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/sci/ - Science & Math


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11333459 No.11333459 [Reply] [Original]

This literally makes no sense

>> No.11333466

>>11333459
1/2 for both

>> No.11333492

>>11333459

Nigga, there's only two kids and the question assumes either one will answer the door.

What else could it be but 50% chance?

>> No.11333502

>>11333459
1/2
1/3

>> No.11333518

>>11333459
The difference is that older boy, younger girl is a possibility in the first case but not the second case.

>> No.11333526

>>11333459
For (ii), we have that the kids are either (g,b) or (g,g) with 1/2 probability. Since a girl answered the door, we have three possibilities, all with equal probability. So the odds are 1/3.

>> No.11333820

>>11333526
>>11333502
>retarded brainlet thinks that sex is influenced by age
lol'd. At what age did you become a girl, anon?

>> No.11333878

>>11333459
Four possibilities, all basically equally likely, for (oldest,youngest)
BB
BG
GB
GG

P(GB or BG | daughter answers)
=[P(daughter answers | GB or BG)*P(GB or BG)]/[P(daughter answers | GB or BG)*P(GB or BG) + P(daughter answers | GG)*P(GG) + P(daughter answers | BB)*P(BB)]
=[1/2 * 1/2] / [1/2*1/2 + 1*1/4 + 0*1/4]
=[1/4]/[1/4+1/4]
=[1/4]/[1/2]
=1/2

If a daughter answers the door, then "B,B" is not correct. Therefore three are three equally likely remaining possibilities: BG, GB, and GG. So the probability in (i) is 2/3.

If you know that the eldest of the two children is a daughter, then the two equally likely possibilities are GB and GG.

P(GB | daughter answers)
=[P(daughter answers | GB) * P(GB)] / [P(daughter answers | GB)*P(GB) + P(daughter answers | GG)*P(GG)]
=[1/2 * 1/2] / [1/2*1/2 + 1*1/2]
=[1/4]/[3/4]
=[1/3]

In sum,
i) 1/2
ii) 1/3

>> No.11333913

>>11333878
>If you know that the eldest of the two children is a daughter, then the two equally likely possibilities are GB and GG.
That means 1/2, not 1/3.

>> No.11333926

>>11333878
>So the probability in (i) is 2/3.
>If you know that the eldest of the two children is a daughter, then the two equally likely possibilities are GB and GG.
>In sum,
>i) 1/2
>ii) 1/3

wat

>> No.11333935
File: 276 KB, 1066x600, d1a.png [View same] [iqdb] [saucenao] [google]
11333935

>>11333459
https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

>>11333492
>>11333820
>>11333913
subhumans

>> No.11333938

>>11333935
>literal inbred literally didn't read his own links
https://www.sciencedirect.com/science/article/abs/pii/001002778290021X?via%3Dihub
Correct answer is 1/2.

>> No.11333946

>>11333878
BG and GB are the same thing

>> No.11333953

>>11333946
Not for birth order which is relevant to the second question.

>> No.11333958

>>11333878
am i the only one for whom the bayesian theorem is completely unreadable when applied to problems?
I can easily follow the reasoning myself and end up at 1/3 but reading this bunch of random letters has always been unreadable to me

>> No.11333961

>>11333953
It's not relevant to the outcome. It just turns the question into "a boy or girl could exist. what are the odds of either?" in which case it's still a 50/50

>> No.11334076

>there are people here who can't see why it's 1/2 and 1/3
What are you even doing on a science board if you can't understand basic probability? You clearly can't have a real understanding of anything else discussed here.

>> No.11334078

>>11334076
See >>11333938
>>11333935
inbred

>> No.11334086

>>11334078
You got it backwards retard

>> No.11334088

>>11334086
It literally says the answer is 1/2 and 1/2. It's even in peer-reviewed format, you know, the format that someone on /sci/ should prefer?

>> No.11334117

>>11334088
Not the same problems.

>> No.11334122

>>11334117
Literally the same problem.

>> No.11334126
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11334126

>>11333459
It is 50%, either the other child is a boy or it is not.

>> No.11334125

>>11334122
Retard

>> No.11334129

>>11334125
Inbred.

>> No.11334212

>>11334129
You are wrong and retarded. In the problem in the paper, we know that the child we have encountered is the eldest. In the OP problem, we don't know that. It changes the answer.

>> No.11334216

>>11334212
... are you OK, retard? Forgot to take your pills today or what?

>> No.11334222

>>11333459
Grab all the nouns and count them.

>> No.11334310
File: 77 KB, 1300x867, happyasianman.jpg [View same] [iqdb] [saucenao] [google]
11334310

>>11333459
Sample space: MM MF FM FF
M = male, F = female
(i) The only options are FM or FF since we know the first child is a female. So only 1 of 2 options fit the criteria. 50%
(ii) This cuts out MM from the sample space. So the new sample space is MF FM FF. Only 2 of 3 fit the criteria. 66%

>> No.11334363

>>11334310
You mong, if a daughter answers the door how can MM be an option?

>> No.11334393

>>11334216
>no argument
Did you finally realize you're wrong but are too pussy to admit ti?

