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The proportion of girls in one family is a biased estimator of the proportion of girls in a population consisting of many families because you are underweighting the families with a large number of children.

If there were just 1 family, then your formula would be wrong, but the average of the percentage of girls you would observe would be

$$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \bigg(\frac{n}{n+1}\bigg) = 1-\log2 = 30.69\%.$$

Half of the time, you would observe $0\%$ girls.

If you have multiple families, the average of the observed percentage of girls in the population will increase.

For 2 families, the average percentage of girls would be

$$\sum_{n=0}^\infty \frac{n+1}{2^{n+2}} \bigg(\frac{n}{n+2}\bigg) = \log 4 - 1 = 38.63\%.$$

More generally, the average percentage for $k$ families is

$$\sum_{n=0}^\infty \frac{n+k-1 \choose k-1}{2^{n+k}} \bigg(\frac{n}{n+k}\bigg) = \frac{k}{2}\bigg(\psi(\frac{k+2}2)-\psi(\frac{k+1}2)\bigg)$$

where $\psi$ is the digamma function which satisfies

$$ \psi(m) = -\gamma + \sum_{i=1}^{m-1} \frac1i = -\gamma + H_{m-1}$$ $$ \psi(m+\frac12) = -\gamma -2\log 2 + \sum_{i=1}^m \frac{2}{2i-1}.$$

With a little work, one can verify that this goes to $1/2$ as $k\to \infty$. So, for a large population such as a country, the official answer of $1/2$ is approximately correct, although the explanation is misleading. In particular, for $10$ couples, the expected percentage of girls is $10 \log 2 - 1627/252 = 47.51\%$ contrary to what the official answer suggests. With $k$ families, the expected proportion is about $1/2 - 1/4k$.

It is not enough to argue that the expected number of boys equals the expected number of girls, since we want $E[G/(G+B)] \ne E[G]/E[G+B].$ Expectation is linear, but not multiplicative for dependent variables, and $G$ and $G+B$ are not independent even though $G$ and $B$ are.

The proportion of girls in one family is a biased estimator of the proportion of girls in a population consisting of many families because you are underweighting the families with a large number of children.

If there were just 1 family, then your formula would be wrong, but the average of the percentage of girls you would observe would be

$$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \bigg(\frac{n}{n+1}\bigg) = 1-\log2 = 30.69\%.$$

Half of the time, you would observe $0\%$ girls.

If you have multiple families, the average of the observed percentage of girls in the population will increase.

For 2 families, the average percentage of girls would be

$$\sum_{n=0}^\infty \frac{n+1}{2^{n+2}} \bigg(\frac{n}{n+2}\bigg) = \log 4 - 1 = 38.63\%.$$

More generally, the average percentage for $k$ families is

$$\sum_{n=0}^\infty \frac{n+k-1 \choose k-1}{2^{n+k}} \bigg(\frac{n}{n+k}\bigg) = \frac{k}{2}\bigg(\psi\left(\frac{k+2}2\right)-\psi\left(\frac{k+1}2\right)\bigg)$$

where $\psi$ is the digamma function which satisfies

$$ \psi(m) = -\gamma + \sum_{i=1}^{m-1} \frac1i = -\gamma + H_{m-1}$$ $$ \psi\left(m+\frac12\right) = -\gamma -2\log 2 + \sum_{i=1}^m \frac{2}{2i-1}.$$

With a little work, one can verify that this goes to $1/2$ as $k\to \infty$. So, for a large population such as a country, the official answer of $1/2$ is approximately correct, although the explanation is misleading. In particular, for $10$ couples, the expected percentage of girls is $10 \log 2 - 1627/252 = 47.51\%$ contrary to what the official answer suggests. With $k$ families, the expected proportion is about $1/2 - 1/(4k)$.

It is not enough to argue that the expected number of boys equals the expected number of girls, since we want $E[G/(G+B)] \ne E[G]/E[G+B].$ Expectation is linear, but not multiplicative for dependent variables, and $G$ and $G+B$ are not independent even though $G$ and $B$ are.

The proportion of girls in one family is a biased estimator of the proportion of girls in a population consisting of many families because you are underweighting the families with a large number of children.

If there were just 1 family, then your formula would be wrong, but the average of the percentage of girls you would observe would be

$$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \bigg(\frac{n}{n+1}\bigg) = 1-\log2 = 30.69\%.$$

Half of the time, you would observe $0\%$ girls.

If you have multiple families, the average of the observed percentage of girls in the population will increase.

For 2 families, the average percentage of girls would be

$$\sum_{n=0}^\infty \frac{n+1}{2^{n+2}} \bigg(\frac{n}{n+2}\bigg) = \log 4 - 1 = 38.63\%.$$

More generally, the average percentage for $k$ families is

$$\sum_{n=0}^\infty \frac{n+k-1 \choose k-1}{2^{n+k}} \bigg(\frac{n}{n+k}\bigg) = \frac{k}{2}\bigg(\psi(\frac{k+2}2)-\psi(\frac{k+1}2)\bigg)$$

where $\psi$ is the digamma function which satisfies

$$ \psi(m) = -\gamma + \sum_{i=1}^m \frac1i = -\gamma + H_{m-1}$$ $$ \psi(m+\frac12) = -\gamma -2\log 2 + \sum_{i=1}^m \frac{2}{2i-1}.$$

With a little work, one can verify that this goes to $1/2$ as $k\to \infty$. So, for a large population such as a country, the official answer of $1/2$ is approximately correct, although the explanation is misleading. In particular, for $10$ couples, the expected percentage of girls is $10 \log 2 - 1627/252 = 47.51\%$ contrary to what the official answer suggests.

