Poll: A women has two kids, one is a boy, what are the odds the other is also a boy?

[TZer0]

Also this is flawed:

'If we remove 0.25 we must distribute evenly.'

Why? You are removing it because of zero boys being in the G-G case. But why must you distribute it evenly? Answer that. 'Just because' isn't an answer.

But regardless, in your algorithm, as much as you hate to admit it, 50% of your boys have a brother, and 50% of your boys have a sister. Do the math. Really -do the math.-

Answer to the first question: this involves convergence, so if that makes your head wobble, don't read it and just accept it.

First, if we get a G-G-occurrence, we're inside the 25% which tell us to re-run the test due to a illegal occurrence. Within these 25% there'll be four intervals B-B, G-B, B-G and G-G, therefore, we must split 25% into four pieces and add those to the other probabilities (because they give us the same result), however, we might get another G-G-occurrence inside this interval. Therefore we must split that interval (which is 6.25% of the original probability) into four pieces and add three of these to the other probabilities and then split the fourth one etc. This sums up to adding 0.25/3 to every one of the other intervals as that is the limit of this sum.

I can admit that 50% of "my" boys have a brother and the other 50% have a sister. However, you forget that if your first child is a girl, she can either have a brother or a sister. If you analyze what I just wrote there, you'll have your answer. Four occurrences, one illegal, two false, one true.

Sec, I'll even draw you a probability tree.

 

[NickIsCool]

*snipping double quote*

The odds:
2 girls - 0%, because we're told.
2 boys - 33%
boy AND girl - 66%.

where does it tell you that either child came first?

... oh wait

i just realized my mistake

so you where right about your question

but the answer to the OPs question is still 50%

Both are the same question


"A woman has two kids, (atleast) one is a boy, what are the odds the other is also a boy?"

EQUALS

"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?"

The odds of getting the same uncertain thing twice in a row are lower.
Lower than the odds of getting [first A and then B OR first B and then A] combined.

wrong

they are not the same question

with the original question we KNOW that one of them is a boy

You mean to say, that you would not be able to figure out that:
(atleast) "one of them is a boy"

when you are given the following information:

"A woman has two kids, she does NOT have TWO girls"?

obviously you dont understand what im trying to get at here

so im done arguing with you about this

its 4 A.M.

im goin to sleep

 

[TZer0]

*snip*

Here you go, hope this makes sense. 0.33 with 3 recurring before anyone comments that.

 

[DracoSuave]

Also this is flawed:

'If we remove 0.25 we must distribute evenly.'

Why? You are removing it because of zero boys being in the G-G case. But why must you distribute it evenly? Answer that. 'Just because' isn't an answer.

But regardless, in your algorithm, as much as you hate to admit it, 50% of your boys have a brother, and 50% of your boys have a sister. Do the math. Really -do the math.-

Answer to the first question: this involves convergence, so if that makes your head wobble, don't read it and just accept it.

First, if we get a G-G-occurrence, we're inside the 25% which tell us to re-run the test due to a illegal occurrence. Within these 25% there'll be four intervals B-B, G-B, B-G and G-G, therefore, we must split 25% into four pieces and add those to the other probabilities (because they give us the same result), however, we might get another G-G-occurrence inside this interval. Therefore we must split that interval (which is 6.25% of the original probability) into four pieces and add three of these to the other probabilities and then split the fourth one etc. This sums up to adding 0.25/3 to every one of the other intervals as that is the limit of this sum.

I can admit that 50% of "my" boys have a brother and the other 50% have a sister. However, you forget that if your first child is a girl, she can either have a brother or a sister. If you analyze what I just wrote there, you'll have your answer. Four occurrences, one illegal, two false, one true.

The question is this.

"A women has two kids, one is a boy, what are the odds the other is also a boy?"

Examining the case of where a girl has a brother or sister is applying -that- statistic to the question of the gender of the sibling of a boy. That's a fallacy.

