Distribution of ages of 3 children in a family Please consider Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 30, p 88.



*A family has 3 children, creatively named A,B, and C.

(a) Discuss intuitively (but clearly) whether the event “A is older than B” is independent
of the event “A is older than C”.
(b) Find the probability that A is older than B, given that A is older than C.

Let X, Y, Z be the ages of A, B, C respectively.
I assumed X, Y, Z are i.i.d. random variables with each following uniform distribution U(0,18). (18 can be changed to any constant, obviously it doesn't affect the probabilities in question here.)
By this logic the event (X>Y) and (X>Z) are independent events, each having probability 0.5. $P(X>[Y \cap Z])$ is just $P((X>Y)∩(X>Z))=P(X>Y)\times P(X>Z)=1/4$.
However the author's solution takes a discrete approach, assuming the 6 possible birth orders to be equally likely events with probability 1/6 each. If we make this assumption we can see that A is older than B means A is not the youngest, which takes out 2 of these 6 possibilities. Making the event "A is the oldest" 50% likely instead of 1/3 likely originally. Hence they're not independent. Given (A>B) increases the probability of (A>C).
This is completely logical and understandable, but my question is why are the two approaches not producing the same conclusions? What am I missing? I can spot one obvious difference that the birth order approach ignores the possibilities X=Y, Y=Z etc., but even under the uniform distribution assumption these events have zero probability (by standard continuous distribution properties).

The ages of $A,B,$ and $C$ may be independent†, but the events of pairwise orders are not.   If you are told that $A$ is one of the two oldest children (because $A$ is older than $C$) it should raise your anticipation that $A$ is also older than $B$.
(† though, actually, they are not independent if they have the same mother; but that is beside the point.)
Assuming that even in the case of twins and triplets the children will still be described as older and younger, there are $6$ equi-plausible arrangements of age. $$\begin{array}{|c|c|} & C<A & B<A & \text{Both} \\ A<B<C \\A<C<B\\B<A<C & & \checkmark \\B<C<A & \checkmark& \checkmark& \checkmark\\C<A<B & \checkmark\\C<B<A & \checkmark& \checkmark& \checkmark\end{array}$$  
The events $C<A$ and $B<A$ are not independent because they are codependent on the age of $A$.
