Boy Born on a Tuesday - is it just a language trick? The following probability question appeared in an earlier thread:

I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?

The claim was that it is not actually a mathematical problem and it is only a language problem.

If one wanted to restate this problem formally the obvious way would be like so:
Definition: Sex is defined as an element of the set $\\{\text{boy},\text{girl}\\}$.
Definition: Birthday is defined as an element of the set $\\{\text{Monday},\text{Tuesday},\text{Wednesday},\text{Thursday},\text{Friday},\text{Saturday},\text{Sunday}\\}$
Definition: A Child is defined to be an ordered pair: (sex $\times$ birthday).
Let $(x,y)$ be a pair of children,
Define an auxiliary predicate $H(s,b) :\\!\\!\iff s = \text{boy} \text{ and } b = \text{Tuesday}$.
Calculate $P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y))$
I don't see any other sensible way to formalize this question.

To actually solve this problem now requires no thought (infact it is thinking which leads us to guess incorrect answers), we just compute
$$
\begin{align*}
& P(x \text{ is a boy and } y \text{ is a boy}|H(x) \text{ or } H(y)) \\\\
=& \frac{P(x\text{ is a boy and }y\text{ is a boy and }(H(x)\text{ or }H(y)))}
        {P(H(x)\text{ or }H(y))} \\\\
=& \frac{P((x\text{ is a boy and }y\text{ is a boy and }H(x))\text{ or }(x\text{ is a boy and }y\text{ is a boy and }H(y)))}
        {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\\\
=& \frac{\begin{align*} &P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday}) \\\\
   + &P(x\text{ is a boy and }y\text{ is a boy and }y\text{ born on Tuesday}) \\\\
   - &P(x\text{ is a boy and }y\text{ is a boy and }x\text{ born on Tuesday and }y\text{ born on Tuesday}) \\\\
   \end{align*}}
   {P(H(x)) + P(H(y)) - P(H(x))P(H(y))} \\\\
=& \frac{1/2 \cdot 1/2 \cdot 1/7 + 1/2 \cdot 1/2 \cdot 1/7 - 1/2 \cdot 1/2 \cdot 1/7 \cdot 1/7}
        {1/2 \cdot 1/7 + 1/2 \cdot 1/7 - 1/2 \cdot 1/7 \cdot 1/2 \cdot 1/7} \\\\
=& 13/27
\end{align*}
$$

Now what I am wondering is, does this refute the claim that this puzzle is just a language problem or add to it? Was there a lot of room for misinterpreting the questions which I just missed? 

Well, given the unstated assumption that the writer is a mathematician and therefore not using regular english, then I agree with the 13/27 answer. 
But in everyday english, from "there are two fleems, one is a glarp" we all infer that the other is not a glarp. 
From "there are two fleems, one is a glarp, which is snibble" we would still infer that the other is not a glarp. Whereas from "there are two fleems, one is a glarp which is snibble" (absence of comma, or when spoken, difference in intonation) we would infer that the other is not a snibble glarp, but it could still be an unsnibble glarp.
