# To which game-based security definition is Perfect Secrecy equivalent?

We all know the classic definitions of perfect secrecy, being $$\Pr[M=m|C=c]=\Pr[M=m]$$ and $$H(M|C)=H(M)$$

But now what I've asked myself:
If we were to remove the polynomial restriction on the attacker, which game-based security definition would be equivalent to perfect secrecy and how would one show this?

Of course I've thought about this myself and it's obvious that known- and chosen-plaintext security is a given as we still can't learn anything new about the plaintext if a scheme is perfectly secret.
Even with chosen-ciphertext attacks I don't see how one could deduce more information on a perfectly secret scheme, given that the challenge ciphertexts would still be independently encrypted.
What I kinda can't grasp though is the reverse direction and of course the formal proof of equivalence.

• I don't see how they could be comparable. There are perfect-secure schemes (e.g., OTP) that would fail catastrophically in a typical CPA/CCA setting, where there is an encryption oracle available, and therefore, there is key reuse. – cygnusv Oct 31 '16 at 21:49

In other words, consider a game between an adversary and a simulator. The simulator chooses its key material. The adversary may submit messages and get their encryption in return. He then chooses a distribution over plaintext messages (all of equal length) and a binary predicate $f$, such that the probability that $f(m)=0$ is $1/2$. The simulator chooses a message according to the distribution, encrypts it and sends the challenge to the adversary. The adversary must now determine $f$. His advantage is the difference between his probability of guessing $f(m)$ correctly and $1/2$.
With a small amount of work, it now follows from perfect secrecy that any adversary in this game will have advantage exactly $1$. Also, if any unlimited adversary has advantage exactly $1$, it must follow that you have perfect security.