There are even trickier aspects to this question. For example, what is the strategy of the guy telling you about his family? If he always mentions a boy first and not a daughter, we get one probability; if he talks about the sex of the first born child, we get a different probability. Your calculation makes a choice in this issue - you choose the version of "if the father has a boy and a girl, he'll mention the boy".
What I'm aiming to is this: the question is not well-defined mathematically. It has several possible interpretations, and as such the "problem" here is indeed of the language; or more correctly, the fact that a simple statement in English does not convey enough information to specify the precise model for the problem.
Let's look at a simplified version without days. The probability space for the make-up of the family is {BB, GB, BG, GG} (GB means "an older girl and a small boy", etc). We want to know what is $P(BB|A)$ where A is determined by the way we interpret the statement about the boys. Now let's look at different possible interpretations.
1) If there is a boy in the family, the statement will mention him. In this case A={BB,BG,GB} and so the probability is $1/3$.
2) If there is a girl in the family, the statement will mention her. In this case, since the statement talked about a boy, there are NO girls in the family. So A={BB} and so the probability is 1.
3) The statement talks about the sex of the firstborn. In this case A={BB,BG} and so the probability is $1/2$.
The bottom line: The statement about the family looks "constant" to us, but it must be looked as a function from the random state of the family - and there are several different possible functions, from which you must choose one otherwise no probabilistic analysis of the situation will make sense.
The overlap (possibility that both children are of the specific kind mentioned) is the issue, I think. Here’s a picture that helped me understand. Consider a random point $(x,y)$ in the unit square.
If you know that one of the two coordinates is bigger than $1/2$ (in set $A$), you have narrowed down the possibilities to the solid red region in the left picture. The chance that the other coordinate is also bigger than $1/2$ (you’re in $A\times A$) is about $1/3$ (blue-spotted part of the red region). It’s significantly less than the size of $A$ ($1/2$), because $A\times A$ is a significant part of $A\times I\cup I\times A$, the red region you know you’re in.
However, if you know that one coordinate is not only in $A$, but also in $B$ (say between $0.67$ and $0.71$), you are in the cross-shaped region on the right ($B\times I\cup I\times B$). The blue part of that region ($A\times B\cup B\times A$) is pretty close to $1/2$ of the cross, because the overlap is not so significant.
Best Answer
The problem with your proposed answer is the sentence "Why can't I just assume that that boy was born on some Xday ..." There is no "that boy," because it is possible that there are two boys.
Here is a more detailed explanation. Let $A$ be the event that both children are boys. Let $B_1$ be the event that at least one child is a boy born on a Sunday, $B_2$ the event that at least one child is a boy born on a Monday, and so on for $B_3$ through $B_7$. We have $\bigcup_{i=1}^7 B_i = B = $ the event that at least one child is a boy.
As you have observed, the reasoning for the Tuesday problem would apply to any day of the week, so $P(A \mid B_1) = P(A \mid B_2) = \cdots = P(A \mid B_7) = 13/27$. It is tempting now to reason as follows: The sets $B_1$, ..., $B_7$ are disjoint, since a child can't be born on two different days of the week. And using this disjointness, it is not hard to show that $P(A \mid B)$ is the average of $P(A \mid B_1)$ through $P(A \mid B_7)$ (just use the definition of conditional probability), so it is also 13/27. But we know that $P(A \mid B) = 1/3$, so something is wrong.
The mistake is that the sets $B_1$ through $B_7$ are not disjoint, because there could be two boys born on different days. So the argument that $P(A \mid B)$ is the average of the $P(A \mid B_i)$ doesn't work.
For more on this, see my book with Stan Wagon, Bicycle or Unicycle, problem 49.