Tuesday Two Boys puzzle - Intuition for why the answer changes from $\frac{1}{3}$ to $\frac{13}{27}$?

Question

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

$$\frac{\frac{g}{2}-\frac{g^2}{4}}{g-\frac{g^2}{4}}=\frac{2-g}{4-g}=\frac{13}{27}.$$

Answer 1

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.

Answer 2

As far as your "old puzzle", "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?" is concerned, imagine 1000 people. Of those 1000, 1/4, 250, get both tails, 1/2, 500, get one head and one tail, and 1/4, 250, get both heads. We eliminate those who got both tails so we are left with 750 people, 250 of whom got both heads. The probability of "both heads" is 250/750= 1/3.

For the "new puzzle", imagine 49000 people. 1000 of them got two "gold coins", 12000 got one "gold coin" and 36000 got no gold coin. Each flips their coins. Of the first group, 250 get both heads, 500 get one head and one tail, 250 get both tails so 750 get at least one "gold head". Of the second group, 3000 get both heads, 3000 get heads on the gold coin and tails on the non-gold coin, 3000 get heads on the non-gold coin and tails on the gold coin and 3000 get tails on both coins so 6000 get a "gold head". Of the last group, since none had a gold coin none got a "gold head". That is, we eliminate all but 6750 of whom 3250 got "both heads". The probability is 3250/6750.