Well, Jow Forums?

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If you chose the gold-gold box you are left with:
1gold, 1gold+1silver

If you chose the gold-silver box, you are left with
1silver, 1gold-1gold

Either way its aa 2/3 chance. The confusion arises when you consider the gold+gold box to be a single unit, when its actually not. This would leave you with a 50/50 chance.

you picked a gold ball.
Which box are you in now?
Which is more likely?

fpbp. if you don't understand conditional probability, read a book, don't try to argue away your ignorance on Jow Forums. if you still want to say 50% without reading up, unironically kill yourself

Your intuition is incorrect, and you should take a statistics class.

>How isn't it 50/50?
Because conditional probability. Question asks what is the probability you picked box 1 given that you drew gold. Consider:

Prior probability of picking gold: P(G) = 1/2
Prior probability of picking box 1: P(b1) = 1/3
Conditional probability of picking gold given box 1: P(G|b1) = 1

Substitute A=b1, B=G, and apply Bayes Theorem.

P(A|B) = P(B|A) * P(A) / P(B) = 1 * (1/3) / (1/2) = 2/3.

the absolute state of burger education
>it's a gold ball
this immediately takes the box with 2 silver balls out of the picture. You've either got the left or middle box. And one of those boxes has 2 gold balls. It's 50/50.

Grab gold ball 1 from box 1: remaining ball is gold.
Grab gold ball 2 from box 1: remaining ball is gold.
Grab gold ball from box 2: remaining ball is silver.

2/3, you brainlets

The problem with your math is that it isn't a 1/3 chance you picked any one box. The question begins AFTER you've drawn a gold ball, you the silver-silver box was never a factor to begin with.

all the boxes are single units. the balls are unified within the boxes.

Hate to break it to you, but Thomas Bayes, the man whose theorem can be used to irrefutably prove that the answer is 2/3, was British, not American.

The game was rigged from the start.

Yeah I explained it weird. This user's explanation is more cohesive:

This being my first time seeing this and coming from someone who until very recently never browsed Jow Forums, I would say 2/3. Logic mandates that you exclude the third box with 2 silvers right off the bat, the fact that you grabbed a gold already means that for this problem, the third box may not as well exist. With that you are left with four balls, one of which you already pulled and it is gold. 2 gold 1 silver left as potentials for the next pull. 2/3 chance of getting gold, IMO.

These. See: en.wikipedia.org/wiki/Bertrand's_box_paradox

50% cause it's one of the first two boxes, so it remains either one gold ball or one silver ball

This obviously. Conditional probability clearly doesn't apply here.

but there are actually 3 possibilities.
You can draw the first gold ball and be left with a gold ball.
you can draw the second gold ball and be left with a gold ball/
You can draw the 3rd gold ball and be left with a silver ball.

2/3

Yes, but if you picked the box with the silver ball then your probability of picking gold is only 50%, whereas if you picked the box with two gold balls your probability is 100%.

Consider the following:
>box 3 is out of the picture
>there are three balls remaining: two gold and one silver
>there are three possible outcomes for your next ball: gold ball 1, gold ball 2, and silver ball 1
>2/3 of the time you will pick gold

0.499999...%

You are confusing prior and posterior probability.

In Bayes theorem, we seek to compute a posterior probability. That is, the probability of event A occurring in light of the knowledge that event B occurred. We do this by using the prior probability of both events (that is, the probability of the event occurring WITHOUT THE KNOWLEDGE INHERENT IN THE PROBLEM), and the likelihood of the event occurring (which is, of course, where we reverse the event and its conditional, and compute P(B|A)).

You cannot use the posterior probability to represent the prior probability, because we are computing the posterior probability. You are jumping the gun and misunderstanding the theorem entirely.

This is a conditional probability problem by its very definition, you ignoramus!

It doesn't matter which gold ball you draw; gold is gold

>conditional probability
there is no conditional probability, you fool. it is stated that you already picked a gold ball (100% probability).

there is fallacy in the question.

The first ball drawn is irrelavent other than narrowing it down to the left and middle boxes.

At that point, you have one box with a gold ball and another box with a silver ball.

So it's 50%.

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How retarded can people be?

You have lifted a gold ball, that means you are in box 1, in which you have a 100% chance of picking up another gold ball - Or you are in box 2, in which you you have a 0% chance of lifting up a gold ball. Average is sum / 2, that is, 0,5 or 50%.

The answer is 50%

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This is the correct interpretation. Anyone else is trying to conflate the monty hall problem with this problem and can't read english.

this is so wrong it hurts

you ALREADY DREW a ball, it's gold, this doesn't apply

you ALREADY DREW a ball, so it's either box 1 or 2, there is a 50/50 chance it's 1

anyone saying 2/3 is either trolling hard or a pseudo-intelectual mathfag wannabe spouting shit they don't understand about bayes theorem

You already know it's a gold ball. It doesn't ask you "IF the first ball is gold, which is the probability?". It states the ball is gold and asks you what the probability of the next ball being gold is. This removes the "knowing" part required for a conditional probability.

It's a trick question.

I have an equal chance of choosing the second gold ball in box one as the others. There are three total choices.