>> No.11334430

>>11334363
Shit nigga, you have negative IQ don't you?

>> No.11334461

>>11334310
why are you treating FM and MF as different situations

>> No.11334524

>two sentences to ask what sex the child would be if there are two children in a household

why do academics to this

>> No.11334554

>>11334310
>This cuts out MM from the sample space. So the new sample space is MF FM FF. Only 2 of 3 fit the criteria. 66%

>Now **WHEN A DAUGHTER ANSWERS THE DOOR** what is the probability that the other child is a boy.

You want the probability of the non-answer being male given that the answer is female and you know at least one female exists.

>> No.11334564

the chance is actually 51% for a boy and 49% for a girl or somesuch

>> No.11334571

>>11333953
Birth order isn't relevant to this question, it's just window dressing. 'Eldest daughter' should be interpreted as 'at least one daughter exists'.

>> No.11334576

>>11333459
Ifs obviously 1/2 for both cases, but ill prove it.

Sample space for first case:
As a daughter answered the door then [math]E = \{ F,M \}[/math] as [math]\mathcal{N} (E) = 2[/math] and youre only picking one element, [math]p = 1/2[/math]

The sample space for the second case is exactly the same as for the first, as the daughter answered the door, sure, now you know that the one that answered the door is the eldest, but that doesnt change the probability of the second child having either sex, so the probability for both cases is 1/2
[eqn]Q.E.D[/eqn]

>> No.11334595

>>11333878
wrong, in the second case you know the eldest is a girl, so that leaves only BG and GG

>> No.11334672

>>11334524

Thirsty pedophiles

>> No.11335265

>>11334576
>now you know that the one that answered the door is the eldest
You don't know this. It's only 1/2 if you do.

>> No.11335280

>>11334310
See >>11333946
MF and FM in this scenario are the same thing

>> No.11335388

It's 1/2 geometrically, but 1/3 linguistically. It's not a paradox, just people unable to separate geometry and linguistics as per usual. This is pretty much the root of every problem with the human species.

>> No.11335402

i) is 2/3
ii) I don’t know but not 1/2

First you are informed that they have two children and no info is given of sex. there is a 50/50 per child and there are two unknown children then it’s BB BG GB GG at 1/4 odds of each. GB AND BG are the same but this is unnecessary to point out so I’ll consider each 1/4.

At a later moment you discover one is a girl. This means that there could only exist GB BG or GG. So the odds of the other child being a boy are 2/3 and girl 1/3.

I don’t know about the second one. I feel like 1/2 isn’t right because there is a 1/2 chance of each answering, so for it to be 1/2 you would have to know for certain that the youngest is a boy. I feel like 1/4 may be correct but I don’t know why. Like there is a 50/50 that either (G) or (G/B) answers, clearly B can’t be 1/2 because there are two independent chances here.

I read the wiki after thinking about it but I don’t believe the second is the same question as the boy girl paradox.

>> No.11335423

>>11335402
Me here, nvm it says when a daughter answers, so it is 1/2.

i) 2/3
ii) 1/2

This is correct

>> No.11335439

>>11335402
> it’s BB BG GB GG at 1/4 odds of each. GB AND BG are the same but this is unnecessary to point out so I’ll consider each 1/4.
Then how did you get 2/3? 2/4 of the choices are boy, 2/4 of the choices are girl. It's 50-50, 1/2.
> for it to be 1/2 you would have to know for certain that the youngest is a boy.
The age doesn't actually matter to the question, or even giving the gender of one the children. If the daughter answers, then the other child is still either a girl or a boy. 1/2.

>> No.11335449

>>11335439
>literally and unironically using the "50%, it either happens or it doesn't" meme

>> No.11335460

>>11335449
There's two kids. One of them is a girl. What's the other one? You already know what one of them is so just take that out of the equation. The question can be reformatted as "there is a child. What are the odds that it's a boy?"

It's one of two options. 50-50 shot of boy or girl.

The second question is the same question with an extra unnecessary detail that doesn't have to do with the answer it's asking for.

>> No.11335472

>>11335460
Hint: Define your sample space carefully!

>> No.11335478

>>11335472
You've been duped into using a formula that doesn't work and is illogical

>> No.11335481

>>11335439
The second one I reread and understand it’s 1/2.

The first there is a universe of four possible combinations of sexes the kids could have. When a girl answers the number of possible combinations of the children collapses to 3 from 4. That’s why the denominator is 3. GB BG GG. G answers and then the question is answer what are the odds that the other child is a boy. So of the 3 possible choices in the universe you check to see how many the other is a boy. GB AND BG. So 2/3. Sorry if my terms aren’t right I don’t know the stats well.

>> No.11335489

>>11335481
>The first there is a universe of four possible combinations of sexes the kids could have.
3

>> No.11335502

>>11335481
It already gives you one of the genders and the question really is about the second child. "What is the probability that one of the children is a boy?" In that case, it's 1/2. You don't need the full scope of possibilities of both children because they're not both part of the question

>> No.11335505

>>11335489
You have two kids, the first can be a boy or girl and the second can be a boy or girl. Four possible different choices. It is not we wait to see if both are boys, both are girls, or one is a girl and one is a boy but it can’t be determined which is which. U best be trolling.

