Since AM = BP.NP, it seems the go to "reduction" to AM relies on randomized reductions to 3SAT rather than the Karp reductions we use for deterministic complexity classes.
This is a wrong intuition. Regardless of how you define your complexity class $\mathbb{C}$, if there exists any problem $\mathrm{A}\in \mathbb{C}$ such that for every problem $\mathrm{B}\in \mathbb{C}$, you have $\mathrm{B} \leq_p \mathrm{A}$, then $\mathrm{A}$ is a many-one complete problem of $\mathbb{C}$.
In fact, even a problem that is complete by randomized reductions for $\mathrm{AM}$ is not known. Put in other word, it seems very hard to just pin down any particular decision problem in $\mathrm{AM}$ so that we can have some non-trivial reduction from other problems known to be in $\mathrm{AM}$.
See mathoverflow.net/questions/34469 and cstheory.stackexchange.com/questions/1233; in short, the definition of AM relies on a promise, and this makes it tricky to define a reduction. – sdcvvc
That is one of the obstacles on the way to find a complete problem for $\mathrm{AM}$. This is also applicable to $\mathrm{BPP}$, $\mathrm{RP}$, $\mathrm{co}$-$\mathrm{RP}$, $\mathrm{ZPP}$. These classes requires the poly-time probabilistic Turing machine to have bounded error probability on all instances. The situation is much easier for $\mathrm{PP}$, this class does not put any requirement on the error probability, whichever result has the higher probability is the answer of the machine so we can easily catch a complete problem for it, namely $\mathrm{MAJ}$-$\mathrm{SAT}$.
