A basic birthday probability question The problem: Suppose the members of a family of 3 were either born on Saturday or Sunday. What are the chances of this family having birthday parties on both Saturday and Sunday?
First approach (from the textbook):
With $x_{i}$ as the number of birthdays on the day $i$:
$n(S)=\binom{4}{3}=4$, as the number of the solutions of $x_{1}+x_{2}=3, x\geqslant0$.
$n(A)=\binom{2}{1}=2$, as the number of the solutions of $x_{1}+x_{2}=3, x\geqslant1$.
Thus $P(E)=\frac{n(A)}{n(S)}=\frac{1}{2}.$
Second approach (mine):
Since the events are independent, the chances of all 3 being born on Saturday would be $(\frac{1}{2})^3$, same with Sunday. So the chances of having birthdays on both days would be $1-2\times \frac{1}{8}=\frac{3}{4}$.

So which one of these approaches are correct? And what is the reason of the other giving the wrong result?
 A: Of course we assume that any given person is equally likely to be born on Saturday as Sunday, particularly given that no other explicit statement to the contrary is given.  This is the so-called "non-informative" prior, based on the "principle of indifference."  Also, on average there are as many Saturdays as Sundays within a year.  (As "meta-reasoning," I would point to the fact that the OP has a reputation of $31$, and so would likely have made this assumption but not voiced it explicitly.)
The second is correct.  Person $A$ is on one of the weekend days.  Call it Saturday (without loss of generality).  What is the probability that at either $B$ or $C$ (or both) are on Sunday?  Well, it is 1 minus the probability both are on Saturday, i.e., $1 - (1/2)^2 = 3/4$.
Another way:  What is the probability all three are on Saturday?  $(1/2)^3$. What is the probability all three are on Sunday?  $(1/2)^3$.  So what is the probability all three on one same day?  $1/8 + 1/8 =1/4$.  What is the probability they are not all on the same day?  $1 - 1/4 = 3/4$.
