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According to this answer, 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 (which is a necessary condition for planarity) is: $$2^{-n(n-1)/2}\sum_{k=0}^{3n-6}{n(n-1)/2\choose k}$$

Therefore, the plot for this equation is:

enter image description here

If I choose a random graph with 10 vertices, what is the probability that it will remain planar after adding an edge between two randomly chosen vertices?

To be more clear:

Consider event $P_{1}$, the event of choosing a planar graph $G_{1}$ in the universe $U_{10}$ of graphs with 10 vertices.

Consider a subset $S$ of the universe of graphs with 11 vertices $U_{11}$ conditionally defined by the graph $G_{1}$ we got in event $P_{1}$ in this way: $S$ is formed by all possible graphs that can result from adding one edge to $G_{1}$.

And then finally consider event $P_{2}$, the event of choosing a planar graph in the universe $S$ defined in the previous step.

Therefore I want to know what is the probability of event $P_{2}$.

Ruan
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  • When you say "remain planar", are you implying that you're conditioning on the original graph being planar? Also, why do you bring up the $3n-6$ condition? Are you interested in the probability that a graph continues to fulfil that condition when you add an edge -- or only in the exact probability that it remains planar? – joriki Aug 10 '18 at 10:12
  • I have edited the question, I hope it's clear now. The $3n - 6$ condition was brought up only as a suggestion of a way to measure the probability of planarity, but I'm open to other solutions. – Ruan Aug 10 '18 at 10:38

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