The answer to the second question is that the ratio becomes undefined because the society has a 100% chance of dying out. The reason why this happens is fascinating: statistical fluctuations in the balance of males and females (a "second order effect") in a population that otherwise would appear to be stable in size (a "first order effect") will cause it to decrease.
Analysis
To make it clear what this statement means, here are the assumptions I use:
The "society" (let's just call it a population because it acts in a very non-human way) consists of males and females, all of which might potentially breed.
Breeding is performed by associating males and females permanently into as many pairs as possible. Each pair produces one offspring sequentially.
A pair produces males with 50% probability, independently with each birth and independently of all other pairs.
Breeding for a pair stops when either (a) a male is produced or (b) a predetermined finite limit $l$ of offspring is reached. (The latter is a necessary nod to realism: no individual is immortal!)
After all pairs stop breeding, the new generation commences breeding in the same way. (That is, there is no overlap of generations and no pairing between generations.)
Under these assumptions, all relevant information is conveyed by the number of males $M$ and number of females $F$ in the current generation. The number of breeding pairs equals $\min(M,F)$. Each pair produces one male (unless it reaches the limit) and a number of females according to a geometric distribution. This is a truncated geometric distribution for the number of females. If there were no limit, it would be a geometric distribution where the probability of $F$ females is
$$\begin{align}
\Pr(F=0) &= 1/2 \\
\Pr(F=1) &= 1/4 \\
\cdots \\
\Pr(F=k) &= 2^{-k-1} \\
\cdots
\end{align}
$$
Because there is a limit, this list stops with $\Pr(F=l) = 2^{-l-1} + 2^{-l-2} + \cdots = 2^{-l}$. Consequently the expected number of males per breeding pair is
$$
\mathbb{E}[M] = \left(1/2 + 1/4 + \cdots + 2^{-l} + 0 \times 2^{-l}\right) =1 - 2^{-l}.
$$
and the expected number of females can similarly be computed. It is clear that in a large population, because the expected number of males is less than one per breeding pair, the expected number of breeding pairs is less in each generation.
Furthermore--regardless of whether we include a limit in the model--variation in the numbers of males and females will cause the population eventually to die out anyway. This is because the number of breeding pairs can only equal the smaller of the numbers of males or females produced in the preceding generation and fluctations in both of those (which are up or down from the expected value with roughly equal probability) will cause the smaller of the two to fluctuate downwards much more often than it fluctuates upwards. This mechanism further forces the population down to zero.
Experiments
Rather than carry out a full and rigorous analysis of the effects of fluctuation, let's do a simulation and look at what happens to small populations of, say, $500$ males and $500$ females. I did this calculation $10,000$ separate times for $81$ new generations (setting the limit to an infinite value, to be conservative):
Clearly the continuation of this population is hopeless: although initially there is some chance that it increases, inevitably the population shrinks. The shrinking quickly is evident: within $12$ new generations, there is almost no chance the latest generation exceeds the size of the first generation (of $1000$). There is almost no chance this particular population will last more than $100$ generations.
(Larger populations will take more than $12$ generations to shrink with near certainty, but the same things eventually happen to them.)
Here is R code to perform the simulation.
next.gen <- function(pop, p=1/2, limit=30) {
n.females <- rgeom(n <- min(pop), p)
has.male <- n.females <= limit
return(c(males=sum(has.male),
females=sum(n.females[has.male]) + sum(rep(limit, n)[!has.male])))
}
simulate <- function(pop, n=1, ...) {
result <- matrix(0, nrow=n, ncol=2)
colnames(result) <- c("males", "females")
n.gen <- 1; k <- 0
while(n.gen <= n && sum(pop <- result[n.gen,] <- next.gen(pop, ...)) > 0)
{n.gen <- n.gen+1}
return(result)
}
pop <- c(males=500, females=500)
set.seed(17)
system.time(d <- replicate(10^4, rowSums(simulate(pop, 81, limit=Inf))))
par(mfrow=c(3,3))
tmp <- sapply(seq(1,81,10), function(i) hist(d[i,], breaks=seq(0,1200,25),
main=paste("Generation", i+1),
xlab="Total population"))
For (a), a simple way to look at is that you've reduced your probability space to only the combinations that have at least one boy:
BB = 1/3
BG = 1/3
GB = 1/3
GG is no longer a possibility based on the fact that your neighbor said he had at least one boy. Of the possibilities remaining, you're left with a 2/3 probability that he has a girl. The information he gave you reduced the probability of him having a girl from 3/4 to 2/3. Formally, this can be shown as follows: $$P(At\ least\ one\ girl|At\ least\ one\ boy) = \frac{P(At\ least\ one\ girl\ \cap At\ least\ one\ boy)}{P(At\ least\ one\ boy)} $$
From your original box, we can see the probability of having at least one boy and at least one girl is BG + GB = 0.25 + 0.25 = 0.5, but we need to divide by the probability of at least one boy, which is BB + BG + GB = 0.25 + 0.25 + 0.25 = 0.75, so we get $\frac{\tfrac{1}{2}}{\tfrac{3}{4}} = \frac{2}{3}$.
For (b), now that we've seen a boy, the only uncertainty remaining is the gender of the other child, and given no other information, the probability of the other child being female is 1/2, which is the answer.
Best Answer
Start with no children
repeat step
{
Every couple who is still having children has a child. Half the couples have males and half the couples have females.
Those couples that have females stop having children
}
At each step you get an even number of males and females and the number of couples having children reduces by half (ie those that had females won't have any children in the next step)
So, at any given time you have an equal number of males and females and from step to step the number of couples having children is falling by half. As more couples are created the same situation reoccurs and all other things being equal, the population will contain the same number of male and females