Does a random action have probability? Imagine I have a coin and I throw it into the air.  It has side a and side b. It is absolutely random which side is going to be facing up.
Knowing that,

What probability does the coin have to fall and facing up the side (a)?

*

*a) 50%

*b) 100%

*c) unknown

*d) 0%


How can it be calculated?
 A: First of all, please note that when we assign probabilities to a physical events, we are making a mathematical model. Some mathematical models seem to fit reality well. Some mathematical models, which may still be useful, fit not so well.  But whether we have good fit or not, we should remember that the mathematical model is not the same thing as the reality.
Let's produce a probabilisitc model of our coin tossing. If there is good reason to believe, because of symmetry, that the two sides are equally likely, then in our model of the situation, we let the probability the coin lands showing $a$ be $p$. Then the probability that it shows $b$ is also $p$. We have $p+p=1$, and therefore $p=1/2$ ($50\%$).
However, that is not the whole story!  "Random" in ordinary speech sometimes carries the connotation of "equally likely." But often it doesn't. For example, when you throw two fair dice, the event "sum of the pips is $7$" is a random event. for emphasis, you could say it is absolutely random.  However, a sum of $7$ and a sum other than $7$ are definitely not equally likely in the model that experience shows best fits the actual tossing of dice. 
When the term "random" is used carefully in a mathematical discussion of probability, it does not carry the additional meaning of "equally likely." If we mean equally likely, we say so.
A: Quick answer is that, yes, it is random. That is, either side is equally likely. Probability of either is 50%. Now, this is not exact. Coins are all slightly different and it is possible that the odds of one side is slightly higher, maybe 40.999% versus 50.001% etc but for the purposes of sports, it is good enough. If you make a weird enough coin, you can probably get something quite different. 
There is no way to calculate it from mathematics. First you assume a model and then you do mathematics with your model to see what other facts will logically follow from your previous assumptions. You could figure it out with physics. That is, specify the coins properties and how it is tossed and you can calculate (via computer simulation) exactly how it will land. Well, exactly only in theory because sometimes it will land right on the edge and the outcome will be infinitely sensitive to the initial conditions.  
You may also be wondering about the probability of say three heads in a row. If you assume that either side is equally likely (or with some other odds), you can use mathematics to calculate the odds of 3 heads. THAT is in fact just mathematics because it is determined from your initial assumption of odds and that they throws are independent
A: You can calculate the probability empirically using statistics: perform a large number of coin toss experiments and tabulate the results.
You can also construct an ideal abstract model of the coin and argue that it is symmetric: the coin is indistinguishable from a 180 degree rotation around any diameter, except for the marking which distinguishes one side from the other. 
From that you can argue that the marking hardly makes a difference to the outcome (and in the ideal model, we can make the marking as light and small as we want: we can have our abstract coin's head side be marked by a nearly massless, vanishingly small, yet faintly detectable particle of matter).
If the coin is symmetric, and the behavior of the toss is random, we can argue that there is no reason for it to land on one side or the other. This is a variation on reductio ad absurdum. If the coin does have a greater probability of landing on one side than the other, then its sides are distinguishable which contradicts the model which makes them (almost) perfectly indistinguishable.
