Tuesday Two Boys puzzle - Intuition for why the answer changes from $\frac{1}{3}$ to $\frac{13}{27}$? In either the old or the new puzzle (see below), the "wrong answer" (that most people initially give) goes something like this:

Well we already know one coin is heads ("one child is a boy"). So $$\Pr(\text{Both coins are heads})=\Pr(\text{The other coin is heads})=\frac{1}{2}.$$

Let $g$ be the proportion of coins that are Golden ("proportion of boys that are born on Tuesday").
As $g\rightarrow1$, the answer to the new puzzle approaches the answer of the old puzzle: $\frac{2-g}{4-g}\rightarrow\frac{1}{3}$. To me there is a satisfying intuitive explanation for this: As $g\rightarrow1$, Golden heads becomes the same thing as heads, so we are back to the old puzzle.
As $g\rightarrow0$, the answer to the new puzzle approaches the "wrong answer": $\frac{2-g}{4-g}\rightarrow\frac{1}{2}$. What's the intuition for this? (I can't think of any and haven't come across any satisfying explanation.)
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Context for my question
I am avoiding answers that focus on criticizing the original wording of the puzzles (see e.g. this or this). So I've tried to reword the puzzles so that there is no ambiguity (feel free to suggest improvements).
Old puzzle (1959)

Everyone in the world randomly picks two coins from a huge pile of fair coins. Each person flips his two coins.
We kill everyone who didn't get at least one heads.
What is the probability that a randomly-chosen survivor got both heads?

Solution: $\frac{3}{4}$ (of the original population) are survivors and $\frac{1}{4}$ (of the original population) got both heads. Hence the answer is $\frac{1}{4}$ ÷ $\frac{3}{4}$ = $\frac{1}{3}$.
New puzzle (2010)

Everyone in the world randomly picks two coins from a huge pile of fair coins. Each person flips his two coins.
It turns out that a proportion $g=1/7$ of the coins are Golden, i.e. contains one milligram of gold (though this is undetectable except with special equipment).
We kill everyone who didn't get at least one Golden heads.
What is the probability that a randomly-chosen survivor got both heads?

Solution: Let $\hat{H}$ denote Golden heads and $X$ denote any flip of any coin.
By the inclusion-exclusion principle, the proportion of survivors (out of the original population) is:
$$\Pr(\hat{H}X)+\Pr(X\hat{H})-\Pr(\hat{H}\hat{H})=\frac{g}{2}1+1\frac{g}{2}-\frac{g}{2}\frac{g}{2}=g-\frac{g^2}{4}=\frac{27}{196}.$$
And the proportion (out of the original population) who got both heads with at least one Golden, is:
$$\Pr(\hat{H}H)+Pr(H\hat{H})-\Pr(\hat{H}\hat{H})=\frac{g}{2}\frac{1}{2}+\frac{1}{2}\frac{g}{2}-\frac{g}{2}\frac{g}{2}=\frac{g}{2}-\frac{g^2}{4}=\frac{13}{196}.$$
Hence the answer is
$$\frac{\frac{g}{2}-\frac{g^2}{4}}{g-\frac{g^2}{4}}=\frac{2-g}{4-g}=\frac{13}{27}.$$

As gold gets rarer, the probability that anyone has two golden heads becomes very small. In that case "pick a random survivor" becomes indistinguishable from "pick a random golden coin among those that came up heads, and see how the other coin picked up by the same person landed". And the result of that is obviously $\frac12$.
The reason for the difference from the native answer $\frac12$ is that people with two winning coins aren't proportionally more likely to be chosen than people with only one winning coin. However, when $g\to 0$, then double-wins become very rare compared to single wins, and so they don't skew the probability very much.