Your scenario counts the following outcomes faithfully:

The woman's known boy is the eldest, and the youngest is a boy. That is accounted for by B-B in your algorithm.
The woman's known boy is the eldest, and the youngest is a girl. That is accounted for by B-G in your algorithm.
The woman's known boy is the youngest, and the eldest is a girl. That is accounted for by G-B in your algorithm.

Which in your algorithm accounts for: The woman's known boy is the youngest, and the eldest is a boy?

It does not. It -excludes- an outcome, systematicly. Which means it fails probability science AND statistical analysis.

Or are you going to then say that the chances that the woman is talking about an eldest boy with a younger brother is only 12.5%?

Well, to disprove that:

B-B
B-G
G-B
G-G are the only possibilities.

Half of all scenarios have an older brother. Therefore, if I am the younger child, I am 50% likely to have an older brother. Conversely if I have am the elder, I am 50% likely to have a younger brother.

As well, brothers are equally distributed on both sides of the equation, so if I am a boy, I am equally likely to be older or younger. There is a 50% probability distribution.

Therefore, there is a 50% chance I am an older brother, and if I am older, there is a 50% chance I have a younger brother.

Therefore, I have a 25% chance of being a boy with a younger brother.

Therefore, an algorithm that would claim I have a 12.5% chance of such is fundamentally flawed.



Also, I understand how your damn algorithm works; that's how I -know- how -exactly- it is flawed.

 

[TZer0]

Which in your algorithm accounts for: The woman's known boy is the youngest, and the eldest is a boy. It does not. It -excludes- an outcome, systematicly. Which means it fails probability science AND statistical analysis.

I'm sorry, but it does exclude. It excludes G-G by re-running the test if that should happen.
Let's quote wikipedia!

Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?"

In this problem, a random family is selected. In this sample space, there are four equally probable events:Older child Younger child
Girl Girl
Girl Boy
Boy Girl (invalid option, removed with lines through this option)
Boy Boy (invalid option, removed with lines through this option)


Only two of these possible events meets the criteria specified in the question (e.g., GB, GG). Since both of the two possibilities in the new sample space {GB, GG} are equally likely, and only one of the two, GG, includes two girls, the probability that the younger child is also a girl is 1/2.
(..)
Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

This question is identical to question one, except that instead of specifying that the older child is a boy, it is specified that at least one of them is a boy. If it is assumed that this information was obtained by considering both children[15], then there are four equally probable events for a two-child family as seen in the sample space above. Three of these families meet the necessary and sufficient condition of having at least one boy. The set of possibilities (possible combinations of children that meet the given criteria) is:Older child Younger child
Girl Girl (invalid option, removed with lines through this option)
Girl Boy
Boy Girl
Boy Boy


Thus, if it is assumed that both children were considered, the answer to question 2 is 1/3.

 

[DracoSuave]

obviously you dont understand what im trying to get at here

so im done arguing with you about this

its 4 A.M.

im goin to sleep

It is the same question logically:

A or B is the same logically as NOT(NOT A AND NOT B)

 

[DracoSuave]

Which in your algorithm accounts for: The woman's known boy is the youngest, and the eldest is a boy. It does not. It -excludes- an outcome, systematicly. Which means it fails probability science AND statistical analysis.

1. We don't know if the boy is the older one or not.
2. I'm sorry, but it does exclude. It excludes G-G by re-running the test if that should happen.

It excludes the case where the boy she speaks of is the youngest and not the oldest. Yours treats this as the same outcome as where she speaks of the eldest and not the youngest. They are NOT the same outcome, and do not share the same probability space.

That is the flaw in the algorithm.

I know it excludes G-G, -and it should- but it also -systematically- excludes one of: (Elder Boy is known, Younger Boy is not) or (Younger boy is known, Elder boy is not).

And if it excludes a scenario -which meets the qualifying criteria- then it is flawed.

Hense why I say: Don't count the family. Count the -boy-.


