TITLE: Probability of a graph having at least 1 k-clique
QUESTION [5 upvotes]: I need to estimate the probability $P(\text{Graph G has at least 1 k-clique})$, any precision will do. I know the edge probability, say $p$, so the average number of the edges, $EK$, is $pm(m - 1)/2$, where $m$ is a number of vertices. I tried to use the Turan theorem with the Markov inequality, but Turan condition is too strong, so I ended up with the estimation $P(\text{Graph G has at least 1 k-clique}) \ge 0$.
I tried to use more precise Chebyshev inequality instead, but to calculate $E(K)^2$ is really hard in my case.
Does anyone know what should I do? Is there maybe any estimates for the probability of a k-clique in a random graph, so I could use it instead of precise Turan condition?

REPLY [5 votes]: If you fix the size of the graph, and vary $k$, then the probability is close to $1$ for small $k$, and close to $0$ for large $k$.  There is a sharp drop at $k \approx r(p,n) = 2(\log n)/\log(1/p)$, which is the size of the largest clique in almost every random graph with $n$ vertices and edge probability $p$: the probability tends to be close to $1$ for $k < r(p,n)$, and close to $0$ for $k > r(p,n)$.
I don't think that generic inequalities are enough to establish this threshold; Bollobás in Section 11.1 of his Random Graphs textbook uses a pure counting argument (and is one source where the facts above can be found).
Another good source is the early article

B. Bollobás and P. Erdös, Cliques in random graphs, Math. Proc. Camb. Phil.Soc. 80, 1976, 419–427. doi:10.1017/S0305004100053056 (PDF)

which deals with the distribution in more detail.  The Matula technical report (referenced by another answer) is also worth reading for its Example 1.