I am studying Alon-Spencer's The Probabilistic Method, and theorem 2.3.2 reads: $\exists$ a 2-coloring of $K_{m, n}$ with atmost ${m \choose a}{n \choose b}2^{1-ab}$ monochromatic $K_{a,b}$
Now while trying to think about this, once I select a random $K_{a,b}$, the probability of it being monochromatic should be $2^{1-(a+b)}$ instead of $2^{1-ab}$ since (for the denominator of the probability) we are choosing colors for $a+b$ vertices independently. Where am I going wrong?
We're coloring the edges; $K_{a,b}$ has $ab$ edges, so there is a $2^{-ab}$ chance they're all red (say) and $2^{-ab}$ chance they're all blue. In total, $2^{1-ab}$.
Alon and Spencer skip a bit of the setup for the sake of having a very "quick" application of the probabilistic method. But we know that this has to be an edge-coloring problem, because the vertex-coloring version of the problem is very silly. If you try to color the vertices of $K_{m,n}$ to avoid a monochromatic $K_{a,b}$, just color all the vertices on one side red and all vertices on the other side blue.
(Vertex-coloring versions of the monochromatic subgraph problem are usually silly, though they might be silly in other ways - think about what happens with $K_n$, for instance - so we never consider them. It probably did not occur to Alon and Spencer that you could be coloring anything other than the edges.)