Family with a daugter/odds they have a son

[C_K_Dexter_Haven]

The difficulty with probability is that the question is VERY dependent on the phrasing and on the knowledge you have. In this case, the ORDER makes a difference, so as soon as you say “my first child is a son” you’ve made the probability of the second child being a girl 50%. If you say “one child is a son” without specifying whether the first or the second in any way, then the chance of the other one being a girl is 2/3.

By analogy, an old joke question: I have two coins that total 15 cents. One of the coins is not a nickel. What are the two coins? If you said it’s impossible, then you’re ignoring the hidden ordering of the coins. The (admittedly trick) answer is that one of the coins is not a nickel, it’s a dime; the other coin is a nickel.The point here is that wording and ordering makes a huge diff in the answer.

[bump]

This is completely stupid. What about this situation?

Let’s say you’re at work and one of your co-workers says he has 2 kids. He goes on to let slip that his daughter was going to a slumber party this weekend. He doesn’t mention the other child at all. What are the odds that his other child is a son?

[Chronos]

50%. He has two kids, the one who’s going to a slumber party and the one who isn’t. The one who’s going to a slumber party is a girl, the other could be either.

Like I’ve said before, it’s really hard to come up with a real-world situation where you have exactly the information that leads to the 2/3 answer, which is why that answer is so unintuitive.

[tim-n-va]

@bump

This is completely stupid. What about this situation?

Let’s say you’re at work and one of your co-workers says he has 2 kids. He goes on to let slip that his daughter was going to a slumber party this weekend. He doesn’t mention the other child at all. What are the odds that his other child is a son?

2/3. If you had information that the daughter was the oldest child or the youngest, it would be 1/2.

[Heart_of_Dorkness]

@bump

This is completely stupid. What about this situation?

Let’s say you’re at work and one of your co-workers says he has 2 kids. He goes on to let slip that his daughter was going to a slumber party this weekend. He doesn’t mention the other child at all. What are the odds that his other child is a son?

@Chronos

50%. He has two kids, the one who’s going to a slumber party and the one who isn’t. The one who’s going to a slumber party is a girl, the other could be either.

Like I’ve said before, it’s really hard to come up with a real-world situation where you have exactly the information that leads to the 2/3 answer, which is why that answer is so unintuitive.

No, Chronos, I disagree. This is exactly the situation that leads to the 2/3 answer.

1. Coworker says, “I have 2 kids.”
   What are the possible ways this could be true?
   They had a son, and then had another son.
   They had a son, and then had a daughter.
   They had a daughter, and then had a son.
   They had a daughter, and then had another daughter.
   These are all the ways in which he could have two kids, and each is equally likely, a 1/4 chance.

2. Coworker says, “My daughter Claire is going to a slumber party this weekend.”
   So, of his two kids, one is a girl. What are the possible ways this could be true?
   [del]They had a son, and then had another son.[/del]
   They had a son, and then had a daughter.
   They had a daughter, and then had a son.
   They had a daughter, and then had another daughter.
   These are all the ways in which he could have two kids AND have one of them be a girl, and each is equally likely, a 1/3 chance.

3. What are the odds that the unmentioned kid is a boy?
   They had a son, and then had a daughter. = **1/3 chance **
   They had a daughter, and then had a son. = 1/3 chance
   They had a daughter, and then had another daughter. = 1/3 chance
   So there are two 1/3 chances for the other kid to be a boy, and only one 1/3 chance for it to be a girl.

[bump]

No, this isn’t the 2/3 chance situation.

The fact that the guy has a daughter doesn’t have any bearing on whether or not he also has a son. It’s the same thing as saying that he has a daughter and his wife’s pregnant- what are the chances of him having another son?

Obviously in that case, it’s 50/50. It’s 50/50 in my example as well, and if you carefully read the OP’s question, it’s a variation on the same question and is also 50/50.

[Find_Friends]

@bump

Ok… real world example. I have a wife and a son. Let’s hypothetically say my wife’s currently pregnant.

Are you telling me we have 66% chance of having a daughter for our second child?

Hey, bump, can you at least understand that those who do “get it” are NOT saying that the gender of one child causes the probability of the other to change?

They are saying it’s a matter of (my choice of expression) “information framework” not causation.

Sigh. It seem that the same misunderstanding crops up every single time the question appears here.

