the first moment method
The first moment method is the workhorse refinement of the basic probabilistic method that uses the expectation of a counting random variable to prove existence or non-existence. It answers questions of the form 'does there exist a configuration with no bad substructures?' and 'is some random quantity typically large?' by computing a single number — the first moment E[X], where X counts the offending substructures. It is called the first moment method because it uses only E[X], the first moment, with no information about the variance or higher moments.
Two complementary deductions follow from the expectation alone. First, if X is a non-negative integer-valued random variable counting bad events and E[X] < 1, then P(X = 0) > 0, because E[X] >= 1 * P(X >= 1) means P(X >= 1) <= E[X] < 1; so a configuration with no bad substructures exists. This is precisely a union bound dressed as an expectation, since by linearity E[X] = sum of probabilities of the individual bad events regardless of any dependence between them — linearity of expectation requires no independence, which is the secret of its power. Second, the same one-sided control gives existence of a value at least the average: there is always a point of the sample space where X >= E[X] (and one where X <= E[X]), so to prove some object has X large it suffices to show the average is large.
Linearity of expectation working without independence is the crucial feature: even when the indicators of the bad events are wildly correlated, E[sum of indicators] = sum E[indicators] exactly. This is what lets the method handle overlapping cliques, intersecting sets, or correlated edges with no extra work. The honest limitation is that the first moment is one-sided: E[X] small forces P(X = 0) > 0, but E[X] large does not by itself force P(X > 0) > 0, because all the mass could sit on a rare event with a huge value. To prove a count is positive with high probability you must control the variance — that is the job of the second moment method.
In a random graph G(n,p), let X count the number of triangles. Each of the C(n,3) triples is a triangle with probability p^3, so by linearity E[X] = C(n,3) p^3 ~ (np)^3 / 6, with no need to worry that triangles share edges. If p = c/n with c < 1 then E[X] -> c^3/6 stays bounded; if instead np -> 0 then E[X] -> 0, so P(X >= 1) <= E[X] -> 0 and the graph is triangle-free with probability tending to 1.
Linearity of expectation needs no independence — the first moment is computed term by term even when terms overlap.
E[X] -> 0 proves X = 0 with high probability, but E[X] -> infinity does NOT prove X >= 1 with high probability: the mass may concentrate on a rare outcome of huge value. Proving positivity requires the second moment.