Geometric distribution with increasing chance of success I have the following problem (perhaps it could be seen as a variant of the geometric distribution):
I'd like to know the expected number of trials needed to get the (single) special ticket among n tickets. Tickets are drawn without replacement. 
The distribution of a generalized version of this problem is here: https://math.stackexchange.com/a/436247
The given distribution is: $p_i \prod_{k=1}^{i-1}(1 - p_k)$, where $p_i$ is the probability of success at i-th trial. This looks like a generalized version of the geometric random variable. 
In my specific case, $p_i = \frac{1}{n-i+1}$, because on each trial, the number of tickets to draw reduces by 1. 
It might seem trivial, but I'm having trouble to calculate the mean of "my" distribution, which would give the expected number of trials until success... This could be a well-known problem already, but I was not able to find a reference for it (I found it's "inverse" here: A Diminishing Geometric Distribution). 
So, any help is appreciated.
EDIT: took out the sentence "until I'll certainly get the special ticket on the last trial" because one could infer that $p_n = 1$, implying $p_i = 0$ for $i < n$, which is not the case.
EDIT: took out the sentence "on the 1st trial the chance of selecting the special ticket is $1/n$, on 2nd it is $1/(n-1)$ and so on." because it can be misleading. See @Karn Watcharasupat's answer and comments below.
 A: Let $N$ denote the number of trials needed.
Think of this as placing the selected ticket on one of numbered spots $1,2,\dots,n$ such that every spot has equal chances to be elected for that.
Let spot $i$ be the place of the $i$-th drawn ticket.
Then $\mathsf P(N=i)=\frac1{n}$ for $i=1,2,\dots,n$.
So the expectation is: $$\mathsf EN=\sum_{i=1}^nP(N=i)i=\frac1{n}(1+2+3+\cdots+n)=\frac12(n+1)$$
Also we have $P(N=i\mid N\geq i)=\frac1{n-i+1}$ but that is irrelevant.
A: Thanks to @drhab for pointing out my mistake.
So this is from what I understand...
Let $X$ denotes the number of trial up to and including the trial where the special ticket is successfully obtained.
\begin{align}
  & P(X=r)={{p}_{r-1}}\prod\limits_{i=1}^{i=r-1}{\left( 1-{{p}_{i}} \right)} \\ 
 & =\frac{1}{n-r+1}\prod\limits_{i=1}^{i=r-1}{\left( 1-\frac{1}{n-i+1} \right)} \\ 
 & =\frac{1}{n-r+1}\prod\limits_{i=1}^{i=r-1}{\left( \frac{n-i+1-1}{n-i+1} \right)} \\ 
 & =\frac{1}{n-r+1}\prod\limits_{i=1}^{i=r-1}{\left( \frac{n-i}{n-i+1} \right)} \\ 
 & =\frac{1}{n-r+1}\left( \frac{n-1}{n-1+1}\times \frac{n-2}{n-2+1}\times \frac{n-3}{n-3+1}\times \cdots \times \frac{n-r}{n-r+1} \right) \\ 
 & =\frac{1}{n-r+1}\left( \frac{\require{enclose} \enclose{horizontalstrike}{n-1}}{n}\times \frac{\enclose{horizontalstrike}{n-2}}{\enclose{horizontalstrike}{n-1}}\times \frac{\enclose{horizontalstrike}{n-3}}{\enclose{horizontalstrike}{n-2}}\times \cdots \times \frac{n-(r-1)}{\enclose{horizontalstrike}{n-(r-1)+1}} \right) \\ 
 & =\frac{n-r+1}{n\left( n-r+1 \right)}  \\
& =\frac{1}{n}
\end{align}
Now, for the expectation $\operatorname{E}(X)$:
 \begin{align}
  & \operatorname{E}(X)=\sum\limits_{r=1}^{n}{rP(X=r)} \\ 
 & =\sum\limits_{r=1}^{n}{\frac{r}{n}}  \\
& = \frac{1}{n} \frac{n}{2} (n+1)\\
& =\frac{n+1}{2}
\end{align}
