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

[veloper]

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.

which one are you talking about? the boy...who is a boy? or the other..boy?

 

[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.

which one are you talking about? the boy...who is a boy? or the other..boy?

Sorry, I didn't really understand your question.

But my point is: can't we just forget the child who is known to be a boy?

 

[veloper]

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.

which one are you talking about? the boy...who is a boy? or the other..boy?

Sorry, I didn't really understand your question.

But my point is: can't we just forget the child who is known to be a boy?

I think we never knew him at all!

but enough silliness for now. This had been fun.

 

[DracoSuave]

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.

Thus, if it is assumed that both children were considered, the answer to question 2 is 1/3.
However, if it is assumed that the information was obtained by considering only one child, then the problem is an isomorphism of question one, and the answer is 1/2

If you are going to quote, do NOT quote out of context, please. The article, and, in fact, that statement does not stop where you quoted. Doing so is bad form, and in some circles is considered intellectually dishonest.

Please read, Ambiguous Problem Statements [http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Ambiguous_problem_statements] as well, regarding this issue.

Like I said, I -read the article-. You did not, clearly.

Also, do not appeal to authority in a mathematical debate. It shows a lack of rigor on your own part, and deconstructs your argument. Especially when said authority has an -edit- button.

But, regarding the original implication:

Two boys - 1/4
One boy, one girl - 2/4
Two girls - 1/4

And to continue:

Total number of boys: 4.

Number of boys per outcome in Two Boys: 2; chance of a boy being from each individual outcome: 50%
Number of boys per outcome in one boy, one girl: 1; chance of a boy being from each individual outcome: 25%
Number of boys per outcome in two girls: 0; chance of a boy being from each individual outcome: 0%

So: Chance of a boy being from two boys= 1 outcome, times 2 boys, equals 2 in 4.
Chance of a boy being from one boy, one girl = 2 outcomes, times 1 boy each, equals 2 in 4.
Chance of a boy being from two girls = 1 outcome, times 0 boys, equals 0 in 4.

Thusly, it is mathematically vindicated that you exclude two girls, because there are zero boys in it. It is not mathematically vindicated that you only include two boys once, because there are two boys in it.

The question here then, is the the woman selected at random, or is she not? If she is selected from a random group of women, then yes, the answer is 1/3. However, we do not know this, that is not part of the question. Therefore the only random element we have is the boy, and therefore -that- must be the standpoint of our examination; we are not dealing with a random woman, we are dealing with a random boy.

"Or are you going to argue with wikipedia?"

 

[ace_of_something]

The woman doesn't determine the gender of the child gender is passed on by the father. And I dunno, is this a riddle? If so the paramiters are weak

 

[Hamster at Dawn]

But you don't know that the first one is boy, you only know that one of them is boy. That's a key difference.

Yes, it's true that if you had a brother it would be irrelevant to the probability of your gender, but if all we knew was that you had a sibling, and at least one of you was male, but not knowing which one, that would effect the probability of your gender.

Actually that makes more sense although I still find it a bit confusing to think about. Thing is, you are trying to find the gender of the OTHER child. You already know that the one you're not trying to find the gender of is a boy. Back to the situation with my fictional brother, you don't know if my brother is older or younger than me so you could argue the same thing... or maybe I'm just being a bit retarded. I guess it's too much for me to get my head around.

 

[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.

Thus, if it is assumed that both children were considered, the answer to question 2 is 1/3.
However, if it is assumed that the information was obtained by considering only one child, then the problem is an isomorphism of question one, and the answer is 1/2

If you are going to quote, do NOT quote out of context, please. The article, and, in fact, that statement does not stop where you quoted. Doing so is bad form, and in some circles is considered intellectually dishonest.

Please read, Ambiguous Problem Statements [http://en.wikipedia.org/wiki/Boy_or_Girl_paradox#Ambiguous_problem_statements] as well, regarding this issue.

Like I said, I -read the article-. You did not, clearly.

Also, do not appeal to authority in a mathematical debate. It shows a lack of rigor on your own part, and deconstructs your argument. Especially when said authority has an -edit- button.

But, regarding the original implication:

Two boys - 1/4
One boy, one girl - 2/4
Two girls - 1/4

And to continue:

Total number of boys: 4.

Number of boys per outcome in Two Boys: 2; chance of a boy being from each individual outcome: 50%
Number of boys per outcome in one boy, one girl: 1; chance of a boy being from each individual outcome: 25%
Number of boys per outcome in two girls: 0; chance of a boy being from each individual outcome: 0%

So: Chance of a boy being from two boys= 1 outcome, times 2 boys, equals 2 in 4.
Chance of a boy being from one boy, one girl = 2 outcomes, times 1 boy each, equals 2 in 4.
Chance of a boy being from two girls = 1 outcome, times 0 boys, equals 0 in 4.

