anyone else hate our generation? >>>/b/318600495

File: 19 KB, 444x384, 1300855214669.png [View same] [iqdb] [saucenao] [google]

Anonymous Sun Mar 27 03:39:37 2011 No.2781144 [Reply] [Original]

anyone else hate our generation?

>>>/b/318600495

>>	Anonymous Sun Mar 27 03:40:37 2011 No.2781146 I'm not an americunt at all

>>	Anonymous Sun Mar 27 03:42:37 2011 No.2781150 I am a retarded American. The answer is 200% probability.

>>	Anonymous Sun Mar 27 03:42:37 2011 No.2781151 >>2781144 50% I'm an americunt and i admit most of our population is genuinely retarded, however most nations these days have a vast majority of uneducated masses.

Scientist Sun Mar 27 03:43:37 2011 No.2781155

>>2781144
The simple answer is 50-50. The probability of the second child being a boy or a girl is largely independent of the first child.

Note that they're not perfectly independent in the real world due to fun medical quirks, and the 50-50 rate is also an approximation. There are more of one gender born by a slight offset.

>>	Anonymous Sun Mar 27 03:45:37 2011 No.2781161 I wish physical harm upon you, OP. I mean it.

Anonymous Sun Mar 27 03:45:37 2011 No.2781163

>>2781155

as you say, some women have more x gender than y (not ref. to chromosomes). and there are more men in the world than women, but thats probably socio economic.

anyway its ~50% and lol at the people in that thread saying "duh you said ones a girl the other has to be a boy"

Anonymous Sun Mar 27 03:47:37 2011 No.2781170

It depends on how you take in the grammar of the sentence.

If you take the sentence "1 child is a girl" as a stated rule, then it would be a 0% chance.

If you take the same sentence and acknowledge it as a trick statement, then it would be a 50% chance.

Coming to any conclusion that leads to either 25% or 33% is just representative of a person that thinks they are a genius and thinks they have come up with some ingenious paradoxical answer to the question when in reality they aren't.

>>	Anonymous Sun Mar 27 03:48:37 2011 No.2781173 >>2781163 inatinfanticide I would wager that this could account for much of the disparity between male and female populations.

>>	Anonymous Sun Mar 27 03:49:37 2011 No.2781177 P(GG) =1/3

>>	Scientist Sun Mar 27 03:50:37 2011 No.2781183 >>2781177 How goes the trolling sir? ?

Anonymous Sun Mar 27 03:51:37 2011 No.2781185

>>2781155
it doesn't say the first child is the girl, could be first or second.

3 possibilities, and as boy/girl is 50/50, all equally likely

1st born girl, 2nd born girl
1st born boy, 2nd born girl
1st born girl, 2nd born boy

only 1 of these is two girls so answer is 1/3

1/3 motherfuckers.

britfag btw

>>	Anonymous Sun Mar 27 03:51:37 2011 No.2781187 lrn 2 grammar, the words check out

>>	Anonymous Sun Mar 27 03:51:37 2011 No.2781189 >>2781177 the question is one is a girl (independent of the other) not both are girls hth

>>	Anonymous Sun Mar 27 03:51:37 2011 No.2781191 I cannot make up a probability until I know the genetics of the mother and father.

>>	Anonymous Sun Mar 27 03:52:37 2011 No.2781199 50%

Anonymous Sun Mar 27 03:54:37 2011 No.2781205

>>2781185
And then you realize that

1st born boy, 2nd born girl
1st born girl, 2nd born boy

are the same and you would not count them separately. The problem says nothing about what order the children were born. Just like what >2781170 said you're trying to act smarter than you really are by adding to the problem what isn't there to begin with.

>>	Anonymous Sun Mar 27 03:54:37 2011 No.2781208 Depends on the grammar of the sentence. Jesus, you are retarded! 50%

>>	Anonymous Sun Mar 27 03:55:37 2011 No.2781213 >>2781189 addendum: saying one is a girl in this case is a red herring (because the gender of the other child doesnt matter the way the questions being asked)

>>	Anonymous Sun Mar 27 03:58:37 2011 No.2781226 The first sentance is NON EXCLUSIVE meaning it does not include the second sentance in its meaning. tldr 1/3

>>	Anonymous Sun Mar 27 03:59:37 2011 No.2781231 File: 122 KB, 740x538, words_that_end_in_gry.png [View same] [iqdb] [saucenao] [google]

>>	Scientist Sun Mar 27 03:59:37 2011 No.2781232 The OP picture is clearly asking "What is the probability that child 2 is a girl given child 1 is a girl." If you want to debate the finer points of English grammar, whatever. Take it somewhere else. Not /sci/ence.

>>	Anonymous Sun Mar 27 04:00:37 2011 No.2781233 >>2781177 >>2781185 >>2781226 americaaaaaaaaaaaaaaaaaaaaaaaaaaaaa >:[

>>	Anonymous Sun Mar 27 04:01:37 2011 No.2781236 File: 18 KB, 400x400, basicniggermathfact0.jpg [View same] [iqdb] [saucenao] [google] >>>/b/318613584 oh lawdy..

>>	Scientist Sun Mar 27 04:01:37 2011 No.2781237 >>2781233 It's the same troll, probably the same as OP, probably the same as you. Anyone else who posts in this thread is a troll, who I shall report. Otherwise saged, nuked from orbit, etc.

Anonymous Sun Mar 27 04:02:37 2011 No.2781240

the question is stupid because its actually a complex question.

for one thing, i think female children are more common, and if you know the first child was a girl, that probably changes the odds the second will be a girl, something to do with hormones and shit.

Anonymous Sun Mar 27 04:04:37 2011 No.2781252

This is a standard 6th grade question that the teacher who asks it uses to seems really smart, the answer is 1/3 deal with it.

It is not obvious in any sense unless you have taken probability, ie you need to realize the question asks:

P(child2 = girl | child1 = girl) = ?

not:

P(child2 = girl) = ?

OP is a piece of shit btw

>>	Anonymous Sun Mar 27 04:05:37 2011 No.2781254 >>2781240 Actually, males are more common, but there are more living females because males die earlier and in greater numbers.

>>	Anonymous Sun Mar 27 04:06:37 2011 No.2781258 >>2781252 it can be interpreted either way, unless you add some extra specification, which OP did not do.

>>	Anonymous Sun Mar 27 04:08:37 2011 No.2781266 >>2781240 they literally aren't more common http://www.indexmundi.com/world/sex_ratio.html and hormones do influence gender selection

Anonymous Sun Mar 27 04:08:37 2011 No.2781267

>>2781205
sure, my eldest son and youngest daughter are exactly the same as if they'd been born other way around

not sure if troll or stupid

do you think the three outcomes from tow coin throws
heads heads, heads tails, tails tails are equally likely?

>>	Anonymous Sun Mar 27 04:09:37 2011 No.2781273 >>2781205 the very fact that it says nothing about order means 1/3 is correct wiki boy girl paradox if you are too retarded to work it out

>>	Anonymous Sun Mar 27 04:11:37 2011 No.2781279 >>2781232 depends if your numbering is significant the pic could say child 1 or 2 is a girl, what is the chance other child is a girl

>>	Anonymous Sun Mar 27 04:11:37 2011 No.2781280 >>2781252 Actually, the question asks about the latter. Learn semantics. Or rather, learn them back, because you apparently decided at some point that misinterpreting the meaning makes you smarter. It doesn't.

>>	Anonymous Sun Mar 27 04:14:37 2011 No.2781293 >>2781273 wiking that just proves 50% is the right answer thank you :)

Scientist Sun Mar 27 04:15:37 2011 No.2781295

Damnit. I hate sucking at stats. Brain lapse. Under the obvious reading, I am wrong. Answer is 1/3.

Work it out. Pretty simple. The sample space is { bb, bg, gb, gg}.

We're asked to find the probability
P(both children are girls | at least one child is a girl)

Which is fundamentally different than
P(child 2 is a girl | child 1 is a girl)

The difficulty comes in formalizing the English question into math.

P(both children are girls | at least one child is a girl)
= P(both children are girls & at least one child is a girl) / P(at least one child is a girl)
= P(both children are girls) / P(at least one child is a girl)
= (1/4) / (3/4)
= 1/3

Still, I blame my sucking on parsing English as opposed to sucking at stats, though they're almost the same thing in this case.

