# Probability that a random graph is planar

I've been attempting to solve the following challenge problem from a combinatorics class but am getting absolutely nowhere.

Prove: For sufficiently large $n$, the probability a random graph $G=(V,E)$ with $n$ vertices is planar is less than $2^{-0.49n^2}$.

I've ruled out using Kuratowski's characterization of planarity for proving this (since I know it's hard to determine whether or not a subgraph $G' \subset G$ is a topological $K_{3,3}$ or $K_5$. The other characterization I'm aware of is if a graph is planar, then $\exists v \in V$ such that $deg[v] \leq 5$. So the probability a random graph is planar is equivalent to $1-\mathbb{P}$ ,where $\mathbb{P} =$ the probability that a random graph has no vertices of degree less than or equal to 5.

However, I can't see how to proceed from this observation. Solutions/hints would be appreciated.

• It's disingenuous to say that "the probability a random graph is planar is ... $1-\mathbb{P}$"; you said that IF a graph is planar, then it has such a vertex -- not that the existence of such a vertex is equivalent to planarity. – Nick Peterson Mar 23 '14 at 3:46
• What is your random model for the graph? Each of the $\binom{n}{2}$ is either present or not present, with equal probability $1/2$? – ShreevatsaR Mar 23 '14 at 7:04
• Yes, @ShreevatsaR, each of the $\binom{n}{2}$ edges has a probability of $1/2$ of being included (or more generally probability $p$). Also Nicholas, it's a necessary but not sufficient condition if I'm not mistaken - the existence of such a vertex is not equivalent to planarity, but rather, if no such vertex exists then the graph is NOT planar (so if all vertices have degree 6 or more, then there is no way for it to be planar). – rm95 Mar 24 '14 at 11:18
• If $e\gt3v-6$, the graph is not planar. – Gerry Myerson Mar 30 '14 at 11:31

A random graph on $n$ vertices is a graph which has each of the $n\choose2$ edges independently with probability $1/2$ each. The probability of at most $3n-6$ edges (that's a necessary condition for planarity) is $$2^{-n(n-1)/2}\sum_{k=0}^{3n-6}{n(n-1)/2\choose k}$$ There are standard ways to estimate the sum.
• Vith a very crude bound, $$\sum_{k=0}^{3n-6}\binom{n(n-1)/2}{k}< e\cdot (n^2/2)^{3n} < e^{6n\log n} < 2^{9n\log n}.$$ – Jack D'Aurizio Apr 1 '14 at 23:35
• @JackD'Aurizio: Just for completeness, that turns out to be sufficient to prove the claim: $\dfrac{2^{9n \log n}}{2^{n(n-1)/2}} = 2^{9n\log n - n(n-1)/2}$, and so we want to say that $9n\log n - n(n-1)/2 < -0.49n^2$ or that $9n\log n < n(n-1)/2 - 0.49n^2 = 0.01n^2 - n/2$ for sufficiently large $n$. And this is clearly the case, as the LHS grows as $\Theta(n \log n)$ but the RHS as $\Theta(n^2)$. – ShreevatsaR Apr 6 '14 at 7:16