I am currently working with a statistical distribution, and I'm wondering if any exploration has been done on this.
The distribution is denoted $\xi$. To construct $\xi$ we use auxillary random variables $X_1,X_2, \ldots, X_k$
For indexing purposes we let the random variable
$X_1 = 1$
$X_2 \sim \mathrm{Bin}(4, \frac{1}{2})$
$X_3 \sim \mathrm{Bin}(4X_2, \frac{1}{2})$
and in general
$X_k \sim \mathrm{Bin}(4X_{k-1}, \frac{1}{2})$
So that the number of trials of the binomial distribution is itself a random variable.
One can show that the sum of $P(X_k = 0)$ over all $k$ is about $0.08737$.
The distribution $\xi$ is given by $P(\xi=k) := \dfrac{P(X_{k+1} = 0) – P(X_k = 0)}{0.08737}$ (we divide because probabilities have to add up to $1$).
Best Answer
I think it would be illustrative to rephrase the question more precisely. Let $X_0 = 1$, and define the sequence of random variables $$X_k \mid X_{k-1} \sim \operatorname{Binomial}(n = 4X_{k-1}, p = \tfrac{1}{2}), \quad k = 1, 2, \ldots.$$ Then define a random variable $\xi$ by $$\Pr[\xi = k] = \frac{\Pr[X_{k+1} = 0] - \Pr[X_k = 0]}{\lim_{n \to \infty} \Pr[X_n = 0]}, \quad k = 0, 1, 2, \ldots$$
It is worth noting that $$\begin{array}{c|c|c|c}k & \Pr[X_k = 0] & \Pr[\xi = k] & \Pr[\xi = k-1]/\Pr[\xi = k] \\ \hline 0 & 0 & 0.715283 & -- \\ 1 & 0.0625 & 0.196295 & 3.64393 \\ 2 & 0.0796518 & 0.0603027 & 3.25515 \\ 3 & 0.084921 & 0.019112 & 3.15523 \\ 4 & 0.0865909 & 0.00611566 & 3.12509 \\ 5 & 0.0871253 & 0.00196292 & 3.1156 \\ 6 & 0.0872968 & 0.000630643 & 3.11256 \\ 7 & 0.0873519 & 0.000202675 & 3.11159 \\ 8 & 0.0873696 & 0.0000651421 & 3.11128 \\ 9 & 0.0873753 & 0.0000209381 & 3.11118 \\ 10 & 0.0873772 & 6.73002 \times 10^{-6} & 3.11115 \\ 11 & 0.0873777 & 2.1632 \times 10^{-6} & 3.11114 \\ 12 & 0.0873779 & 6.9531 \times 10^{-7} & 3.11113 \end{array} $$ so that it looks like $\xi$ asymptotically behaves much like a geometric distribution. But other than this, I doubt that there is a nice closed form for $\xi$.