You are looking at a much more manageable version of the famous birthday problem, which has 365 choices for each "die". The link describes the solution as well. In short, the trick for counting duplicates is to rather counting the cases without a duplicate - that is much simpler.
Imagine that you threw your fair six-sided die and you got ⚀. The
result was so fascinating that you called your friend Dave and told
him about it. Since he was curious what he'd get when
throwing his fair six-sided die, he threw it and got ⚁.
A standard die has six sides. If you are not cheating then it lands on each side with equal probability, i.e. $1$ in $6$ times. The probability that you throw ⚀, the same as with the other sides, is $\tfrac{1}{6}$. The probability that you throw ⚀, and your friend throws ⚁, is $\tfrac{1}{6} \times \tfrac{1}{6} = \tfrac{1}{36}$ since the two events are independent and we multiply independent probabilities. Saying it differently, there are $36$ arrangements of such pairs that can be easily listed (as you already did). The probability of the opposite event (you throw ⚁ and your friend throws ⚀) is also $\tfrac{1}{36}$. The probabilities that you throw ⚀, and your friend throws ⚁, or that you throw ⚁, and your friend throws ⚀, are exclusive, so we add them $\tfrac{1}{36} + \tfrac{1}{36} = \tfrac{2}{36}$. Among all the possible arrangements, there are two meeting this condition.
How do we know all of this? Well, on the grounds of probability, combinatorics and logic, but those three need some factual knowledge to rely on. We know on the basis of the experience of thousands of gamblers and some physics, that there is no reason to believe that a fair six-sided die has other than an equiprobable chance of landing on each side. Similarly, we have no reason to suspect that two independent throws are somehow related and influence each other.
You can imagine a box with tickets labeled using all the $2$-combinations (with repetition) of numbers from $1$ to $6$. That would limit the number of possible outcomes to $21$ and change the probabilities. However if you think of such a definition in term of dice, then you would have to imagine two dice that are somehow glued together. This is something very different than two dice that can function independently and can be thrown alone landing on each side with equal probability without affecting each other.
All that said, one needs to comment that such models are possible, but not for things like dice. For example, in particle physics based on empirical observations it appeared that Bose-Einstein statistic of non-distinguishable particles (see also the stars-and-bars problem) is more appropriate than the distinguishable-particles model. You can find some remarks about those models in Probability or Probability via Expectation by Peter Whittle, or in volume one of An introduction to probability theory and its applications by William Feller.
Best Answer
Probability of a die not turning up $n$ is $1-1/n$. Probability of not turning up $n$ in any of the $n$ dice is $(1-1/n)^n$. If you subtract this from $1$, it'll be probability of at least one $n$ turning up when you throw $n$ dice, i.e.
$$p=1-(1-1/n)^n\rightarrow 1-e^{-1}$$ as $n$ goes to $\infty.$