3 coins interpretation of the boy and girl paradox Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

I think I have come up with a new interpretation for this problem. If children where coins, boys were heads and girls were tails, we would have $3$ coins: $2$ of them can be both heads and tails, and the other one is loaded so it can only be heads. We start throwing one of the $2$ normal coins. If it is heads, we throw the other normal coin. If it is tails, we throw the loaded coin. This way, the chance of both coins being heads ends up being ${1 \over 2}\cdot {1 \over 2}$, which is ${1 \over 4}$. This is neither ${1 \over 3}$ nor ${1 \over 2}$, which are the classic answers to this problem. 

Is my interpretation wrong?
This would be easier with two coins, both fair.

Version (a): toss both coins.  If both are tails, start again.  Otherwise  record what happened.  About $\frac{1}{3}$ of records will show two heads.    

Version (b): toss the first coin.  If it is tails, start again.  Otherwise toss the second coin and record what happened.  About $\frac{1}{2}$ of records will show two heads.   

Your version is closer to "There is a family who wants at least one son.  If their first child is a boy they will have another child, but if it is a girl then they intend to use adoption or gender-selective abortion to make sure their second is a boy." That indeed would give a probability of $\frac14$ for the probability of two boys, but it is not either of the standard forms of the question.
In your method you are calculating based on the chance that if the first throw of normal coin is a head, however in the original problem one head is always guaranteed