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

[Avatar Roku]

You already know the first one is a boy, so the probability of the second being a boy is 50%. I see what people are saying about the 33% and it makes sense but it also doesn't. If you didn't know what gender I was then you would have a 50/50 chance of guessing correctly. If I then tell you that I have a brother, that doesn't change the odds.

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.

No, that would only be if it said "there are two kids, one is a boy, what is the probability that the first/second (doesn't matter which) is boy?" In this case, it's "Two kids, one's a boy, what's the probability the other is a boy?" Since that's the case, it may as well just be asking "what's the probability that, if a woman has one child, (s)he's a boy?"

So it's 50%. The first child is irrelevant to the calculations.

That's the thing, it's not the first child that's a boy, it's simply one of the the children that's a boy. It could be the first child, or it could be the second. Therefore both the first and the second child are relevant.

As one guy said, don't think of it as "One is a boy." think of it as "They are not both girls." The two are equivalent, but with the second statement it is obvious that there is only a 1/3 chance of them both being boys.

Not really. The known child is irrelevant, whether he's the first or second, as there's a 50% chance that the unknown is a boy, completely independent of his/her brother.

Look at it like this: a woman has two children. That means there's four options:

GG
GB
BG
BB

Obviously, given the parameters of the problem, GG is out. On the surface, it would seem that there's 3 options (BG, BB, GB) left, but, as that isn't true, since the order doesn't matter. Therefore, GB and BG are the same, meaning there's only two options, BB and BG/GB.

For any given family with two children, they are twice as likely to have a boy and a girl as two boys. If you eliminate the possibility that they have two girls that does not change the fact that it they are twice as likely to have a girl and a boy as they are two boys.

Interesting argument, and I can see where you're coming from, but you're wrong, at least by my interpretation of this awfully worded problem. You see, since we're told that one is a boy, that's set, making the other child's sex independent. That wouldn't matter if it was asking us about both of them, but it's not, as the question portion of the problem says "what are the odds the other is also a boy?"

At this point, though, I'm willing to just chalk up our differences here as the problem being very poorly worded, as your arguments were completely valid, if for a different possible interpretation than mine.

 

[Paragon Fury]

Wait, what is the question?

Is it "A woman hs two children. Given that the first is a boy, what are the chances of the second one being a boy?" or "A woman has two children. What are the odds that both the first and second child will be a boy?"

Because those are two totally different questions with totally different answers.

 

[Avatar Roku]

Wait, what is the question?

Is it "A woman hs two children. Given that the first is a boy, what are the chances of the second one being a boy?" or "A woman has two children. What are the odds that both the first and second child will be a boy?"

Because those are two totally different questions with totally different answers.

That's the general consensus, this is a horridly worded question that's more or less open to interpretation.

 

[stone0042]

What exactly are we trying to prove here?

 

[Pimppeter2]

Ahh, but what are the chances of it being Godzilla?
[sub]Honestly...I want to know[/sub]

 

[Avatar Roku]

Ahh, but what are the chances of it being Godzilla?
[sub]Honestly...I want to know[/sub]

...0%? -5%? Just a couple hunches.

 

[Seldon2639]

50% since the chances of getting a boy or girl in general is based on whether the egg receives either the X or the Y chromosome.

So unless the second child is a fraternal twin, then that is my answer.

You've got it backwards. The egg is always an X chromosome. The sperm is what determines, since it can be either X or Y

I know, I meant that the egg would receive either a second X chromosome resulting in a girl, or a Y chromosome resulting in a boy

Ah, I misunderstood what you wrote. I thought you meant what the egg itself receives from the mitosis which creates the eggs in the first place.

 

[Maze1125]

You see, since we're told that one is a boy, that's set, making the other child's sex independent.

That's the mistake everyone's making.
All we know is that one of the children is boy, we don't know which one is.
And it does matter because, although the order doesn't matter, an ordering must still be chosen. That is necessary for probability calculations. As we don't know which is a boy, no matter what ordering we pick, it could still be either of them, and so the remaining child can never be independent of the other.

And I'm not just pulling stuff out of my arse here, this is a well known probability paradox, and the wording is key. The fact that we know one of the children is male, but not which, gives a unique case where the probability of the other child being male is 1/3.

Wait, what is the question?

Is it "A woman hs two children. Given that the first is a boy, what are the chances of the second one being a boy?" or "A woman has two children. What are the odds that both the first and second child will be a boy?"

Because those are two totally different questions with totally different answers.

In fact, the question is neither of those. My previous paragraph should explain it well enough.

 

[Sejs Cube]

Answering the question as posed, the genders of the two kids have no interrelation.