---

But don’t get me wrong. I once had a non-conversations with a seemingly intelligent person trying to get him to understand that clockwise and counter-clockwise are not absolute directions, and that a wheel or disk exhibits each depending on which side you view. THAT still boggles my mind. I at least understand how this matter can be difficult to get. I myself might well have had a WTF? reaction the first time on SDMB had I not come across it in my preteen years, in a Time-Life magazine.

BTW, this article seem to be comprehensive and address ambiguities.

[ZenBeam]

@bump

Let’s say you’re at work and one of your co-workers says he has 2 kids. He goes on to let slip that his daughter was going to a slumber party this weekend. He doesn’t mention the other child at all. What are the odds that his other child is a son?

As always, the critical factor is whether you are told information about a particular child or not. When you have information about a particular child, the other child is a well-defined person as well.

Claire is a particular child. The question is about the sex of the other child, her sibling. We know nothing about this other child, so the chance of the other child being a boy is 50%.

@Heart_of_Dorkness

No, Chronos, I disagree. This is exactly the situation that leads to the 2/3 answer.

If you insist on listing equal-probability situations in terms of older/younger shildren, there are four possibilities here (older child listed first):
1 Claire, Boy
2 Claire, Girl
3 Boy, Claire
4 Girl, Claire

You have no information whether Claire is older or younger, so it’s 50/50 whether Claire is older or younger. If Claire is the older child, there’s no reason for a boy or a girl to be preferred, so cases 1 and 2 have the same probability. Likewise if Claire is the younger child. So all four cases 1 - 4 have to be equally probable, and the chance of the other child being a boy is 50%.

@bump

This is completely stupid.

What exactly are you referring to? Could you explain which of these statements from post 40 you agree with and which you disagree with?:

@ZenBeam

Look, just make a sample of four families:
1 Girl, Boy
2 Boy, Girl
3 Boy, Boy
4 Girl, Girl

This is simple, right? For families with two children, those four cases are all equally likely, and there are no other cases to consider.

@ZenBeam

“everyone who’s first child is a daughter”: 1 and 4 raise their hand. Half have a boy.
"Everyone who has a second child that is a daughter ": 2 and 4 raise their hand. Half have a boy
“they have at least one daughter”: 1, 2, and 4 raise their hand. 2/3rds have a boy.

Aren’t all three of these statements true? If not, which don’t you agree with?

@ZenBeam

Cases 1 and 2 are in no way equivalent for you. You have zero chance of ending up with Case 1, with a girl for your first child and a boy for your second, and a 50% chance of ending up with an older boy and a younger girl.

Isn’t this true?

@ZenBeam

But in the OP’s question, when all we know is that at least one child a girl, Case 1 is possible, and just as likely as Case 2 or Case 4.

Do you disagree that in this situation, Case 1 is possible?

[Heart_of_Dorkness]

@bump

The fact that the guy has a daughter doesn’t have any bearing on whether or not he also has a son. It’s the same thing as saying that he has a daughter and his wife’s pregnant- what are the chances of him having another son?

No, it’s not the same thing. As shown above, there are, in reality, four equally likely ways to end up with two kids. And of those, there are three equally likely ways to end up with two kids, one of which is a daughter. But if you say he has a daughter and his wife’s pregnant, then there are only two equally likely ways to end up with two kids, one of which is a daughter - you’ve eliminated the possibility of him already having a son and his wife being pregnant with a daughter.

You’re absolutely right - the fact that someone has a daughter has no bearing on whether or not they also have a son. It seems like that’s what is being claimed here. But it’s not.

The problem is, odds are never absolute, only relative. The odds of anything happening depend on how you define the context in which it happens. In my line of work, data analysis, we call this context a “universe”. If you alter your universe, you alter the odds. So in the universe of “all two-kid families”, you have a 50% chance of having “one of each”, and a 50% chance of having “two of the same”. Or, you have a 25% chance of having each of “boy-girl”, “girl-boy”, “girl-girl”, and “boy-boy”. But if you narrow your universe to include only “all two-kid families where one kid is a girl”, you eliminate one of the above possibilities, and so the odds of the three remaining possibilities all go up to 33%.

And if you further narrow your universe to include only “all two-kid families where the first kid is a girl”, you again eliminate one of the above possibilities, and the odds of the two remaining possibilities go up to 50%. And this is what you’re doing when you say, “They have a girl and are pregnant.” You’ve eliminated the “boy-girl” possibility.