>>	Anonymous Sun Mar 27 04:17:37 2011 No.2781305 >>2781252 no, it asks P(child2 = girl \| child1 = girl) or P(child1 = girl \| child2 = girl)

>>	Anonymous Sun Mar 27 04:17:37 2011 No.2781308 >>2781170 I´d support this, but i´m not an expert at genetics, so if anyone comes up with some other value for probability, mabye he just used Mendelian inheritance laws to determine the chance.

>>	Anonymous Sun Mar 27 04:18:37 2011 No.2781311 >>2781295 it's called "probability" not "stats". also, your long ass post seems to imply that we all care SO much that you FINALLY figured it out. fucking namefaggots think everyone needs to hear their every thought.

>>	Scientist Sun Mar 27 04:19:37 2011 No.2781320 >>2781311 Also posting for the benefit of others, as no one had yet posted a complete correct answer.

>>	Anonymous Sun Mar 27 04:20:37 2011 No.2781324 >>2781293 >not sure if troll or stupid no it says both depending on interpretation

>>	Anonymous Sun Mar 27 04:20:37 2011 No.2781329 >>2781295 > implying each coupling has the same probability The calculation is correct though. Someone good at genetics?

Scientist Sun Mar 27 04:22:37 2011 No.2781337

>>2781329
>implying each coupling has the same probability
Covered that here bro:
>>2781155
>Note that they're not perfectly independent in the real world due to fun medical quirks, and the 50-50 rate is also an approximation. There are more of one gender born by a slight offset.

>>	Anonymous Sun Mar 27 04:23:37 2011 No.2781339 >>2781320 haven't read thread properly, one was posted based on enumerating the space of equally likely outcomes here >>2781185

>>	Anonymous Sun Mar 27 04:24:37 2011 No.2781347 >>2781144 IT'S ALL ABOUT THE WORDING OF THE QUESTION

>>	Anonymous Sun Mar 27 04:25:37 2011 No.2781352 Doesn't the concept of transgender rule out the possibility of 50% boy-girl split. I know the probability might be smaller, but there isn't any other information to warrant not giving it 33%. 33% boy-girl-it

>>	Anonymous Sun Mar 27 04:27:37 2011 No.2781357 i love it when that pompous ass tripfag scientist gets shit wrong

>>	Anonymous Sun Mar 27 04:27:37 2011 No.2781358 >>2781295 NOPE FAGGOT. ANSWER IS 0%. IT SAYS ONE CHILD IS A GIRL. THAT IMPLIES THE OTHER OF THE TWO IS A BOY. YOU LOSE MOTHERFUCKER!

Scientist Sun Mar 27 04:29:37 2011 No.2781364

>>2781358
Yes, that part is the reading. Does "1 child is a girl" mean "Exactly 1 child is a girl" or "at least one child is a girl"? That's a question of English, which I don't care too much about. I would hazard that the default reading, especially in context, is that it's saying "at least one child is a girl". Otherwise it's a very silly question - the questioner asks "One child is a girl, and one is boy. Are both children girls?".

>>	Anonymous Sun Mar 27 04:32:37 2011 No.2781379 >>2781280 Ya misinterpreting stuff makes me a genius thanks. The point is only an anus asks this question. And only an anus speaks of semantics.

>>	Anonymous Sun Mar 27 04:32:37 2011 No.2781380 This is just a poorly setup probability question. Genetics isn't just a fucking tossup or random or some shit. There are factors that have to be taken into account. And that's beyond the grammatical ambiguity of the question. Shit's dumb, man.

>>	m1ckey !dkJI/P7Y32 Sun Mar 27 04:32:37 2011 No.2781381 We have this retarded thread about the boy/girl paradox everyday, with the same ambiguous wording, trolling and butthurt. y u do dis OP

>>	Anonymous Sun Mar 27 04:34:37 2011 No.2781397 >>2781305 mb

Anonymous Sun Mar 27 04:37:37 2011 No.2781418

All right, fags, I'll explain this to you.

We have a two-element set {x1, x2} of variables that can have values of "b" or "g" (we're assuming equal probability for both).

One variable is chosen from the set, we're given its value, and asked about the value of the other one, so it's:

P(x1 was chosen)*P(x2 =g | x1 = g) + P(x2 was chosen)*P(x1 = g | x2 = g)

Which equals:

(P(x1 was chosen) + P(x2 was chosen))/2

Anyone who gives a different answer doesn't know math and cannot into abstract thinking.

>>	Anonymous Sun Mar 27 04:38:37 2011 No.2781424 File: 206 KB, 409x383, 1293563347700.png [View same] [iqdb] [saucenao] [google] >mfw when they don't take in the probability of hermaphrodite

>>	Anonymous Sun Mar 27 04:42:37 2011 No.2781445 Context Context Context Context Context Has the woman had no children yet? 33% chance of girl/girl Has the woman already had children and we know one is a girl? 50% chance of second child being a girl

Scientist Sun Mar 27 04:46:37 2011 No.2781466

>>2781418
>A woman has 2 children. 1 child is a girl. What are the odds that the other child is also female?

Let's assume further that it means:
>A woman has 2 children. (At least) 1 child is a girl. What are the odds that the other child is also female?

That's equivalent to:
>A woman has 2 children. (At least) 1 child is a girl. What are the odds that both children are female?

That's also equivalent to:
>A woman has 2 children. What are the odds that both children are female /given/ that (at least) 1 child is a girl?

= P(both children girls | at least one child is a girl)

The mistake is assuming the English is equivalent to:
P(child 1 is a girl | child 2 is a girl)

In English, that would be:
>A woman has 2 children, Ann and Sue. What is the probability that Ann is a girl given that Sue is a girl?

And that is a very different question. Using names, we can (wrongly) manipulate the English as follows:

>A woman has 2 children. 1 child is a girl. What are the odds that the other child is also female?

>A woman has 2 children. (At least) 1 child is a girl. What are the odds that the other child is also female?

>A woman has 2 children, Ann and Sue. Ann or Sue is a girl, and possibly both. What is the probability that Ann and Sue are both girls?

>A woman has 2 children, Ann and Sue. Ann is a girl. What is the probability that Ann and Sue are both girls?

That last step is an incorrect changing of the English. It changes from "At least one of Ann or Sue are girls" to "Ann is a girl". Once you isolate that piece of data to a single child as opposed to a statement of both children, you change the question.

>>	Anonymous Sun Mar 27 04:46:37 2011 No.2781468 >>2781424 Then you would have to factor in the probability of any given fetus developing traits of both sexes. You do it mister I'm so fucking outside the box.

Anonymous Sun Mar 27 04:55:37 2011 No.2781510

>>2781466
>Let's assume further that it means
>assume

Yes, that's what it is. Assuming things that weren't stated.

>Once you isolate that piece of data to a single child as opposed to a statement of both children, you change the question.

Only the question does explicitely isolate the data to a single child. You're the one changing the question.

>>	Scientist Sun Mar 27 04:57:37 2011 No.2781525 >>2781510 lolwut? I think we're on different pages. The answer is 1/3. (It's 0 if you take the silly reading that "1 child is a girl" means "exactly 1 child is a girl and the other child is not".)

>>	Anonymous Sun Mar 27 04:59:37 2011 No.2781534 I´m sure you all have read this: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox ... right?

>>	Anonymous Sun Mar 27 05:01:37 2011 No.2781544 File: 44 KB, 387x375, 1300718036928.jpg [View same] [iqdb] [saucenao] [google] >>2781445 >Has the woman had no children yet? >one child is a girl

>>	Anonymous Sun Mar 27 05:01:37 2011 No.2781546 >>2781525 >I refuse to accept I'm wrong. Okay, then, I won't force you to.

Scientist Sun Mar 27 05:05:37 2011 No.2781555

>>2781546
Could you point out where you disagree with my transformations? I'd be curious.

1) A woman has 2 children. 1 child is a girl. What are the odds that the other child is also female?
2) A woman has 2 children. At least 1 child is a girl. What are the odds that the other child is also female?
3) A woman has 2 children. At least 1 child is a girl. What are the odds that both children are girls?
4) A woman has 2 children. What are the odds that both children are girls /given/ that at least 1 child is a girl?
5) A woman has 2 children. Find P(both children are girls | at least 1 child is a girl)

P(both children are girls | at least 1 child is a girl)
6** = P(both children are girls & at least 1 child is a girl) / P(at least 1 child is a girl)
7** = P(both children are girls) / P(at least 1 child is a girl)
8** = (1/4) / (3/4)
9** = 1/3

>>	Anonymous Sun Mar 27 05:06:37 2011 No.2781564 File: 27 KB, 450x286, 1300418327631.jpg [View same] [iqdb] [saucenao] [google] >>2781555 trips = you are correct, sir

>>	Reiiama Kotsu, Inevitable Sun Mar 27 05:12:37 2011 No.2781577 Answer is 'it depends' because this troll didn't include the definitive statement(s). Depending on interpretation it is either 1/3 or 1/2. /thread.

