For the probability function p(n) which is recursive, how to obtain the maximum n under the condition p(n)≤0.1?

Question

here is the recursive function
p(1)=0$p(1)=0$
p(n+1)=(1−10−8n)p(n)+10−8n$p(n+1)=(1-{10}^{-8}n)p(n)+{10}^{-8}n$
I want to get the maximum n that makes p(n)≤0.1$p(n)\le 0.1$　
I tried to get the solution of the function but it failed. It seems possible to use something like Chernoff Bounds,Chevyshev’s Inequality,Markov’s Inequality to calculate the bound of n, but I don't know how to start.
The original question is to choose 8 numbers from 0−9$0-9$ arbitrarily, arrange and combine them in order. Find the largest n that after repeating n times, probability of duplicates in n numbers≤0.1$\le 0.1$
(I want to find largest n as possible,under the premise of computability, without computer calculation. The answer may not be the number exactly at the critical point, but the number close to the critical point in the calculable range)
Can anyone help me?

Answer 1

Using a computer, the maximum value of n$n$ for which p(n)≤0.1$p(n)\le 0.1$ is

p(4590)≈0.0999628p(4591)≈0.100004.$$\begin{gathered} p(4590)\approx 0.0999628 \\ p(4591)\approx \mathrm{0.100004.} \end{gathered}$$

Let c=10−8$c={10}^{-8}$. Note that the given recursion may be written as

1−p(n+1)=(1−cn)(1−p(n)).$$1-p(n+1)=(1-cn)(1-p(n)).$$

This suggests defining an auxiliary sequence b(n)=log(1−p(n))$b(n)=\log (1-p(n))$, hence

log(1−p(n+1))=log(1−cn)+log(1−p(n))$$\log (1-p(n+1))=\log (1-cn)+\log (1-p(n))$$

or

b(n+1)=log(1−cn)+b(n)$$b(n+1)=\log (1-cn)+b(n)$$

and we get the telescoping sum

b(n)=∑k=0nlog(1−cn)=log∏k=0n(1−ck),$$b(n)=\sum _{k=0}^n\log (1-cn)=\log \prod _{k=0}^n(1-ck),$$

consequently

p(n)=1−∏k=1n−1(1−ck).$$p(n)=1-\prod _{k=1}^{n-1}(1-ck).$$