Consider a totally different example: health statistics. We hear stuff all the time like, “The average person has X% odds of getting cancer.” In this case, without any other context, we can only assume that the universe includes everyone in the entire world. But we also know that your personal chances of getting cancer can vary greatly depending on things like health, family history, exposure to toxins, etc. So if you don’t smoke or drink, you eat well and exercise, you avoid too much sun exposure, you have no family history of cancer, etc. then your personal odds of getting cancer are much lower than the average, say 1/3 X%. That makes sense, right? But what are your odds of getting cancer, then? Are they X, or 1/3 X? Is X wrong? No, not if we’re including you as one person in the universe of “everybody”. But if we narrow the universe down to “you”, then your odds change.

Now let’s say you move to another part of the world, where for those who can’t afford it (which is most people) nutrition is poor and medical care is terrible. People in this community are therefore likely to die fairly young of causes other than cancer. You, however, can afford good food and good care, exactly like when you were living here. Great, right? Except that now - in the universe of your community - your odds of getting cancer go way, way up. You are much more likely than most people in your community to get cancer, because you’ll be living long enough to get it! Now, does the fact that you moved, or that these other people are dying have any bearing on your actual health? Obviously not. But describing a universe that includes only them and you does have bearing on your odds. *Relative *to their odds, yours increase.

So to come back to the OP, what is the universe described? “There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.)”

“There is a family with two children. You have been told this family has a daughter.” = the universe of families with two children, one of which is a daughter.

This universe includes all families that had a boy and then a girl, all families that had a girl and then a boy, and all families that had a girl and then a girl. For the one-of-eaches, the order doesn’t matter; whether they had a girl and then a boy OR a boy and then a girl, we’re including them in the universe. But in doing so, we’re including twice as many one-of-eaches as girl-girls, and so having one of each is twice as likely - in this particular universe.

[RedSwinglineOne]

@Chronos

Yes, exactly.

Consider this scenario: You’re in a room with hundreds of couples, each of them parents to two children. You say “Everyone in the room whose first child is a daughter, raise your hand”, … Of these families, half will have two girls, and half will have a boy and a girl.

Now, though, suppose you’re in the same room, and you say “Everyone in the room who has at least one daughter, raise your hand”, … Of these families, 1/3 will have two girls, and 2/3 will have a boy and a girl.

The point I would like to add to this is that if you ask the first question, and then ask the second question, all of the people who raised their hands only when asked the second question will be people with a boy (and a girl). If they had two girls they would have raised their hand the first time. If they had two boys they would not have raised their hand at all. It is this addition to the group of people who all have a boy that change the answer from 1/2 to 2/3.

[Heart_of_Dorkness]

@bump

The fact that the guy has a daughter doesn’t have any bearing on whether or not he also has a son. It’s the same thing as saying that he has a daughter and his wife’s pregnant- what are the chances of him having another son?

Wait, maybe this well help.

“I have two kids, one is a girl.” is NOT the same universe as “I have a girl and my wife’s pregnant.”
“I have two kids, one is a girl.” IS the same universe as “I have a girl and my wife’s pregnant with either a boy or a girl… or wait, maybe it’s that my wife’s definitely pregnant with a girl, and the existing child could be either. Man… we never should have named the poor kid Pat.”

You see? If you say that “I already have a girl and my wife is pregnant”, you’re eliminating the possibility of having a boy and your wife being pregnant with a girl. You’d still have two kids, one would still be a girl, but you’re not including them anymore.

[Irishman]

@bump

Here’s the problem with your reasoning.

1 and 2 are equivalent. For the purposes of the question, it doesn’t matter if it’s Girl/Boy or Boy/Girl.

So there are actually 3 possible options for the family in this question: one child of each gender, both boys, and both girls.

The problem with this analysis is that each birth is a separate event with a separate probability. So the first child has a 50/50 chance and the second child has a 50/50 chance. When you lump the “one child of each gender” categories, you are making unequal sample sizes.

boy boy
boy girl
girl boy
girl girl

Look at those options. 50% of the first children are boys. 50% of the second children are boys. Two independent events, so those sample sizes are equal. Now we go in and eliminate part of the sample - the boy boy set. That leaves only 75% of our initial total sample set to work with. From that 75%, we still have three equal sized groups.

