This is going to get a bit long for a comment.
Suppose you are asking three questions in sequence:
1 Do you have exactly two children?
If the answer is yes, you include the person on your sample, and ask
2 Do you have a child who is a boy born on a Tuesday?
If the answer is yes, the person continues in the sample - call this $BT$ and you ask
3 Are both of your children boys?
Say the yes answers to this are $BB$, you are looking for $\frac {BB}{BT}$
Note that (given reasonable assumptions) one fourteenth of all children are tuesday-born boys. However, at stage 2, there will be $0$ or $1$ or $2$ Tuesday-born boys in the family. Case $0$ is discarded, and the other two are not equally likely.
Taking the day and the sex into account, there are 14 possibilities for each child - 196 altogether. In one of these, both are Tuesday-boys. There are 26 more possibilities containing a Tuesday-boy, making 27 in all (the $2$ case takes two of the expected 28) - twelve of which have two boys (two of the fourteen having been taken into account already), and fourteen (as expected) with a boy and a girl.
Counting, one sees the proportion with two boys is $\frac {13}{27}$.
I noted this, because it was not explained in the answers to the closely related question Zev Chonoles linked.