A changing probability relating to primes The probability that a process succeeds at step $t$ is
$$\left( \prod_{k=1}^{t-1}{ 1-\text{Prime}(p+k)^{-1/c} } \right)(\text{Prime}(p+t))^{-1/c}$$
Here $p$, $t$ and $c$ are naturals, and $\text{Prime}(x)$ is simply the $x$th prime.
How many steps, on average, does the proccess take until it succeeds, for a given $p$ and $c$?
MY ATTEMPTS
The first thing I tried was to try to convert the prime in the function to a power of $x$, hoping that I could test it in my math software and get close to the same asymptotics:
$$\prod_{k=1}^{t-1}{ \left(1-(\text{Prime}(p)^{d \cdot k})^{-1/c} \right) }\text{Prime}(p+t)$$
The idea was to approximate $\text{Prime}(p+k)$ with $x^{d \cdot  k}$ for $d \approx 1$, say $d=1.1$.  But by my calculations, this wasn't exacting enough.  Perhaps I have to rethink $d$.
That was probably my best attempt so far.  I'm still thinking that maybe we can approximate the prime values with a power of $x$, or some suitable function.  I'm not sure what else to try.  I feel like I'm way off here.
Anyways, the idea seems to be to get the product equal to around $1/2$, which would be an average number of steps or trials.
 A: *

*For a r.v. $X$ on $\mathbb{Z}_{>0}$ with $\mathbb{P}(X=n)=p_n$, the expected value is $\mathbb{E}X=\sum_{n=1}^{\infty}np_n$ (of course).

*If $p_n=q_n-q_{n+1}$, where $1=q_1>q_2>q_3>\ldots$ and $S=\sum_{n=1}^{\infty}q_n$ converges, then $\mathbb{E}X=S$.

*Now let $a_n\in(0,1)$ with $\prod_{n=1}^\infty(1-a_n)=0$. If we put $q_n=\prod_{k=1}^{n-1}(1-a_k)$, assuming $q_1=1$, then $p_n[{}=a_n q_n]$ has the needed form. With $a_k=\text{Prime}(p+k)^{-1/c}$, $\prod_{n=1}^{\infty}(1-a_n)=0$ does hold.

*In the case $c>1$, $S=\sum_{n=1}^{\infty}q_n$ converges. This is shown using $1-a_n<e^{-a_n}$ and estimating $\sum_{n=1}^{N}a_n$ from below, using $a_n\sim(n\log n)^{-1/c}$; it grows at least as fast as $N^a$ for any $a<1-1/c$, which is sufficient. Thus, the answer is $S$. It doesn't have a closed form - you have to leave it as is.

*Finally, suppose $c=1$. Then, from Mertens' third theorem, $\lim_{n\to\infty}q_n\log n$ exists and is positive. Together with $a_n\sim(n\log n)^{-1}$, this easily implies that $\sum_{n=1}^{\infty}np_n$ diverges. Thus, the answer is $\infty$.

