/sci/ - Science & Math

50%
derp

>>4528749
Do you want a sticker?

>>4528751
nice big gold one please

100% since you chose the "same box" again.

>>4528753

>>4528767

only one of those 2 boxes has 2 gold.

75%

l2 Bayes, /sci/

The scenario would only be possible if you chose the box with two gold, so yes, the answer is indeed 100%, since the question is "the odds that you get gold from the same box"

I want to say 66%.

>>4528868

What if you happened to pick the box with the one gold and one silver?

>>4528883
Huh. You're right. 50% then.

>>4528859
Care to explain how Mr Bayes helps?

I too wish to say 2/3, from my pathetic understanding of conditional probability.

the answer is 50%, there is nothing to debate about high school statistics.

>>4528893
>says there is nothing to debate about high school statistics
>gets the answer wrong

>>4528890

You got gold the first time from the box, so you know you didn't get the box with the two silvers.

Therefore, you either have the 2 gold box or the 1 gold and 1 silver box. There's a 50% chance that it's either, and there's a 100% chance you'll draw gold again from the 2 gold box and only a 50% chance that you'll draw gold again from the 1 gold and 1 silver box.

0.5*1+0.5*0.5 = 75%

>>4528899

there is only 2 boxes with golds in them and you already know your box contains at least one gold. The third silver box means nothing. It might as well not even exist.

>>4528901
But the question is only asking about the second pull, so Bayes doesn't apply.

>>4528911
>There's a 50% chance that it's either
>>4528901
>The third silver box means nothing

Hmm. And where did Bayes' theorem even come in?

>>4528913

I'm assuming this is with replacement.

>>4528918

Full retard.

>>4528925
Oh you. I don't see any statements in that post that even begin to relate to Bayes' theorem. Perhaps your explanation was just shit?

http://en.wikipedia.org/wiki/Bayes'_theorem#Statement_and_interpretation

>>4528901
that's retarded, there's a zero percent chance from pulling gold again from the 1-gold-1-silver since you already pulled out its only gold

I wanna say 2/3. The way I think of it there's 6 coins to choose from initially, all equally likely. The fact that there's three boxes is irrelevant at the start, it's just artificially grouping the 6 coins into 3 sets. There's still 6 coins each equally likely.

Since there's two gold coins in one box and one gold coin in another box, if you pull a gold coin you know it had a 2/3 chance coming from the double box and a 1/3 chance coming from the 1 and 1 box. If you drew from the 2-gold box, there's a 100% chance of drawing another gold. If you drew from the 1-gold-1-silver box, there's a 0% chance of drawing another gold. So in total, there's a 2/3 chance of drawing another gold, 1/3 chance of drawing silver. Please tell me how this is wrong.

>>4528933

HURR DURR If you didn't use the equation explicitly in a way that I can just plug and chug you suck at explaining.

Derp.

>>4528936
>Implying you don't suck at explaining.

You didn't say what you're prior was, you didn't say what was conditional on what. It isn't my fault you can't explain your workings out for shit.

>>4528935

the 2 silver box is irrelevant, you already know the box contains a gold. It's not a single set of 6 coins either, it's 3 sets of 2 with one of those sets thrown in the trash.

one set grants a 0% chance and the other grants a 100% chance, so it's 50%.

>>4528935
You are not supposed to account for all that stuff. The question is quite clear; AFTER you have pulled a gold, what is the chance to get another? I.e. the problem starts when the 2 silver box is already excluded. Which means the probability is ofcourse 50%.

Like all retarded statistics troll thread problems its mostly a matter of how you interpret the original question.

If you look at the combined probability to first draw a gold and then draw another gold we have a completely different scenario.

>>4528935

The grouping into three boxes is NOT arbitrary since you have to pick the second coin from the same box.

Also, I'm assuming that you put the coin back into the box, otherwise the problem would be trivial.

>>4528950
How would not replacing the coin make it trivial?

>>4528942

I was trying to explain it a way that exposed the intuition behind the theorem so you can actually understand what you're doing rather than just plugging numbers into a meaningless equation.

