An extension to this question:
If a function, on average, reduces its input by $1/4$ (round down for integers), and terminates when input is less than a threshold, say $1$, what is the expected number of rounds for it to terminate?
More formally, we have: given initial size $n_i$ and the threshold size $n_t$, the number of rounds is $R\equiv \min\{k: (\prod_k^i X_i)<n_t/n_i\}$, where $X_i$ are iid with support $[0, 1]$ and $\mathbb{E}[X_i]=3/4$. The goal is to find $\mathbb{E}[R]$.
I have run some quick scripts in python, and here are the results compared to the intuitive (but probably incorrect), answer of $\log_{4/3}(n_i/n_t)$:
When $n_i = 10^7$, and experiment repeated $10^5$ times, we get the rounds to termination $54.85193$, while $\log_{4/3}(10^7) = 56.027459457557676$. I have been consistently getting results lower.
Also, this is only for triangular distribution between $0.5$ and $1$ with mode $0.75$ (It's even lower for uniform $[0.5, 1]$, which results to $53.24309$ rounds). Is there a way to approach this for any given support and expectation (distribution potentially known)?