Also, I did read the wikipedia article. Also continue to read where it says that the problem comes from naivety in determining valid outcomes. And continue to read where it says that 50% is mathematically correct, assuming of course, that the birth of one child doesn't actually effect the gender of another child or that gender distribution is equal.

I told you, I read the article. I also did not -stop- reading until the end.

 

[Clyde]

If this thread teaches anyone anything, let it be that math and language translate roughly.

 

[TZer0]

Which in your algorithm accounts for: The woman's known boy is the youngest, and the eldest is a boy. It does not. It -excludes- an outcome, systematicly. Which means it fails probability science AND statistical analysis.

1. We don't know if the boy is the older one or not.
2. I'm sorry, but it does exclude. It excludes G-G by re-running the test if that should happen.

It excludes the case where the boy she speaks of is the youngest and not the oldest. Yours treats this as the same outcome as where she speaks of the eldest and not the youngest. They are NOT the same outcome, and do not share the same probability space.

That is the flaw in the algorithm.

I know it excludes G-G, -and it should- but it also -systematically- excludes one of: (Elder Boy is known, Younger Boy is not) or (Younger boy is known, Elder boy is not).

And if it excludes a scenario -which meets the qualifying criteria- then it is flawed.

Hense why I say: Don't count the family. Count the -boy-.


Also, I did read the wikipedia article. Also continue to read where it says that the problem comes from naivety in determining valid outcomes.

I told you, I read the article. I also did not -stop- reading until the end.

Nope, you didn't understand the code. It only disqualifies a result if it hits the "no boys"-option.

Look, in this question it doesn't matter if we define one of them as younger or older, we've got two instances of child. One impossible outcome, three possible of which two false, one true. I've given you a probability tree. Isn't that enough? http://www.escapistmagazine.com/forums/read/18.157846?page=5#3928958

If you're oh-so-right, then make your own tree, your own algorithm and your own spreadsheets and prove it to me. Because to now, I've been the only one here showing the most solid evidence, nothing of which has been disproved.

Also, it isn't too late to admit fault

 

[veloper]

If this thread teaches anyone anything, let it be that math and language translate roughly.

Exactly.

Escapists are parsing the english question in different ways.

 

[veloper]

If this thread teaches anyone anything, let it be that math and language translate roughly.

Exactly.

Escapists are parsing the english question in different ways.

People got very angry at me for doing that the last time this question came up. I still don't understand why it bothers some people to point out that the words in a, you know WORD PROBLEM matter.

I think that's part of the fun.
It would be a boring question, if the wording was very careful and exact and left no room for different interpretation.
It's a tease.

 

[DracoSuave]

Look, in this question it doesn't matter if we define one of them as younger or older, we've got two instances of child. One impossible outcome, three possible of which two false, one true. I've given you a probability tree. Isn't that enough? http://www.escapistmagazine.com/forums/read/18.157846?page=5#3928958

It's enough that I can point out the obvious flaw.

You have a boy.

There are four instances of boy in the probability chart. 2 of those instances have a brother, 2 do not. Therefore, your probability distribution is 2/4, or 50%.

It -does not get simpler than that.-

Your chart only -qualifies- three instances of boy. But there are four boys in all possible two-sibling pairings. 3 != 4.

Therefore you will exclude 25% of all boys with a single sibling.

For every three instances of a boy-outcome in your distribution, you will ignore and exclude one boy-outcome. That is why your algorithm is flawed. Saying that I don't know it doesn't change this fact. Because I -do- understand it.

I can't make it any plainer than that.

However, your analysis would be correct in THIS scenario:

'A woman has a boy. She tells you her boy does not have an older brother, and that he has only one sibling. What are the chances that the other sibling is a boy or a girl?'

However, that is NOT the case, and that is why your scenario fails. This is what you are doing:

'3 out of 4 two-sibling pairings have boys. Therefore, if you have a pairing with a boy, there is a 2/3 chance the other is a girl.'