>>4528950

you DON'T put the coin back in the box. the question is trivial. This thread is full of idiots.

>>4528950
>He thinks the problem is any less trivial if you include replacement

>>4528954

Because then it's too easy. It's just 50%.

>>4528955
>I was trying to explain it a way that exposed the intuition behind the theorem

Yes, by not using the theorem.

>>4528954

the question is still trivial, just not easy enough for an 8 year old to get the answer. Which this question is by the way, stop over thinking it for god's sake.

Without replacement: 2/3
With replacement: 5/6

>>4528960

Oh I'm sorry, I simply assumed you had an elementary understanding of probability theory.

How wrong I was.

>>4528967

but that's wrong you fucking retard. holy shit.

THERE ARE ONLY 2 BOXES TO CONSIDER. YOU ALREADY KNOW YOU DON'T HAVE THE THIRD SILVER BOX. TAKING THE SILVER BOX INTO ACCOUNT IS LIKE TAKING INTO ACCOUNT THE APPLE TREE IN THE BACKYARD WHEN COUNTING THE NUMBER OF APPLES IN A BOWL.

>>4528967
>didn't understand the question.jpeg
how does it feel that if you got this on a test you would get the wrong answer, because you cannot understand basic english?

its 100% clear from the wording that you do not replace coins. AND its also 100% clear that the problem starts in a scenario where you only have 3 coins left to chose from, among which 2 are golds.


>fukken aspies overthinking at 500mph.J-pegg

>>4528969
I simply assumed you knew what Bayes' theorem even was. My mistake.

>>4528975
>its 100% clear from the wording that you do not replace coins

I didn't realize it was a psychic examination, and that additional information was being beamed into my head. Please accept my apologies.

The answer is 2/3.

>>4528978

it doesn't matter if he knows what baye's theorem is. because it doesn't apply.

>>4528975
>its also 100% clear that the problem starts in a scenario where you only have 3 coins left to chose from, among which 2 are golds.

Exactly. So the probability is 2/3. I don't see what there is to get so worked up about.

>>4528982

the answer is 1/2, you either pull out silver or gold. There is no magical third coin in the box. How fucking hard is this to understand.

>>4528983
I know it doesn't. If you turn your attention to >>4528859
You will see that he thinks it does. I simply pressed him for an explanation of why. He still has not provided one.

>>4528984
no, because you are either picking from the box with 1 gold left, or a box with 1 silver left. fukken retardos.

>>4528978

>implying that at this point you aren't just trying to justify your stupidity

>>4528990
But there were two possible ways of picking the box that will have one gold left in it.

I think you guys might be over thinking it a tad bit. In order to get another gold out of a box, there still needs to be one left, meaning only one box could be chosen. The probability of that box being picked twice is 1/3.

>>4528984
you have to pick twice from the same box you imbecile.

>>4528994
no, because you already know you have picked a box with one gold. you should not account for the initial pick. its 1/2. If you didnt know that you already picked a box with one gold your answer would be right.

>>4528988

And I never will because you're too stupid to understand it anyway, making the whole endeavor a complete waste of time.

I'd much rather have some fun by insulting you until you decide to get off your lazy ass and figure it out for yourself.

>>4528994

>>4529000
Fascinating story, friend.

>>4528994

you aren't picking a box. you never pick a box. the box is chosen for you. this question is specifically designed to confuse.

the question is no different than: "you flip a coin and get heads, what are the chances you're next coin flip will be heads". The answer is 50%, you have all been successfully trolled, goodnight.

Case by case...
Recall that gold WAS DRAWN!
So...
1/2(1/3+1/2) = 5/12

>>4529011

>checking that gold was drawn.

...what? I don't think you understand the question.

Misleading question. Are you picking a box at random again with the other boxes included in the set or are you just reaching into the same box?

