Family with a daugter/odds they have a son

[Skinny]

From the Monty Hall problem
http://www.straightdope.com/columns/read/916/on-lets-make-a-deal-you-pick-door-1-monty-opens-door-2-no-prize-do-you-stay-with-door-1-or-switch-to-3

Now, I don’t have a problem with the Monty Hall question and please don’t add more on that here, as it’s been gone over well enough. The problem is with this little nugget.

“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.)”

He goes on to explain it like this

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.

This doesn’t seem to make sense, but the data there supports the findings, except the data isn’t correct. If you claim the order of female and male siblings counts, then the order of female and female siblings must count as well, same goes for the two males.

It should look like this. We will name our female child Claire to clear it up a bit.

(1) Child A is Claire and Child B is male.

(2) Child A is Claire and Child B is female.

(3) Child A is female and Child B is Claire.

(4) Child A is male and Child B is Claire.

(5) Child A is male(x) and Child B is male(y).

(6) Child A is male(y) and Child B is male(x).

We know one child is Claire, eliminating choice #5-6. In 2 cases the remaining child is male, in 2 cases female. If order matters in one case, it matters in all.

[Chronos]

It should look like this. We will name our female child Claire to clear it up a bit.

This is the problem. We can’t name “our female child”, because “our female child” isn’t necessarily a specific individual. If the family has two daughters, then “our female child” can be either one of them.

The problem is counterintuitive because in real life, it’s very rare to learn something like “the family has at least one daughter” without that information being about some specific individual (say, “the eldest child is a daughter”, or “the favorite child is a daughter”, or “the child I just met is a daughter”). If you do have information about a specific individual, then the other child is equally likely to be a boy or a girl, just as you’d expect. If, however, you somehow gain the information that at least one is a girl, without gaining information about a specific child, then it’s 2/3 that the other one is a boy.

[bump]

This isn’t the Monty Hall problem though; there’s no role of Monty here, which is what makes that problem 2/3 rather than 1/2.

Contestants who switch have a 2/3 chance of winning the car, while contestants who stick have only a 1/3 chance. One way to see this is to notice that there is a 2/3 chance that the initial choice of the player is a door hiding a goat. When that is the case, the host is forced to open the other goat door, and the remaining closed door hides the car. “Switching” only fails to give the car when the player had initially picked the door hiding the car, which only happens one third of the time.

In the OP question there is no Monty choosing a door and narrowing things down- the question of the children is the same thing as saying “A couple has a daughter and the wife is already pregnant with their second child. What are the chances of it being a boy?”.

Without a Monty, this is a 50/50 problem.

[Terr]

@Chronos

The problem is counterintuitive because in real life, it’s very rare to learn something like “the family has at least one daughter” without that information being about some specific individual (say, “the eldest child is a daughter”, or “the favorite child is a daughter”, or “the child I just met is a daughter”).

For the purposes of the problem here, what is the difference between the statement “the family has two kids, at least one of which is a daughter” and “the family has two kids, and the one I just met is a daughter”?

[Skinny]

@Chronos

This is the problem. We can’t name “our female child”, because “our female child” isn’t necessarily a specific individual. If the family has two daughters, then “our female child” can be either one of them.

Yes, the female child could be the first or the second female child of two female children, that’s two separate scenarios.

The female child is a specific child since we know they have it, if you think it matters that the brother might be younger or older, then it matters if her sister is younger or older too. You can’t have it matter only if the other is male.

Are you trying to say that if the question was “You have been told this family has a daughter named Claire, what are the odds they also have a son?” That the answer would be different?

If order doesn’t matter, then it’s just

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is Male and Child B is male.

We don’t know if the daughter is first or second, so if order counts, it counts for the two females as well, either way it’s 50%. 2/3 is not counterintuitive, it’s just omitting data.

[Chronos]

Are you trying to say that if the question was “You have been told this family has a daughter named Claire, what are the odds they also have a son?” That the answer would be different?

