Suppose we have a probability distribution $p$ over $\{0, 1, 2\}$, with probabilities $p_0$, $p_1$ and $p_2$, and $p_0 + p_1 + p_2 = 1$. Now suppose we repeatedly choose an element randomly from this set until we get a $2$, with the probabilities above. E.g. we might get the sequence $0, 0, 0, 1, 0, 1, 1, 0, 0, 2$ with probability ${p_0}^6{p_1}^3p_2$.
Now, we can "parse" this sequence into a compressed form of integers, with each integer except the last denoting the number of $0$s preceding each $1$. The last integer is the number of remaining terms in the sequence i.e. the remaining $0$s and the $2$. So the above would be $3, 1, 0, 3$. Now suppose we generated a number from the compressed sequence as follows: raise $2$ to the power of the first number, $3$ to the power of the second, $5$ to the power of the third, and so on, to give a prime factorisation. Then we get, for the above example, the number $2^3 \times 3 \times 7^3$. This process leads to a probability distribution over all numbers except $1$. Denote $P(n)$ as the probability of generating the number $n$ by this process.
Question: What is the minimum $k_n$ such that $P(n + k_n) < P(n)$?
It is easy to show that $k_n \leq n$. Because $n + n = 2n$ and the prime factorisation of $2n$ is greater than $n$. Is there a better upper bound than this, in terms of $p_0$, $p_1$ and $p_2$?