Whats the answer b?

It doesn't matter.
In the beginning you picked a box with gold in it (which was 2 in 3), thus eliminating the all silver box.
You're still left with only two boxes, so 1 in 2 you picked the all gold one

If you randomly choose the right box, you have a 0% chance of drawing a gold ball and therefore counting it.

If you randomly choose the middle box, you'll have a 50% chance of drawing a gold ball and therefore counting it.

If you randomly choose the right box, you have a 100% chance of drawing a gold ball, and therefore counting it.

Since you count the gold box every time and the mixed box 50% of the time, 2/3rds of counted events will be from the all gold box.

Was supposed to be right, middle, left obviously

You do one attempt, as per the problem description, and each round is independent, as per... math, except you know you eliminated the only-silver box by drawing a gold ball from your chosen box.
Round two has two potential boxes left, 1 in 2
quod erat demonstrandum

Whats the equation with the original question and whats the equation for the one you replied to?

It can be a thousand attempts or one attempt. It doesnt change the answer which isnt 50% by the way

Guy asked if it was more likely to have picked the gold-only box if there was 1000 gold and a 1 gold 999 silver one. The answer then depends on what specific question you're trying to answer.
Is it more likely you drew the gold ball from the 1000 gold box? Yes, the gold box had a 1 in 1 chance, the other had a 1 in 1000 chance.
Does it influence the probability of the overall outcome? No, still 1 in 2.

I really hope you're not asian. Your parents would kill you for you displaying such a lack of basic mathematical knowledge

>Is it more likely you drew the gold ball from the 1000 gold box? Yes
So you do understand this on some level. Youre also more likely to have drawn a gold ball from the first box in the original question

Ligma

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>Youre also more likely to have drawn a gold ball from the first box in the original question

In the original question chance of drawing a gold ball from the gold box is 1 in 1, from the mixed one 1 in 2 and from the only silver box 0, sure.
The chance of picking a gold ball is 3 in 6, however, or 1 in 2. Even if you picked a silver ball (chance of 3 in 6 or 1 in 2 again), it would not change the outcome. 1 in 2 chance to pick a gold ball in your second attempt (since one of the boxes was eliminated by previous knowledge about the distributions/contents)

The first box has more gold balls in it. Therefore the probability that we took it from the first box is higher than if we took it from the second, therefore the probability of getting a second gold ball is higher

It's not higher that we actually took it in a one attempt experiment. Picking any box has a chance of 1 in 3. The only thing we know is that we did not pick the only-silver one. Leaving two candidate boxes that we could have picked in the first round - we do not know which. 1 in 2

>The only thing we know is that we did not pick the only-silver one
You also know you picked a gold ball. Which means you can bet on the fact that it more than likely came from the first box. The math doesnt change based on the number of attempts. The probability of getting any given number on a six sided die is 1 in 6 regardless if you roll it 1 time or a million times

0.5%? bravo

it's a very famous math problem. you are wrong. you can perform OP's problem in real life if you want to check for yourself, it has been done.
It's not 50/50. Google Monty python door problem if you wanna learn more.

Dont feel dumb. the problem is very famous for defying intuition.

The answer is 2/3 faggots and anybody who claims otherwise is assuming the balls are ordered and the first ball is always selected

Proof for the retards at the back of the bus;
Let x be the ball you took, G for gold S for silver

Every possible state
[X, G] [G, S] [S, S]
[G, X] [G, S] [S, S]
[G, G] [X, S] [S, S]

In 2/3 of these states the next ball you take is gold

All other states are invalid as you would have selected silver and discarded your test

Now, like smol brain idiots let's examine every possible state if you assume that you always pick the 1st ball from each box

[X, G] [G, S] [S, S]
[G, G] [X, S] [S, S]

Oh would you look at that we arrive at 2 possible states. Thank you for your time you can close this thread now (and go back to school)

in other words.... you have no clue how to solve it?

Listing out all the possible scenarios like he did is exactly how you solve problems like this

sorry wrong person

41.5%
I won’t explain

This post converted me, I finally get that I was being a 50/50 retard. I've acended to the 2/3s master race. Thanks user.

en.m.wikipedia.org/wiki/Bertrand's_box_paradox
You are ready my son

gg - 2 gold box
gs - gold and silver box
ss - 2 silver box

33.(3)% each chance for each box.
and 50% chance that first ball will be gold

You take first a gold ball
you chose gg and you have 100% of take gold
you chose gs and you have 0% of take gold

so is 50%-50% yes? no...

is more probably that you chose gg, because your first ball was gold!

