No.
Throwing two dice is the same as throwing a dice twice.
Let's think naturally:
What is the set of outcomes when a die is thrown twice?
It is just the set of all ordered pairs such that, the first of the two entries tell you the outcome on the first throw and the second of the entries tell you the outcome on the second throw.
What is the set of outcomes when two dice are thrown?
Without loss of generality, I can label the dice as die A and die B. So, now, the set of outcomes will be the set of ordered pairs such that he first of the two entries tell you the outcome on die A and the second of the entries tell you the outcome on the die B.
So, radically, the set of outcomes, technically called the sample space, is the same.
To conduct an experiment with $100$ dice, if you have the means, you could buy $100$ dice and use a method that is reasonably unbiased in generating outcomes, save time or throw a single die $100$ times, the outcomes will be unbiased but you'll waste time.
The point is both methods have their own practical merits and demerits.
You have computed the probability that (for example) the first and second throws of
the pair of dice produce pairs (doublets) and the others do not.
It could also be the case that the first throw is a pair, the second is not,
but the third throw is (and the remaining two are not).
This is also an event where you get a pair exactly twice, but it is
disjoint from the other event.
So you must also consider how many possible different events you could be looking for.
The two pairs could come anywhere within the sequence of throws.
Best Answer
Yes, it is. You can visualize it quite nicely with a tree.
The sum of all probabilities must be $1$. Not getting at least one $6$ is equivalent to getting a non-$6$ on every roll. The probability for rolling "not a six" is $1-\frac{1}{6} = \frac{5}{6}$. As the outcome "not six" would have to happen 4 times in a row, we get that the probability of rolling "not six" on every roll is $(\frac{5}{6})^4$, therefore all other possibilities must be $1-(\frac{5}{6})^4$ so that the sum of all possibilities is $1$.