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 $1/2$"?

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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

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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