>>	Anonymous Sun Mar 27 05:17:37 2011 No.2781588 Probability is bullshit mumbo jumbo math.

Scientist Sun Mar 27 05:17:37 2011 No.2781589

>>2781577
I disagree. It requires a rather strained interpretation to get 1/2. You need a pretty big rewrite to go from

"What is the probability that both children are girls given that at least child is a girl?"

to

"What is the probability that both children are girls given that Ann is a girl?"

The knowledge "at least one child is a girl" is fundamentally different than "the oldest is a girl" or "Ann is a girl" or any other knowledge that first distinguishes the children then says one is a girl. It eliminates the sample set in different ways.

As I myself am evidence of, humans quickly wrongly conflate these two. Humans suck at intuitively understanding stats.

>>	Anonymous Sun Mar 27 05:19:37 2011 No.2781592 >>2781588 To a certain extent, you are correct. When it comes to repeating an experiment very often, it is the shit though.

>>	Anonymous Sun Mar 27 05:21:37 2011 No.2781596 >>2781588 Not sure if it's bullshit, but it is the biggest pain in the ass that I've ever come across in my mathematical explorations. Maybe I'm just not wired for that kind of thing...

Anonymous Sun Mar 27 05:22:37 2011 No.2781601

>>2781555
Actually, most of those transformations are wrong, which becomes apparent when you convert the phrases into mathematical notation.

"Do not solve mathematical problems in natural language, it's inherently illogical" was among the first thing that I was taught in my university courses. How can you call yourself a scientist if you don't know that? Of course the foremost problem here is that the question was stated in natural language in the first place. Trolling at its finest.

Scientist Sun Mar 27 05:23:37 2011 No.2781605

>>2781596
In short, the question is asking
"Sample set is {bb, bg, gb, gg}. What are the odds that both are g, given that at least one is g."

For some reason, lots of people want to reinterpret the OP question as:
"Sample set is {bb, bg, gb, gg}. What are the odds that both are g, given that the first is g."

I was wrong initially. I'm trying to see if I can put this into words better. Once you formalize it, it's easy as pie, but the really damned annoying part is changing it from English to formalized symbolic form.

>>	Scientist Sun Mar 27 05:24:37 2011 No.2781610 >>2781601 Would you be so kind as to present your steps and your answer? I did try to convert to formalized symbolic form as soon as possible, but that required first clearing up the English.

>>	Anonymous Sun Mar 27 05:27:37 2011 No.2781618 File: 11 KB, 250x325, 1301165157871.jpg [View same] [iqdb] [saucenao] [google] >>2781610 i think you are trolled, sir

>>	Scientist Sun Mar 27 05:28:37 2011 No.2781620 >>2781618 Dunno. I've been particularly stupid in this thread. I want to at least double check myself.

>>	Anonymous Sun Mar 27 05:35:37 2011 No.2781638 File: 85 KB, 480x600, 477_son_i_am_disappoint.gif [View same] [iqdb] [saucenao] [google] wow. you scifags got trolled even worse than /b/.

Anonymous Sun Mar 27 05:37:37 2011 No.2781645

>>2781610
I already did.

>A woman has two children.

S - set of children
|S| = 2.

>1 child is a girl

x ∈ S
x = g

>What are the odds that the other child is also female?

y ∈ S - {x}
y = ?

We're assuming that ∃(x ∈ S) P(x = g) = P(x = b) = 50%, of course.

>>	Anonymous Sun Mar 27 05:39:37 2011 No.2781650 >>2781645 >y = ? Ugh, that should of course be: P(y = g) = ?

>>	Scientist Sun Mar 27 05:40:37 2011 No.2781655 >>2781645 I disagree with your use of sets. For example, if both children were girls, then x member S y member S x = g y = g Thus \|S\| = 1. But you also said that \|S\| = 2. Thus you're pretty mistaken somewhere around here.

Anonymous Sun Mar 27 05:41:52 2011 No.2781660 [DELETED]

Once again, too many errors.

>>2781610
>A woman has two children.

S - set of children
|S| = 2.

>1 child is a girl

x ∈ S
x = g

>What are the odds that the other child is also female?

y ∈ S - {x}
P(g = y) = ?

We're assuming that ∀(x ∈ S) P(x = g) = P(x = b) = 50%, of course

Anonymous Sun Mar 27 05:45:37 2011 No.2781668

>>2781655
Okay, you're right, this was unclear, once again:

>A woman has two children.

S - set of children
|S| = 2.

>1 child is a girl

x ∈ S
gender(x) = g

>What are the odds that the other child is also female?

y ∈ S - {x}
P(gender(y) = g) = ?

We're assuming that ∀(x ∈ S) P(gender(x) = g) = P(gender(x) = b) = 50%, of course

>>	Anonymous Sun Mar 27 05:46:37 2011 No.2781671 >>2781645 >>2781660 err, i don't get it. The sample space should be the set of all possible outcomes, the outcomes are genders of the 2 children, right ? what is S in your answers ?

Scientist Sun Mar 27 05:46:37 2011 No.2781672

>>2781655
>>2781660
Now, if you wanted to formalize it as:

x, y member S
|S| = 2

"1 child is a girl"
girl(x)
aka girl(x) = true

"What are the odds that the other child is also female?"
What is P(girl(y)) ?

--

Ok, I think I disagree with your English to symbolic logic step
"1 child is a girl" => girl(x)
The English "One child is a girl" is equivalent to "At least one child is a girl", not "Child x is a girl". That remains the source of our contention.

>>	Scientist Sun Mar 27 05:48:37 2011 No.2781676 >>2781672 To beat this dead horse, I argue that the English "one child is a girl", using your notation, means: girl(x) OR girl(y) I disagree that it means girl(x)

Scientist Sun Mar 27 05:51:37 2011 No.2781686

>>2781676
Hmm, one might be able to phrase it more as:

x, y are the children of the woman.
S = { x, y }
|S| = 2

"One child is a girl"
implies
(there exists a child f in S) [girl(f)]
which of course is equivalent to
girl(x) OR girl(y)

To do it a longer way:
"One child is a girl"
implies "there exists a child of the women which is a girl"
implies (there exists a child f in S) [girl(f)]
implies girl(x) OR girl(y)

>>	Anonymous Sun Mar 27 05:52:37 2011 No.2781692 >>2781672 Actually, it does not. I could rewrite this to ∃(x ∈ S) gender(x) = g (or girl(x), if you prefer) which is pretty much equivalent to your "at least", and it would still not change anything.

>>	Anonymous Sun Mar 27 05:56:37 2011 No.2781702 >>2781676 >girl(x) or girl(y) Of course not. The y variable wasn't even introduced at that point yet.

Scientist Sun Mar 27 05:59:37 2011 No.2781712

>>2781692
Oh, but it would.

"A woman has 2 children. 1 child is a girl. What is the probability that the other child is also female?"

Call x and y the children of the woman.
Let S = {x, y}
Thus |S| = 2

"1 child is a girl."
implies (there exists f in S) [gender(f) = g]

"What is the probability that the other child is also female?"
asks "What is P(gender(??) = g".

Well shit, we don't know what the "other" child's name is, x or y.

We can rephrase it as:
"What is the probability that the other child is also female?"
** What is the probability that both children are female?
** What is P(gender(x) = g AND gender(y) = g) ?

Now, we are given something, which means the question actually means:
"What is the probability that the other child is also female?"
** What is the probability that both children are female /given/ that 1 child is female?
** What is P(gender(x) = g AND gender(y) = g | (there exists f in S) (gender(f) = g))

P(gender(x) = g AND gender(y) = g | (there exists f in S) (gender(f) = g))
= P(gender(x) = g AND gender(y) = g AND (there exists f in S) (gender(f) = g)) / P((there exists f in S) (gender(f) = g))
= P(gender(x) = g AND gender(y) = g) / P((there exists f in S) (gender(f) = g))
= (1/4) / (3/4)
= 1/3

Any questions?

