Probability of a family name surviving based on number of offspring If a family patriarch (e.g name = "Adams") has x number of children, and the probability of having a son or daughter is 50%, then what is the probability of his family name surviving for 3 or 4 or more generations?
(Assume all children marry, and only sons carry on the Adams name.)
E.g. If he has 1 child , probability  to survive 1 generation is 50%.
If he has 2 children, who each have 2 children,
75% (?) chance to survive 1 generation, then ? % chance to survive 2 generations, 3 generations, and so on?
Are there any models based on probability that can present this, e.g. in a simple program with 2 inputs, # of children per generation, and # of generations?
If this could be turned into a formula that can be charted, would the line approach an asymptote of 100% and resemble a mirrored exponential line?
 A: Each male child has (passes the family name to) $n$ children and each child has probability $p=1/2$ of being male. That is, each male child is modelled by a binomial random variable $X\sim B(n,p=1/2)$.
To investigate extinction (survival), we model a branching stochastic process. That is, we have a Galton–Watson process - one of the most common formulations of a branching process. If $Z_t$ is the state in generation $t$, we have the recurrence
$$
Z_{t+1}=\sum_{i=1}^{Z_{t}} X_{t, i}.
$$
Let $P_t$ be the extinction probability by the generation $t$. This extinction process can be analyzed using the method of probability generating function  $G$. We have that
$$
P_{t}=G\left(P_{t-1}\right), P(0)=0.
$$
In our case, we have random variables $X_{n,i}$ with binomial distribution $X_{n,i}\sim B(n,p=1/2)$ and we know that the PGF of a binomial random variable is $G(z)=[(1-p)+p z]^{n}$. It follows that
$$
P_{t+1}=\left(\frac{P_t+1}{2}\right)^n, P(0)=0.
$$
Finally, probability of the family name surviving for at least first $t$ generations is the complementary probability of the extinction probability $P_t$ by the generation $t$, which is $1-P_t$.
When $n\le 2$, the survival probability goes to zero $(1-P_t)\to0$. This is not hard to find, as seen in "Calculating the extinction probability for $X_{n,i}\sim B(2,p)$".
When $n=3$, the survival probability goes to  $(1-P_t)\to [1-(\sqrt{5}-2)]\approx 0.763932\dots$ as already calculated in "Branching process and calculating the probability of extinction  for $X_{n,i}\sim B(3,1/2)$".
For larger $n$, the survival probability $(1-P_t)$ indeed approaches larger values.
For example, for $n=3,4,5,6$ you can plot $1-P_t$ for $t=1,2,\dots,10$,