If you put the gold ball you took back in the box, the answer is 75%. If you don't replace the ball, the answer is 50%.

>you ALREADY DREW a ball, it's gold

There are three gold balls, pajeet

The 'paradox' is only confusing because it's set up to confuse the person at first. Only reason there is 3 boxes is to fuck with you when you find the real answer.
Get rid of the 2 silvers, and it is easier to realize why the answer is true. There are three possible outcomes.

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thanks for proving my point people in this thread saying 2/3 are trolling, dumbass

It's still 1/2. Notice how it says
>from the same box

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this is correct. monty hall has no bearing on this one

You draw a ball. It's gold. There are three possible gold balls to draw, and you chose one of them.

You are twice as likely to chose a ball in box 1 than box 2, because there are three gold balls and two of them are in box 1. What is the chance you will choose a gold ball from box 1 and not box 2?

Two out of three times, you fucking mongoloid.

>There is no conditional probability
You don't know what conditional probability is. P(A|B) means "A given B", as in "under the event that B occurs, what is the probability that both A and B occur?". And I don't know why people are talking about Monty Hall because nobody brought it up.

TWO BOXES. TWO POSSIBILITIES YES OR NO.
how the fuck is anyone able to say this is 2/3 1/3. this a 50/50 shot.
someone explain how this isn't true. you can't. it's simple logic.

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1/F

You have to draw from the same box you drew the first ball, you complete imbecile

imagine being this stupid

It's MIT niggers overcomplicating and trying to change reality for their retarded mathematics. I bet if you ran a simulation of this it would end up as 50/50.

Correction, I was not including the fact that it said the next ball drawn would be FROM THE SAME BOX. That means that it would be 50/50 as TWO of the four total balls in box 1 or two would be accounted for, if you took a ball from one of the two boxes with a gold ball, chances are 50/50 the next ball is gold.

Imagine that. Drawing the other ball from the same box. Lets try it.

Draw ball 1. It's gold. Companion ball 2 is gold.
Draw ball 2. It's gold. Companion ball 1 is gold.
Draw ball 3. It's gold. Companion ball 4 is silver.

The companion ball being gold is probably some really weird fraction like 48/57

Oh sweetie. Does it hurt being that retarded?

google bertrand's box paradox and then kill yourself

>this immediately takes the box with 2 silver balls out of the picture
Yes, but it also takes the silver/gold box out of the picture half the time. It takes the gold/gold box out of the picture none of the time. So the chosen box is twice as likely to be the gold/gold rather than the silver/gold.

Yes, two possibilities, but those two possibilities aren't equally likely. You're more likely to have picked the gold/gold box, since half the times when you "pick" the gold/silver box, you end up picking the silver ball, and you have to throw that attempt out.

But you are forgetting, we've already drawn the ball. You are left with:

You drew from box 1: 1 gold ball left
You draw from box 2: 1 silver ball left

The actual act of drawing was never a factor. It was ALREADY drawn. SO it doesn't matter if it was gold ball 1 or 2 that was originally drawn, because it's already happened. All you know is that you have a gold ball in you hand and a box that has either 1 remaining gold ball or a silver ball.

wrong

but why?

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...

>All you know is that you have a gold ball in you hand and a box that has either 1 remaining gold ball or a silver ball.

I had a higher chance of drawing a gold ball from box 1 than box 2. Which means there is a higher chance that the other ball is gold.

No, cute frogposter, if I picked box two, I could have als gotten a silver ball. So if I have a gold ball the probability that it is from box 1 is greater than that it is from box 2.
Thus it can not be 50%, and it is 2/3th.

Alright anyone who really believes it's 1/2 (50%) please write some code that produces that result.

If you get an algorithm that produces that result, post it here. I'll correct any bugs and get it to output 2/3 (67%).

Remember when writing your implementation to be careful what must happen for you to "discard" a trial: you must first select a box randomly, then select a ball randomly. Just because you've selected the gold/silver box doesn't mean that you can magically grab the gold ball from that box just because it exists. (You can also grab a ball randomly out of the six, and derive the box it's in based on that.)

I did. the wiki page for the paradox . HOWEVER Bertrand's paradox is worded differently from this one. From the wiki:
>"The 'paradox' is in the probability, after choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, of the next coin drawn from the same box also being a gold coin."

THIS problem:
>"You take a ball from the box at random. It's a gold ball. What is the probability that the next gold ball you rake from the same box will also be gold."

Bertrand's box says "IF" its gold, what is the possibility, while this meme is saying "IT IS" gold, so what is the possibility. Bertrand's paradox does take the last silver-silver box into account, while OP's does not.

You're missing a choice.

There are three gold balls total. What's the chance of drawing a gold ball from box one?

It's 1/3, there are three boxes and only one of them has two gold balls, i.e. it is one third.

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2/3

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no fuckwit.

the ball IS gold. You don't need to factor in the double silver one at all.

So you've got one of the first two boxes. The remaining ball in that box is either gold or silver.

50% chance.

Kill yourself

50%. It's either a gold ball or not.

This image is a bait for people who already knew this paradox and don't bother to actually read the text. Here the propability is 50%, while in the original paradox it's 2/3

>So you've got one of the first two boxes. The remaining ball in that box is either gold or silver.

