AGH! a smiley! try again:
“A couple has two children, one of them a girl. What are the odds the other is a boy?” (from a letter to Cecil Adams)
If the questioner knows the answer, according to the customs of English semantics, he has just told us the answer: “one of them” is a girl. The other is a boy. Easy.
Oh! You meant to say, “A couple has two children, at least one of them a girl. What are the odds the other is a boy?” Well, that’s… interesting.
OK, let’s phrase it this way:
Case A: The Johnsons, we are told, have a child. This normally appears as something like, “They have a child in the second grade.”
Knowing nothing about any other children that they may or may not have, we will assume roughly a 50% chance that the child we know of is a boy. (I could guess 50% chance male, 49% female, 1% hermaphrodite, but without more information, that’s making some highly specific assumptions about actual biological probabilities.)
Solution (by definition): {chance male}:{chance female}=1:1
Case B: The Smiths have a daughter. You’ve met her. (She’s blonde, her name is Sally, she’s between eight and eleven years old.) You are told they have one other child. You are not told the child’s age.
There are two obvious questions we can ask here, each with two possible answers:
[ul][li]One is the relative age of the child–whether this unknown child is older or younger than Sally. The actual probabilities here are dependent upon Sally’s age as well as upon the ages of the parents, and may be unequal. Let’s name the (unknown) probability that the child is older than Sally o:y.[/li][li]The other is the sex of the unknown child. Once we define the child’s age, we can agree that this probability is, by definition, as in Case A, an even split, or 1:1.[/ul][/li]So we can list the relative probabilities of different answers as to both the child’s age and sex as follows:
{Sally has an older sister}:{Sally has an older brother}:{Sally has a younger sister}:{Sally has a younger brother}=o:o:y:y
We divide out the o:y ratio, and find that Sally’s existence has no bearing on the actual probabilities of her sibling’s sex. (We will assume normal reproductive systems and random copulation patterns in her parents.)
However, this is an ideal construction: Sally’s parents could have some biological reason that they are more likely to produce daughters, and it is more likely that Sally was born to such parents than that she was born to parents biologically more likely to produce sons.
So, is the real answer that Sally is (however slightly) more likely to have a sister than a brother? Probably. Not so fast! We have to overcome an objection first.
The Objection: If we do a statistical sampling (in an ideal world) of families with exactly two children, it will appear that the following ratio obtains: {older boy, younger boy}:{older boy, younger girl}:{older girl, younger boy}:{older girl, younger girl}=1:1:1:1
Thus, of two-child families “with at least one daughter,” 2/3 have a son, 1/3 have two girls. From this, my fellow Dopers (and Cecil, and Marilyn vos Savant) appear to argue, there is a 2:1 likelihood that a girl with one sibling will have a brother. This is counter-intuitive, and I will show it to be incorrect. I was tempted to say that this is one of those paradoxes which make probabilistic reasoning suspect (see note at end), but in fact the confusion arises from an oversimplification.
Those who espouse the “2/3 answer” are attempting to compute the probabilities of different families–which is fine, if you’re asking the following question: “Assuming equal likelihood of a woman giving birth either to a son or to a daughter; of families with two children, at least one of which is a girl, what proportion shall have sons?” That question asks the statistical incidence of a particular family composition among other families.
But the given question was, “A couple has two children, at least one of them a girl. What are the odds the other is a boy?”
Or as Cecil phrased it, “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?”
This is a different question, dealing with odds within this family–and it has a different answer.
In answering this question, those who espouse the “2/3 answer” are not sufficiently considering the family with two daughters. In the statistical sample given in the Objection, there are four possibilities for Sally. Let’s call the family types E, F, G, & H:
Type E: older boy, younger boy --no possibility for Sally
Type F: older boy, younger girl --one possibility for Sally
Type G: older girl, younger boy --one possibility for Sally
Type H: older girl, younger girl–two possibilities for Sally
Doubtful? Let’s frame it again, like this:
Case C: You have just met the Smythes (let’s say, in a public place). A girl appears, clearly their daughter. They call her “Sally.” Mrs. Smythe says, “We have one other child.” --STOP–
What are the possibilities?
I: Sally is the younger child–the daughter–of family type F.
II: Sally is the elder child–the daughter–of family type G.
III: Sally is the younger child–the second daughter–of family type H.
IV: Sally is the elder child–the first daughter–of family type H.
These are the same possibilities (in a different order) as in Case B, and they have the same odds: in this order o:y:o:y, as defined above. Remember, we’re asking about the unseen child’s probabilities within this family. Having seen Sally, we know that she could occupy two different positions in family type H, but she could occupy only one position in family type F and only one position in family type G. Thus, we must weight the likelihood of her being in family type H as twice that of her being in family type F and as twice that of her being in family type G.
{Smythes are Type F}:{Smythes are Type G}:{Smythes are Type H}=1:1:2
Since in types F & G, Sally would have a brother, and in type H she would have a sister, there are equal chances that Sally’s sibling is a boy or a girl.
Thus, given the question as stated, the likelihood that the unknown child is a boy is as in Case A, that is, 1:1, or 1/2.
Quod erat demonstratum.
Note: There are certain cases where probability problems will not work, but will reveal an multitude–even an infinity, I suppose–of differing results, depending on the “solution’s” approach. The class of problems I think of is the “continuous probabilities.” Fortunately, the above problem is not one if those.