Anonymous Sun Mar 27 06:05:37 2011 No.2781736

>>2781712
>Call x and y the children of the woman.
>Well shit, we don't know what the "other" child's name is, x or y.

Duh.

Don't do this. It's unnecessary, it's not stated in the question, and it complicates your reasoning and makes you reach an incorrect conclusion in the process.

I'm pretty sure >>2781668 is correct already, read it carefully.

>>	Anonymous Sun Mar 27 06:06:37 2011 No.2781739 BB GG BG GB 3 possible variations when known one is girl 1 possible variation with the other child being girl too chance is 1/3

Scientist Sun Mar 27 06:07:37 2011 No.2781742

>>2781736
That's who I'm arguing with. Again, it boils down to what I think is the rather clear translation from English to propositional logic. Namely,

"1 child is a girl"
implies (there exists f in S) (gender(f) = g)

or
"1 child is a girl"
implies gender(x) = f

I think that those are very different translations. I think that the "(there exists f in S) (gender(f) = g)" translation is the correct one. With that out of the way, the answer 1/3 follows quite readily.

>>	Anonymous Sun Mar 27 06:10:37 2011 No.2781756 If the woman has one child a girl the other must be a boy so the probability is 0%.

Scientist Sun Mar 27 06:12:37 2011 No.2781763

>>2781756
And again, we've covered that. That's a possible, though I think silly, interpretation of the English. I think that "1 child is a girl. What is the probability that the other child is a girl" does not mean "Exactly 1 child is a girl. What is the probability that the other child is a girl?" based on context. If you want to interpret it that way, fine, though I think equally plausible is the interpretation "At least one child is a girl."

>>	Anonymous Sun Mar 27 06:14:37 2011 No.2781769 >>2781742 No, it does not, because no matter how you note it, you still have >the other child As in, the child that isn't x: y ∈ S - {x} \|S-{x}\| = 1, so we don't have to worry about choice, possible results are b and g, both with 50% probability.

Scientist Sun Mar 27 06:15:37 2011 No.2781776

>>2781742
To be complete, the 3 possible translations are:

gender(x) = girl
gender(x) = girl AND NOT gender(y) = girl
(there exists f in S) (gender(f) = girl)

The first leads to probability 1/2. I think that this is a rather bad reading of the English.

The second leads to probability 0. Plausible reading, but silly.

The last reading leads to probability 1/3. I think this is the clearest reading, though the reading giving 0 is at least sensible and a correct translation of the English.

Scientist Sun Mar 27 06:17:37 2011 No.2781785

>>2781769
>No, it does not, because no matter how you note it, you still have

Can we agree that the two claims:
** gender(x) = girl
** (there exists f in S) (gender(f) = girl)
are two distinct and different claims?

The first creates a sample space of
{ (x = girl, y = boy), (x = girl, y = girl) }

The second creates a sample space of
{ (x = boy, y = girl), (x = girl, y = boy), (x = girl, y = girl) }

That leads to very different probabilities.

>>	Anonymous Sun Mar 27 06:19:37 2011 No.2781797 I can;t believe how this is still going. The only right answer is 50%. Well done America.

>>	Scientist Sun Mar 27 06:19:37 2011 No.2781801 Let's put it to you like this. I flip a quarter and a penny. I tell you that at least one of them landed heads. What is the probability that the other landed heads? The answer is 1/3. Go ahead, perform the experiment repeatedly, and tally the results.

>>	Anonymous Sun Mar 27 06:24:37 2011 No.2781816 >>2781785 Yes, they're different. No, the first one is a correct reading. No, both lead to the same result.

Scientist Sun Mar 27 06:25:37 2011 No.2781820

Here's another way of thinking about it.

Randomly roll 2 boy/girl children. Remove the case that there are no girls.

That leaves sample space { bg, gb, gg }.

Now, someone else who knows the results picks a girl and labels that x.

Case bg:
There's only one choice for x. The other is a boy. Thus the "P is 0".

Case gb:
As above. "P is 0".

Case gg:
We're not sure which was labeled as x. However, it doesn't matter, the "other" child is also a girl. Thus "P is 1".

Add up the numbers, and you get P = 1/3.

>>	Anonymous Sun Mar 27 06:25:37 2011 No.2781822 >>2781801 No, the answer is 1/2, and I already pointed out why. Let's just say that the question you're actually asking is not the question you think you're asking.

Scientist Sun Mar 27 06:27:37 2011 No.2781827

>>2781816
>Yes, they're different.
Good.

>No, the first one is a correct reading.
Meh, politely disagreed.

>No, both lead to the same result.
... just went full retard.

"... given gender(x) = girl" will give a different answer than "... given gender(x) = girl or gender(y) = girl".

>>	Anonymous Sun Mar 27 06:30:37 2011 No.2781836 100 replies and on average 56& of 4chans users are americans which means so far in this thread is only 1 american who has got the answer correct

>>	Anonymous Sun Mar 27 06:31:37 2011 No.2781842 >>2781827 Well, I tried, I really did. You're hopeless.

>>	Anonymous Sun Mar 27 06:32:37 2011 No.2781844 Answer is 1/3 All those who say otherwise are either trolling/retards and can't read, mostly both.

>>	Anonymous Sun Mar 27 06:33:37 2011 No.2781848 >>2781820 are you fucking serious? you must have a low iq if you need to think this much about such a simple problem. no one tell this fag about the monty hall problem, or he will lose 5 weeks of his life trying to figure it out.

>>	None !ShaanKGPQ6 Sun Mar 27 06:33:37 2011 No.2781849 >>2781801 So you're saying that if I flip two coins, the result of the second coin toss is dependent on the first? Wat.

>>	Anonymous Sun Mar 27 06:33:37 2011 No.2781851 >>2781801 In this situation one coin is being flipped twice. Good god you people are dumb.

>>	Anonymous Sun Mar 27 06:34:37 2011 No.2781853 >>2781842 girl x is girl: U={GB,GG},the probability for child y to be girl is 1/2 girl x is girl or girl y is girl: U={GB,BG,GG}, the answer then is 1/3 you are a retard i am not the one you were talking to by the way, just a bystander.

>>	Anonymous Sun Mar 27 06:38:37 2011 No.2781863 let x be child 1 y be child 2 the question can be framed as: p(y=g\|x=g) or p(y=g or x=g \| x=g or y=g) different questions, different answers >>2781853 this is correct

Anonymous Sun Mar 27 06:38:37 2011 No.2781867

>>2781851
Exactly, that's a good example.
In this situation we are looking for the outcome where both throws are for instance tails(T) and we know one the throws ended up tails up.
Our possible variations with 2 throws when one was tails are:
TT
TH
HT

therefore the chance of getting 2 tails is 1/3

>>	None !ShaanKGPQ6 Sun Mar 27 06:41:37 2011 No.2781879 >>2781867 That's implying order matters. It doesn't.

>>	Anonymous Sun Mar 27 06:42:36 2011 No.2781881 you are all stupid niggers just read the wiki fags

>>	Anonymous Sun Mar 27 06:42:37 2011 No.2781880 >ITT: child gender is a Bernoulli trial...

>>	Anonymous Sun Mar 27 06:44:37 2011 No.2781884 >>2781879 Order doesn't matter in that? It's entirely correct and valid analogy.

>>	Anonymous Sun Mar 27 06:44:37 2011 No.2781885 >>2781867 If I toss a coin there is a 50% chance it will land heads and a 50% chance it will land tails. If I toss it again there is still a 50% chance it will land heads and a 50% chance it will land tails. Where the fuck are you getting 33% from?

>>	Anonymous Sun Mar 27 06:45:37 2011 No.2781889 >>2781879 Oh but it does. If we don't know which throw was tails there's 2 possible wrong outcomes and only 1 right outcome

>>	Anonymous Sun Mar 27 06:45:37 2011 No.2781890 >>2781885 If your small mind can't comprehend it, do the experiment. hundred times shoud be enough but more is better.

>>	Scientist Sun Mar 27 06:46:37 2011 No.2781893 >>2781879 I don't know what you're trying to say. I do know the answer is the same whether I'm using one quarter and one dime, or two dimes, or two kids's genders. It's the same thing.

>>	Anonymous Sun Mar 27 06:49:37 2011 No.2781899 >>2781890 I saw your math, but it has nothing to do with the topic at hand. You seem to think the first coin toss has an effect on the second toss or some shit, when both tosses are in fact separate events. Go be retarded somewhere else.

