I read the Wikipedia Page on ciphertext indistinguishability. Here it gives the following outline of the IND-CPA game:
- The challenger generates a key pair PK, SK based on some security parameter k (e.g., a key size in bits), and publishes PK to the adversary. The challenger retains SK.
- The adversary may perform a polynomially bounded number of encryptions or other operations.
- Eventually, the adversary submits two distinct chosen plaintexts $M_0,M_1$ to the challenger.
- The challenger selects a bit $b\in \{0, 1\}$ uniformly at random, and sends the challenge ciphertext $C = E(PK, M_b)$ back to the adversary.
- The adversary is free to perform any number of additional computations or encryptions.
- Finally, the adversary outputs a guess for the value of b.
From the first step, it seems that the challenger only generates a secret key once. That means that the same secret key is used throughout the entire game.
However, then it says this:
Although the adversary knows $M_0,M_1$ and PK, the probabilistic nature of E means that the encryption of $M_{b}$ will be only one of many valid ciphertexts, and therefore encrypting $M_0,M_1$ and comparing the resulting ciphertexts with the challenge ciphertext does not afford any non-negligible advantage to the adversary.
I think that 'the probabilistic nature of E' means that $E$ depends implicitly on the secret key. Then if the secret key is different every time, we could view $E$ as a probabilistic algorithm. However, we earlier saw that the secret key should be fixed for the entire game. Then how can $E$ be probabilistic?
I also looked at this question. The answer doesn't say anything about the probabilistic nature of $E$.
As a follow-up, I would also like to know if (and how) the decryption function $D$ is deterministic or probabilistic.