>>Case by case...
>>Recall that gold WAS DRAWN!
>>So...
>>1/2(1/3+1/2) = 5/12
Sorry, I read four boxes, let me fix that:
1/2(1/3+2/3) = 1/2

>>4529011
none of your numbers make any kind of sense.

initial chance to pick a gold box: 2/3

chance to pick gold from two gold box: 1

chance to pick gold again from one gold box: 0

these are the only numbers one should possibly see in your "equation". What the FUCK are you even trying to do?

you cannot just ignore that the coins are grouped into boxes, and you MUST draw from the same box twice.

>>4529018

>what are the odds i pull out gold again from the same box
>from the same box

>>4529030
FAIL, I did not adjust for the fact that you could draw a silver from the box with two on the second turn:
1/2(1/3+1/3(1+1/2))=5/12

Alright, riddle-me-this you fucking retards, What if the gold coins were slightly heavier than the silver ones, and it affects how it flows through the air. This is happening on a windy day, in the middle of a snowstorm, during a lunar eclipse, while the sun is launching a solar flare burst towards the earth, knocking out all electronics so it cannot be recorded using tools, and the earths rotation shifts by 0.0000000000000000023%, and your professor is on a sick leave, the coins also have a 0.14% copper impurity, and meanwhile an earthquake is occurring, and japan is flooding again.

Also, the man holding the box has Parkinsons.

>>4529041

you can't, you are pulling a coin from the box you already pulled a gold out of. the 2 silver box might as well be in the garbage pail next to the table at this point.

>>4529041

refer to >>4529033

>>4529046

what if I am also colour blind and I left the room to take a piss.

If he puts it back in: 75%
If he doesn't put it back in : 50%

let's rephrase the question a way you clusterfucks can understand:

you decide to flip a coin 2 times. On the first flip you get heads. What is the chance that the second coin flip will also be heads.

>>4529065
that question is not equivalent. What you wrote should (in prob. theory context) be interperted as what is the probability to flip heads twice in a row... which is not 50%. But 1/4.

>>4529073

you are either the OP trying to stir the fire, or an idiot.

>>4529078
No, just stating the obvious. Of course on each throw there is 50% to get heads/tails. But if you start to combine flips then you are dealing with another problem.

In the OP post it is 100% clear that you should NOT work with combined probabilities. Answer is 50%. Answer to your question is 1/4. umad?

>>4529085

the point of my question is that the first heads flip is already accounted for, it's asking the chances of heads for a single coin flip.

The question I asked was a troll question from months ago and the answer IS 50%.

<div class="math"> P(X_2=g|X_1=g) = \frac 2 3 </div>

>>4529116

you're equation is faulty as it takes into account the first coin pick when it's already accounted for.

>>4529125
Which line is wrong, and what is wrong with it?

>>4529133

The first coin pull is already pulled as part of the question. It is gold. The question simply wants to know if the second coin pull will be gold or silver.

>>4529144 cont.

your equation discarding due to the first coin pull, when we already know the first pull to be 100% gold, skews the results.

remove the silver-silver box and retest, as we already know it isn't chosen.

>>4529144
>>4529157
http://en.wikipedia.org/wiki/Conditional_probability

>>4529144

Looking back at this thread it seems nobody has noticed that the issue of the first coin being selected is where all the disagreements have sprang from. The question states that we have chosen a gold coin to begin with, but it does NOT tell us the process that led to it being chosen. In my program I assumed all possible coin selections to be equally likely (thus you are twice as likely to pick the 'gg' box), which gives a probability of 2/3 to get two golds.

If you suppose that the 'sg' and the 'gg' box are chosen with equal likelihood then you will get a different answer. 1/2 to be exact. But this scenario seems unintuitive since most people would presume the player to be choosing coins at random from the very beginning.

Glad that's sorted.

>>4529157
>your equation discarding due to the first coin pull, when we already know the first pull to be 100% gold, skews the results.

I don't think it should, since the player gets a free retry each time he picks silver. But I'll whack it into MATLAB and see what happens.

>>4529178

but the question already states that you picked gold as your first coin. One of the main points of probability is that you don't take previous attempts into account. That's gambler's fallacy.

it does not matter if he picked a million boxes before the one presented in the question, as that's the only one that matters and he picked gold.