>>	Anonymous Sun Mar 27 06:49:37 2011 No.2781902 What I'm saying is that the text clearly states that one toss was already done for us. It turned out to be a girl. So atm you have G x. Now what's the chance of x being a girl? 50%.

>>	Anonymous Sun Mar 27 06:51:37 2011 No.2781910 >>2781899 you do know that you are discarding the results where the first child is not a girl right? of course they are dependent, read the fucking question. retard sure is retarded

>>	Anonymous Sun Mar 27 06:51:37 2011 No.2781913 If the question would be: "1st child was a girl. What are the odds that second child is girl too?" Then the answer would be 50%. Then our pool of possible variations would be {GG, GB} with one being correct

>>	Anonymous Sun Mar 27 06:51:37 2011 No.2781914 >>2781899 Must you make me repeat my words If your small mind can't comprehend it, do the experiment. hundred times shoud be enough but more is better. Do the experiment. Then come back if you have problem

Scientist Sun Mar 27 06:51:37 2011 No.2781916

>>2781885
Let me try to formalize.

Someone approaches you. A woman. She says that she has two children. She further says that one of the children is female (or both are female). She asks what is the probability that both are female if she's not lying.

So, P(both female | one or both are female)

Let's look at the sample space
{ bb, bg, gb, gg }, each with equal probability.

P(both female | one or both are female)
= P(both female & one or both are female) / P(one or both are female)
= P(both female) / P(one or both are female)
= (1/4) / (3/4)
= 1/3

Any questions?

>>	Anonymous Sun Mar 27 06:53:37 2011 No.2781920 You flip a coin. I now flip a coin. What are the chance my toss is the same as yours?

>>	Anonymous Sun Mar 27 06:53:37 2011 No.2781923 >>2781910 Nope. Both children are a separate coin toss. The first one was heads(girl). The second coin toss has a 50% chance of being either heads(girl) or tails(boy). I really can make this any simpler.

>>	Anonymous Sun Mar 27 06:54:37 2011 No.2781927 >>2781923 no it's not, stop trying to change the question.

>>	Anonymous Sun Mar 27 06:54:37 2011 No.2781928 I flip a coin once and get tails. What are the odds I get tails if I flip again. 1/2 Retards.

>>	Scientist Sun Mar 27 06:54:37 2011 No.2781929 >>2781923 >The first one was heads(girl). Reread OP pic. That's not what was given. It was given one of the two flips was heads(girl), not that either particular one was heads(girl).

>>	Anonymous Sun Mar 27 06:54:37 2011 No.2781930 >>2781923 *Really can't make this any simpler

>>	Anonymous Sun Mar 27 06:56:37 2011 No.2781935 >>2781916 there you go retards, you guys are either too stupid or trolls.

>>	Anonymous Sun Mar 27 06:56:37 2011 No.2781937 >>2781929 Goddamn it, it doesn't matter. Either way each child still has a 50% chance of being a boy or a girl.

Scientist Sun Mar 27 06:57:37 2011 No.2781943

>>2781937
You're confusing and conflating
gender(x) = girl
with
gender(x) = girl or gender(y) = girl
They're fundamentally different. Look at the resulting change to the sample space:

{ gb, gg }
vs
{ bg, gb, gg }

They are different pieces of information.

>>	Anonymous Sun Mar 27 06:58:37 2011 No.2781946 >>2781937 ahahahaha you retards sure are retarded >>2781916 math, disprove it or gtfo

>>	Anonymous Sun Mar 27 06:58:37 2011 No.2781947 >>2781916 >Any questions? Why did you change the question's phrasing? You don't have to answer, it was rhetorical.

>>	Anonymous Sun Mar 27 06:59:37 2011 No.2781948 >>2781928 >>2781920 You are not asking same question as OP is. you are asking this >>2781913 which is not same

Scientist Sun Mar 27 07:02:37 2011 No.2781961

>>2781947
>Changed the question
Don't think I did.

"Question: A woman has two children. 1 child is a girl. What are the odds that the other child is also female?"

So,
P(other child female | 1 child is female)
= P(both children female | 1 child is female)
= P(both children female | one or both of the two children is female)
= P(both children female & one or both of the two children is female) / P(one or both of the two children is female)
= P(both children female) / P(one or both of the two children is female)
= (1/4) / (3/4)
= 1/3

>>	Anonymous Sun Mar 27 07:02:37 2011 No.2781964 >>2781943 The lady is saying " I have 2 children, one is a girl. what is the chance of my other one also being a girl." It can only be a boy or a girl. 50% motherfucker.

Scientist Sun Mar 27 07:04:37 2011 No.2781976

>>2781964
The probabilities are not independent. You cannot apply that rule. She did not say "My daughter Ann is a girl. What is the probability that my other child is a girl?". She said "One of my children (or both) is a girl. What is the probability that both of my children is a girl?". That is a fundamentally different question.

>>	fucking sage'd Anonymous Sun Mar 27 07:05:37 2011 No.2781981 >>2781964 now i know you are a troll because no one is this stupid, gtfo

>>	Anonymous Sun Mar 27 07:08:37 2011 No.2781998 >>2781976 >She said "One of my children (or both) is a girl. What is the probability that both of my children is a girl?" Did she? Let's read OP again... no, apparently not.

>>	Anonymous Sun Mar 27 07:10:37 2011 No.2782005 50% chance the second child will be a girl.

Scientist Sun Mar 27 07:10:37 2011 No.2782008

>>2781998
What does "1 child is a girl" mean?
Your options are:
1- My daughter Ann is a girl.
2- Exactly one of my children is a girl.
3- One or both of my children are girls.

1 is the stupid and wrong reading.
2 is I think not the intended reading.
3 is clearly the intended reading.

>>	Anonymous Sun Mar 27 07:10:37 2011 No.2782011 File: 8 KB, 225x224, images.jpg [View same] [iqdb] [saucenao] [google] >find: "Scientist" >1 of 35 matches found >Scientist confirmed for faggot

>>	Anonymous Sun Mar 27 07:12:37 2011 No.2782020 >>2782008 4. "One child is a girl." You really should know you have no point when you have to rephrase the question for it to fit your "solution".

Scientist Sun Mar 27 07:13:37 2011 No.2782024

>>2782020
What does that mean? How is that option different than the other 3 posted? How does it affect the sample space of

{
(oldest girl, youngest girl),
(oldest girl, youngest boy),
(oldest boy, youngest girl),
(oldest boy, youngest boy)
}

?

How do you interpret that to affect the available sample space? Which of the above pairs is no longer allowed?

>>	Anonymous Sun Mar 27 07:14:37 2011 No.2782027 >>2781144 50,000ppm

>>	Anonymous Sun Mar 27 07:15:37 2011 No.2782030 >>2782024 >oldest >youngest Don't introduce shit that wasn't stated. It wasn't stated, period.

>>	Scientist Sun Mar 27 07:15:37 2011 No.2782031 >>2782011 That's nice.jpg. Stand back now. I'm practicing MATHS.

Scientist Sun Mar 27 07:16:37 2011 No.2782034

>>2782030
I know it's not stated. However, there are two children. Whatever label you want to give them.

The fact remains that the sample space before the knowledge "1 child is a girl" is: { gg, gb, bg, bb }

It is not { gg, gb, bb }

So, how does the knowledge "1 child is a girl" affect the sample space of { gg, gb, bg, bb } ? Specifically which elements of this set become disallowed?

>>	Anonymous Sun Mar 27 07:18:37 2011 No.2782039 >>2782030 The age is in there to separate the 4 different possibilities. There is nothing wrong when they are labeled like that. This is the start pool: GG GB BG BB Then we start to analyze it to find the solution.

>>	Anonymous Sun Mar 27 07:20:37 2011 No.2782050 trick question says 1 is a girl, so: 0 chance

>>	Anonymous Sun Mar 27 07:21:37 2011 No.2782052 >>2781236 1 right? my friend is sitting right next to me and swears it's 0

Scientist Sun Mar 27 07:22:37 2011 No.2782059

>>2781236
>>2782052
There is no good answer. When approached one way, the limit is 1.
lim as x->0+ of x^0 = 1

When approached another way, the limit is 0.
lim as x->0+ of 0^x = 1

There is no obvious definition to that. It might be defined a particular way for a particular class or domain, but there is no one "sensible" way, just like whether Natural Numbers includes 0 or not is rather arbitrary and it depends.