However, what you've not done is account for the fact that 50% of all boys come from a two-boy pairing. Therefore, taking a run of all families and then excluding all girls does not -actually give you a fair distribution of all pairings with boys-.

So, if you have a pairing with a boy, there is a 50% chance that boy comes from a two-boy pairing. Otherwise, it comes from a boy-girl or girl-boy pairing.

So, 50% of the time, it has a 100% chance of having a brother, and 50% of the time, it has a 0% chance of having a brother.

That adds up to 50/50.

 

[Scary_Bob]

The answer is zero, "A woman has two kids, one is a boy" Therefore the odds of the other one being a boy is zero ^_^

Really though, the question is worded to ambiguously so as to validate either 50% or 33%.
If the OP had stated that at least one was a boy then 33% would be correct... mostly.

 

[Arkhangelsk]

It isn't 33% We already have the fact that one boy is born. The chances of another one being born is a different conundrum. The chances are still 50% that it's either a boy or a girl. Or if we want to be really petty, we could try to figure out the odds for hermaphrodites and twins.

And if this question is a trick question, as in you've already said the answer through saying that she has two kids, and "one is a boy, which is the other one" (as in you've already stated a fact on how many boys she has), I will do this to you:

Spoiler: I will do this to everyone who act like that

 

[Clyde]

So why does the poll include 25% and 100%?
I'm glad few are picking them, but ...

 

[Trotgar]

The answer is 50. Many people have explained it earlier, if one is known to be a boy, that doesn't reduce the chance of the other one being a boy as well.

And if it's a trick question, see the second post above this.

 

[TZer0]

It's enough that I can point out the obvious flaw.

You have a boy.

There are four instances of boy in the probability chart. 2 of those instances have a brother, 2 do not. Therefore, your probability distribution is 2/4, or 50%.

It -does not get simpler than that.-

Your chart only -qualifies- three instances of boy. But there are four boys in all possible two-sibling pairings. 3 != 4.

Therefore you will exclude 25% of all boys with a single sibling.

For every three instances of a boy-outcome in your distribution, you will ignore and exclude one boy-outcome. That is why your algorithm is flawed. Saying that I don't know it doesn't change this fact. Because I -do- understand it.

I can't make it any plainer than that.

However, your analysis would be correct in THIS scenario:

'A woman has a boy. She tells you her boy does not have an older brother, and that he has only one sibling. What are the chances that the other sibling is a boy or a girl?'

However, that is NOT the case, and that is why your scenario fails. This is what you are doing:

'3 out of 4 two-sibling pairings have boys. Therefore, if you have a pairing with a boy, there is a 2/3 chance the other is a girl.'

However, what you've not done is account for the fact that 50% of all boys come from a two-boy pairing. Therefore, taking a run of all families and then excluding all girls does not -actually give you a fair distribution of all pairings with boys-.

So, if you have a pairing with a boy, there is a 50% chance that boy comes from a two-boy pairing. Otherwise, it comes from a boy-girl or girl-boy pairing.

So, 50% of the time, it has a 100% chance of having a brother, and 50% of the time, it has a 0% chance of having a brother.

That adds up to 50/50.

We have four instances of boy, B-B (2), B-G (1), G-B(1), sum: 4.

Fine then, we'll start on the basics. Care to tell me what the probability of getting the following is if we allow G-G and boy and girl at random?
Two boys -
One boy, one girl -
Two girls -

Also, while you're at it, care to prove wikipedia wrong as well?

http://en.wikipedia.org/wiki/Boy_or_Girl_paradox
We have this question: Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
Wikipedia gives this answer: Thus, if it is assumed that both children were considered, the answer to question 2 is 1/3.

Let me illustrate this:
You have no boys or girls. You pick two random children, get two girls? try again. If not, boy + boy = success. You're not supposed to define one of them as a boy or girl, then draw the other one.