This situation is not analogous to the question posed by the OP.

It's 1 in a million that you picked from the box with only 1 speck of gold so 999,999/1,000,000 that it is full of gold.

It is, we're dealing with a extremely-low-odds event which already happened (picking the solitary speck of gold, analogous to getting 999 heads in a row)

>the chance I grabbed the one gold speck is amazingly low
Imagine not knowing what probability means.

That's not the same problem... You forgot that we are comparing two boxes, not a singular coin. You are flipping two coins. Your analogy oversimplifies the problem.

fucking retarded high school drop outs think it's 50%. The answer is just under 100%. Let's say we pick a random human being and they turn out to browse r9k, and we ask the chance of them being a fembot. That shit wouldn't be 50%.

>its 50% cuz theres only 2 possibilities, fembot or not fembot

Attached: brainlets.jpg (900x900, 83K)

There are two elements that are unlike your coin flip example.
1. We are ignorant of whether or not the unlikely event happened.

2. More crucially, whether or not the unlikely event happened actually has an impact on what we're trying to predict.

Let's say we have 10 boxes. All of them have 9 silver coins and 1 gold coin, except for box 10, which has 10 gold coins.

1) You DID draw one random coin from one random box, and it was a gold one. What's the chance you drew it from box 1? Since you chose the box at random, and theoretically you could draw gold from every one of them, (with different odds per box, but that doesn't matter since the coin is already drawn) the chance is 1/10

2) You WILL draw one random coin from a random box. What are the chances it will be gold? there are 19 gold coins and 100 total coins so 19%. What are the chances of it being from box 1 AND gold? 10% chances of picking box 1 and IN THAT PARTICULAR BOX 10% chances of the coin being gold, so 1%

Both scenarios 1) and 2) ask about the odds of a gold coin coming out of box 1, the difference being in 1) the coin is already drawn hence the chance of it being gold is 100% (a given), in 2) the coin will be drawn so we must consider the individual odds of each box.

Yep, it's zero percent.

Because the dust is coloured, not the boxes.

>1) the chance is 1/10
BEFORE drawing, the chance is 1/10. But, given that you have drawn gold, the chance that you HAVE DRAWN from box one is 1/19. Probability is about information and ignorance.

OK I see it now...makes sense

I am going to say 1/2. The question is only interested if the remainder of a box is full of gold after withdrawing a gold speck. That rules out 1 box since it's all silver. Two boxes remain containing any gold. The scenario implies you'll have a 100% chance of picking up a single gold speck from among the 2 last boxes, so you just have to worry about if the remainder of the contents in the boxes is gold or not. So, 50%.