>>	Anonymous Sun Mar 27 07:23:37 2011 No.2782063 >>2781742 >propositional logic call it propositional calculus or just logic

>>	Anonymous Sun Mar 27 07:27:37 2011 No.2782075 >>2782034 It should be easier for you to understand if you order the children by the time we learned about them. First is the child in sentence 2, the second is the "other" child from sentence 3. What now?

>>	Scientist Sun Mar 27 07:29:37 2011 No.2782078 >>2782075 You haven't answered my question. Can you please draw out a sample space before "1 child is a girl", and draw out a separate sample space given "1 child is a girl". I await your answer.

>>	Anonymous Sun Mar 27 07:30:37 2011 No.2782084 >>2782075 You can do that but then it turns a bit harder to understant. If we go by that then the pool is: GG GB The trick here is that GB is twice as probable as in: GG 2*GB From this we get 1/3

>>	Anonymous Sun Mar 27 07:30:37 2011 No.2782086 lol are you stupid? >the other child >also female directly implying the first is female and not at all asking for the possible results of both childrens' genders.

>>	Anonymous Sun Mar 27 07:32:37 2011 No.2782093 >>2781236 http://mathforum.org/dr.math/faq/faq.0.to.0.power.html

>>	Anonymous Sun Mar 27 07:35:37 2011 No.2782111 >>2782059 Hast I checked natural numbers doesn't include zero

>>	Scientist Sun Mar 27 07:37:37 2011 No.2782120 >>2782111 In a couple of my math classes, they did. True story.

>>	Anonymous Sun Mar 27 07:42:37 2011 No.2782134 >>2782120 why? that's whole numbers (nonnegative). If zero is included in natural numbers what is the point of making it a distinct term?

>>	Scientist Sun Mar 27 07:43:37 2011 No.2782141 >>2782134 Dunno. It really doesn't affect much though. I was similarly miffed too.

Anonymous Sun Mar 27 07:45:37 2011 No.2782143

From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2

>>	Scientist Sun Mar 27 07:46:37 2011 No.2782147 >>2782143 Bravo. That is the best way of putting it. I was thinking of a good way while running to the store, and that's also the best I could do.

>>	Anonymous Sun Mar 27 07:46:37 2011 No.2782150 >>2782141 I hate these kinds of arbitrary inconsistencies.

Anonymous Sun Mar 27 07:47:37 2011 No.2782155

>>2781144

>>2781144

I hate mathematical problems posed in the English language!

They can be interpreted in any number of ways...

Example 1: A woman has 2 children. 1 is a girl.

This statement could be interpreted to mean 1 is a girl therefore the other is a boy/transexual/other creature.

Example 2: A woman has 2 children. 1 child is a girl. How do we know for certain the child mentioned in the second sentence is related in any way to the first sentence?

Example 3: The problem is solved by common sense and the probability is 0.5. Assuming odds of a boy/girl birth are even.

>>	Anonymous Sun Mar 27 07:47:37 2011 No.2782156 >95% of Americans are retarded and can't do simple probability Evidently OP is one of them.

>>	Anonymous Sun Mar 27 07:48:37 2011 No.2782158 >>2782143 trouble is with the second interpretation, you can't guarantee there is a girl, so the question isn't asked. wikipedia is retarded on this.

Scientist Sun Mar 27 07:51:37 2011 No.2782165

>>2782158
I don't understand what you mean "You can't guarantee there is a girl".

When a random woman approaches you and informs you that one of her two children (or both) is a girl, that alone gives you additional information. It allows you to dismiss the possibility that you live in the universes where she no girl children, and dismiss the universes where she doesn't have two children.

Anonymous Sun Mar 27 07:56:37 2011 No.2782177

>>2782165
i suggest you read the post i linked to before weighing in.

the second interpretation, which i was criticising, results in probability of 1/2. in it a family is chosen at random from all two children families, so there may not even be a girl.

Anonymous Sun Mar 27 07:59:37 2011 No.2782186
File: 39 KB, 512x512, fools.jpg [View same] [iqdb] [saucenao] [google]

ITT: people who think that statistics = odds

simple plain wrong

you idiots who say it is 1/3 to 2/3 because you "take away" the event of two boys and argue that the ratio is the chance for the respective result

it is not

the chance for the sex of the other child is not in any way affected by the "girl" who is set as a girl

>>	Anonymous Sun Mar 27 08:03:37 2011 No.2782200 >>2782186 problem is no single girl is "set as a girl" that's why people have trouble with this problem

>>	Anonymous Sun Mar 27 08:05:37 2011 No.2782203 >>2782186 2:1 ratio = 1/3 chance 0/10 Troll

>>	Anonymous Sun Mar 27 08:05:37 2011 No.2782204 >>2782200 there is no problem and yes at least one child is a girl, thus one child is set as a girl

Anonymous Sun Mar 27 08:06:37 2011 No.2782206

>>2782186
but there are two ways to get 1 girl and one boy, but only one way to get 2 girls. Hence we have 3 possible outcomes. The sum of the probabilities of all of the outcomes is equal to 1. Hence getting one girl and one boy happens 2/3 of the time.

>>	Anonymous Sun Mar 27 08:06:37 2011 No.2782209 >>2782203 has /b/ asked to troll? guess people still roll

>>	Anonymous Sun Mar 27 08:07:37 2011 No.2782210 >>2782204 "set as" suggest a particular child all we know is one child is a girl. not that any child is set as a girl. i know i will die on one day, but no one day is set as my day of death

>>	Anonymous Sun Mar 27 08:08:37 2011 No.2782212 >>2782204 >can't tell if retarded or trolling

Anonymous Sun Mar 27 08:09:37 2011 No.2782215

>>2782206

wrong

there is one way to get boy/girl (which is the same as girl/boy) and one way to get girl/girl

you do not want to take the odds of all possible combinations and exclude the results you don't want, in this case boy/boy as you only get the statistic ratio

last post for today, I know I got trolled btw

>>	Anonymous Sun Mar 27 08:09:37 2011 No.2782217 >>2782204 it is not known which child is set as a girl though? 1st or 2nd? so, in order of giving birth, we have three distinct ways her family could have arrived GG. GB, BG

>>	Anonymous Sun Mar 27 08:10:37 2011 No.2782219 >>2782215 see >>2782217

>>	Indexer !t1.kaRtHUs Sun Mar 27 08:11:37 2011 No.2782220 Both 2/3 and 1/2 are correct Prove me wrong.

>>	Anonymous Sun Mar 27 08:11:37 2011 No.2782221 >>2782215 i am about to start having kids, and i can tell you me and my gf think there are two ways we could have a boy and a girl 1st born a son, or second born a son

>>	Anonymous Sun Mar 27 08:11:37 2011 No.2782222 >>2782219 for ways, one did not happen being two boys

>>	Anonymous Sun Mar 27 08:12:37 2011 No.2782225 >>2782215 boy/girl and girl/boy are not the same events.

>>	Anonymous Sun Mar 27 08:13:37 2011 No.2782226 >>2782222 >for yes, four ways of getting a family with a girl in having two boys you are right not herping at all because forced transexual operation upon children counts

>>	Anonymous Sun Mar 27 08:13:37 2011 No.2782227 >I know that i'm a troll Fixed that for you. 1/3 is correct. Anything else and you are retard and troll. There is no way around it. To all: If you you don't understand this then it's your problem that you are so stupid.

>>	Anonymous Sun Mar 27 08:14:37 2011 No.2782232 >>2782221 you could have dizygotic twins

Anonymous Sun Mar 27 08:15:37 2011 No.2782234

>>2782225
>>2782221

they are in this case as you only look out for those cases with at least one girl

still probability for the cases that apply to the question are not connected with the absolut ratio of cases that can happen but actually are excluded

where are you from?

Anonymous Sun Mar 27 08:19:37 2011 No.2782244

this isn't a probability problem, it's a semantics problem.

You can look at it as probability of A given B = Probaility of A intersect B over probability of B and get 1/3

or we can instead consider the population of all the families with tw young children in the world who have at least one little girl and consider how many of those families have two little girls, and how many of them only have one litle girl.
Of course the answer is that there will be half of one and half of the other.

Anonymous Sun Mar 27 08:20:37 2011 No.2782250

>>2782234
you seem to be saying two things

first line

>one child is a girl, could be first born or second born, it is the same event anyway

when of course GB and BG (in order of birth) are distinct events

second line is

>well if we don't exclude what the question asks us to exclude then the answer becomes 1/2.

which while correct, isn't very helpful

Anonymous Sun Mar 27 08:22:37 2011 No.2782253

>>2782234
If you must go by that then GB situation is just twice as probable as GG. There's no way around the facts.