It is not enough to argue that the expected number of boys equals the expected number of girls, since we want $E[G/(G+B)] \ne E[G]/E[G+B].$ Expectation is linear, but not multiplicative for dependent variables, and $G$ and $G+B$ are not independent even though $G$ and $B$ are.

The proportion of girls in one family is a biased estimator of the proportion of girls in a population consisting of many families because you are underweighting the families with a large number of children.

If there were just 1 family, then your formula would be wrong, but the average of the percentage of girls you would observe would be

$$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \bigg(\frac{n}{n+1}\bigg) = 1-\log2 = 30.69\%.$$

Half of the time, you would observe $0\%$ girls.

If you have multiple families, the average of the observed percentage of girls in the population will increase.

For 2 families, the average percentage of girls would be

$$\sum_{n=0}^\infty \frac{n+1}{2^{n+2}} \bigg(\frac{n}{n+2}\bigg) = \log 4 - 1 = 38.63\%.$$

More generally, the average percentage for $k$ families is

$$\sum_{n=0}^\infty \frac{n+k-1 \choose k-1}{2^{n+k}} \bigg(\frac{n}{n+k}\bigg) = \frac{k}{2}\bigg(\psi(\frac{k+2}2)-\psi(\frac{k+1}2)\bigg)$$

where $\psi$ is the digamma function which satisfies

$$ \psi(m) = -\gamma + \sum_{i=1}^{m-1} \frac1i = -\gamma + H_{m-1}$$ $$ \psi(m+\frac12) = -\gamma -2\log 2 + \sum_{i=1}^m \frac{2}{2i-1}.$$

With a little work, one can verify that this goes to $1/2$ as $k\to \infty$. So, for a large population such as a country, the official answer of $1/2$ is approximately correct, although the explanation is misleading. In particular, for $10$ couples, the expected percentage of girls is $10 \log 2 - 1627/252 = 47.51\%$ contrary to what the official answer suggests. With $k$ families, the expected proportion is about $1/2 - 1/4k$.

It is not enough to argue that the expected number of boys equals the expected number of girls, since we want $E[G/(G+B)] \ne E[G]/E[G+B].$ Expectation is linear, but not multiplicative for dependent variables, and $G$ and $G+B$ are not independent even though $G$ and $B$ are.

If there were just one such family, then your formula would be wrong, but the average of the proportion of girls you would observe would be

$$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \bigg(\frac{n}{n+1}\bigg) = 1-\log2 = 0.306...$$

Half of the time, you would observe a proportion of $0$.

If you have multiple families, the average proportion of girls in the population will increase.

The proportion of girls in one family is a biased estimator of the proportion of girls in a population consisting of many families because you are underweighting the families with a large number of children. For a large population such as a country, the official explanation is essentially correct.

The proportion of girls in one family is a biased estimator of the proportion of girls in a population consisting of many families because you are underweighting the families with a large number of children.

If there were just 1 family, then your formula would be wrong, but the average of the percentage of girls you would observe would be

$$\sum_{n=0}^\infty \frac{1}{2^{n+1}} \bigg(\frac{n}{n+1}\bigg) = 1-\log2 = 30.69\%.$$

Half of the time, you would observe $0\%$ girls.

If you have multiple families, the average of the observed percentage of girls in the population will increase.

For 2 families, the average percentage of girls would be

$$\sum_{n=0}^\infty \frac{n+1}{2^{n+2}} \bigg(\frac{n}{n+2}\bigg) = \log 4 - 1 = 38.63\%.$$

More generally, the average percentage for $k$ families is

$$\sum_{n=0}^\infty \frac{n+k-1 \choose k-1}{2^{n+k}} \bigg(\frac{n}{n+k}\bigg) = \frac{k}{2}\bigg(\psi(\frac{k+2}2)-\psi(\frac{k+1}2)\bigg)$$

where $\psi$ is the digamma function which satisfies

$$ \psi(m) = -\gamma + \sum_{i=1}^m \frac1i = -\gamma + H_{m-1}$$ $$ \psi(m+\frac12) = -\gamma -2\log 2 + \sum_{i=1}^m \frac{2}{2i-1}.$$

With a little work, one can verify that this goes to $1/2$ as $k\to \infty$. So, for a large population such as a country, the official answer of $1/2$ is approximately correct, although the explanation is misleading. In particular, for $10$ couples, the expected percentage of girls is $10 \log 2 - 1627/252 = 47.51\%$ contrary to what the official answer suggests.

It is not enough to argue that the expected number of boys equals the expected number of girls, since we want $E[G/(G+B)] \ne E[G]/E[G+B].$ Expectation is linear, but not multiplicative for dependent variables, and $G$ and $G+B$ are not independent even though $G$ and $B$ are.