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”, then you pick one of the couples with their hand raised. If you then ask one of those couples “What is your first child’s name?”, they’ll be able to tell you (and maybe the name they’ll tell you is Claire). 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”, then pick one of the couples. If you ask that couple “What is your daughter’s name?”, they might tell you, or they might say “Which one?”. To a family with two daughters, the question “What is your daughter’s name?” doesn’t make sense. Of these families, 1/3 will have two girls, and 2/3 will have a boy and a girl.

[Terr]

@Chronos

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”, then pick one of the couples. If you ask that couple “What is your daughter’s name?”, they might tell you, or they might say “Which one?”. To a family with two daughters, the question “What is your daughter’s name?” doesn’t make sense. Of these families, 1/3 will have two girls, and 2/3 will have a boy and a girl.

You’re changing the terms of the problem. Those don’t include the families answering the question by asking “which one”? In fact, the terms do not include families being asked or answering any questions.

And the answer to “You have been told this family has a daughter named Claire, what are the odds they also have a son?” is 1/2. The “Claire” part is irrelevant.

[Skinny]

@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”, then you pick one of the couples with their hand raised. If you then ask one of those couples “What is your first child’s name?”, they’ll be able to tell you (and maybe the name they’ll tell you is Claire). 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”, then pick one of the couples. If you ask that couple “What is your daughter’s name?”, they might tell you, or they might say “Which one?”. To a family with two daughters, the question “What is your daughter’s name?” doesn’t make sense. Of these families, 1/3 will have two girls, and 2/3 will have a boy and a girl.

So, everyone who’s first child is a daughter and raises their hand has 50% chance of having a son also. Everyone who has a second child that is a daughter has a 50% chance to have a son also. But you think when these two groups are asked if they have at least one daughter, the 50% chance averages out to 2/3 in favor of boys.

The problem is you are using possible gender combinations and claiming it doesn’t matter what order the female came in because they’re both female. Female and female is actually two possibilities like I first stated, just because the female being first of two girls or second is the same result, doesn’t mean it’s not two possible outcomes.

If you put money down at those odds I’ll take the bet.

[ZenBeam]

@Skinny

So, everyone who’s first child is a daughter and raises their hand has 50% chance of having a son also. Everyone who has a second child that is a daughter has a 50% chance to have a son also. But you think when these two groups are asked if they have at least one daughter, the 50% chance averages out to 2/3 in favor of boys.

Those two groups are not independent. The group with two girls is getting counted twice if you just average the first two cases.
Look, just make a sample of four families:
1 Girl, Boy
2 Boy, Girl
3 Boy, Boy
4 Girl, Girl

“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.

[Powers]

@Terr

For the purposes of the problem here, what is the difference between the statement “the family has two kids, at least one of which is a daughter” and “the family has two kids, and the one I just met is a daughter”?

Because in the first statement, the one you just met could be the /other/ daughter, not the one being referred to with “one of our kids is a girl”. In the second statement, the one you just met and the one who is a girl are stipulated as the same person.
Powers &8^]

[Cartoonacy]

Here’s a way to prove it experimentally:

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. Now look at each pair, and discard the ones that have two black cards, since we know that the family has at least one girl. Of the pairs that remain, how many of them have two red cards? (Repeat a few times to get a decent sample size.)

[tim-n-va]

@Skinny

From the Monty Hall problem
On “Let’s Make a Deal,” you pick Door #1. Monty opens Door #2 — no prize. Do you stay with Door #1 or switch to #3? - The Straight Dope

Now, I don’t have a problem with the Monty Hall question and please don’t add more on that here, as it’s been gone over well enough. The problem is with this little nugget.

“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.)”

He goes on to explain it like this

(1) Child A is female and Child B is male.

(2) Child A is female and Child B is female.

(3) Child A is male and Child B is female.

(4) Child A is male and Child B is male.

We know one child is female, eliminating choice #4. In 2 of the remaining 3 cases, the female child’s sibling is male. QED.

This doesn’t seem to make sense, but the data there supports the findings, except the data isn’t correct. If you claim the order of female and male siblings counts, then the order of female and female siblings must count as well, same goes for the two males.

It should look like this. We will name our female child Claire to clear it up a bit.

(1) Child A is Claire and Child B is male.