>> No.11335511

>>11335502
I explained that when you make assumptions matters. Can’t help any further.

>> No.11335516

>>11335423
The first one says a daughter answers too.

Question 1 and Question 2 are the same question. You can't get different answers for both of them.

>> No.11335520

>>11333459
In case one, there are 4 equally likely possibilities:
1) Older girl, younger girl. Older girl answers door.
2) Older girl, younger girl. Younger girl answers door.
3) Older girl, younger boy. Older girl answers door.
4) Older boy, younger girl. Younger girl answers door.
In 2 out of 4 of these scenarios the other child is a boy, so the answer to (i) is 1/2.

In case two, there are 3 equally likely possibilities. This is not the same as in the first scenario because we know that the eldest child is a girl, eliminating possibility 4. This leaves:
1) Older girl, younger girl. Older girl answers door.
2) Older girl, younger girl. Younger girl answers door.
3) Older girl, younger boy. Older girl answers door.
In 1 out of 3 of these scenarios the other child is a boy, so the answer to (ii) is 1/3.

>> No.11335523

ii) is equivalent to i) due to the fact that you can't know if the child that answered the door is the older one.

>> No.11335525

>>11335516
Yes you can, because the different sexes are independent events. You cannot combine BG and GB. you are not understanding the premise and answering a different question.

>> No.11335529
File: 92 KB, 1300x1300, 2CAD497C-DCAA-4E4B-8359-1D2D2D32F5CA.jpg [View same] [iqdb] [saucenao] [google]
11335529

>>11333459
Its 50 50 buuuutttt the girl is older so shes more likely to be older enough to answer the door over the boy

>> No.11335542

>>11335520
The child who answers the door isn't even part of the equation

>> No.11335555

>>11335516
I’m replying again to explain further. The first and second question are not the same. In the first you do not know the girl will answer until after the universe is established. In the second question it specifically tells you that you need to reestablish the universe you’re starting from. The premise is part of the questions and the second question changes the premise. They are only the same question if you use some lossy compression to rewrite the question like you’re doing.

>> No.11335566

>>11335542
It has to be, because it's part of the problem statement that a daughter answers the door.

>> No.11335571

>>11335566
That bit leads up the question, which is asking the probability that the second child is a boy. You already know the first one, so you're not thinking of two children anymore. You're thinking of one child, and what the odds of it being a boy are.

>> No.11335580

>>11335571
The odds of the second child being a boy depends on the gender of the child that answered the door.. If the child that answers the door is a boy in both questions, would you say by your logic that the answer is still 1/2 in both cases?

>> No.11335590

>>11335580
>The odds of the second child being a boy depends on the gender of the child that answered the door
No
>If the child that answers the door is a boy in both questions, would you say by your logic that the answer is still 1/2 in both cases?
Yes

>> No.11335600

>Suppose you know before arriving that the eldest of the two children is a girl. Now, when a son answers the door, what is the probability that the other child is a boy?
You are saying the answer to this is 1/2?

>> No.11335615

>>11335600
In that case, it's 0%, both genders are revealed already and the second child is a girl

>> No.11335625

>>11335615
So if in that case the gender of the child that opens the door reveals information that influences the answer, why do you think this isn't the case in the original question?

>> No.11335631

>>11333459
Poorly written question, in a mathematical field that takes poorly wording for granted

>> No.11335636

>>11335625
Because in what you just described, there's nothing to guess. Both children are revealed. "There is a boy and a girl. Is there a boy?" Yes, the question itself gave the answer. In the original question, you don't know if the second child is a boy or not, and that's what you need to work out. There's two children. One is a girl. Okay, sorted.

Now there's one child. What are the odds he's a boy? If there's two options, boy or girl, then it's 50% chance of boy

>> No.11335638
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11335638

I find this whole question highly problematic, how is this question going to make non-binary students feel?
This teacher should lose his/her/xer/their job.

>> No.11335738

1) 1/4. Since statistically you have a 1/2 chance of bearing a girl each time, so two would be 1/22
2) 3/4 inverse of the above.

>> No.11336131

If fat, both children are neither and shot on the spot.
The question is whether with sanctioned weapons.
However, if said children identify as weapons themselves they may be blanks or loaded with hollow jackets. Then it's a matter. of whether these are male or female jackets.
Unless they identify as children. The jackets, that is.
If the age-binary so-called children identity as jackets then they are unmistakably Republicans.

