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Q&A

How can "information about the birth season" bring "at least one is a girl" closer to "a specific one is a girl"?

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Please see the sentences beside my red highlighted words. I don't understand how "Conditioning on more and more specific information brings the probability closer and closer to $1/2$"?


Example $2.2 .7$ (A girl born in winter). A family has two children. Find the probability that both children are girls, given that at least one of the two is a girl who was born in winter. In addition to the assumptions from Example $2.2 .5$, assume that the four seasons are equally likely and that gender is independent of season. (This means that knowing the gender gives no information about the probabilities of the seasons, and vice versa; see Section $2.5$ for much more about independence.)

Solution: By definition of conditional probability, $$P(\text{both girls} \mid \text{at least one winter girl} )=\frac{P(\text { both girls, at least one winter girl })}{P(\text { at least one winter girl })}$$ Since the probability that a specific child is a winter-born girl is $1 / 8$, the denominator equals $$P( \text{at least one winter girl} )=1-(7 / 8)^{2}$$ To compute the numerator, use the fact that "both girls, at least one winter girl" is the same event as "both girls, at least one winter child"; then use the assumption that gender and season are independent:

$$ P \text { (both girls, at least one winter girl })=P \text { (both girls, at least one winter child) }$$ $$=(1 / 4) P(\text { at least one winter child }) $$ $$=(1 / 4)(1-P(\text { both are non-winter }))$$ $$=(1 / 4)\left(1-(3 / 4)^{2}\right) $$

$$ \text { Thus, } P \text { (both girls|at least one winter girl })=\frac{(1 / 4)\left(1-(3 / 4)^{2}\right)}{1-(7 / 8)^{2}}=\frac{7 / 64}{15 / 64}=7 / 15 \text { . } $$

At first this result seems absurd! In Example 2.2.5, the result was that the conditional probability of both children being girls, given that at least one is a girl, is $1 / 3 ;$ why should it be any different when we learn that at least one is a winter-born girl? The point is that information about the birth season brings "$\color{red}{\text{at least one is a girl}}$" closer to "$\color{red}{\text{a specific one is a girl}}$". Conditioning on more and more specific information brings the probability closer and closer to $1 / 2$.

For example, conditioning on "at least one is a girl who was born on a March 31 at $8: 20 \mathrm{pm}$ " comes very close to specifying a child, and learning information about a specific child does not give us information about the other child. The seemingly irrelevant information such as season of birth interpolates between the two parts of Example 2.2.5. Exercise 29 generalizes this example to an arbitrary characteristic that is independent of gender.

Blitzstein, Introduction to Probability (2019 2 ed) p 52. Original image


Example $2.2 .5$ (Two children). Martin Gardner posed the following puzzle in the $1950 \mathrm{~s}$, in his column in Scientific American. Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls? Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

At first glance this problem seems like it should be a simple application of conditional probability, but for decades there have been controversies about whether or why the two parts of the problem should have different answers, and the extent to which the problem is ambiguous. Gardner gave the answers $1 / 2$ and $1 / 3$ to the two parts, respectively, which may seem paradoxical: why should it matter whether we learn the older child's gender, as opposed to just learning one child's gender?

It is important to clarify the assumptions of the problem. Several implicit assumptions are being made to obtain the answers that Gardner gave.