>>4529178
>The question states that we have chosen a gold coin to begin with, but it does NOT tell us the process that led to it being chosen.

>I pick a box at random
>and get gold
What other interpretations of this could there be besides picking a box at random and drawing a coin from it?

The only ambiguity I can see is whether or not there's replacement.

>>4529171

conditional probability doesn't take into account rolls you've already done. that's gambler's fallacy.

if it did, the chance of me rolling a 6 on a 6-sided die after one million dice rolls that didn't come up 6 would be almost 100%.

>>4529178
>If you suppose that the 'sg' and the 'gg' box are chosen with equal likelihood then you will get a different answer. 1/2 to be exact.

No, you don't:

P(picking sg box) = 1/2
P(drawing g from sg) = (1/2)*(1/2) = 1/4
P(drawing g, g from sg) = 0

P(picking gg box) = 1/2
P(drawing g from gg) = (1/2) * 1 = 1/2
P(drawing g, g from gg) = (1/2) * 1 * 1 = 1/2

P(drawing g) = 1/4 + 1/2 = 3/4
P(drawing g, g) = 0 + 1/2 = 1/2
P(drawing g, g | drawing g) = 2/3

>>4529171

>If two events A and B are statistically independent, the occurrence of A does not affect the probability of B, and vice versa.

>>4529200
Yes you do. If you choose the 'sg' box then your probability of a second gold is zero. If you picked the 'gg' box then it is 1. The only thing that matters here is how you choose the box in the first place.

the problem is people are considering the first coin in their pull when it is irrelevant, all that matters is the second coin pull. This same question was posted here like 2 months ago with the only difference being that the word "coin" was excluded this time.

>>4529205

>This same question was posted here like 2 months ago with the only difference being that the word "coin" was excluded this time.

but the box was already chosen for you. It was part of the question. It's a coin toss probability question presented in the most convoluted method possible.

>>4529205

>The only thing that matters here is how you choose the box in the first place.

but the box was already chosen for you. It was part of the question. It's a coin toss probability question presented in the most convoluted method possible.

>>4529205
Disregard that, I didn't follow your maths. This is all starting to look unintuitive as fuck.

>>4529201
The second coin drawn is not statistically independent from the first coin drawn:

P(draw g,g) = 1/3
P(draw g,s) = 1/6
P(draw s,g) = 1/6
P(draw s,s) = 1/3

Drawing gold the first time biases you to draw gold the second time, and the same with silver.

The reason it is not statistically independent is because both draws depend on which box was chosen.

>>4529223

it's a troll question. The fact so many people got it wrong on a /sci/ board is embarrassing.

>>4529224

yes, it wouldn't be independant if you were drawing two coins. but you are only drawing one coin, and that coin can either be gold or silver. it's a heads or tails question with a bunch of shit thrown on top to make it look hard.

>>4529231
>If I pick a box at random and get gold, what are the odds that I pull out gold again from the same box?
>pull out gold again
>again

>>4529217
>>4529223
Okay I think I've got a handle on it. Your maths states: "P(drawing g from sg) = (1/2)*(1/2) = 1/4", which is different from what I was intimating. What I meant was if that you get the gold coin with equal probability from either box, then the probability of a second gold is 1/2. Your working is pretty much the first case again since you allow the 'gg' box to have twice the likelihood of being picked in the first place.

>>4529228
Would you rather we went back to the days of magnets and geusses? We could do better stuff with our time, but I'd rather have troll maths than troll-nothing-to-do-with-either-science-or-maths.

>>4529237
Box gg doesn't have twice the probability to be picked in the first place:

P(box gg) = P(box gs)

Box gg only has twice the probability to be picked given that the first draw is gold:

P(box gg | gold drawn) = 2*P(box gs | gold drawn)

>>4529235

>If I flip a coin and get heads, if I flip another coin what are the chances i will get heads again.
>what are the chances i will get heads again.
>get heads again.

>>4529253
Yes, there are two coin flips in that problem. The reason we can only consider one coin flip is because the two flips are statistically independent. The same is not true of the two coin draws.

