If population keep trying till they have girl child, what will be the probability of population having more girls than boys and vice versa? I was solving this problem:

In a world where everyone wants a girl child, each family continues having babies till they have a girl. What do you think will the boy to girl ratio be eventually? (Assuming probability of having a boy or a girl is the same)

The solution given was:

Suppose there are N couples. First time, N/2 girls and N/2 boys are born (ignoring aberrations). N/2 couples retire, and rest half try another child. Next time, N/4 couples give birth to N/4 girls and rest N/4 boys. Thus, even in second iteration, ratio is 1:1. It can now be seen that this ratio always remain same, no matter how many times people try to give birth to a favored gender.

My doubt is that will following be the case:
P(population will have more girls)= P(population will have equal number of boys and girls)= 1/2
Consider there are 16 couples

*

*8 give birth to girls and hence stop. 8 give birth to boys, so they give another chance.

*4 give birth to girls and hence stop. 4 give birth to boys, so they give another chance.

*2 give birth to girls and hence stop. 2 give birth to boys, so they give another chance.

*1 give birth to a girl and hence stop. 1 give birth to a boy, so they give another chance.

*Note till now there Number of boys = Number of girls

*Now probability that a single remaining couple give birth to a girl is 1/2. In that case they will stop and there will be one more girl than boys in the population. Probability that a single remaining couple give birth to a boy is 1/2, in which case they will give another chance in which they will again have a 1/2 probability of giving birth to boys, thus again balancing girl-boy ratio.

So am I correct with two facts:
Fact 1:
P(population will have more girls than girls)= P(population will have equal number of boys and girls)= 1/2
Fact 2:
P(population will have more boys than girls) = 0
 A: I'll give an alternative solution to the original problem.
For each $i\in \{1,\ldots,N\},\ $ let $X_i\ $ be the random variable, "number of babies a couple has, up to and including their first girl." Then, for each $i,\ X_i\sim \text{Geo}\left(\frac{1}{2}\right),\ $ where $\text{Geo}\ $ denotes the Geometric distribution. We know that $\text{Exp}(X_i) = \frac{1}{\frac{1}{2}} = 2\ $ for every $i.$
By the algebra of linear combination of random variables, the expected total number of births (in the population) such that every couple in the population has a girl, is given by $\text{Exp}(X_1 + X_2 + \ldots + X_N) = \text{Exp}(X_1) + \text{Exp}(X_2) + \ldots + \text{Exp}(X_N) = 2 + 2 + \ldots + 2 = 2N.$ But each couple has exactly one girl, so that's $N$ girls in total, and therefore $N$ boys in total also. So the expected boy to girl ratio is $1:1.$
As for your "Facts", they are both clearly wrong, and to me show a lack of understanding of discrete probability distributions (Like Geometric or Binomial), but I'm tired and I'll let someone else have the burden of explaining why your facts don't make sense.
A: If there are N couples, the probability that  couple 1 has $k$ boys and all the other couples have no boys is given by ...
$$P_{k, 0, 0,0...}= \frac{1}{2^{N+k} }$$
A moment's reflection should convince you that this same probability applies to any possible way of distributing the $k$ boys among the $N$ couples.
By stars and bars, the number of possible $N$-tuples of whole numbers summing to $k$ is given by...
$$ N_k = \binom{N+k-1 }{N-1  }$$
So $$P(k \text { boys })  = \frac{\binom{N+k-1 }{N-1}}{  2^{N+k}}$$
In particular, if there are $N=16$ couples and there are equal numbers of girls and boys
$$P(16 \text{ boys and} 16 \text{ girls })=\binom{31}{15}2^{-32}\approx 0.07$$