>> No.11336160

>>11334461
Due to the introduction of older/elder

>> No.11336245
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11336245

I like to think I have sortof a good grip on stuff like this but I am brainfarting hard right now

There are the 4 possible outcomes GG BG GB BB
and you are in a situation of 1 of them, and the one of the two who answered the door is clearly G
and my initial thought is that in two of the cases you are in BG or GB, and in one of the cases you are in GG
but is it really only one of the cases? do you not have to account for that you could have met either the first or the second of GG?

and if you are given 10 of each of GG,BG,GB then you would have to discard half of BG GB since you meet B half the time so you are left with 10 GG and 5 BG and 5 GB
so with this thinking then half the time you are in GG and half the time in BG/GB


while in the second case you discard all of GB so that you are left with 10 GG and 5 BG and the other will be B 1/3 of the time?

jeeze

>> No.11336266 [DELETED] 

>>11336245
Yes, 4 possibilities: GG, BG, GB, and BB. (notation: eldest youngest. so BG means boy is the eldest and girl is youngest)
i) a girl answers, so BB can't be possible. then GG, BG, and GB are left. the question is "what is the probability that the other will be a boy?". 2 out of 3 are boys in GG, BG, and GB, so 2/3.
ii) the eldest is a girl, so there are only two possibilities: GG and GB. the question is "what is the probability that the other is a boy?". 1 out of 2 are boys in GG and GB, so 1/2.

>> No.11336272

>>11336245
Yes, 4 possibilities: GG, BG, GB, and BB.
(notation: eldest youngest. so BG means boy is the eldest and girl is youngest)

i) a girl answers, so BB can't be possible.
then GG, BG, and GB are left.
the question is "what is the probability that the other will be a boy?".
2 out of 3 are boys in GG, BG, and GB, so 2/3.

ii) the eldest is a girl, so there are only two possibilities: GG and GB.
a girl answers, so GG and GB are both still possible.
the question is "what is the probability that the other is a boy?".
1 out of 2 are boys in GG and GB, so 1/2.

maybe I'm wrong though

>> No.11336278

>>11336272
>2 out of 3 are boys in GG, BG, and GB, so 2/3.
but in reiterated visits you would always accept GG situations while BG and GB would be rejected half the time so that you are left with
more of GG than of BG
and more of GG than of GB

>> No.11336345

>>11336278
you're not accepting/selecting any situations. the question doesn't say that you will leave if a boy answers. if the question was "you're going by a 1000 houses and you will only stay if a daughter answers, what then is the probability that the other child is a boy?", then you'd be right.

>> No.11336364

>>11336345
>if the question was "you're going by a 1000 houses and you will only stay if a daughter answers, what then is the probability that the other child is a boy?", then you'd be right.
is not that practically what the question is?
in "sampling" for outcomes to use in "estimating" the probability in question you certainly have to reject all the houses where a boy answers, leaving you with a larger proportion of GG houses than BG or GB houses

>> No.11336370

>>11333459
Surely the answer to each is 50%

Although if you wanted to be super technical, the ratio of males to females in nature is not exactly 1:1. It's estimated to be 1.05 - 1.06 to 1.

In which case, the chance of a boy would be 51.2 - 51.5%, according to my calculations.

>> No.11336386

>>11336364
You're confusing where the events that determine the probabilities take place.
The actual events are the birth of the children. The chances of getting a boy or girl are the same, so for 2 children there are 4 equally likely possibilities: GG, BG, GB, and BB.
You can't say anything about rejecting a house, because that implies that the probabilities are affected by something else besides the birth of the children: namely the possibility that you can reject/leave a house.

I understand where you're coming from, but I don't know if I can explain it well enough to clear the confusion.

>> No.11336395

>>11336364
No, in terms of sampling, you are given 1000 houses which are guaranteed to have a girl answer the door, you are never rejecting houses. That's the "ambiguity" that rises the the "debate" of 1/2 vs 1/3 (both are valid answer but require slightly different interpretation).

>> No.11336411

>>11336395
>No, in terms of sampling, you are given 1000 houses which are guaranteed to have a girl answer the door,
I know that, but do you know for certain the proportion of houses of each type is the same, or does GG make up twice as large a proportion as the two others?
>>11336395
>you are never rejecting houses.
I mean, you want 1000 houses so you rejection sample houses until you only have houses where G answers
you reject all the B houses and pretend they never happened

>> No.11336453

>>11333878
But GB and BG are the same

>> No.11336477

>>11333935
This is not the BG paradox since both questions ask about "the other child" instead of the pair.

>> No.11336531

>>11336160
elder DAUGHTER

>> No.11336639

>>11333459
ITT: People misreading the question

You're looking at the probability of one child being a boy or girl. In both questions, the 1st child being a girl or being older is no more important than the detail of the parents existing. The question begins where it asks what the probability of the child that didn't answer the door is. It can either be a boy or a girl.

>> No.11336699

Brainlets, everyone of you. The answer is clearly 1/58.

Agender
Androgyne
Androgynous
Bigender
Cis
Cisgender
Cis Female
Cis Male
Cis Man
Cis Woman
Cisgender Female
Cisgender Male
Cisgender Man
Cisgender Woman
Female to Male
FTM
Gender Fluid
Gender Nonconforming
Gender Questioning
Gender Variant
Genderqueer
Intersex
Male to Female
MTF
Neither
Neutrois
Non-binary
Other
Pangender
Trans
Trans*
Trans Female
Trans* Female
Trans Male
Trans* Male
Trans Man
Trans* Man
Trans Person
Trans* Person
Trans Woman
Trans* Woman
Transfeminine
Transgender
Transgender Female
Transgender Male
Transgender Man
Transgender Person
Transgender Woman
Transmasculine
Transsexual
Transsexual Female
Transsexual Male
Transsexual Man
Transsexual Person
Transsexual Woman
Two-Spirit

>> No.11336725

it's actually 51% or 52%

>> No.11336749

>>11336639
>In both questions, the 1st child being a girl or being older is no more important than the detail of the parents existing.
Wrong. In the second question there are two equally likely possible states of the children, ordered by age:

GG
GB

1 out of 3 times a daughter answers the door, the other child will be a boy.

