In the probabilistic method, I often see we define a sample space and sample the elements of the sample space uniformly at random and use it to prove something exists. For example, in the proof of a lower bound of a Ramsey number, if $\binom{n}{k}2^{1-\binom{k}{2}}<1, R(k,k)<n.$ We start by assuming each edge is colored with probability $0.5$ for each color (blue or red).
What if we do not use the uniform distribution? We'll get a new valid result or results that are unrealizable in practice? Is it necessary to stick with the uniform distribution?
I expect that a non-uniform distribution should be as fine if we try to show the probability of something is non-zero, as long as you do not give too many zero probabilities to realizable objects (e.g. a graph with all red/blue edges). But for the expectation argument in the probabilistic method, I think if you use non-uniform distributions, all expected values could be changed and we could get a weaker or stronger statement than its counterpart theorem that assumes a uniform distribution.
I try to find the discussion on this issue on various resources, but couldn't find it yet.
In fact the second example of the Wikipedia article on the probabilistic method uses a non-uniform distribution. By including each potential edge in a graph on $n$ vertices with probability $p=n^{1/g−1}$, it proves that for large $n$ the probabilites for the graph to contain at most $n/2$ cycles of length less than $g$ and to contain no independent set of size $\left\lceil\frac n{2k}\right\rceil$ are both greater than $\frac12$, and thus there must be a graph that has both of these properties.