This is barely a probability question, but I needed to check to make sure the solution is as simple as I believe it to be.
What is the the expected number $n$ of independent trials needed to have $x$ success (not necessary to be consecutive) given probability $p$?
I would assume since each trial is independent the solution would be $n = x/p$, but perhaps I am overlooking something here.
Best Answer
If you know that the expectation of a geometric variable is $1/p$, where $p$ is the success factor for the variable, you can do the following (which seems to be the way you are thinking about it, and is a very nice method for computing expectations of complicated r.v.'s that can be written as a sum of simpler r.v.'s):
Let
$\ \ \ X_1$ be the number of trials to the first success,
$\ \ \ X_2$ be the number of additional trials to the second success,
$\ \ \ X_3$ be the number of additional trials to the third success
$\ \ \ \ \ \ \vdots$
Each $X_i$ is a geometric variable with success factor $p$; so, $\Bbb E(X_i)={1\over p}$ for each $i$.
Now let $Y$ be the number of trials to the $x^{\text{th}}$ success. Note that $Y=\sum\limits_{i=1}^x X_i$. ($Y$ is, as Robert Israel observes, a negative binomial random variable.)
Then, recalling that the expectation of a sum of random variables is the sum of their expectations: $$\Bbb E(Y)= \Bbb E\Bigl(\sum_{i=1}^x X_i\Bigr)=\sum\limits_{i=1}^x \ \Bbb E(X_i)={x\over p}.$$