In the first question there are three equally likely states, ordered by age:

GG
GB
BG
BB

2 out of 4 times a daughter answers the door, the other child will be a boy.

>> No.11336760

>>11336749
>In the second question there are two equally likely possible states of the children, ordered by age:

>GG
>GB

>1 out of 3 times a daughter answers the door, the other child will be a boy.

i am trying to understand how the fuck you reached your conclusion here, but i just can't

is this a troll

>> No.11336764

>>11336749
>GG
>GB
>1 out of 3 times a daughter answers the door
2 out of 2 times a daughter answers the door. What you're trying to find is the one that didn't answer the door. 1 option is girl, 1 option is boy.

>> No.11336826

>>11336760
>elder daughter with a sister opens the door -> other child is a girl

>younger daughter with a sister opens the door -> other child is a girl

>>elder daughter with a brother opens the door -> other child is a boy

>>11336764
>2 out of 2 times a daughter answers the door.
In what context?

>What you're trying to find is the one that didn't answer the door.
Yeah, what don't you understand about "1 out of 3 times a daughter answers the door, the other child will be a boy?"

>1 option is girl, 1 option is boy.
Yes, but they're not equally likely if you know a specific child is a girl and then see a girl opening the door. The other child maybe the elder, meaning it's automatically a girl. Or it may be the younger meaning it's a boy or girl. So overall it's 1/3.

>> No.11336830

>>11336826
why are you counting "elder daughter who has younger sister" and "elder daughter who has younger brother" as two separate entities? it's just an elder daughter, she's only the elder daughter no matter what her sibling is

>> No.11336838

>>11336826
The question starts at the second child. Don't even think about the first child, it's the most basic logic trap they could have possibly included. The second child can be one of two genders. Boy or girl. Nothing else factors into this besides the most basic probability. Whatever you're plotting, you're entering the wrong information into the wrong theorem.

>> No.11336859

>>11336838
A family has 10 children, of which 9 girls and 1 boy. You ring the bell. A girl answers. Her sibling hides behind her. What is the probability the sibling is a boy?
Using your logic, it's 50%. After all, it could either be a boy or a girl.

>> No.11336864

>>11336859
1 child out of 1 [child you don't know the sex of] could be either a boy or girl

1 child out of [8 girls and 1 boy] is much likelier to be a girl

>> No.11336867

>>11336864
there's only one child hiding behind the girl that opened, not 9:
1 child out 1 [child you don't know the sex of] could be either a boy or girl.

>> No.11336868

>>11336859
In this case there's 2 children total and one of them is given to you. Somehow you're coming up with 2/3 as an answer, imaging extra things about the question that aren't pertinent, and now creating completely different scenarios as example that don't even relate.

>> No.11336870

>>11336867
yes, but that one child is a child randomly selected from a pool of [8 girls and 1 boy]

in the first case it's one child randomly selected from a pool of [1 child of unknown sex]

>> No.11336873

>>11336868
I'm not the anon you originally responded to.
I'm just trying to show how you're conflating situations.

>>11336870
it's not randomly selected from a pool of [1 child of unknown sex].
the pool is: BB, BG, GB, GG.

>> No.11336876

>>11336873
>the pool is: BB, BG, GB, GG.
no it isn't, as you already know that the eldest is a girl, thus the pool is only GB and GG

>> No.11336877

>>11336876
no, that's the second case (ii).
in the first case (i), you don't know that the eldest is a girl.

>> No.11336880

>>11336877
the first case it's simpler, the pool is just G and B

>> No.11336882

>>11336880
the first case is simpler, yes, but the initial pool is BB, BG, GB, and GG.

>> No.11336885

>>11336882
you have no reason for treating GB and BG as two separate items

>> No.11336892

>>11336882
If it were a question of both children, it could be two boys, two girls, or a girl and a boy

You've already been given that one of them is a girl. In trying to guess what the second child is, it could only be a boy or a girl

>> No.11336897

>>11336885
they're separate because they both have a 25% chance of occurring. otherwise, you'd incorrectly assume that these situations would all have a 1/3 chance of occurring: 2 boys, 2 girls, and a girl + boy. While in reality, 2 boys would be 1/4, 2 girls would be 1/4, and girl + boy would be 1/2.

>>11336892
Just because there are two options to pick from, doesn't mean they have the same rate/chance of occurring.
You can either win the lottery or don't, so there are only 2 options. But your chance of winning isn't 50%.

>> No.11336902

>>11336897
>You can either win the lottery or don't, so there are only 2 options
That would be one number out of a million. This is one child out of two.