The situation when you chose first gold ball for each box is:
gg - 100%
gs - 50%
ss - 0%

We have a 100 situation where user takes a ball.
If user chose the ss we don't discuss it anymore 0 from 33.(3) possibles
if user chose the gs we have a 50% chance that we take silver or gold if it is silver we don't discuss it. 33.(3)/2 times from 33 possibles.
If user chose the gg we always discuss it. 33.(3) times from 33.(3) possibles

That means the situation where user takes the first gold ball is true for 33(.3)/2+33.(3) times (first gold from gs + gg). This means we have 50 situations where the gold ball was first. The author asks about the second ball. 2/3 of 50 situations was a gg box and 1/3 gs.
(2*100%+1*0%)/3= 66.(6)% it will be the second ball gold.

50%

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It's not about intuition, or the Monty *Hall problem (which I know well by the way).

The Monty Hall problem is about game theory and strategy, i.e. would it be wise to switch (it would).

It has no relation here however, as switching isn't on the table, and with 2 remaining boxes, one of which we know is the gold-gold one (due to elimination of the silver one), 1 in 2.
Without the option to switch, initial chance of 1 in 3 and the 2nd level chance of 1 in 2 (after elimination of silver-silver) remains the same.

Whether or not to you're allowed to switch makes all the difference here for optional strategy.

you're just autistic

I like how you change the question to justify your retarded answer

See:

why do people always fall for these threads lmao, the 50% posters are baiting

In monty hall you should switch because youre more likely to be intially wrong. Here its 2/3 because youre more likely to have taken a gold from the first box

Yep, it's wrong, you pick one golden ball, by assumption one golden ball will be removed from the other box aswell
2/3 is pea brain answer

Bruh it's a Bernards box paradox, a literal hundred years before the Monty hall problem

Cool story

The proof you are wrong was written in the late 1800s

nope
You randomly selected a box out of three in the beginning and have to stick to it. Your random chance is 1 in 3, and since you cannot switch, it remains that way. After elimination that kicked out one of the duds, your chance is therefore 1 in 2.
I know it's hard to cope with this simple explanation when you felt to big brains for knowing about Monty Hall - just not recognizing it does not actually apply here because you cannot switch

Everybody makes mistakes, not every theory is right

Please do not ruin this thread for me by being right. I was having fun

So you believe the answer to monty hall is 50/50 too then. Since you got that wrong its no wonder why you got this one wrong as well

Only one probability matters: whether ball two will be golden. Given the first ball was gold the picked box either had two gold or one gold to begin with.

it's
50/50

You either pick a silver ball or a gold ball. The third box is out of the equation. The question asks nothing of the probability of picking particular boxes or doing all the steps "correctly". The only thing the question wants to know is what is the probability the next ball is gold, given you have already pick a box with at least one gold ball.

It's a 50/50 chance you have the two gold box in front of you or the gold and silver box in front of you. Given one gold is already removed you have a 50/50 chance of pulling out a gold or silver ball because there are only two possible balls you can pick.

But it's not a theory. It has been tested to death, including by retards who double down on their claims because they are so afraid to admit they are wrong

Still a cool story doe

No, Monty Hall has the option of switching, making it 2/3 in favor of the box you did not pick and that remains unrevealed

He wasn't moving the goalpost, just helping to show why it's higher than 50%.
Which it is.

But it's not

Well, you can admit you're wrong, 2/3 makes zero sense
You're tasked with only picking one more ball
Because the boxes contain either one gold or one silver, it's 50/50

The chance that you randomly pick a box with 2 of the same coins is 2/3. That doesn't change after revealing that you didn't pick the silver-silver one

The probability of you being in box 1 or 2 given you already have a gold ball in your hand isnt equally likely

no i am actually that retarded/full of myself

The riddle doesn't ask you to pick a box, you're told that you already got a gold ball, so the silver/silver box is removed from start

it actually is equally likely

it literally says "you pick a box at random, and get a gold ball"

2/3 only makes zero sense if you're uneducated and rely on intuition and that's ok, some people get jobs and live fulfilling lives, other people are niggers

Cool story doe

Yes it, it's two boxes that contain gold, the boxes' position relative to one another doesn't matter

It isnt. Box 1 has more gold balls. Thats the simplest way to put it. Since it has more gold balls you can reasonably assume that your first gold ball came from the first one

But what if niggers already stole the other gold balls?

But you can feel the weight difference.