>>4529252
Yes that is what I mean.

>>4529260
>/sci/
>having a life

>>4529263

BUT THERE IS ONLY ONE COIN DRAW, OUT OF ONE OF TWO BOXES. THE BOX CAN EITHER BE BOTH GOLD (1) OR ONE GOLD ONE SILVER (0). 50/50.

>pick gold box
>chance of getting gold holding the same box with the gold in it
100%
what.

Doesn't it depend if you put the coin back in or not? It'd be 75% if you did and 50% if you didn't, right?

>>4529283
[spoiler]THERE ARE NO COINS

>>4529276
>THERE IS ONLY ONE COIN DRAW

What the fuck? I just pointed out very clearly where the problem explicitly mentions two coin draws.

>>4529283

yes, but that's not what these autists are arguing about. They literally lack the ability to answer written questions

>>4529288
I just presumed the "gold" and "silver" where coins.

Deal with it.

>>4529289

THE FIRST COIN DRAW HAS ALREADY HAPPENED. IT IS GOLD. THE QUESTION IS ABOUT THE SECOND COIN DRAW. IT'S A GAMBLER'S FALLACY TROLL QUESTION.

.66 percent chance of getting a box with at least 1 gold piece in it
.5 percent of the boxes with at least 1 gold piece in it have 2 gold pieces
.66 * .5 = .33
33% chance

>>4529295
oh nvm, its just .5 as we aren't asking about the probability from picking a box
oopsies

>>4529283
Nope. It's 2/3 if you don't and 5/6 if you do.

>>4529295

YOU ALREADY GOT A BOX WITH A GOLD PIECE IN IT. IT WAS YOUR FIRST DRAW.

>>4529299
>>4529297

>>4529300

DIDN'T SEE IT.

>>4529305
now you do
these questions are always tricky because you read too quickly and don't know where to start

>>4529294
You are assuming that the events
draw 2 = gold
and
draw 1 = gold
are statistically independent. They are not.

It is
draw 2 = gold, box = GG
draw 1 = gold, box = GG
etc.
that are statistically independent.

The former are not statistically independent because the first draw gives you information about which box you're drawing from.

statistics is awesome

>>4529312

but the first draw being gold guarantees that you are only drawing from 2 boxes; the third box is statistically irrelevant in this question. it's impossible to get a non 50% answer from a single pull in a 2 option question.

>>4529316

statistics makes me want to bash my skull in because apparently nobody knows how to apply it outside of dice rolls and coin tosses.

>>4529317
The third box is irrelevant, correct. However, that is not the only information you obtain from the draw. Given that you have drawn a coin and obtained gold, it is twice as likely that you have the GG box as it is that you have the GS box:

P(box = GG | draw 1 = gold) = 2/3
P(box = GS | draw 1 = gold) = 1/3

>>4529312
Correction:
>It is
>draw 2 = gold, box = GG
>draw 1 = gold, box = GG
>etc.
>that are statistically independent.

This is only true if you're drawing with replacement. Without replacement, statistical independence doesn't apply here at all. The second draw is completely determined by the box and the first draw.

here is a better question:

you have 200 million boxes, each containing 1000 coins.

if you pull 300 million coins from random boxes, what are the chances the next box you open will be empty.

'bout tree fiddy

> I pick a box at RANDOM and get gold

if we picked a box at random to get gold, its twice as likely to come from GG, so there is a 2/3 chance it came from GG and a 1/3 chance it came from GS. Therefore, there is a 2/3 chance the next coin you pick is gold.

if the problem said:
>a friend looks inside and randomly picks a box with a gold coin, and gives it to you, what are the odds you will pull out gold

then its 1/2 since the odds of having GG and GS are equal

This conditional probability stuff always confused the shit out of me ;(

2/3

>>4530808
Nevermind that, the chances are 50%, we don't pick the ingod, we pick the box and there are only two boxes with gold; although one of them only has one gold ingot and one silver, in which case is purely dependant of WHICH box we picked.