>> No.11336916

>>11333459
case 1: 2/3
reason: after door opens, possibilities are BG, GB, GG. 2 of 3 have boys.

case 2: 1/2
reason: GG, GB. 1 of 2 has a boy.

>> No.11336917

>>11336885
So you believe if you flip a coin twice, you have a 1/3 chance of getting one heads and one tails?

>> No.11336931

>>11336897
But there aren't 2 boys or two girls. There are 2 children of indeterminate sex. Once you determine the sex of one, that has no bearing on the other. Both have a 50% chance to be either sex.

>> No.11336935

>>11336917
no, and the post you responded to is the reason why i think it's 1/2

it's either GB or GG, can't be anything else (GB=BG)

>> No.11336937

>>11336902
You're ignoring where the actual distribution of probabilities take place: birth. (with the lottery you don't ignore where the probabilities are distributed: the amount of different numbers)

You can see for yourself that you're wrong by either doing some programming or flipping a coin a bunch of times:
Flip a coin a bunch of times. On a little piece of paper, write boy for heads, and girl for tails.
Put all the pieces of paper in a bowl and shake/shuffle them.
Take out 2 pieces of paper and place them somewhere. Repeat this until the bowl is empty and you only have pairs of children.
Pick any pair. If one of them is a girl, write an X if the other is a boy and a O otherwise.
When you've gone through all pairs, look at how many X's and O's you have.

>> No.11336941

>>11336830
>why are you counting "elder daughter who has younger sister" and "elder daughter who has younger brother" as two separate entities?
Because they are?

>it's just an elder daughter, she's only the elder daughter no matter what her sibling is
A coin that flips heads and a coin that flips tails can be the same coin, but they are separate events with separate probabilities. What is your point?

>> No.11336942

>>11336931
see for yourself: >>11336937

>> No.11336945

>>11336942
>>11336937
Are you saying you expect a 1/3 2/3 distribution on a fucking coin toss? Get a proper coin lmao.

>> No.11336946

>>11336838
>The question starts at the second child. Don't even think about the first child, it's the most basic logic trap they could have possibly included.
Why? Including that information changes the answer as I showed, so why should it be ignored? You're confusing your faulty intuition for how to answer the question with an argument for how to answer the question. I've already proven it's relevant information, so nothing you're saying is relevant.

>> No.11336956

>>11336699
ok this is based
i can’t believe such a bigoted question is allowed to exist in the current year (2020).

transphobic shitlords

>> No.11336961

>>11336937
>You're ignoring where the actual distribution of probabilities take place: birth.
It makes no difference if it's birth or the opening of the door. You're given that the first one that opens the door is a girl, so you no longer have to guess what the odds of both children together. That one is out of the equation. What are the odds that the single unknown child is a boy, versus the one other option of girl? 50-50. What are the odds that a coin flipped will land on heads? 50-50. It's a one coin flip question, not a two flip question.

>> No.11336968

>>11336946
The information doesn't change the answer, you're trying to answer a question that wasn't asked. You're falling for logical traps that are plain and obvious and claiming that not intentionally being wrong is faulty intuition. You have to be clever.

>> No.11336997

reminds me of the monty hall problem, but that one has an unintuitive solution because a guy who knows the answers fucks with it while you're choosing

this is imo unambiguously 50/50 for both questions

>> No.11337007

>>11335388
elaborate

>> No.11337034

>>11336961
you're incorrectly assuming that the information of "a girl answers" doesn't change the sample space / pool of options out of which you're randomly picking.

do what I wrote about flipping a coin and writing it on paper. you'll see for yourself.

>> No.11337036

>>11336945
no, idiot. actually read what I wrote.

>> No.11337038

>>11337036
You wrote, in response to me saying it's 1/2, to go do a million coin flips like a monkey.

Go fuck yourself.

>> No.11337043

>>11337038
>your reasoning is faulty, and you can see it for yourself by trying it out in real life
>REEEEEEEEEEE

>> No.11337044

>>11337034
Here's my version of it.

Take a piece of paper, write an X for girl, and leave that paper outside the bowl

Now put a piece of paper with an X and a piece of paper with an O in the bowl. Tell me what the odds are that you'll pull out the piece of paper with an O on it

>> No.11337048

>>11337043
My reasoning is obviously correct, since coin flips are 50/50.

Go flip a coin until you see the distribution. I recommend a minimum of 200 flips, but more also works.

>> No.11337050

>>11337044
that has nothing to do with the question in the OP.

>> No.11337054

>>11337048
yeah, no shit. I didn't say coin flips are 1/3 or 2/3. Coin flips are 50/50%, so they can be used to come up with 100 random pairs of 2 kids.

>> No.11337057

>>11337050
That IS the question in the OP. One of the children has nothing to do with chance, it's pre-determined. You're determining the probability of one child, now in isolation.

>> No.11337058

>>11337054
Exactly. Get to flipping.

>> No.11337063

>>11337057
Wrong. You're changing the question to something more simple, but you're losing information. The extra information is that a girl answers the door, which means that the sample space / pool of possibilities is reduced from "BB, BG, GB, GG" to "BG, GB, GG".