Thusly, it is mathematically vindicated that you exclude two girls, because there are zero boys in it. It is not mathematically vindicated that you only include two boys once, because there are two boys in it.

The question here then, is the the woman selected at random, or is she not? If she is selected from a random group of women, then yes, the answer is 1/3. However, we do not know this, that is not part of the question. Therefore the only random element we have is the boy, and therefore -that- must be the standpoint of our examination; we are not dealing with a random woman, we are dealing with a random boy.

"Or are you going to argue with wikipedia?"

I just used wikipedia because I know it is correct (at the moment) in this matter.

We're not dealing with an isomorphism of question 1. We're dealing with information regarding any of the two children, but we don't know which one. We're dealing with a random woman, not a random boy. The question isn't: "we have a random boy with a sibling, what is the chance that he has a brother?"
Therefore, we're dealing with this:
Two boys - 1/4
One boy, one girl - 1/2
Two girls - 1/4 - redraw
Here you have the probability-table. You can tell me what the results will be (yes, I know you did).

 

[Ace of Spades]

Make a punnet square.

 

[Ildecia]

It's variable probablity, but I'm not sure the question is phrased right,

The oldest kid is a boy?

anyway

Boy Boy is in
Boy Girl is in
Girl Girl is out
Girl Boy is in

So it's 33%

so... that still doesn't make sense... i understand the probability logic... but every birth has a 50/50 probability.

births aren't conditional probability so idk what your looking at

 

[Escapist_V]

As I see it the Boy | Girl paradox serves to point out the flaw in probability theory. It's what happens when you leave out essential information or use terms that are ambiguous, not constraining the question enough. People interpret the question differently, adding in factors like order and age in which changes the outcome. A definitiv answer can't be found because the question itself is flawed.

 

[Trotgar]

It's variable probablity, but I'm not sure the question is phrased right,

The oldest kid is a boy?

anyway

Boy Boy is in
Boy Girl is in
Girl Girl is out
Girl Boy is in

So it's 33%

so... that still doesn't make sense... i understand the probability logic... but every birth has a 50/50 probability.

births aren't conditional probability so idk what your looking at

Yeah, we ain't talking about boy boys and boy girls etc, the child who we know to be a boy isn't important. We are just talking is the one child a Boy or a Girl. The existence of the another child doesn't have anything to do with the other being a boy or a girl.

 

[teisjm]

If she already has one boy, and gets pregnant the chance of the second one turnign out to be a boy when he's born is 50 %. Theres 2 possible outcomes (not ocunting hermaphrodite baby)

Girl or
Boy

both have a 50% chance of happening (if we assume theres not some genetic thingy that makes one genders birthrate dominant)

If she has no children and plans to have two children, the chances of both beeing boys are 25%
theres four possible outcomes (not counting twins, hermaphrodites or the antichrist)

a girl and then another girl.
a girl and then a boy.
a boy and then a girl.
a boy and then anotehr boy.

Again assuming that theres no genetic predisposition to get one gender more than the otehr each of the 4 options has a 25% chance of happening, since only one is boy + boy the chance of gettign 2 boys is 25%, if you have none already.

I'm not gonna vote, cause i can't read clearly from your question which of the 2 situations you're tlaking about.

 

[TZer0]

The question also isn't "we're dealing with a random mother."

Correct. We're dealing with a random mother with at least one son. Therefore we must split up the G-G interval evenly on the other ones in order to remove it/calculate this numerically. The whole mother-thing is there to visualize the problem in another way. It could've been coins or whatever, still the same problem. Got two tails? success. Got one tail, one head? failure. Got two heads? retry.

It is all just a task with probability-spaces.

 

[Maze1125]

Now, our possible outcomes are:

The first child is the known boy, and the second child is also a boy. [BOY]
The first child is the known boy, and the second child is a girl. [NOT A BOY]
The second child is the known boy, and the first child is also a boy. [BOY]
The second child is the known boy, and the first child is a girl. [NOT A BOY]

All these outcomes are equally probable. Therefore, 2 out of 4 outcomes have the unknown child as a boy.

Except that that is a fallacious probability table.

Let FB be the event that the first child is a boy, let OB be the event that at least one of the children is a boy, etc.

P(FB|OB) = P(FB intersect OB)/P(OB) = P(FB)/P(OB) = (1/2)/(3/4) = 2/3
Therefore, necessarily, P(FG|OB) = 1/3

Conditioning on the first child being a boy:

P(SB|(FB intersect OB)) = P(SB intersect FB intersect OB)/P(FB intersect OB) = P(SB intersect FB)/P(FB) = (1/4)/(1/2) = 1/2
Therefore, P(SG|(FB intersect OB)) = 1/2
P(SB|(FG intersect OB)) = P(SB intersect FG intersect OB)/P(FG intersect OB) = P(SB without FB)/P(SB without FB) = 1
Therefore P(SG|(FG intersect OB)) = 0, but that was obvious anyway.