If the world was like people that suggest that 1/2 is the correct answer then we would have equal amounts of families with GG GB and BB.
This in turn means that the first child would have massive impact on the gender of the second child. As in, you already have one girl so the next child will most likelybe boy.

That is clear bullshit. The anwer is 1/3 to this question.

>>	Anonymous Sun Mar 27 08:23:37 2011 No.2782254 >>2782234 so if i threw two coins, you are saying there are only three outcomes. two heads, heads tails, and two tails, and these have equally likely probability i would very much like to gamble with you, i would take all your cash.

>>	Indexer !t1.kaRtHUs Sun Mar 27 08:24:37 2011 No.2782258 File: 87 KB, 800x600, herpderpimtrollingsci.jpg [View same] [iqdb] [saucenao] [google]

Anonymous Sun Mar 27 08:25:37 2011 No.2782259

>>2782250

it is

ratio is the absolut number

chance/odds/probability has nothing to do with the absolut cases WHENEVER you exclude some cases.

exclude one case - ratio becomes 2:1

but the chance for the event A being either girl/boy or girl/girl are the same es for the event B boy/girl or girl/girl

chance and ratio are about the same if there are no defining conditions

>>	Anonymous Sun Mar 27 08:26:37 2011 No.2782263 >>2782258 1/2, and gtfo my /sci/ also applies to everyone who answered other than 1/3 for OP's question /thread

>>	Anonymous Sun Mar 27 08:26:37 2011 No.2782266 File: 24 KB, 400x300, poker_cards.jpg [View same] [iqdb] [saucenao] [google] >>2782254 this is the answer to all the 1/2 flat earthers gambling they would soon change their odds once the money started leaving their pockets

>>	Anonymous Sun Mar 27 08:28:37 2011 No.2782274 >>2782263 >head buried in sand one day you will accept that 1/3 is a valid answer until then you must remain a mere acolyte

>>	Anonymous Sun Mar 27 08:28:37 2011 No.2782276 >>2782254 >>2782253 I'm laughing actually because it has got to be so hard to stay in your role without rofl

>>	Anonymous Sun Mar 27 08:28:37 2011 No.2782277 File: 1.09 MB, 1787x2688, 1301050180608.jpg [View same] [iqdb] [saucenao] [google] >>2782266

Anonymous Sun Mar 27 08:29:37 2011 No.2782278

>>2782259
>These two unrelated situations have the same chance
>Because of this the actual question has also the same chance
Because fish and whales swim birds also swim

What if you tried to aswer the question presented instead of going to entirely different things.

>>	Anonymous Sun Mar 27 08:29:37 2011 No.2782282 >>2782259 >must be troll as everything you are saying confirms my 1/3 answer

>>	Anonymous Sun Mar 27 08:30:37 2011 No.2782286 >>2782274 the question >>2782258 is different from OP's. read carefully.

>>	Anonymous Sun Mar 27 08:31:37 2011 No.2782289 >>2782276 Considering that your role is based on some retard biology that first born child influences the next childs gender.

>>	Anonymous Sun Mar 27 08:32:37 2011 No.2782291 >>2782286 my bad. didn't read it

>>	Anonymous Sun Mar 27 08:32:37 2011 No.2782294 trolls trolling trolls while being trolled

>>	Anonymous Sun Mar 27 08:33:37 2011 No.2782300 >>2782289 and as it does not the chance for OP to have two girls when one definatly is a girl is exactly 50%

>>	Anonymous Sun Mar 27 08:37:37 2011 No.2782322 >>2782277 what odds would you want for two tails. if i'd thrown two coins, examined them, and told you at least one was a tail? i'm offering $1.50 + your stake back, for each dollar you stake good deal?

Anonymous Sun Mar 27 08:38:37 2011 No.2782327

>>2782300
No it's not. Chance is 1/3 this can be proven with evidence. That is what we love in /sci/
Do the coin test.
Flip two coins
If TT flip again (two boys)
If both are H (HH/girlgirl) mark that result to paper
If only one is H then mark it also but to different column

Repeat hundred times

You should see the result by now.

>>	Anonymous Sun Mar 27 08:39:37 2011 No.2782334 50%, anyone saying otherwise is a massive retard

>>	Anonymous Sun Mar 27 08:39:37 2011 No.2782335 File: 115 KB, 1024x939, 1233506310735.jpg [View same] [iqdb] [saucenao] [google] >>2782327 tails are obviously girls heads are boys

>>	Anonymous Sun Mar 27 08:40:37 2011 No.2782340 >>2782334 so you would leap at the chance for these odds >>2782322

>>	Anonymous Sun Mar 27 08:40:37 2011 No.2782341 >>2782335 Well if that pleases you then use those. It doesn't chance the result.

>>	Anonymous Sun Mar 27 08:42:37 2011 No.2782353 >>2782341 heads a girls as they give head tails are boys as they have little tails infront

>>	Anonymous Sun Mar 27 08:43:37 2011 No.2782356 >>2782327 so TT (boys/boys) will never happen? it will but you exclude it and whenever at least one is girl you ask (what is the other?) course you get 2 to 1 that it is a boy, but in every individual case the chance is 1 to 1

Anonymous Sun Mar 27 08:45:37 2011 No.2782367

>>2782327
I have a better trial.
Flip 2 coins. If they are both tails re flip.
Now pretend there is a second person with you. Tell them that one of the coins is heads. Then pretend they said they think the other coin is heads. They will be right about 50% of the time.

>>	Anonymous Sun Mar 27 08:46:37 2011 No.2782371 >>2782322 no. But if you examined them told me one coin was tails, and then asked me what I thought the other coin was, I would take those odds, because I would be right 50% of the time

>>	Anonymous Sun Mar 27 08:47:37 2011 No.2782374 >>2782356 but those individual cases where there are two boys HAVE been excluded by the question but at least you are beginning to get it

>>	Anonymous Sun Mar 27 08:48:37 2011 No.2782381 >>2782371 >essentially the same bet okay i'll do it i suggest you try it for yourself though, this being the internet

Anonymous Sun Mar 27 08:49:37 2011 No.2782385

>>2782356
Did anyone talk about individuals. This whole thread is about that situation in the op. BB is dismissed because it is known that there is atleast one girl.
If one of the coins is T then othe other is H, considering than you will be marking HH in different spot. Just do the damn thing and you undestand.

>>2782367
That is because you are a massive retard. I will be right in 2/3 cases.

>>	Anonymous Sun Mar 27 08:49:37 2011 No.2782386 >>2782371 no, that is the same thing try it with two coins flip 2, whenever one is a tail record the other one it will be 1/3 of the time

>>	Anonymous Sun Mar 27 08:50:37 2011 No.2782390 >>2782381 I have done. It's not the same bet. When you mention the other coin, you are suddenly asking a question about a singular coin. Just one, not both. And as the coin flips are independant, the odds of a coin coming up tails is about 50%.

Anonymous Sun Mar 27 08:51:37 2011 No.2782393

>>2782374

not sure if trolling

the so called problem ain't problem of finding odds
problem can be made very basic if A and B are 50% likely, how likely is it that if A happens the second case is not A? it's 50%

now you don't need to care which is a girl and which is a boy, OP only asked if A, how likely A again? the same as if B how likely B again? 50/50

>>	Anonymous Sun Mar 27 08:51:37 2011 No.2782394 ITT there's a sucker born every minute no wonder casinos are rich if even /sci/can't do two flip of a coin odds 1/3 suckers

>>	Anonymous Sun Mar 27 08:51:37 2011 No.2782395 >>2782385 Did you do it? Because I've performed this trial.

Anonymous Sun Mar 27 08:54:37 2011 No.2782402

>>2782390
you haven't done it. properly

you are probably throwing one coin, seeing if a tail, then throwing the other.

if you threw two coins together and examined one tail and two tail occurrences you'd find the former occurred twice as often as the latter

HH, HT, TH, TT

>>	Anonymous Sun Mar 27 08:55:37 2011 No.2782405 >>2782394 Lol retard, how about you use simple logic ? Its 1/2.

>>	Anonymous Sun Mar 27 08:55:37 2011 No.2782411 >>2782390 herped so hard he derped

>>	Anonymous Sun Mar 27 08:56:37 2011 No.2782413 >>2782402 you're still examining two coins. Why? We aren't asking about 2 coins, we are asking about a single coin. The other coin as it were.

