Probability of a random graph being triangle-free 
Let $S$ be the set of all graphs with vertex set $\{1,2,\ldots,n\}$. A random graph $G\in S$ has probability $2^{-{n \choose 2}}$. Show that a random graph almost surely contains a triangle.

My attempt so far: we want to show that as $n$ goes to infinity, the probability of a triangle-free graph goes to zero. I tried to find a general formula for the probability of a triangle-free graph with $n$ vertices, but it seems impossible. I know that for any three vertices, the probability of not being a triangle is $\frac{7}{8}$, but I can't calculate the probability of the union of all such three vertices. Is there any other way to approach this problem?

In this situation, you can greatly overestimate the probability of no triangle and still get it to approach zero.  Divide your set of $n$ vertices into disjoint sets of size $3$ (with perhaps one or two vertices left over if $n$ isn't divisible by $3$). So you've got at least $(n-2)/3$ pairwise disjoint sets of size $3$. What's the probability that none of these sets forms a triangle? Well, each one of these sets fails to form a triangle with probability $7/8$ (as you said in the question). And, because the three-element sets under consideration are disjoint, these probabilities, for different $3$-element sets, are independent.  So the probability that none of these $(n-2)/3$ (or more) sets forms a triangle is $(7/8)^{(n-2)/3}$ (or less).  This approaches zero as $n\to\infty$.  
The probability that there are no triangles at all is, of course, $\leq$ the probability that none of our particular $3$-element sets forms a triangle, so it also approaches zero.