Ergo it's a straight 50/50 shot at the second kid being a boy.

 

[Aesir23]

50% since the chances of getting a boy or girl in general is based on whether the egg receives either the X or the Y chromosome.

So unless the second child is a fraternal twin, then that is my answer.

You've got it backwards. The egg is always an X chromosome. The sperm is what determines, since it can be either X or Y

I know, I meant that the egg would receive either a second X chromosome resulting in a girl, or a Y chromosome resulting in a boy

Ah, I misunderstood what you wrote. I thought you meant what the egg itself receives from the mitosis which creates the eggs in the first place.

No problem, I should probably be more clear in what I type anyway. Not exactly my strong suit since I type almost exactly how I think, and let's face it. ADD is not-hey, shiny! (Yes, I do have ADD. I'm just joking at my own expense, not making fun of others who have it *hides behind a flame shield from people who may get offended*)

 

[DracoSuave]

Both 33.3% and 50% are correct. It all depends on if we look at the one boy specified as unique occurrence which must be included or not.

http://en.wikipedia.org/wiki/Boy_or_Girl

Probability does not work that way.

You see, since we're told that one is a boy, that's set, making the other child's sex independent.

That's the mistake everyone's making.
All we know is that one of the children is boy, we don't know which one is.
And it does matter because, although the order doesn't matter, an ordering must still be chosen. That is necessary for probability calculations. As we don't know which is a boy, no matter what ordering we pick, it could still be either of them, and so the remaining child can never be independent of the other.

And I'm not just pulling stuff out of my arse here, this is a well known probability paradox, and the wording is key. The fact that we know one of the children is male, but not which, gives a unique case where the probability of the other child being male is 1/3.

In fact, the question is neither of those. My previous paragraph should explain it well enough.

Then your math is wrong. Using your logic, there are four possibilities.

First, we must establish the probability that the known boy is the first or the second child.

For the first child: BB or BG are the two outcomes, both equally probable at 2 out of 4, or 50%.

For the second child: BB or GB are the two outcomes, both equally probable at 2 out of 4, or 50%

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.

The problem with the 33% argument is that it fails to count BB twice, because there are -two- occurances of a boy in there, not one. The fact that she says she has a boy means that outcome is -twice as likely- to apply in this case, rather than equally so.

With completely unknown information, yes, it would be equally likely, however with -known- information you must wieght the outcomes different based on their likelihood to fulfill her criteria. BB is twice as likely to fulfill her criteria as any other outcome, therefore it must be weight twice as much. This argument already permits the weighting of criteria based on this; it uses the weighting of the GG scenario at 0 as part of it. What it fails to do is apply this correct procedure to any other outcome. And thusly, it arrives at the wrong answer.

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.

MACM 101, bitches.

-----------------------------

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%.

 

[veloper]

"A women has two kids, one is a boy, what are the odds the other is also a boy?" ~33%

"A women has two kids, the oldest a boy, what are the odds the other is also a boy?" ~50%

This is a trick with language & logic and has little to do with probability.

 

[TZer0]

Both 33.3% and 50% are correct. It all depends on if we look at the one boy specified as unique occurrence which must be included or not.

http://en.wikipedia.org/wiki/Boy_or_Girl

The problem with the 33% argument is that it fails to count BB twice, because there are -two- occurances of a boy in there, not one. The fact that she says she has a boy means that outcome is -twice as likely- to apply in this case, rather than equally so.

Did you even click the link? No? Not at all? You didn't scroll down the solution of the question? Ok then. I'll explain it then.

I don't think you understood what kind of experiment this is. The BB-option isn't supposed to be counted twice. This experiment can be solved the following way: You've got two coins, you flip them both. If you get tails-tails, you add one to your counter, if you get tails-heads or heads-tails you don't add one, if you get heads-heads, you flip again. Obviously, the tails-tails-option will happen 25% of the time, or 33.3% due to the re-flips.

Also, I've been that nice that I've solved this task for you.. numerically, it runs the test 10000 times, should be accurate enough? http://snipt.org/plpp - Python required.

The reason this works is because neither boys have a unique quality. If we however named one of them Simon, the younger one, then we're allowed to "lock" him to a position, say, the left one, and only change whatever is in the right position. Then we'll get SB and SG as options, two boys having 50% chance to occur.

.3 means recurring before anyone tries picking on it.

This guy got it right:

"A women has two kids, one is a boy, what are the odds the other is also a boy?" ~33%

"A women has two kids, the oldest a boy, what are the odds the other is also a boy?" ~50%

This is a trick with language & logic and has little to do with probability.

 

[NickIsCool]

The options at the start can be organised like this.