Box one has twice the chance of being picked over box two. If I picked a gold ball, it is more likely for me to have picked it from box one, therefore it is more likely for the other ball to be gold too.

there's no such thing as gold ball1 and gold ball 2, it's either gold or silver 50%

/thread

every fucking time there is a nigger like you, who is not only ignorant, but also insolent.
i used to be you.
you are wrong.

>there is no conditional probability

Box one has a higher chance of being picked over box two, because it has more gold balls in it.

Lets not get anti-semetic here

Go on then
Make a program that simulates this and check if you have a 50% chance

no it doesn't you brainlet cunt.

You can't see into the box. The silver ball in the gold/silver box doesn't come into the equation until after the first selection which IS gold.

It's not asking the probability of picking a gold ball from all the boxes or the probability of picking the double gold box.

It's already saying you've picked a gold ball, what are the chances of it being the box that now contains either a silver ball or a gold ball.

50/50

The mistake people make here is that they're counting macrostates (boxes) instead of microstates (balls).
There are 2 macrostates, which makes it looks like it's 50%, but there are 3 microstates, actually making it 67%.

Half the balls are gold.
Half the boxes that contain a gold ball contain another gold ball.
Half of the time you will pick a gold ball.
Half of the time that you've picked a gold ball you'll pick another gold ball.

>The silver ball in the gold/silver box doesn't come into the equation until after the first selection which IS gold.

And there are three possible gold selections, two of which are in box one.

You're forgetting that the problem states that you picked a box and grabbed a ball both at random.

but you've ALREADY drawn a gold ball and don't put it back in. You begin the problem with 1 box with 1 gold ball left and 1 box with 1 silver ball left.

2/3 of the gold balls are in one box.
HMMMMMMM

50/50
You either get a gold one or don't.
You guys are overthinking it.

Not this shit again. Fuck off 1/2 high school dropouts.

Box 1 is all gold
Box 2 is half gold
Box 3 is no gold but is irrelevant because you picked a gold ball already
If you picked box 1, you got a gold and get another gold
If you picked box 2, you got a gold and then a silver

When all the factors involved are 2's, only a retard can come up with 2/3 from it

I think at this point, people that say its 50-50 are just taking the piss and trolling for a reaction.
This shit ain't effort.

Based retard who assumed it was Bertrand's box paradox without actually reading it. Enjoying your high horse?

>but you've ALREADY drawn a gold ball

Yep. And I had a 2/3 chance of drawing that gold ball from box one.

that only applies if you didn't have the information that a box was picked at random and a ball was grabbed at random.

not him, but it is the same problem. Saying "if" doesn't change anything, because you know the box was picked at random.
If you didn't know the box was picked at random, your argument holds.

>If you picked box 2, you got a gold and then a silver
If you picked box 2 you had a fifty percent chance of getting a gold. There is a 1/6 chance you pick each ball. If you've picked gold, it increases to 1/3. So now we have balls 1, 2, and 3. Balls 1 and 2 are together, ball 3 is with a silver. If you picked ball 1, you get another gold. If you picked ball 2, you get another gold. If you pick ball 3, you get a silver. 3 possibilities, two outcomes result in a second gold ball, 2/3.

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But none of those choice factor in to the outcome. The question doesn't ask "if you chose a box and then a ball, what are the chances", its says" you got a gold ball, what are the chances of what is left".

prove this wrong. protip: you can't. written in Matlab btw.

% There are three boxes:
box1 = {'silver','gold'};
box2 = {'silver','silver'};
box3 = {'gold','gold'};
boxes = {box1,box2,box3};
n = 1e6;
bothGold = 0;
oneGold = 0;
count = 0;
for ii = 1:n
% pick a box at random (uniform dist. integer in range 1-3)
randBox = randi([1 3],1,1);
% put your hand in and pick a ball at random
firstDraw = randi([1 2],1,1);
firstBall = boxes{randBox}{firstDraw};

switch firstBall
case 'gold'
count = count + 1;
% the ball IS gold, draw second ball
if firstDraw == 1;
secondDraw = 2;
else
secondDraw = 1;
end
secondBall = boxes{randBox}{secondDraw};

switch secondBall
case 'silver'
%log if 1 silver 1 gold
oneGold = oneGold + 1;
case 'gold'
%log if both gold
bothGold = bothGold +1;
end

case 'silver'
% ignore and start over, this is irrelevant. the problem states
% the ball IS gold.
end
end

%display answer :^)
bothGold/count

Box 1: 1,000,000 gold balls
Box 2: 1 gold ball, 999,999 silver balls
Box 3: 1,000,000 silver balls

So if I draw a gold ball, is the chance of getting another one 50/50?
The number of gold balls in each box matters.
Pic related is you.

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Illiterate. OPs image clearly states that we have already chosen a gold ball. WE never made the choice between the boxes, the question did for us. Therefore the last box (silver-silver) was never a factor at all. Stop assuming this question is worded like Bertrand's paradox.

>pseudorandom number generator
Pfffft, enjoy your inaccurate model you fucking plebe