Number of trials rolling six 6-sided dice to get 6 unique values I have six 6-sided dice, and I want to roll them until I have all six distinct values (assume them to be naturals 1-6), in no particular order.
The naive strategy would be to roll all 6 dice until I have all 6 values. The expected number of trials until success is $\frac{6^6}{6!} = 64.8$.
A smarter strategy would be to roll a die until its value is distinct from the values of other dice already rolled. Eg. the first die gets rolled once, its value is 3, and then I roll the second one until its value is something else than 3, say 4. Then I roll the 3rd one until its value is neither 3 nor 4, etc. This way, the expected number of trials is $\sum_{i=1}^{6}\frac{6}{i} = 14.7$.
An even smarter strategy would be to first roll all 6 dice, then pick the duplicates ‒ all except one ‒ and roll these again. Repeat untill I have all 6 values. For example:


*

*first roll: 1 1 2 4 4 4 (the order doesn't matter)

*pick 1 4 4, leave 1 2 4 on the table

*roll the 3 dice (which had values 1 4 4) again

*repeat this until there are values 1-6 on the table


Question: what is the expected number of trials with this strategy?
 A: As a back-of-the-envelope calculation, I expect around $$\frac{\pi^2}6n$$ rolls are needed.
Most rolls are spent tidying up the final few numbers.
It takes $n$ rolls to get the final number.
With two numbers to go, success is four times as likely, as two dice each have two successful rolls, so around $n/4$ rolls are needed to advance.  With three to go, around $n/9$ are needed.  The chance of advancing more than one step at a time is relatively small.  So my leading-order estimate is $n+(n/4)+(n/9)+...$ which is the number at the top of this answer.
EDIT  A better fit seems to be
$$\frac{\pi^2}6n-\frac12\sum_{k=1}^n\frac1k-0.75$$
I got the first correction from a more precise version of the argument above, but the $0.75$ is taken from simulations, a million trials at each of $n=2$ to $20$.
The following graph shows the difference between the average and $\pi^2n/6$.  One curve is simulations, the other curve is from the formula above.

EDIT: I want to record where the first correction term comes from.
I change variables, so that $s_k=t_{n-k}$, and $k$ is the number of dice being rolled.  Following the accepted answer,  out of $n^k$ possible rolls, most roll no new numbers; sometimes one of the $k$ dice rolls one of $k$ new numbers; or two roll the same new number; or two roll different new numbers.  The rest will be $O(n^{k-3})s_k$, and negligible for this calculation.
$$n^k s_k=n^k+(n-k)^k s_k + \\
k^2 (n-k)^{k-1} s_{k-1} + \\
{k\choose2} k (n-k)^{k-2} s_{k-1} + \\
{k\choose2} (k^2-k) (n-k)^{k-2}s_{k-2}+...$$
To leading order, $k^2s_k=n+k^2s_{k-1}$, which leads to the $n\pi^2/6$.
Now bring in the next order, let $s_k=n×a_k+b_k$.  We know $a_k=\sum^k_{i=1}(1/i^2)$.  The $b_{k-2}$ term is negligible, but by combining the known $a_{k-1}$ and $a_{k-2}$, it simplifies to $b_k=b_{k-1}-1/(2k)$