>> No.11337065

>>11337058
After doing the steps I outlined, there won't be a 50/50 distribution of X's and O's. You're claiming there will be, so do it yourself and how you're wrong.

>> No.11337068

>>11337063
And I'm saying that the question IS simple and you're complicating it to the point where you're getting the answer wrong.

>> No.11337088

>>11337065
Because your steps are wrong. You're treating BG and GB as if they're separate occerances. They're not. Whether the girl who answers the door is older or younger than her sibling has no bearing on that sibling's sex.

All of the possible sex combinations for two random children are BB, BG, and GG. BB is automatically discounted, because you know at least one of them is a girl. Therefore the possible combinations that are left are BG and GG. Therefore the probabiltiy of the other child being male is 1/2.

>> No.11337125

>>11337088
No, the possible combinations that are left are GB, BG, and GG. So the probability is 2/3.

If you flip a coin twice, you have 4 possibilities: HH, TH, HT, TT. (H = heads, T = tails)
You're assuming that TH and HT are the same, so that you only have HH, TH, and TT.
Then you say that if you discount all HH, there's a 50% chance it's TH.
Go try that out in practice and see how that theory works out for you.

>> No.11337131

>>11337125
You're not flipping a coin twice. You're only flipping it once. The first is already a G, you don't flip that one.

>> No.11337141

>>11337131
the couple has 2 children, so 2 coins have been flipped.

>> No.11337143

>>11337068
If it is so simple, why are you so afraid of testing it in the real world? Oh wait, because you don't want to deal with the possibility of being wrong.

>> No.11337153

>>11337141
Not from your perspective. You are determining what the other child is while already knowing that the first is a girl. Therefore it's 50/50, you're only determining the sex of one child.

>> No.11337164

>>11337143
nobody is saying that performing the procedure you're suggesting wouldn't bring us to the same conclusion as you

we are telling you that the procedure you're suggesting is wrong, not that your result itself is wrong (though it obviously is as a result of the result of a faulty procedure, but that's secondary)

>> No.11337169

>>11337153
so if you have 1000 families with 2 children, from the families that have at least 1 girl, in 50% of the cases the other is a boy and in the other 50% the other is a girl?

>> No.11337171

>>11337164
it clearly seemed like you were implying that performing the procedure would give a different answer. but good that you cleared that up.

so what would you answer to >>11337169 ?

>> No.11337180

>>11337169
what the first child is irrelevant, in human biology what the sex of a previous child was has no bearing on what the next one will be

adding an imaginary brainfart-full of dismissed households that you discarded because a boy answered the door is leading you astray

>> No.11337193

>>11337180
>what the first child is irrelevant
false. it changes the way you calculate probabilities.

>in human biology what the sex of a previous child was has no bearing on what the next one will be
true. I never denied this.

>adding an imaginary brainfart-full of dismissed households that you discarded because a boy answered the door is leading you astray
it's a girl that answered. anyway, adding up imaginary situations is how you calculate chance. you don't actually have to flip a coin 3 times to calculate that the chance to get 3 successive heads is 0.5*0.5*0.5.

>> No.11337196

>>11337193
No, you don't have 1000 families with 2 children. You have 1000 families always with 1 girl and 1 child of indeterminate sex.

>> No.11337201

>>11337196
>>11337169

>> No.11337210

>>11337193
>false. it changes the way you calculate probabilities.
Yes, if you choose to do it wrong. Stop that.

>it's a girl that answered. anyway, adding up imaginary situations is how you calculate chance. you don't actually have to flip a coin 3 times to calculate that the chance to get 3 successive heads is 0.5*0.5*0.5.
You're adding up imaginary situations that can't happen and then only discounting HALF of them. That's the problem. The first flip you're doing is not a flip. It's not "flip only take the girls, discard the boys", it's "this is always a girl, there is no first flip".

>> No.11337222

>>11337201
>>11337210
no, that would be the case if you knew you were going to a house that had at least one girl. you don't. so you're not selecting only the houses that have 1 girl and 1 unknown child.
you're selecting a house at random, and then you learn that 1 is a girl.

>> No.11337233

>>11337222
that would be the case if you started your scenario before you learned what the first child is

but upon starting, the door has been answered and you already know that one is a girl

so you just do 1 coin flip to find the sex of the other child

>> No.11337236

>>11337233
the scenario does start before you learned what the first child is.
it starts when you assume that there's a 50/50 chance a child is born male/female. then it follows that families will have a 1/4 distribution of BB, BG, GB, GG each. etc.

>> No.11337238

>>11337236
no, it starts when the question mark hits

>> No.11337246

>>11337238
¿kek?

>> No.11337248

>>11337246
no habla upside down

>> No.11337320

>>11336968
I answered the question that was asked. Given the information that the eldest child is a girl and that the child that answered the door is a girl, what is the chance the other child is a boy. The answer is that 1 out of 3 times a girl answers the door, the other child will be a younger brother. What do you think is the difference?

>> No.11337333

>>11333459
P(A) = 1/2, for all A. As Stephen Wolfram proved.

