Security of AES counter mode vs. CBC mode
The requirement for the AES-CTR input blocks is, that they should be unique during the lifetime of a key. In most of the cases a random 96bit nonce is used with a 32bit counter that starts from 0. If the same input block for AES-CTR occurs twice, AES-CTR is not CPA secure any more. In this case, this can be due to a counter-overflow after $2^{32}$ blocks or due to colliding randomly chosen 96bit nonces (birthday paradox: 50% chance after $\sqrt{2^{96}}$ messages. Consider the following case:

Two distinct 1-Block messages $P$ and $P'$ are sent under the same key $K$ (that might be negotiated beforehand) and with the same nonce $N$. The attacker knows that the related cipher texts $C$ and $C'$ where calculated by XORing them with the keystream (which is based on the nonce and the counter):

$C = P \oplus E_K(N,0)$

$C' = P' \oplus E_K(N,0)$

Then the attacker can simply xor the cipher texts

$C\oplus C' =  P \oplus E_K(N,0) \oplus P' \oplus E_K(N,0) = P \oplus P'$

and he obtains the ''distance'' between the two plain texts. Due to redundancies in the English language, he might be able to determine $P$ and $P'$.

This problem is also known as the "two-time-pad". Once the same keystream is XORed with the plaintext, we get into trouble. Therefore, it is important, that the input for the AES encryption is unique during the lifetime of a key. It does not have to be unpredictable, just unique.