It depends on the number of families.

Let $X$ be the number of children in a family, it is geometric random variable with $p=0.5$, i.e., $$ P(X = x) = 0.5^x, x=1,2,3... $$ which implies (E(X) = 2)

For $N$ families, the ratio is $$ \frac{\sum (X_i - 1)}{ \sum X_i} = \frac{\sum X_i - N}{\sum X_i} = 1 - \frac{N}{ \sum X_i} $$

Since $\sum X_i /N \longrightarrow E(X) = 2$ (law of large number), the ratio coverages to 1/2 if $N \longrightarrow \infty$.

If there are only finite families, let $T$ be the total number of children of the country: $T = \sum X_i$, then

 $T$ has a negative binomial distribution with pmf

 $$ P(T=t) = C^{t-1}_{N-1} 0.5^t, t = N, N+1... $$

therefore $$ E\left[ \frac{N}{\sum X_i} \right] = E\left[ \frac{N}{T} \right] = \sum_{t=N}^{\infty} \frac{N}{t} C^{t-1}_{N-1} 0.5^t = {_2F_1} (N, 1, N+1, -1) $$ where $_2F_1$ is the hypergeometric function.

Therefore the expected ratio is $ 1- {_2F_1} (N, 1, N+1, -1) $.

It depends on the number of families.

Let $X$ be the number of children in a family, it is geometric random variable with $p=0.5$, i.e., $$ P(X = x) = 0.5^x, x=1,2,3... $$ which implies $E(X) = 2$

Suppose there are $N$ families in the country, the girl ratio is $$ \frac{N}{ \sum X_i} $$

Since $\sum X_i /N \rightarrow E(X) = 2$ (law of large number), the ratio coverages to 1/2 if $N \rightarrow \infty$.

If there are only finite families, let $T$ be the total number of children of the country: $T = \sum X_i$, then  $T$ has a negative binomial distribution with pmf  $$ P(T=t) = C^{t-1}_{N-1} 0.5^t, t = N, N+1... $$

It implies $$ E\left[ \frac{N}{\sum X_i} \right] = E\left[ \frac{N}{T} \right] = \sum_{t=N}^{\infty} \frac{N}{t} C^{t-1}_{N-1} 0.5^t = {_2F_1} (N, 1, N+1, -1) $$ where $_2F_1$ is the hypergeometric function.

Therefore the expected girl ratio is ${_2F_1} (N, 1, N+1, -1) $.

It depends on the number of families.

Let $X$ be the number of children in a family, it is geometric random variable with $p=0.5$, i.e., $$ P(X = x) = 0.5^x, x=1,2,3... $$ which implies (E(X) = 2)

For $N$ families, the ratio is $$ \frac{\sum (X_i - 1)}{ \sum X_i} = \frac{\sum X_i - N}{\sum X_i} = 1 - \frac{N}{ \sum X_i} $$

Since $\sum X_i /N \longrightarrow E(X) = 2$ (law of large number), the ratio coverages to 1/2 if $N \longrightarrow \infty$.

If there are only finite families, let $T$ be the total number of children of the country: $T = \sum X_i$, then

$T$ has a negative binomial distribution with pmf

$$ P(T=t) = C^{t-1}_{N-1} 0.5^t, t = N, N+1... $$

therefore $$ E\left[ \frac{N}{\sum X_i} \right] = E\left[ \frac{N}{T} \right] = \sum_{t=N}^{\infty} \frac{N}{t} C^{t-1}_{N-1} 0.5^t = {_2F_1} (N, 1, N+1, -1) $$ where $_2F_1$ is the hypergeometric function.

Therefore the ratio is $ 1- {_2F_1} (N, 1, N+1, -1) $.

It depends on the number of families.

Let $X$ be the number of children in a family, it is geometric random variable with $p=0.5$, i.e., $$ P(X = x) = 0.5^x, x=1,2,3... $$ which implies (E(X) = 2)

For $N$ families, the ratio is $$ \frac{\sum (X_i - 1)}{ \sum X_i} = \frac{\sum X_i - N}{\sum X_i} = 1 - \frac{N}{ \sum X_i} $$

Since $\sum X_i /N \longrightarrow E(X) = 2$ (law of large number), the ratio coverages to 1/2 if $N \longrightarrow \infty$.

If there are only finite families, let $T$ be the total number of children of the country: $T = \sum X_i$, then

$T$ has a negative binomial distribution with pmf

$$ P(T=t) = C^{t-1}_{N-1} 0.5^t, t = N, N+1... $$

therefore $$ E\left[ \frac{N}{\sum X_i} \right] = E\left[ \frac{N}{T} \right] = \sum_{t=N}^{\infty} \frac{N}{t} C^{t-1}_{N-1} 0.5^t = {_2F_1} (N, 1, N+1, -1) $$ where $_2F_1$ is the hypergeometric function.

Therefore the expected ratio is $ 1- {_2F_1} (N, 1, N+1, -1) $.