You select 10 numbers from the set $\{2,3,\dots,12\}$. You then continually roll 2 fair dice and sum them up, until your selection of 10 numbers come up.
For example if your selection was 7,7,7,7,8,8,8,6,6,6 (4 7's, 3 8's and 3 6's), and you roll the dice repeatedly and get 7,7,6,5,8,7,7,9,8,3,5,10,12,6,6,3,2,5,7,9,8, you may stop now because 4 7's have come up, 3 8's and 3 6's.
What is the best choice of 10 numbers so as to minimise the number of rolls ?
On
Using a brute-force exhaustive search, and a recursion for each case, I get the following answer:
There are two equivalent optimal strategies, namely $$4,5,6,6,7,7,7,8,8,9$$ $$5,6,6,7,7,7,8,8,9,10$$ yielding, for the expected number of rounds, $$ e=\frac{a}{b} \approx 28.26676327 $$ where $$ \begin{align*} a&=71526610479792733682076713232552201067\\[4pt] b&=2530413892848114144358747803518976000\\[4pt] \end{align*} $$ Remarks:
Intuitively, the multiplicities of the selected numbers should be in proportion to their probability of occurring in a single round.
Thus, for example, for the simpler game where you roll a single die, and get to guess $6$ numbers, the optimal selection is $1,2,3,4,5,6$.
If in each round you roll two dice, and get to guess $36$ numbers, I suspect the optimal selection is $$2, 3,3, 4,4,4, 5,5,5,5, 6,6,6,6,6, 7,7,7,7,7,7, 8,8,8,8,8, 9,9,9,9, 10,10,10, 11,11, 12 $$ For the game in question, where in each round you roll two dice, and get to guess $10$ numbers, then, since the multiplicities for the selected values must be nonnegative integers, it's not possible for the multiplicities to be in proportion to the associated probabilities. Yet, it's intuitive that they should be approximately in those proportions.
(too long for a comment, but not a full answer). I'll work it out for $n=2$, $n$ being the number of values you select.
Let $E[a,b]$ be the expected number of throws it take assuming you chose $a,b$. Similarly, let $E[a]$ be the expected number of throws it takes just to see $a$. Of course $E[a]=\frac 1{P(a)}$.
For $a=b$ clearly choosing $7$ is best. In that case one sees at once that $$\boxed{E[7,7]=\frac 2{P(7)}=12}$$
Now assume that $a\neq b$.
Considering the possible outcomes of the first toss (i.e. "$a$", "$b$", or neither) it is easy to see that $$E[a,b]=P(a)\times \left(E[b]+1\right)+P(b)\times \left(E[a]+1\right)+(1-P(a)-P(b))\times \left(E[a,b]+1\right)$$
This implies that $$\boxed {E[a,b]=\frac {P(a)P(b)+P(a)+P(b)}{P(a)P(b)(P(a)+P(b))}}$$
It is easy to compute that $$E[7,8]=E[7,6]=9.92727\cdots<12$$ and that $$E[6,8]=10.8$$ so as the OP expected, choosing $\{7,6\}$ or $\{7,8\}$ is optimal. (remark: it is not difficult to rule out the other possibilities, by direct computation if nothing else.)
Note I: this method generalizes to more choices but the cases start to multiply badly. My guess for $n=3$ would be $\{6,7,8\}$ but I have not verified this and it could easily be incorrect.
Note II: this method certainly lends itself to automation. If you have computed all the expectations for $n$ choices then you can get them for $n+1$ by a recursion as was done above. This is hardly a pencil and paper method, however. there are $\binom {19}9=92378$ possible selections when $n=9$ and $\binom {20}{10}=184756$ when $n=10$.