Simply put, what does “perfect secrecy” mean? I would like to ask for a clear (but maybe not so deep) explanation of what the term "perfect secrecy" means.

As far as I have researched and understood, it has to do with probabilities of assuming that a certain variable will be the key for a certain cipher text. And unless I'm confusing some concepts, the one-time pad is the only cipher known to have perfect secrecy, since no amount of resources would be enough to break it.

My notions of probability are kind of weak, which is why I don't understand most documents that speak of it.
Perfect Secrecy (or information-theoretic secure) means that the ciphertext conveys no information about the content of the plaintext. In effect this means that, no matter how much ciphertext you have, it does not convey anything about what the plaintext and key were. It can be proved that any such scheme must use at least as much key material as there is plaintext to encrypt. In terms of probabilities, it means that the probability distribution of the possible plaintexts is independent of the ciphertext.
We contrast this with semantic security, which I define by quoting the seminal 1984 paper of Goldwasser&amp;Micali:

Whatever is efficiently computable about the cleartext given the cyphertext, is also efficiently computable without the cyphertext.

For two examples, I quote my answer to this related question:

When used correctly, the One Time Pad (OTP) is information-theoretic secure, which means it can't be broken with cryptanalysis. However, part of being provably secure is that you need as much key material as you have plaintext to encrypt. Such a key needs to be shared between the two communicants, which basically means you have to give it to the other person through a perfectly secure protocol (eg by hand/trusted courier). So, actually it just allows you to have your trusted meeting in advance, rather than at the time of transmitting the secret information.
To illustrate this, consider what happens if one tries to brute force OTP? Since you have allowed an attacker infinite computational resources, he can keep guessing keys and calculating the appropriate plaintext until every key has been tested. Supposing the message was $b$ bits long, this would leave him with $2^b$ possible keys, each of which would generate a unique plaintext, making $2^b$ plaintexts. What is important here is that this means they would have candidate plaintexts corresponding to ever possible bit-string of length $b$. This means, even if you knew the message was &quot;Meet me at the stadium at 2?:15&quot; (where ? is 0,1,2 or 3), you still wouldn't have any idea what the ? was, because the possible plaintexts would contain this string with every possible value of ?.
Most cryptographic methods we use now are computationally secure. There are lots of different ways to do this, and I'll just sketch at a few of them. We might come up with a reduction to a problem conjectured to be hard (eg the Diffie-Hellman Problem or Discrete Log Problem). That is, we prove that &quot;If you can break my cipher, you can solve [hard-problem]&quot;, meaning our problem is at least a difficult to solve as [hard-problem]. So, if the problem is indeed hard to solve, so must cracking our encryption be.