So that means the silver only box is redundant to the riddle

doesn't matter. three boxes, random pick, 1 in 3

No, it's important for the initial probability distribution, 1 in 3 for either combo, but 2 in 3 for same balls.
Once you choose, you eliminate one of the 2 / 3 boxes, so the mixed box is still 1 / 3 and the gold-gold is 2 /3 per the initial setup

If you select a person from the prison population and they are a nigger, what is the likelihood of the next person you pick from the prison population being a nigger?

It's not 2/3 because the riddle only asks me AFTER I get a gold ball, so this means the riddle tells me beforehand that the silver box has no importance

I dont even know what youre arguing but its pointless. This can be tested in real life if this were done 100 times a person would get gold balls the first time 50 times, and of those 50 they would get a gold ball second 33 times, not 25. Thats a fact

wrong. the riddle asked you to pick a box at random, and you yourself then figure out it's not the silver-silver you picked, since you got a gold ball. Nobody told you which one is the silver-silver before you chose

Irrelevant. The GG box was 2x as likely to be selected due go the filtering applied

Yes, so I get a gold ball, thus the silver only box is removed from equation, before the riddle asks me the percentage
So it's 50/50

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Alright mein kinder, I shall explain this.
Consider the following: We have three boxes, and you can't see inside the boxes.
We can ignore the 2 silver's box.
So we have 2 gold box, and one g box. That gives us 4 balls in total to consider.
Case one, we grab the silver ball first.
We ignore this case.
Case two: We grab the first ball of the two gold case.
Case three: We grab the second ball of the two gold case.
Case four: We grab the gold ball of the mixed case.
As you can see, there are three cases that we count, and two of them *are* the gold box.
We are just as likely to get all four cases, but case 2 and 3 look the same to us.

probabilities cannot be tested in real life. you're a fucking moron. At best you can derive an approximated guess, and quantify how confident you're in that guess (p-value).
N=100 furthermore is so fucking low that it's completely meaningless.

so 2 in 4 or 1 in 2

The placement of the boxes is undisclosed and is not mentioned in the riddle, the drawing is just the exemplify, but the riddle doesn't specify the order of the boxes, therefore their placement is not accounted for

We can do it 10,000 times then. We'll get 5000 gold balls first and of those well get 3333 gold balls a second time

All the approximate numbers when testing this are closer to 2/3 than 1/2

N=10000 is little bit better, but my rough estimation is that the confidence level would still very much suck.

Stop with stupid experiments for problems that have exactly one clear and undeniable correct mathematical answer

No, 2/3. Remember, case one has already been discounted. In two of the three cases, we get a second gold ball.
The 4th case is just as likely as 2 and 3.
But we can't tell the difference between two and three.

yes, so?

Could probably get the noise to an acceptable level after just 50 tbh.

you cannot just discount the initial case, you picked before you knew about the outcome

Correct

So in 180 tests you take each ball an equal amount of times, 30

In 60 tests total you get 2x G
In 30 tests you get 1x g
In 90 tests you get Silver and the problem is a non starter.

In any scenario where you select a box and the order of the balls is not known or predetermined the outcome is that the GG box is held twice as often

See:

10,000 is plenty if you were right. The numbers would be close to 50%, and they simply arent. Experiments are worth more than theories

So that means it's 2/3 is the correct answer, not 1/2. Objective reality is pointing towards the actual mathematical answer.

Sure they can. Couldnt we flip a coin to determine the probability of it landing on tails? Cant we roll dice to figure out which total is the most likely?

But case one doesn't satisfy the condition set "Given your first draw is a golden ball"
So case 1 must be removed as a possibility.
In reality, there are six cases total, but cases 5 and 6 are from the double silver box, and are equally ignored

I already did this test, 100 times, and I only got gold each time, therefore the chance of gold is 100%

no, if you only do N=10000 there is a high chance you will get a results that's far off the right result.
Same as flipping a coin 10000 times and getting heads is improbable but not impossible. You can be just "unlucky" in your experiment. But doing the same thing 1 billion times, and the improbable becomes a lot more pronounced. Chances of you being unlucky are slim, but not eliminated

You uh... sure there isn't some bias in there lil' Timmy?

Its 2/3, you can also calculate it with bayes theorem.

Yeah, but the odds of that happening even in a string of 100 is *tremendously* low.

Given enough time, I can roll a 6 with a 6-sided dice 10000 times in a row (eventually). If I get really lucky, I might do it on the first attempt. That doesn't yet mean the chance of rolling a six with a six-sided dice is 100%

Bayes has nothing to do with this

Nope, the exact same setup, gold each time
Therefore the chance of gold is 100%

100 in a row is not tremendously low, it's very much feasible. It's far more likely e.g. than somebody being struck by lightning, but we know that people regularly but rarely are actually struck by lightening