Attacking any one in many public keys
Actually, a well known result is that, for any cryptosystem that relies on the hardness of the DLog problem (including ECDlog), there is no such reduction in strength if you have $k$ keys.  That is, the problem of &quot;here are $k$ keys, break any one&quot; is essentially as hard as the problem &quot;here is one key, break it&quot;.
The proof is straight-forward; first off, we assume that the cryptosystem is based on Dlog, hence breaking the system is equivalent to solving some DLog problem.  Further suppose we did have a method of, given $k$ values $a_1G, a_2G, …, a_kG$, of finding $a_i$ for some $i$, and this takes effort $M$ [1].
Then, given $bG$, we can attempt to recover $b$ by selecting $k$ random values $r_1, r_2, …, r_k$; and compute $r_1(bG), r_2(bG), …, r_k(bG)$.  We then hand this off to our oracle, which gives us the value $c$ s.t. $cG = r_i(bG)$ for some $i$; we then compute $b = cr_i^{-1}$, and thus recovering $b$, at the cost of $M$ plus the time to do $k$ point multiplications (which is quite feasible assuming $k$ isn't ridiculously large).
Similar proofs would apply to DH-based cryptosystems as well.

[1]: I'm using additive notation, mostly because I find it easier to read...