How to verify that the average is really a simple calculation? There are many problems were the calculation of the actual probability is a hard one but evaluating the average (expected value) is easy.
Example
Shuffle two decks of cards and turn over each card of both decks simultaneously: How many times will you find the same card on each deck? Calculating the probability for at least one hit is quite tedious (via the inclusion-exclusion principle: $1-1/e$) but calculating the expected value of this happening is quite easy (via linearity of the expected value operator: 1). 
It is interesting to note that the calculation of the average being easy is not dependent on the random variables being independent of each other. But what is important for this to work is that the marginal probabilities are not being changed by their potential dependence (so e.g. even if two throws of a die are not independent they can still be fair).
My question
How can you be sure that the marginal probabilities are not being changed by their dependence? This is: How can you differentiate situations were the calculation of the average is therefore  easy and were even the average is hard to calculate (like the actual probabilities) because the marginal probabilities change?
 A: OK, I'll try to answer this even though I'm still not sure I've properly understood what you mean by "change".
Linearity of expectation says that the expectation value of the sum of two random variables (defined with respect to the same probability space) is the sum of their expectation values (and analogously for linear combinations). This always holds and is simply a consequence of the definition of the expectation value as an integral with respect to the probability measure. (If you don't know this terminology, you can replace "probability space" by "set of events with probabilities assigned to them" and "integral with respect to the probability measure" by "sum/integral of values times probabilities").
Correctly applied, this holds also in the case of your quantum dice. Let's say we throw the dice one after the other, and the dice work such that the first die has a uniform distribution and the second die shows a number less than or equal to the one on the first die, with the same probability for all such numbers. Then the conditional expectation value for the second die, given that the first die showed $k$, is $(k+1)/2$, and the total expectation value for the second die is therefore
$$\sum_{k=1}^6\frac16\frac{k+1}2=\frac1{12}\left(\frac{6(6+1)}2+6\right)=2\frac14\;.$$
The expectation value for the first die is $3\frac12$, and you can easily verify that the expectation value for the sum is the sum of the expectation values, $5\frac34$.
Here we had to use conditional expectation values because the we had no independent way of knowing the expectation value of the second die; its probabilities are only given dependent on what the first die shows. That doesn't change the linearity of expectation, but it prevents us from using a shortcut to evaluate the expectation value without thinking about conditional probabilities.
In How to calculate the expected value when betting on which pairings will be selected, things were different in that we had an independent way of knowing the expectation value for each match. We could have figured out the expectation value for the second match by imagining some physical process that randomizes the matches, considering the draws in the process sequentially and adding up conditional expectation values for the second match depending on what was drawn in the first. But that wasn't necessary because whatever concrete physical process was chosen to randomize the matches, it was by definition chosen such that all teams and all matches are treated the same and the expectation value for all matches is the same. Thus, if we had unexpectedly found that the expectation value for the second match was different from the first, this would have been not a failure of the linearity of expectation, but a failure to choose an appropriate process to randomize the matches.
Given that the expectation values are all the same by definition, we don't have to go to all the hassle of figuring out conditional probabilities and expectation values; we can just multiply one of them by the number of matches.
Now you might ask: But what if we don't say they're the same by definition, but focus on some particular process of randomizing the matches and want to know if that particular process leads to equal expectation values for all matches, don't we then have to figure out the conditional probabilities? That's a question about symmetry, about how to tell whether a process treats all objects of a certain kind equally. In the present case, you presumably had in mind something like a process where balls bearing the names of the teams are successively drawn from an urn, and the question is then how to tell whether all the possible matches have the same probability of being drawn in the second draw "even though" two balls have already been removed. This is clearly the case, since the balls that were removed were themselves picked randomly with uniform distribution, so there's nothing there that could break the symmetry between the teams.
I hope I've made this slightly clearer; if it's still not clear, I suggest you take some time to come up with an example where what you call the "easy calculation" works even though you wouldn't expect it to, or vice versa.