Say we take 100 families with the equal distribution as above, 25 families with each set. Now drop out the boy boy set. That leaves 75 families, 25 with a boy girl, 25 with a girl boy, and 25 with a girl girl.

You can see that 50 out of 75 have at least 1 boy. That is 2/3. Only 25 out of 75 have no boys. That is 1/3.

The trick here is that the equal 50/50 distributions occur, and then you eliminate a select part of the distribution that skews the remaining distribution. Then you ask a question about that remaining distribution. That is you we achieve the skewed result.

@Cartoonacy

Shuffle a deck of cards and deal them out into 26 piles of two. Each pair of cards represents a family with two children; black cards are boys and red cards are girls.

@John_W.Kennedy

Order doesn’t actually matter. And you don’t need to use playing cards. Just toss two coins 100 times and eliminate all the cases (about 25) with two heads. Of the about 75 remaining, about 25 will be two tails and about 50 will be one head and one tail.

I’m not sure I like the playing card example. Something about the fact that the deck constrains the later options. I like the two coin toss method better. Each toss is independent, each coin is independent.

@Amateur_Barbarian

Okay, I’m going to be doggedly stupid here. No matter how many times I read the “2/3” arguments, they sound like variations of the gambler’s fallacy. Each flip of a coin has a 50/50 chance; each sperm-egg dance has a 50/50 chance. Minor biological possibilities aside, no one flip/dance has any effect on any other in the series.

This is all about sorting the results, not changing the outcome of the next coin toss. Take the equally distributed results, cut out some part skewed to one side, and what is left is skewed.

Take a rectangle of paper. Cut through it on a diagonal. The remaining bit is no longer a rectangle. We’re doing the same thing with the results pool. We’ve cut a diagonal - families that are boy boy. The remaining results pool is no longer a rectangle, it is skewed.

@bump

This isn’t the Monty Hall question, and that’s why I think applying that logic to this one is wrong.

No, it is not the Monty Hall problem. It is not like the Monty Hall problem. The only similarity is that the results are non-intuitive and not 50/50.

You’re right in my example that I’ve ruled out B and C, but in the OP’s example, there’s no difference between B and C because it doesn’t matter who came first.

But it does matter who came first, when you are laying out equal distributions from the probabilities of having occurred.

Just knowing that a couple has a daughter doesn’t change the odds that they may also have a son, assuming that sons and daughters are born with equal frequency.

Right. What is different is that the case where the family has two daughters, they could be thinking of 1 daughter, or they could be thinking of the other daughter. That gives them a pool of daughters twice as large as the family that only has one daughter. The pool of families with two daughters is not twice as large, but the options of which daughter is twice as large. That is the thing that skews the results.

As soon as you specify one of the daughters in any way, you have eliminated that skew, because you have eliminated the possibility you could be talking about the other daughter.

@Heart_of_Dorkness

No, Chronos, I disagree. This is exactly the situation that leads to the 2/3 answer.

[snip]

3. What are the odds that the unmentioned kid is a boy?
   They had a son, and then had a daughter. = **1/3 chance **
   They had a daughter, and then had a son. = 1/3 chance
   They had a daughter, and then had another daughter. = 1/3 chance
   So there are two 1/3 chances for the other kid to be a boy, and only one 1/3 chance for it to be a girl.

No, Chronos is right. Once you’ve specified the daughter is going to a slumber party, the question became about the child who is not going to the slumber party. It doesn’t matter if the daughter going to the slumber party is older or younger, the other child cannot be the child going to the slumber party.

Another way to think about this whole problem…

There is a family with two daughters: Alice and Betty.

Ask that family, do you have at least one daughter? The Dad says “yes” and thinks of Alice, the Mom says “yes” and thinks of Betty. For a family with only one daughter, Claire, they can only think of Claire. Because the family where the Dad thinks of Alice and the Mom thinks of Betty is the same family, that part of the pool of answers for “Do you have at least one daughter” overlaps, it is not independent.

[Heart_of_Dorkness]

@Irishman

@Heart_of_Dorkness

No, Chronos, I disagree. This is exactly the situation that leads to the 2/3 answer.

