Suppose that, when rolled for the first time, a special 6-sided dice shows $1,\ldots, 6$ with probability $\frac{1}{6}$ each, and then, upon rerolling, shows with equal probability a number greater or equal to the previous outcome (so e.g. if you roll a $4$, the subsequent outcome is going to be one of $4, 5, 6$, each with probability $\frac{1}{3}$). Now, what is the probability that the $n$-th roll is a $6$?
This is a homemade question and I'm wondering if there is an elegant way to solve it, avoiding messy computation.
If $p_n$ is the probability that the $n$th roll is a six, the generating function for this sequence is $$\phi(s)=\sum_{n\geq1} p_n s^n={120s\over(6-s)(5-s)(4-s)(3-s)(2-s)(1-s)}.$$ A partial fraction expansion gives us the explicit formula $$p_n=\sum_{j=0}^5 {5\choose j}{(-1)^j\over (j+1)^n}.$$ The first few values are $$p_1={1\over 6},\quad p_2={49\over 120 },\quad p_3={13489\over 21600},\quad p_4={336581\over 432000}.$$