We can perfectly well decide that the outcomes are double $1$, a $1$ and a $2$, double $2$, and so on, as in your proposal. That would give us $21$ different outcomes, not $36$.
However, these $21$ outcomes are not all equally likely. So although they are a legitimate collection of outcomes, they are not easy to work with when we are computing probabilities.
By way of contrast, if we imagine that we are tossing a red die and a blue die, and record as an ordered pair (result on red, result on blue) then, with a fair die fairly tossed, all outcomes are equally likely. Equivalently, we can imagine tossing one die, then the other, and record the results as an ordered pair.
You can compute probabilities using your collection of outcomes, if you keep in mind that for example double $1$ is half as likely as a $1$ and a $2$. The answers will be the same, the computations more messy, and more subject to error.
We have a drawing with replacement of $10$ balls out of an urn containing $6$ red balls, numbered from $1$ to $6$, and 30 white balls.
The number $R$ of red balls drawn is binomially distributed; i.e., the probability $p_r$ that we draw exactly $r$ red balls is given by
$$p_r={10\choose r}\left({1\over6}\right)^r\left({5\over 6}\right)^{10-r}\qquad(0\leq r\leq 10)\ .$$
Assuming that we draw exactly $r$ red balls, the numbers on these $r$ balls define a map $f:\ [r]\to[6]$, and all $6^r$ such maps are equally likely. Out of these maps $6!\left\{\matrix{r \cr 6\cr}\right\}$ are surjective, where $\left\{\matrix{r \cr 6\cr}\right\}$ (called a Stirling number of the second kind) denotes the number of ways to partition $[r]$ into $6$ nonempty blocks. Therefore the probability $q_r$ that on the $r$ red balls drawn all $6$ numbers are present, is given by
$$q_r={6!\left\{\matrix{r \cr 6\cr}\right\}\over 6^r}\qquad(6\leq r\leq10)\ .$$
It follows that the overall probability $P$ of the event described in the question is given by
$$P=\sum_{r=6}^{10} p_r\>q_r={416216045\over8463329722368}\doteq0.0000491788\ .$$
Best Answer
It is entirely impossible to know exactly what the author meant. But it seems to me like a relatively safe assumption that they meant that the two dice show the same result.
"But as far as I know its called doublet." And I usually call it a pair. Many things have several different names. That's just the way it is.