A problem on probability concerning distributions of particles I saw this problem in An Introduction to the Theory of Statistics by Mood, Graybill, and Boes (2nd ed.). I am quite intrigued by the problem. Here it is:

Suppose that a particle is equally likely to release one, two or three
  other particles and suppose that these second generation particles
  are, in turn, equally likely to release one, two or three third
  generation particles. Find the pmf of the number of third generation
  particles.

I am thinking of doing this by listing all possible outcomes (which I suppose are from 1 to 27), and then manually compute for the probabilities of each. However, I know that it will be very tedious. So I am wondering if there is an easier way for this (perhaps using discrete distributions?). Any help will be very much appreciated.
 A: Use a generating function.  For one generation (no pun intended), the generating function is
$$
F(z) = \frac{z+z^2+z^3}{3}
$$
since there is a one-third probability of producing either one ($z$), two ($z^2$), or three ($z^3$) offspring.  To determine the generating function for the grandchildren, we simply write
$$
\begin{align}
F(F(z)) & = \frac{F(z)+[F(z)]^2+[F(z)]^3}{3} \\
        & = \frac{z^9+3z^8+6z^7+10z^6+12z^5+12z^4+16z^3+12z^2+9z}{81}
\end{align}
$$
I didn't do that by hand; I used a symbolic math package.  Anyway, at this point, one can merely read off the probabilities: $1/81$ probability of nine grandchildren, $3/81 = 1/27$ probability of eight grandchildren, etc.
ETA: I noticed that you expected a range of $1$ to $27$ descendants.  That would happen at the fourth generation; for that, you could compute $F(F(F(z)))$.  That yields a big long expression that's a bit too large for this space.  But I can say, for instance, that the most common result is six great-grandchildren, for which the probability is
$$
\frac{142182}{1594323} = \frac{5266}{59049} \doteq 0.089180
$$