(2) Child A is Claire and Child B is female.

(3) Child A is female and Child B is Claire.

(4) Child A is male and Child B is Claire.

(5) Child A is male(x) and Child B is male(y).

(6) Child A is male(y) and Child B is male(x).

We know one child is Claire, eliminating choice #5-6. In 2 cases the remaining child is male, in 2 cases female. If order matters in one case, it matters in all.

You get the four options by enumerating over combinations of independent events. Birth A is male or female. Independent of that Birth B is male or female. Four possible outcomes as explained.

Your six choices include two and three which are identical: Birth A female, Birth B female.
It also includes choices five and six which are identical: Birth A male, Birth B male.

[bump]

@ZenBeam

Those two groups are not independent. The group with two girls is getting counted twice if you just average the first two cases.
Look, just make a sample of four families:
1 Girl, Boy
2 Boy, Girl
3 Boy, Boy
4 Girl, Girl

“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.

The OP question:

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?

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.

When we find out that one of the children is a girl, we can rule out the both boys category, and then the chance that it’s a boy is 1 in 2 chance that the family has one of each versus 2 girls.

[Heart_of_Dorkness]

@bump

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.

While it’s arguably true that groups 1 and 2 are equivalent, in the sense that they both represent a family with one boy and one girl, they aren’t literally the same group of people. If you treat them as one group - that is “one of each”, rather than “girl/boy” and “boy/girl”, there will be twice as many families in the “one of each” group as in either the “girl/girl” group or the “boy/boy” group. And therefore, it will be twice as *likely *to have “one of each”. Here’s why:

Start with 100 families. Each of them has a kid. Given 50/50 gender odds, you’ll get:

50 families with a girl
50 families with a boy

Now, each of them has another kid. Again, with 50/50 odds, you’ll get:

25 families with a girl, and then a girl
25 families with a girl, and then a boy
25 families with a boy, and then a girl
25 families with a boy, and then a boy

For the sake of the question, we eliminate those families that don’t have a girl, and we’re left with 75 families; 25 with just girls, and 50 - twice as many - with “one of each”:

25 families with a girl, and then a girl
25 families with a girl, and then a boy
25 families with a boy, and then a girl

So the chances of one of these families also having a son is 50/75 or 2/3.

[Telemark]

@bump

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

As said, if you group them into three options (which is a perfectly reasonable way of looking at the problem) then the probabilities are: 50%, 25%, and 25%.

All options don’t have equivalent probabilities - this is the crux of understanding the question.

[Amateur_Barbarian]

As with so many of these problems, it comes down to semantic wanking. If you carefully construct the query one way AND observe rigid logical rules, the answer is one thing; construct the query another way that most people would see as identical to the first AND again observe rigid logical rules and the answer is different.

I honestly don’t understand the amount of skull time and typing sweat expended on these constructs when in the end they make Clinton’s “the meaning of ‘is’” seem deep and rational.

[Heart_of_Dorkness]

@Telemark

All options don’t have equivalent probabilities - this is the crux of understanding the question.

That’s it in a nutshell. There are two possible outcomes to playing the lottery: either you win, or you lose. But that doesn’t mean you have 50/50 odds.

[Terr]

@Powers

Because in the first statement, the one you just met could be the /other/ daughter, not the one being referred to with “one of our kids is a girl”.

You haven’t met any in the first statement.

[DataX]

@Heart_of_Dorkness

That’s it in a nutshell. There are two possible outcomes to playing the lottery: either you win, or you lose. But that doesn’t mean you have 50/50 odds.

Yes, but that isn’t going to convince the people who are going to say:

“But the chances of winning the lottery are 1 in 1,000 (or whatever). The chances of having a boy (or girl) ARE 50/50”

I agree it is 2/3

[Chronos]

Quoth Amateur Barbarian:

If you carefully construct the query one way AND observe rigid logical rules, the answer is one thing; construct the query another way that most people would see as identical to the first AND again observe rigid logical rules and the answer is different.

Correct. The two constructions that most people think are identical aren’t actually identical, and the subtle differences (that most people overlook) are actually significant and lead to a different answer.