I used the Markov Chain approach to solve this problem. There are six states corresponding to the number of sides of the die already seen and the transition matrix is bidiagonal, with only the current state of the next highest one possible. It is then possible to create the set of difference equations linking the probabilty vector after N rolls of the die to the vector after N-1 rolls. These equations solve to give values of p1(N),p2(N),...,p6(n) which are the probabilities of having seen 1,2,..6 faces after n rolls, respectively. It is more convenient to write the probabilities are (n+1) throws so:

p1(n+1) = (1/6)^n
p2(n+1) = 5 * [ (2/6)^n - (1/6)^n ]
p3(n+1) = 10 * [ (3/6)^n - 2(2/6)^n + (1/6)^n ]
p4(n+1) = 10 * [ (4/6)^n - 3(3/6)^n + 3(2/6)^n - (1/6)^n ]
p5(n+1) = 5 * [ (5/6)^n - 4(4/6)^n + 6(3/6)^n - 4(2/6)^n +(1/6)^n ]
p6(n+1) = 1 - 5(5/6)^n + 10(4/6)^n - 10(3/6)^n + 5(2/6)^n - (1/6)^n ]

The pattern here, for anyone familiar with Pascal's Trinagle and combinatorics is clear, with the sign of the combinations alternating between + and -, and the proportion being summed reducing by 1/6.

The probability that it will need N throws is just the (1/6) times the probability that 5 sides had been seen after N-1 throws, i.e.

P(N rolls before all 6 sides are seen) =
(5/6) * [ (5/6)^(N-2) - 4*(4/6)^(N-2) + 6*(3/6)^(N-2) - 4*(2/6)^(N-2) + (1/6)^(N-2) ]
for N=2,3,4,...
and 0, otherwise.
By the way, it can be quicly seen that for N=2,3,4,5, this formula gives 0, as it should.

The mean of a random varable with thie PDF can be worked out to be 14.7 using basic results of sums of infinite series.

The Median is solved by seeing at which value of N the cumulative probabilities reach 0.5.
The CDF, i.e. P(number of rolls <=N) is, by summing the series:

1 - 5*(5/6)^(N-1) + 10*(4/6)^(N-1) - 10*(3/6)^(N-1) + 5*(2/6)^(N-1) - (1/6)^(N-1)

This cannot be solved explicitly, but by iteration gives a solution of the median as 12.8110473

It's not exactly SAS, but I hope you enjoy anyway!

p1(n+1)