[snip]

3. What are the odds that the unmentioned kid is a boy?
   They had a son, and then had a daughter. = **1/3 chance **
   They had a daughter, and then had a son. = 1/3 chance
   They had a daughter, and then had another daughter. = 1/3 chance
   So there are two 1/3 chances for the other kid to be a boy, and only one 1/3 chance for it to be a girl.

No, Chronos is right. Once you’ve specified the daughter is going to a slumber party, the question became about the child who is not going to the slumber party. It doesn’t matter if the daughter going to the slumber party is older or younger, the other child cannot be the child going to the slumber party.

Wait, why does the slumber party matter? We could also specify that she’s named Becky and is wearing jeggings. All that matters is that we know there are two kids, we know one of them is a girl, and that’s all we know. As long as we’re allowing that the *other *child could be a) either older or younger and b) either a boy or a girl, then we’re still dealing with a 2/3 scenario. And indeed, that’s all we know:

@Chronos

He has two kids, the one who’s going to a slumber party and the one who isn’t. The one who’s going to a slumber party is a girl, the other could be either.

So I agree, it doesn’t matter if the daughter going to to the slumber party is older or younger. That means it still could be that:

They had a son, and then had a daughter. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a boy. = **1/3 chance **
They had a daughter, and then had a son. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a boy. = 1/3 chance
They had a daughter, and then had another daughter. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a girl. = 1/3 chance

Each of these cases can be described by the coworker as “I have two kids. My daughter Becky is going to a sleepover this weekend,” and each is equally likely.

What am I missing?

[tim-n-va]

Standard conditional probability P(B|G) = P(G∩B)/P(G)

That is, probablility of B given G is the probability of G and B divided by the probability of G.

P(G∩B) = .5 (of the four possible combinations, two are boy/girl and girl/boy)
P(G) = .75 (of the four possible combination all have a girl except one)

so the probability of a boy, given that one child is a girl is .5/.75 or 2/3.

[ZenBeam]

@Heart_of_Dorkness

\What am I missing?

Post 48.

[Irishman]

@Heart_of_Dorkness

Wait, why does the slumber party matter? We could also specify that she’s named Becky and is wearing jeggings.

Yes, we could specify her name, or what she is wearing, or if she has tattoos or not. The point is, if we identify the specific kid in any way, then we have narrowed the conditions down to “this kid” and “the other kid”. Once you have done that, the probabilies are back to 50/50.

That is the point of why it is very unlikely to get this situation in any real conversation. You are almost always going to be given a “this kid/other kid” situation, not a “one of whom is, what’s the other?” situation.

All that matters is that we know there are two kids, we know one of them is a girl, and that’s all we know.

No, we know a specific kid is going to a slumber party. This kid is a daughter and is going to a slumber party, the other kid can be a girl or a boy.

If he had let it slip one of his kids was a daughter, and later mentioned one of his kids was going to a slumber party, and assuming that boys have slumber parties as equally as girls, then you would not have enough information to connect the two, so you would be in the 2/3 condition. But the stated example, the daughter is going to the slumber party. Ergo, we’re asking about the kid not going to the slumber party.

So I agree, it doesn’t matter if the daughter going to to the slumber party is older or younger. That means it still could be that:

They had a son, and then had a daughter. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a boy. = **1/3 chance **
They had a daughter, and then had a son. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a boy. = 1/3 chance
They had a daughter, and then had another daughter. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a girl. = 1/3 chance

No. Your third case is misrepresentative. Change it to:

They had a daughter Becky, and then had another daughter. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a girl. = 1/3 chance

They had a daughter, then had another daughter Becky. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a girl. = 1/3 chance

And then realize all your chances add up to 4/3. Oops. Unless you think Becky could be the son.

Each of these cases can be described by the coworker as “I have two kids. My daughter Becky is going to a sleepover this weekend,” and each is equally likely.

What am I missing?

They are not equally likely, because Becky could be the older daughter or the younger daughter in a two daughter family. You have identified a specific child.

Either the family had Becky, and then another child, or the family had a child, and then had Becky. That’s the only two options. Within that set, the other child could be a boy or a girl. But Becky can never be the other child, and Becky can never be a boy.