>>	Anonymous Sun Mar 27 08:56:37 2011 No.2782414 >>2782405 HH HT TH TT count the one tail, count the two tails

Anonymous Sun Mar 27 08:59:37 2011 No.2782422

>>2782413
your original quote here >>2782371 was
>But if you examined THEM told me ONE coin was tails
(my emphasis)

if you want to change the question halfway through a debate, then, well, good luck with that

my version, and your original version, exactly models OP's question

>>	Anonymous Sun Mar 27 09:01:37 2011 No.2782427 >>2782414 You don't even need to be throwing two coins, one is already a girl, what is the chance that the other one is as well ? the chance that the one coin flip is H (girl) is 50 fucking %.

>>	Anonymous Sun Mar 27 09:02:37 2011 No.2782431 >>2782427 don't feed him with anger

>>	Anonymous Sun Mar 27 09:03:37 2011 No.2782434 >>2782427 yes, but which one? 1st born or second born? we don't know, so we must look at all three situations GG, GB, BG (where first born is written first)

>>	Anonymous Sun Mar 27 09:05:37 2011 No.2782444 >>2782427 question doesn't say first coin/child is a tail/girl if it did it would be 50% second was a tail/girl it just says that one of them is.

Anonymous Sun Mar 27 09:06:37 2011 No.2782453

>>2782434

inb4 first/second born does not resample the order of coins tossed

the question does not state if it is first or second and it does not matter, one is a girl, what is the other? if you want to flip coins you will only have to flip one coin at all

>>	Anonymous Sun Mar 27 09:06:37 2011 No.2782455 >>2782431 u mad too at not getting it? problem is the answer to the question you have in your head IS 50% you just have the wrong question in your head and don't know it

Anonymous Sun Mar 27 09:09:37 2011 No.2782459

>>2782453
but assuming one coin is a tail and then flipping a single coin forces order on a situation that does not mention order, thus giving the incorrect result

no lie, if you took all the two children families in the word, 2/3 of them would be boy girl, and one third would be girl girl

>>	Anonymous Sun Mar 27 09:10:37 2011 No.2782462 The answer is 100%, because they are identical twins. Suck it asspies.

>>	Anonymous Sun Mar 27 09:11:37 2011 No.2782465 i'll break down the thread like this 1/3 but allowing 1/2 based on a mangled interpretation - god tier 1/2 only - you will never be smart tier not even trolling http://www.bbc.co.uk/dna/h2g2/A19142246

>>	Anonymous Sun Mar 27 09:14:37 2011 No.2782471 >i hate this generation Oh, grow up. You sound like those faggots 15 year olds who go onto ac/dc videos and say how great they are because they listen to "good music"

Anonymous Sun Mar 27 09:15:37 2011 No.2782472

>>2782459

>three children, one a girl, one a boy
>odds for the child not mentioned to be a boy/girl?

>three children, two girls
>odds for the child not mentioned to be a boy/girl?

>two children, one girl
>odds for the child not mentioned to be a boy/girl?

the question really is not about the actual ratio

however you will be right in more cases if you say that the other child is a boy, still no change in odds for the child itself.

you are affected but the child is not

>>	Anonymous Sun Mar 27 09:16:37 2011 No.2782477 >>2782471 you don't really argue about taste, do you?

>>	Anonymous Sun Mar 27 09:16:37 2011 No.2782478 49% chance, since there are 51% proportion of men on the planet.

>>	Anonymous Sun Mar 27 09:17:37 2011 No.2782479 >>2782471 u mad? >>2782472 cluthcingatstraws.com i love the smell of proving people wrong in the morning smells of victory

>>	Anonymous Sun Mar 27 09:18:37 2011 No.2782484 >>2782471 >arguing about maths is just the same as arguing about opinion

>>	Anonymous Sun Mar 27 09:19:37 2011 No.2782489 What are the odds that the OP is a faggot? 100%

Anonymous Sun Mar 27 09:51:37 2011 No.2782569

assuming every second post in this topic was made by a troll
>chance for a specific sex is 1:1

assuming I am trying to solve the problem, I am not a troll, I am not totally fucked up
>one child is a girl

how likely is the next post going to be a troll?
how likely is the post before written by a troll?

while it is more likely that both posts are not written by trolls (2:1) the chance for each post is 1:1

>what are the odds that the other child is also female?
50% percent chance for the other child

>what would you guess if you were asked?
you'd set your "bet" on a boy as it is more likely in more cases to be right

>>	Anonymous Sun Mar 27 09:57:37 2011 No.2782585 File: 45 KB, 1720x370, boygirl.png [View same] [iqdb] [saucenao] [google] http://en.wikipedia.org/wiki/Boy_or_Girl_paradox /thread

>>	Anonymous Sun Mar 27 09:59:37 2011 No.2782589 >>2782569 >50% percent >facepalm myself as the question is "the odds that the other child is also female" the answer is 1:1, quod erat demonstrandum

>>	Anonymous Sun Mar 27 10:00:37 2011 No.2782592 100%

>>	Anonymous Sun Mar 27 10:04:37 2011 No.2782603 >>2782589 1: 2 http://www.bbc.co.uk/dna/h2g2/A19142246

>>	Anonymous Sun Mar 27 10:05:37 2011 No.2782605 >>2782569 >50% percent chance for the other child >you'd set your "bet" on a boy as it is more likely in more cases to be right walt whitman salutes your herpitude

>>	Anonymous Sun Mar 27 10:07:37 2011 No.2782610 50%. http://en.wikipedia.org/wiki/Gambler%27s_fallacy

>>	Anonymous Sun Mar 27 10:18:37 2011 No.2782642 >>2782605 the question in the article is "how likely is one child a boy or a girl?" repeat odds for one of two children being a boy or girl or whatever the chance for the child itself are 50/50

>>	Anonymous Sun Mar 27 10:19:37 2011 No.2782645 For the retards, this question is exactly the same as OPs; Question: A woman has 2 children. Both are not boys. What is the probability that both are girls?

>>	Anonymous Sun Mar 27 10:21:37 2011 No.2782650 >>2782610 not actually the case here it is genuinely 1/3 binomial distribution BB = 1/4 BG (either order) 1/2 GG = 1/4 as BB is excluded we have (1/4)/(3/4) = 1/3

Anonymous Sun Mar 27 10:24:37 2011 No.2782659

my computer is fucked, im borrowing my brother's computer. When i try to turn it on it comes to the "chose how to start up" screen, then i chose one of em(tried em all). when i chose one, it comes a black screen and a short beep, and it restarts =[ any ideas how to fix it? it is a windows xp.

Anonymous Sun Mar 27 10:25:37 2011 No.2782665

>>2782645
not the same question

the question is "What are the odds that the other child is also female?"

answer lies within the word "also" which can be interpreted as "both at the same time" or as "the chance for a child born before or after the girl"

>>	Anonymous Sun Mar 27 10:27:37 2011 No.2782672 >>2782665 not the guy you are responding to but if you can't see they are the same question i would leave /sci/

>>	Anonymous Sun Mar 27 10:29:37 2011 No.2782675 >>2782659 >>2782659 >>>/g/

>>	Anonymous Sun Mar 27 10:29:37 2011 No.2782676 >>2782665 both are not boys = at least one child is a girl the other child is a girl, given one child is a girl = both children are girls so equivalent, and you are seriously retarded

>>	Anonymous Sun Mar 27 10:31:37 2011 No.2782683 258 posts and 14 image replies omitted. Click Reply to view. I am wondering why do I come here.

Anonymous Sun Mar 27 10:33:37 2011 No.2782692

>>2782672

>both are not boys
>there is no boy, therefore two girls
or
>there can only be one boy

this is not the same question as "one is a girl..." stupid

the problem relies in the chosen words which can indicate several answers that do not exclude each other based on the way you read the question.

but I've explained that

>>	Anonymous Sun Mar 27 10:34:37 2011 No.2782699 its ambiguous (either 1/3 or 1/2 depending on how you think about it) and OP trolled everyoner

>>	Anonymous Sun Mar 27 10:40:37 2011 No.2782723 >>2782699 /no it's an obvious 1/3.

>>	Anonymous Sun Mar 27 11:41:37 2011 No.2782892 Having already had a girl, the historical family birth records of the population at large would imply that the woman will have a slightly greater probability of having another girl.

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