  • It assumes that gender is binary, so that each child can be definitively categorized as a boy or a girl. In fact, many people don't neatly fit into either of the categories "male" or "female", and identify themselves as having a non-binary gender.
  • It assumes that $P($ boy $)=P($ girl $)$, both for the elder child and for the younger child. In fact, in most countries slightly more boys are born than girls. For example, in the United States it is commonly estimated that 105 boys are born for every 100 girls who are born.
  • It assumes that the genders of the two children are independent, i.e., knowing the elder child's gender gives no information about the younger child's gender, and vice versa. This would be unrealistic if, e.g., the children were identical twins. Under these (admittedly problematic) simplifying assumptions, we can solve the problem as follows. Solution: With the assumptions listed above, the definition of conditional probability gives $$P( \text{both girls} \mid \text{elder is a girl} )=\frac{P(\text { both girls, elder is a girl })}{P(\text { elder is a girl })}=\frac{1 / 4}{1 / 2}=1 / 2,$$ $$P( \text{both girls} \mid \text{at least one girl} )=\frac{P(\text { both girls, at least one girl })}{P(\text { at least one girl })}=\frac{1 / 4}{3 / 4}=1 / 3$$ (We solved the second part of the problem in terms of girls rather than boys to make it a bit easier to compare the two parts of the problem.) It may seem counterintuitive that the two results are different, since there is no reason for us to care whether the elder child is a girl as opposed to the younger child. Indeed, by symmetry, $$P(\text{both girls} \mid \text{younger is a girl })=P(\text{ both girls }\mid \text{ elder is a girl })=1 / 2$$ However, there is no such symmetry between the conditional probabilities $P$ (both girls|elder is a girl) and $P($ both girls $\mid$ at least one girl). Saying that the elder child is a girl designates a specific child, and then the other child (the younger child) has a $50 %$ chance of being a girl. "At least one" does not refer to a specific child. Conditioning on a specific child being a girl knocks away 2 of the 4 "pebbles" in the sample space ${G G, G B, B G, B B}$, where, for example, $G B$ means the elder child is a girl and the younger child is a boy. In contrast, conditioning on at least one child being a girl knocks away only $B B$.

Op. cit. p 49. Original image

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‘More specific information’, here, is informally referring to the value of the (positive) likelihood ratio: the ratio between how likely it is to get that information if a proposition is true, and how likely it is to get that information if the proposition is false. (I say ‘informally’ because, as you're aware, specificity is also a technical term in statistics, which is computed somewhat differently. The author here is using ‘specific’ in its general sense, not the technical sense.)

In this example, if the proposition is question is that both children are girls, the likelihood ratio corresponding to the information that the oldest child is a girl is $\frac{P(\text{oldest is girl} \mid \text{both are girls})}{P(\text{oldest is girl} \mid \text{at least one is not a girl})}$, or $\frac{1}{1/3} = 3$. The likelihood ratio corresponding to the information that at least one child is a girl is $\frac{P(\text{at least one is girl} \mid \text{both are girls})}{P(\text{at least one is girl} \mid \text{at least one is not a girl})}$, or $\frac{1}{2/3} = \frac32$. For information like a winter child is a girl, the likelihood ratio will be somewhere between these two values—in this case, $\frac{P(\text{at least one is winter girl} \mid \text{both are girls})}{P(\text{at least one is winter girl} \mid \text{at least one is not a girl})} = \frac{7/16}{1/6} = \frac{21}{8}$.

Note that the difference between ‘girl’ and ‘winter girl’ shows up on both sides of that ratio—the probability that at least one is a winter girl is less than the probability that at least one is a girl, both under the assumption that both are girls and under the assumption that at least one is not a girl. Any additional requirement placed on the girl will do this. But critically, the probability is hit harder in the case where at least one child is not a girl—because the both-girls case has two chances to meet the requirement, and the other case has one or zero. So additional requirements increase the likelihood ratio, and the stricter the requirement, the fewer girls there are who could meet it, and the stronger the effect. ‘The oldest child’ identifies exactly one girl, which is the strictest possible requirement short of impossibility, so we can't expect the likelihood ratio to increase beyond 3.

I'm guessing your book covers the value of likelihood ratios elsewhere; for the purposes of this example, it's enough to understand that a likelihood ratio completely captures the value of information in the context of a particular proposition. If you have an initial assumption for the probability of the proposition, the likelihood ratio is all you need to compute the probability of the proposition given the new information. The larger the likelihood ratio, the greater the conditional probability will be. So a likelihood ratio between two others will result in a conditional probability between the probabilities you'd get from those alternate pieces of information, and everything I just said about likelihood ratios applies to conditional probabilities.

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