A boy born on a Tuesday

Idea. The condition is about a specific, fairly rare type of child. Count the equally likely configurations directly instead of trusting intuition.

Step 1: the condition. Ordered pairs (first child, second child) with at least one Tuesday boy: the first child is a Tuesday boy ($14$ pairs), or the second is ($14$ pairs), minus the pair counted twice where both are. That is $14+14-1=27$ pairs.

Step 2: both boys. Among these, pairs where both are boys: the first is a Tuesday boy and the second any boy ($7$), or the second is a Tuesday boy and the first any boy ($7$), minus the double count: $7+7-1=13$.

Step 3. $$\mathbb{P}(\text{two boys}\mid \text{a Tuesday boy})=\frac{13}{27}\approx \mathrm{0.481.}$$

Why the weekday matters. If you only knew “at least one is a boy”, the answer would be $\frac{1}{3}$. If you knew “the elder is a boy”, it would be $\frac{1}{2}$. A specific detail like the weekday nearly identifies one child, so the answer moves close to $\frac{1}{2}$. What changes the answer is how the information was obtained, which is why these puzzles need a precise statement.