Therefore, the probability table goes like this:

BB 2/3 * 1/2 = 1/3
BG 2/3 * 1/2 = 1/3
GB 1/3 * 1 = 1/3
GG 1/3 * 0 = 0

It is a lazy lack of rigor, and should be rejected by that principle alone.

Failure to understand this concept = failure at discrete mathematics, ergo failure at probability science.

This is the only part of your post I agree with.

MACM 101, bitches.

I could post my credentials, but there is no way to prove I'm not making them up, sufficed to say, they're far better than a single 101 module.

Also, disproof of the 33% via ad absurdum:

A woman has two objects. One is a fish. The other is a child. What is the probability of that child being a boy or a girl?
A woman has two children. One has ADD. What is the probability of the other child being a boy or a girl?
A man has two children. One dislikes him. What is the probability of the other child being a boy or a girl?

The answer in all of those is 50%, because the gender of the second child is a completely independant event. Therefore the following is true:

A woman has two children. One has the quality X, where X is unique to that child and is not a relation to the other child. What is the probability of the other child being a boy or a girl?

The probability is 50% for all X. Therefore, substituting 'being a boy' for X means as well that it is 50%.

You have not shown that the event that one is a boy is "unique to that child and is not a relation to the other child." you're just presuming that fact. Although this does back up your claim that you're failing to be rigorous.

 

[Maze1125]

As I see it the Boy | Girl paradox serves to point out the flaw in probability theory. It's what happens when you leave out essential information or use terms that are ambiguous, not constraining the question enough. People interpret the question differently, adding in factors like order and age in which changes the outcome. A definitiv answer can't be found because the question itself is flawed.

A definitive answer can most definitely be found, just because some people like to exclude information they have or include information they don't have doesn't mean that there isn't a clear definitive answer using precisely the information that we have.

If she already has one boy, and gets pregnant the chance of the second one turnign out to be a boy when he's born is 50 %. Theres 2 possible outcomes (not ocunting hermaphrodite baby)

Girl or
Boy

both have a 50% chance of happening (if we assume theres not some genetic thingy that makes one genders birthrate dominant)

If she has no children and plans to have two children, the chances of both beeing buys are 25%
theres four possible outcomes (not counting twins, hermaphrodites or the antichrist)

a girl and then another girl.
a girl and then a boy.
a boy and then a girl.
a boy and then anotehr boy.

Again assuming that theres no genetic predisposition to get one gender more than the otehr each of the 4 options has a 25% chance of happening, since only one is boy + boy the chance of gettign 2 boys is 25%, if you have none already.

I'm not gonna vote, cause i can't read clearly from your question which of the 2 situations you're tlaking about.

Actually, the question is neither of those.

 

[Maze1125]

Like I said, any 33% answer assumes information that is not given in the question.

No it doesn't.

My proof two posts above yours uses exactly the information given and gives 1/3 as the answer.

 

[teisjm]

Spoiler

As I see it the Boy | Girl paradox serves to point out the flaw in probability theory. It's what happens when you leave out essential information or use terms that are ambiguous, not constraining the question enough. People interpret the question differently, adding in factors like order and age in which changes the outcome. A definitiv answer can't be found because the question itself is flawed.

A definitive answer can most definitely be found, just because some people like to exclude information they have or include information they don't have doesn't mean that there isn't a clear definitive answer using precisely the information that we have.

If she already has one boy, and gets pregnant the chance of the second one turnign out to be a boy when he's born is 50 %. Theres 2 possible outcomes (not ocunting hermaphrodite baby)

Girl or
Boy

both have a 50% chance of happening (if we assume theres not some genetic thingy that makes one genders birthrate dominant)

If she has no children and plans to have two children, the chances of both beeing buys are 25%
theres four possible outcomes (not counting twins, hermaphrodites or the antichrist)

a girl and then another girl.
a girl and then a boy.
a boy and then a girl.
a boy and then anotehr boy.

Again assuming that theres no genetic predisposition to get one gender more than the otehr each of the 4 options has a 25% chance of happening, since only one is boy + boy the chance of gettign 2 boys is 25%, if you have none already.

I'm not gonna vote, cause i can't read clearly from your question which of the 2 situations you're tlaking about.

Actually, the question is neither of those.

Oh so she already have two children, and we know that one of them is a boy?

that leaves us with 3 possible way she could've had them

Boy first then anotehr boy
Girl first then a boy
Boy first then a girl

each has a 33% chance, so it's 33.

(Still not counting twins, hermaphrodites and divine beeings.)