>> No.11337864

>>11333459
1. Define sample space set (2 children, ordered by answering door)
(b = boy, g - girl)
bg
bb
gb
gg
problem subspace set selection (g answers door)
gb
gg
probability space set (has a b)
gb
so one item set of probability space, from two item set of problem subspace

probability is 1/2

2. define sample space set (capital G will represent older female child))
bG
Gb
Gg
gG
problem subspace set selection (g or G answers door)
Gb
Gg
gG
probability space set (has a b)
Gb
so one item set of probability space, from three item set of problem subspace

probability is 1/3

>> No.11338198

>>11337320
The girl who answered the door isn't even part of the equation. She's not a question of probability, she's certain. What is left to probability is the second child being a boy or not, which is unrelated to the gender of the first child.

>> No.11338265

i think we can safely conclude that the question is ass for being ambiguous

>> No.11338393

>>11337864
In 1, bg and bb are never an option. It already says a daughter answers.

In 2, bG and gG aren't options, it says the elder daughter answered the door.

In both scenarios, it's 1/2

>> No.11338574

>>11338265
It's not ambiguous at all. If we assume a 1:1 natural ratio in the world for males and females, then the answer to both questions is 50%.

But if you want to be technical, the natural ratio of males to females is actually estimated to be around 1.05 to 1.06 males for every 1 female.

Source:
>https://en.wikipedia.org/wiki/Human_sex_ratio

Using these figures, the answer to each question would be 51.2% (if the ratio is 1.05:1) or 51.5% (if the ratio is 1.06:1), if my calculations are correct.

However, the question probably doesn't expect you to know this factual figure of 1.05 - 1.06. Especially since that isn't actually a figure, it's a range, since different studies have produced slightly different results.

It's reasonable to assume that the question wants you to assume that the ratio is 1:1. So then the answer to each question is 50%.

>> No.11338857

>>11338574
those ratios are for births and does not take into account that the values heavily adjust themselves due to higher child mortality
this is one of the most autistic "I am so smart" posts in this thread
to only go on to be completely wrong on every account (even the actual answer).

>> No.11338930

>>11338857
>the values heavily adjust themselves due to higher child mortality
Yeah that's fair enough.

The rest of your post is ad hominem garbage though, indicating that you're a moron (that's not ad hominem, that's an insult, which you deserve for being a fucking moron).

>> No.11339321

>>11334461
>>11336160
The introduction of elder/younger has nothing to do with it. The fact that the two children are distinguishable is what causes FM and MF to be different. The fact that you knew ahead of time that one was a girl the second time is what causes the answer to the second to be different. In the first we distinguish the children by "opened the door" and "did not", while in the second we have information on "eldest" and "youngest" but the results come from "opened the door" and "did not" just as in the first. Because "eldest"-"youngest" and "opened door"-"did not" are completely independent the statistical properties change.

Go on, try tossing a labelled pair of coins several times. For the first part, choose any distinguishing property that is independent of the labelling (this could be "landed first" "landed more on the left", etc), and partition your sample space that way. Then partition your sample space based on the labelling. you will get different statistical distributions.

>> No.11339345

First question is simple. You have 4 possibilities:

1 G at the door, G elsewhere
2 G at the door, B elsewhere
3 B at the door, G elsewhere
4 B at the door, B elsewhere

If G opens a door, only 1 and 2 are possible and are equally likely, hence 1/2

>> No.11339411

>>11339321
In both cases you only toss a single coin to determine the second child.

>> No.11339696

>>11338393
>it says the elder daughter answered the door.

No, it is says when *A* daughter opens the door.
It does NOT say when the elder daughter opens the door

>> No.11339765

>>11339696
it doesn't matter if she's older

>> No.11340662

>>11339765
lmao what a retard

>> No.11340913

>>11338393
>it says the elder daughter answered the door.
it actually doesn't....

>> No.11340925

>>11335590
So they tell you their older child is a girl and a boy answers the door what's the chance that the other child is a boy?

>> No.11341155

For the second question;
The original pool of posibilities is:
BB BG GB GG (younger-elder)
If you know the elder is a girl you are only left with:
BG
GG
So therefore probability is 1/2.
I'm still confused and not sure about this, why should being younger or older have anything to do? The question for me is always the same, you have a child what are the probabilities of it being a boy.
Please correct me if i'm wrong.

>> No.11341176

>>11341155
Can´t believe im this retarded, it does not say the elder girl opened the door.

>> No.11341180

>>11339411
*sigh*
just do the experiment, brainlet

>> No.11341182

>>11335265
The text literally says the daughter answered the door

>> No.11341245

>>11333878
What if the girl has a penis it should be BB
BG
GB
BG(B)
G(B)B

>> No.11341257

>>11341182
It could have been the youngest daughter.

>> No.11341261

>>11341257
She's trans

>> No.11341288

>>11341182
no it does not retard

>> No.11341528

>>11341180
>just do the wrong experiment :)

>> No.11342640

>>11336699
I assume * represents self inflicted termination

>> No.11343353

>>11342640
>self inflicted termination
are you trying to sound as autistic as possible

>> No.11343369

>>11336699
This