Expected number of jumps to just exceed $(1-\rho)$ quantile in the line segment $[0, 1]$

Question

Suppose $X$ is random variable from some unknown distribution $f(X)$. I'm given a black-box/algorithm that takes a number $c\;: 0 \leq c < 1$ and outputs a number(randomly generated) using the truncated distribution $f_c(X)$ where the domain of $X = [c, 1]\; \therefore f_0(X) = f(X)$. Now I need to estimate the $E[\text{count}]$ for the values of $count$ returned by the following code:

Input: rho (where $0<\rho <1$)

$Q(1-\rho)$: Value of Quantile Function [$(1-\rho)^{\text{th}}$-quantile] of the distribution function $f_0(X)$

int get_count(float rho) {
  float c = 0.0;
  int count = 0;
  while(c < Q(1-rho)) {
    c = blackbox_algo(c);
    count++;
  }
  return count;
}

I tried to construct a recurrence by conditional expectation(as generally done in case of geometric distribution) but it seems to be not correct as the memoryless property is not valid here. Any help is welcome :)

Disclaimer: It's not a homework problem. I'm a budding researcher in engineering and ran into situation where such kind of problems are arising. This problem has a similarity with this one but these are not identical

Answer 1

You can forget about finding the expected number of steps if you don't know what $f$ is.

Original question

It should be obvious that if $f(x) = 0$ for every $x \in [1-ρ,1]$, then your get_count will run forever. It suggests that making $f$ sufficiently close to zero in that region would make the expected count go to infinity, and in fact this is so.

Present question

If you take two distributions with the same cutoff $t = Q(1-ρ)$ but different distribution over $[0,Q(1-ρ)]$, then you'll get different count distributions. Just try for discrete distributions to see why. Say $t = 0.5$, and in one case $X = 0$ whenever $X < t$, and in the other case $X$ is $0$ or $0.1$ with equal probability whenever $X < t$. Then the second case will obviously give a lower expected count because truncation occurs faster.