B - B 25%
B - G 25%
G - B 25%
G - G 25%

At least one of them is a boy.

B - B 33%
B - G 33%
G - B 33%
G - G 0%

Thus, the probability of the other sibling being a boy is 33%

B - B 50%
B - G 50%
G - B 0%
G - G 0%

the question asks the gender of the other child, not the order in which they where born or who was older

so the 2nd and 3rd possibilities are the same

"A women has two kids, one is a boy, what are the odds the other is also a boy?" ~33%

"A women has two kids, the oldest a boy, what are the odds the other is also a boy?" ~50%

This is a trick with language & logic and has little to do with probability.

the question does not involve the age of the womans children

this is not a trick, it is a very simple question

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

lets break this down

"A women has two kids"

it states our subjects

"one is a boy"

now we know that one child is a boy

"what are the odds the other is also a boy?"

now we are asked this question

no where in the question is the age of the children revealed, so it must be assumed that it is a variable that has no place here

so people who say that there is a 33.3% chance of the other child being a boy is WRONG

let me explain why

their possibilities for the childrens gender was as follows
B - B
B - G
G - B
G - G

the next step they make is correct. they remove the "G - G" result. that is good because none of the results that are relevant to this question have both the children being girls because we KNOW one of the is a boy.

now are possibilities are a bit closer to what we want
B - B
B - G
G - B
G - G

now quiet a few of the poster just stopped and said "i got it!" i guess with out looking at the results

did you notice that with out knowing which came first, and it not being relevant, that we have a duplicate result that some overlooked

without that knowledge our results can be changed around at will and still be the same

like so

B - B B - B
B - G G - B
G - B B - G
G - G G - G

notice any similarities?

if not, then let me help you


B - B B - B
B - G G - B
G - B B - G
G - G G - G

see it now?now lets see what results we need and which ones we dont

B - B This can stay because it is relevant to the question
B - G This can also stay for the same reason
G - B This one cannot because it is the same result as the one above it
G - G This one cannot because it is not relevant to the question

now lets see just the ones we need

B - B
B - G

i dont know about you, but that looks alot like a 50% chance of two boys

edit: hope this makes more sense to you
edit edit: found a typo

 

[TZer0]

The options at the start can be organised like this.

B - B 25%
B - G 25%
G - B 25%
G - G 25%

At least one of them is a boy.

B - B 33%
B - G 33%
G - B 33%
G - G 0%

Thus, the probability of the other sibling being a boy is 33%

B - B 50%
B - G 50%
G - B 0%
G - G 0%

the question asks the gender of the other child, not the order in which they where born or who was older

so the 2nd and 3rd possibilities are the same

Wrong. B-G != G-B.
They're two different occurrences and we need both. In this case we're not allowed to lock a boy in a position.

You may also check my numerical solution: http://snipt.org/plpp
http://www.escapistmagazine.com/forums/jump/18.157846.3928485

 

[Azraellod]

B - B 50%
B - G 50%
G - B 0%
G - G 0%

the question asks the gender of the other child, not the order in which they where born or who was older

so the 2nd and 3rd possibilities are the same

*groans in annoyance*

I was bored of arguing my point, so I deliberately left this thread. But if I must.

Spoiler

They amount to the same thing, but you are forgetting to add the probabilities.

These probabilities start out as equal, so:

(BB) = 1/4
(BG) = 1/4
(GB) = 1/4
(GG) = 1/4

(BB) = 1/3
(BG) = 1/3
(GB) = 1/3
(GG) = 0

(BB) = 1/3
(BG)+(GB) = 2/3

Thus, 33%

 

[DracoSuave]

Both 33.3% and 50% are correct. It all depends on if we look at the one boy specified as unique occurrence which must be included or not.

http://en.wikipedia.org/wiki/Boy_or_Girl

The problem with the 33% argument is that it fails to count BB twice, because there are -two- occurances of a boy in there, not one. The fact that she says she has a boy means that outcome is -twice as likely- to apply in this case, rather than equally so.

Did you even click the link? No? Not at all? You didn't scroll down the solution of the question? Ok then. I'll explain it then.

I don't think you understood what kind of experiment this is. The BB-option isn't supposed to be counted twice. This experiment can be solved the following way: You've got two coins, you flip them both. If you get tails-tails, you add one to your counter, if you get tails-heads or heads-tails you don't add one, if you get heads-heads, you flip again. Obviously, the tails-tails-option will happen 25% of the time, or 33.3% due to the re-flips.

Yes I did read the link. I also didn't -stop- reading the article half-way through.

