How do you calculate probability of rolling all faces of a die after n number of rolls? 
Possible Duplicate:
Expected time to roll all 1 through 6 on a die
Probability of picking all elements in a set 

Im pretty new to the stackexchange, and posted this is statistics, and then discovered this site, and thought it was much more appropriate, so here I go again:
It is fairly easy to figure out what is the average rolls it would take to roll all faces of a die [1 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 14.7], but that got me thinking of a seemingly more complicated problem.
Say you roll a die 1-5 times, the is the odds of ALL faces showing, is obviously 0. If you roll a die 6 times, the odds of all faces showing can easily be calculated like so:
1 * (5/6) * (4/6) * (3/6) * (2/6) * (1/6) = .0154 or 1.54%
Now is where I get stuck. How to do 7, or more times, and calculate it with n.
Any tips is helpful!

The probability of not rolling a 1 in $n$ rolls is $(5/6)^n$, similarly for not rolling a $2,\ldots,6$.  Now, $6(5/6)^n$ would be the probability that we are not rolling a $1,\ldots,6$, but we would be double counting the rolls where we do not roll both a 1 or a 2.  The probability of not rolling two specified numbers in $n$ rolls is $(4/6)^n$ and there are $\binom{6}{2}$ pairs of numbers.  But if we subtract these out we undercount the rolls that avoid three numbers.  This generalizes to the inclusion-exclusion principle, giving us the probability of missing any number as
$$\binom{6}{1}(5/6)^n-\binom{6}{2}(4/6)^n+\binom{6}{3}(3/6)^n-\binom{6}{4}(2/6)^n+\binom{6}{5}(1/6)^n$$
as the probability of missing at least one number in $n$ rolls.  The probability of rolling all of them is just 1 minus this probability.
