Why is not the answer to all probability questions 1/2.
To talk of probability, you need both the sample space and a probability assigned to each element of the sample space.*
So in the case of a fair coin, the probability is given by the data that the sample space is $\{H, T\}$ and that $\Pr(H) = \frac12$ (and therefore $\Pr(T) = \frac12$). It's only because of this that you can say that the probability of heads is $\frac12$; you can't conclude this just from "either it will be heads or not".
For example, you could have a biased coin, where the sample space is still $\{H, T\}$, but $\Pr(H) = 0.9$ (and $\Pr(T) = 0.1$). In that case, you cannot use the reasoning that "either it will be heads or not" to conclude that heads and tails have equal probability.
In the case of a fair coin, the full reasoning you use is "either it will be heads, or not, and both are equally likely", where the final part (in emphasis) comes from the data given to you, namely that $\Pr(H) = \Pr(T) = \frac12$.
Even in the case of dice, you can still say "either it will be $2$ or not", but you're missing the latter part of the reasoning, that both events are equally likely. (Indeed, for fair dice, the probability of rolling $2$ is only $\frac16$, whereas the probability of rolling something other than $2$ is $\frac56$. But you could have loaded dice too.)
Often, in your textbooks, you may be given just the sample space without the probabilities, leaving implicit the fact that each individual element of the sample space has the same probability. But ideally that should be specified too.
[*]: Note: The discussion above is of a discrete probability space. In general, we have a set ("sigma algebra") of events, and probabilities on them, satisfying certain axioms. But let's not worry about that now.
In the case of a fair coin, the two outcomes are equally probable. $P(\text{heads}) = P(\text{tails}) = 1/2$.
In the case of a (fair) dice, the two outcomes $"2"$ and $"\text{not }2"$ are not equally probable. There are five possible ways of not getting 2, all equally probable, and only one possibility to get 2. In total there are six possible outcomes. Simply put: $$\frac{5}{6} = P(\text{not }2) > P(2) = \frac{1}{6}$$.
We cannot calculate a probability without using other probabilities in the calculation.
When we say that a coin has $P(H)=1/2$, or a die has $P(2)=1/6$ that is not something we learn using probability theory, it is an assumption about physics.
We assume that the die can only land with a face up, and we assume that all faces are equal in geometry and weight distribution, and therefore that the sum of probability for the faces is 1, and that the probability is the same for all 6 faces. Pure physics.
If I roll two dice I believe that each of them exert the same probabilities as a single die, and that they do this independently of one another. This is not something I can learn using probability theory either, it is an assumption about physics that the dice do not coordinate their rolls.
Once we have all these assumptions about die physics we can do all the cool die equity calculations using probability theory.
Unlike the rest of the world, dice are made for producing independent stochastic variables, the assumptions about die rolls have a solid foundation in physics that is pretty much undisputed. But when we try to figure the probabilities of something with real world relevance simple physics will not suffice, rather we might have to rely on psychology, sociology, economics or palaeontology. Putting probabilities on different outcomes is often guesswork, and different variables have all sorts of odd correlations, meaning that basic probability theory (which always assume independent variables) won't work.
When the coin argument seems to be universally applicable it is actually because it is a false argument, it just happens to produce the right result for fair coins.