Let us define $M\_0=2^n$ for $n\in\mathbb{N}$. Let $\ell\in\mathbb{N}$ be the number of random variables we are working with. For $1\leqslant i\leqslant\ell$, we define $M\_i$ to be a random variable following a binomial distribution with parameters $M\_{i-1}$ and $2^{-n}$.


I'm interested in computing $\mathbb{P}\left[M\_\ell=1\middle|M\_\ell\geqslant1\right]$. To be fair, I only needed to lower-bound this probability, which is easy to do by computing $\mathbb{P}\left[M\_1=1\middle|M\_1\geqslant1\right]\approx\frac{\mathrm{e}^{-1}}{1-\mathrm{e}^{-1}}>\frac12$, but, out of curiosity, I was wondering if it was possible to get a closed-form for $\mathbb{P}\left[M\_\ell=1\middle|M\_\ell\geqslant1\right]$, and I do not know of any theorem that would help me with this.


Intuitively, this probability should decrease exponentially fast to $1$, and it might be possible to get the closed-form expression of the probability for $\ell\leqslant3$ by hand, but it will quickly become tedious to operate that way. Is there any clever way I'm missing for this?



Distribution of a random variable obtained by chaining distribution
Abbreviate $N=M\_0=2^n$ and $b=1/N$. As described, the conditional distribution of $M\_i$, given that $M\_{i-1}=n$, is binomial with parameters $n$ and $b$. The conditional generating function of $M\_i$ is therefore
$$
E[s^{M\_i}\mid M\_{i-1}]=(1-b+bs)^{M\_{i-1}}.
$$
Using $g\_i$ to denote the generating function of $M\_i$, this leads to the recursion
$$
g\_{i}(s) = g\_{i-1}(1-b+bs),
$$
and, because $g\_0(s) = s^N$, by recursion,
$$
g\_i(s) = (1-b^i+b^is)^N, \qquad0\le s\le 1.
$$
In other words, $M\_i$ has the binomial distribution with parameters $N$ and $b^i$.


In particular,
$$
P[M\_i=1\mid M\_i\ge 1] = {Nb^i(1-b^i)^{N-1}\over 1-(1-b^i)^N}.
$$