From the article in question:

The authors argue that the reason people respond differently to this question (along with other similar problems, such as the Monty Hall Problem and the Bertrand's box paradox) is because of the use of naive heuristics that fail to properly define the number of possible outcomes

Also, I've been that nice that I've solved this task for you.. numerically, it runs the test 10000 times, should be accurate enough? http://snipt.org/plpp - Python required.

Interesting. However it is flawed. She has a boy. So you do not select 'one of the families with any number of boys', you select -A BOY-. So you do not choose 'one of the pairs that succeeds the criteria' you select 'one of the heads.' The thing is, the double pairs of heads have two heads, and thusly you're more likely to choose one of those pairs than a pair of tails/heads or a pair of heads/tails.

You have only two valid ways of looking at it. You can either look at them as independant events, in which case the boy is irrelevant, or you can look at the boy as relevant, in which case you -must- value all outcomes appropriately.

Looking at all outcomes:
John, Fred
Bill, Tracy
Jill, Don
Marge, Tara

She has a boy from one of these outcomes. That boy has an equal probability of being John, Fred, Bill, or Don. Obviously, he cannot be Tracy, Jill, Marge, or Tara.

John and Fred being of equal probability, means that outcome has twice as many occurances as Bill/Tracy or Jill/Don.

Your argument is -not taking a random boy- from the sample. It's taking a random sample and just rejecting ones that don't follow the criteria, it flatly ignores the fact that a random boy will come from some samples more often than others.

And that is its flaw.

The reason this works is because neither boys have a unique quality. If we however named one of them Simon, the younger one, then we're allowed to "lock" him to a position, say, the left one, and only change whatever is in the right position. Then we'll get SB and SG as options, two boys having 50% chance to occur.

Actually, it's the boy's non-uniqueness that allows you to do this the way I describe. You don't know which boy it is, so you have to choose a random -boy-.

Not a random family.

This guy got it right:

"A women has two kids, one is a boy, what are the odds the other is also a boy?" ~33%

"A women has two kids, the oldest a boy, what are the odds the other is also a boy?" ~50%

This is a trick with language & logic and has little to do with probability.

Any question that asks 'What are the odds' is a probability question. That's what 'the odds' means. And probability is math. And math doesn't play with your tricks. Ignorance of how the math works does not excuse this.

Besides, it was disproven, logically.

A woman has an child who is X. What are the chances the other child is a boy?

X can be any quality that does not involve the other child.

'Is a boy' qualifies under such.

This is a logical proof.

 

[Clyde]

At least the majority is right.

 

[IckleMissMayhem]

It's either 50% or 33%, depending on the way your teacher interprets the question. But you can argue both answers are correct.

 

[veloper]

"A women has two kids, one is a boy, what are the odds the other is also a boy?" ~33%

"A women has two kids, the oldest a boy, what are the odds the other is also a boy?" ~50%

This is a trick with language & logic and has little to do with probability.

the question does not involve the age of the womans children

this is not a trick, it is a very simple question

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

lets break this down

"A women has two kids"

it states our subjects

"one is a boy"

now we know that one child is a boy

"what are the odds the other is also a boy?"

now we are asked this question

no where in the question is the age of the children revealed, so it must be assumed that it is a variable that has no place here

so people who say that there is a 33.3% chance of the other child being a boy is WRONG

let me explain why

there possibilities for the childrens gender was as follows
B - B
B - G
G - B
G - G

the next step they make is correct. they remove the "G - G" result. that is good because none of the results that are relevant to this question have both the children being girls because we KNOW one of the is a boy.

now are possibilities are a bit closer to what we want
B - B
B - G
G - B
G - G

now quiet a few of the poster just stopped and said "i got it!" i guess with out looking at the results

did you notice that with out knowing which came first, and it not being relevant, that we have a duplicate result that some overlooked

without that knowledge our results can be changed around at will and still be the same

like so

B - B B - B
B - G G - B
G - B B - G
G - G G - G

notice any similarities?

if not, then let me help you


B - B B - B
B - G G - B
G - B B - G
G - G G - G

see it now?now lets see what results we need and which ones we dont

B - B This can stay because it is relevant to the question
B - G This can also stay for the same reason
G - B This one cannot because it is the same result as the one above it
G - G This one cannot because it is not relevant to the question

now lets see just the ones we need

B - B
B - G

i dont know about you, but that looks alot like a 50% chance of two boys

That's a very elaborate way of completely missing the point. It's all language.

Most people would have no difficulty with solving the following:
"A woman has two kids, she does NOT have TWO girls, what are the odds she has two boys?" (33%)

this question can be paraphrased like the OP

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

but not as

"A woman has two kids, the oldest a boy, what are the odds #2 is also a boy?" This is because it now contains more information: which of the 2 kids the question is about is now defined.