What is the advantage of an attacker over breaking a 4 digits PIN? When a hardware system is protected by a 4 digits password, what is the advantage of the attacker into breaking that system?

Isn't it $10*10*10*10=10^{4}$? If $\frac{1}{10^{4}}$ is the cost of such an attack then is this reasonable computation for the attacker? In my eyes this seems like a simple brute force attack for 4 numbers. I mean is super easy. Isn't it?
Something went wrong with you calculations there :) A single digit has $10$ possible values, not $2^{10}$. What you were probably trying to calculate was $\frac{1}{10^4}=\frac{1}{10000}$.

It is a bit unclear what you mean by advantage. In cryptography, the advantage $\mathsf{Adv}_\mathcal{A}$ of an adversary $\mathcal{A}$ is the probability that $\mathcal{A}$ is successful ($\mathsf{Succ}_\mathcal{A}$) in attacking something minus the success probability for some trivial attack that cannot be prevented.

For example, for a decision problem, such as CPA security, the advantage is $\mathsf{Adv}_\mathcal{A}=\mathsf{Succ}_\mathcal{A}-\frac{1}{2}$. This is because you acannot prevent $\mathcal{A}$ from guessing the bit.

The advantage of guessing a four digit PIN with one random guess is $1/10^4$.
For at most $t$ random (but distinct) guesses, we get $\frac{t}{10^4}$. 
The advantage for an adversary attacking some authentication service using four digit PINs would therefore usually be $\mathsf{Adv}_\mathcal{A}=\mathsf{Succ}_\mathcal{A}-\frac{t}{10^4}$.

EDIT: In the first version of this answer I assumed that PINs would be chosen uniformly for each try. Of course that's not true. They would indeed be chosen to be distinct from each other. (This makes the probabilities quite a bit easier.)