Expected Value Problem with 100 side dice and 100 doors One hundred doors, one dollar behind each door. Roll a  one-hundred dice for one hundred times. You can take the dollar after the door whose number is rolled out but the dollar is not replaced. What's the expectation? 
What I tried: 
I tried to write a general formula taking ideas from the Coupon collector problem. I found that the E[X]= (N-n)/N where N is the total number of doors and n is the amount we've opened but I don't think that works. Any advice would be appreciated.
 A: For $1\leq i\leq100$ let $X_i$ be a random variable if door number $i$ opened and $0$ otherwise.  The we want $E(\sum X_i)=\sum E(X_i)$ by linearity of expectation.  But $E(X_i)$ is just the probability that door $i$ is opened, or $1$ minus the probability that it is never opened.
$E(X_i)$ is the same for all for all $i$ so we have $$100\Pr(\text{Door 1 isn't opened})=100(1-.99^{100})\approx100\left(1-{1\over e}\right)$$ 
A: Here is  a slightly different approach  by way of enrichment.  Say the
die has  $n$ sides and there  are $n$ doors.  The  answer would appear
from first principles to be
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{n^n} \sum_{q=1}^n q {n\choose q} q! {n\brace q}.}$$
Here $q$  gives the  number of  different values  that have  been seen
where $1\le q\le  n.$ We choose these from the  $n$ available ones and
partition the constituents of the sequence of rolls into $q$ non-empty
sets (Stirling number), one for each  type of element, where the order
of the sets matters (factor $q!$ as  we map each element to the places
where it appears in the sequence). Any door contributes one dollar the
first time  it is opened and  zero dollars thereafter, so  we seek the
expectation of the number of different values that have appeared. Note
also that
$$\frac{1}{n^n} {n\choose q} q! {n\brace q}$$
gives the probability of seeing $q$ different values.
We use the EGF of the Stirling numbers to compute the sum:
$$n! [z^n] \frac{1}{n^n}
\sum_{q=1}^n q {n\choose q} q! \frac{(\exp(z)-1)^q}{q!}
\\ = n! [z^n] \frac{1}{n^n}
\sum_{q=1}^n q {n\choose q} (\exp(z)-1)^q
\\ = n\times  n! [z^n] \frac{1}{n^n}
\sum_{q=1}^n {n-1\choose q-1} (\exp(z)-1)^q
\\ = n\times  n! [z^n] (\exp(z)-1) \frac{1}{n^n}
\sum_{q=0}^{n-1} {n-1\choose q} (\exp(z)-1)^q
\\ = n\times  n! [z^n] (\exp(z)-1) \frac{1}{n^n}
\exp((n-1)z)
\\ =  \frac{n}{n^n} \times  n!
[z^n] (\exp(nz)-\exp((n-1)z))
= \frac{n}{n^n} (n^n - (n-1)^n).$$
This gives for the desired answer
$$\bbox[5px,border:2px solid #00A000]{
n \left( 1 - \left(\frac{n-1}{n}\right)^n\right)}$$
which agrees with the response that was first to appear. We also
have
$$\lim_{n\to\infty} \left(1-\frac{1}{n}\right)^n = \frac{1}{e}$$
so this becomes $\sim n \left(1-\frac{1}{e}\right).$