[C_K_Dexter_Haven]

I like the coin-tossing analogy. To emphasis that the wording of the problem is critical to the answer, consider two versions:
(1) I have tossed a coin twice, and there is at least one H. What is the probability that there is another H? Answer: 2/3
(2) I have tossed a coin twice and the first toss was H. What is the probability that the second toss is H? Answer: 1/2

[tim-n-va]

No. Your third case is misrepresentative. Change it to:

They had a daughter Becky, and then had another daughter. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a girl. = 1/3 chance

They had a daughter, then had another daughter Becky. One of the children, Becky, is going to a slumber party. The child who is not going to the slumber party is a girl. = 1/3 chance

Nope. If you care which daughter (in the two girl scenario) is named Becky, going to the slumber party or any other extraneous fact you have.

Two girls (don’t care about names) = 1/3
OR
Two girls with first named Becky = 1/6
Two girls with the second named Becky = 1/6

[bump]

It does depend heavily on the wording- see the “Second Question.”

Boy or Girl paradox

[Heart_of_Dorkness]

@ZenBeam

@Heart_of_Dorkness

No, Chronos, I disagree. This is exactly the situation that leads to the 2/3 answer.

If you insist on listing equal-probability situations in terms of older/younger shildren, there are four possibilities here (older child listed first):
1 Claire, Boy
2 Claire, Girl
3 Boy, Claire
4 Girl, Claire

You have no information whether Claire is older or younger, so it’s 50/50 whether Claire is older or younger. If Claire is the older child, there’s no reason for a boy or a girl to be preferred, so cases 1 and 2 have the same probability. Likewise if Claire is the younger child. So all four cases 1 - 4 have to be equally probable, and the chance of the other child being a boy is 50%.

Okay, I see the problem. We are posing two different questions. bump said:

@bump

Let’s say you’re at work and one of your co-workers says he has 2 kids. He goes on to let slip that his daughter was going to a slumber party this weekend. He doesn’t mention the other child at all. What are the odds that his other child is a son?

I read this as the coworker saying, “I have two kids. One is a daughter. Her name is Claire, incidentally, and she’s going to a slumber party. What are the odds that my other kid is a boy?” The answer is 2/3. Knowing other information about that daughter is completely irrelevant unless it alters the universe in question: families with two kids, one of which is a girl. Knowing the girl’s name or her weekend plans does not change the universe. According to this interpretation, in the explanation above, possibilities 2 and 4 are each only half as likely as 1 and 3. That is, your coworker has two kids, and one is a girl, who happens to be named Claire. How did this happen? He could have had a boy and then a girl (and named it Claire), a girl (and named it Claire) and then a boy, or a girl (which he may or may not have named Claire) and then a girl (which he certainly named Claire if he did not name the first one Claire - and possibly, even if he did).

But you guys are reading this as the coworker saying, “I have two kids. One is a daughter. Her name is Claire, incidentally, and she’s going to a slumber party. What are the odds that Claire’s sibling is a boy?” The answer is 1/2. Because you’re treating the specification of Claire (and/or the mention of her going to a slumber party) as denoting her as the point of reference, rather than the family. Now, the universe we’re discussing is NOT “families with two kids, one of which is a girl”, it’s “girls with one sibling”.

Again - bear with me - here are the children of four different families, listed in birth order:

Girl A and Boy A
Boy B and Girl B
Girl C and Girl D
Boy C and Boy D

Once more with feeling: how many possible two-child families with at least one girl? Three. How many of them also have a boy? Two. Chances of a two-child family with a girl also having a boy? 2/3.

But now, let’s look at the very same group of people, but instead of talking about “two-child *families *that have at least one girl”, let’s talk about “*girls *with one sibling”. And as an example of a girl with one sibling, let’s consider in specific your coworker’s daughter, Claire, who is going to a sleepover. You know that there is another kid in Claire’s family. She has a sibling. So, what are the chances that her sibling is a boy?

Girl A and Boy A

• If Claire is Girl A, her sibling is a boy. = 1/4
  Boy B and Girl B
• If Claire is Girl B, her sibling is a boy. = 1/4
  Girl C and Girl D
• If Claire is Girl C, her sibling is a girl. = 1/4
• If Claire is Girl D, her sibling is a girl. = 1/4
  [del]Boy C and Boy D[/del]

She has a 1/2 chance of having a brother. And of course, this matches ZenBeam’s example above (just in a different order).

So the bottom line is, as we’ve been saying, when you’re looking at statistics, you have to be really clear on what universe you’re talking about.