Editing Probabilistic method (section)

===Second example===
A 1959 paper of Erdős (see reference cited below) addressed the following problem in [[graph theory]]: given positive integers {{mvar|g}} and {{mvar|k}}, does there exist a graph {{mvar|G}} containing  only [[cycle (graph theory)|cycles]] of length at least {{mvar|g}}, such that the [[chromatic number]] of {{mvar|G}} is at least {{mvar|k}}?

It can be shown that such a graph exists for any {{mvar|g}} and {{mvar|k}}, and the proof is reasonably simple.  Let {{mvar|n}} be very large and consider a random graph {{mvar|G}} on {{mvar|n}} vertices, where every edge in {{mvar|G}} exists with probability {{math|''p'' {{=}} ''n''<sup>1/''g''−1</sup>}}.  We show that with positive probability, {{mvar|G}} satisfies the following two properties:

:'''Property 1.''' {{mvar|G}} contains at most {{math|''n''/2}} cycles of length less than {{mvar|g}}.

'''Proof.''' Let {{mvar|X}} be the number cycles of length less than {{mvar|g}}. The number of cycles of length {{mvar|i}} in the complete graph on {{mvar|n}} vertices is

:<math>\frac{n!}{2\cdot i \cdot (n-i)!} \le \frac{n^i}{2}</math>

and each of them is present in {{mvar|G}} with probability {{math|''p<sup>i</sup>''}}. Hence by [[Markov's inequality]] we have

:<math>\Pr \left (X> \tfrac{n}{2} \right )\le \frac{2}{n} E[X] \le \frac{1}{n} \sum_{i=3}^{g-1} p^i n^i = \frac{1}{n} \sum_{i=3}^{g-1} n^{\frac{i}{g}} \le \frac{g}{n} n^{\frac{g-1}{g}} =  gn^{-\frac{1}{g}} = o(1).</math>
: Thus for sufficiently large {{mvar|n}}, property 1 holds with a probability of more than {{math|1/2}}.

:'''Property 2.''' {{mvar|G}} contains no [[Independent set (graph theory)|independent set]] of size <math>\lceil \tfrac{n}{2k} \rceil</math>.

'''Proof.''' Let {{mvar|Y}} be the size of the largest independent set in {{mvar|G}}. Clearly, we have

:<math>\Pr (Y\ge y) \le {n \choose y}(1-p)^{\frac{y(y-1)}{2}} \le n^y e^{-\frac{py(y-1)}{2}} = e^{- \frac{y}{2} \cdot (py -2\ln n - p)} = o(1),</math>
when

:<math>y = \left \lceil \frac{n}{2k} \right \rceil\!.</math> Thus, for sufficiently large {{mvar|n}}, property 2 holds with a probability of more than {{math|1/2}}.

For sufficiently large {{mvar|n}}, the probability that a graph from the distribution has both properties is positive, as the events for these properties cannot be disjoint (if they were, their probabilities would sum up to more than 1).

Here comes the trick: since {{mvar|G}} has these two properties, we can remove at most {{math|''n''/2}} vertices from {{mvar|G}} to obtain a new graph {{math|''G′''}} on <math>n'\geq n/2</math> vertices that contains only cycles of length at least {{mvar|g}}. We can see that this new graph has no independent set of size <math>\left \lceil \frac{n'}{k}\right\rceil</math>. {{math|''G′''}} can only be partitioned into at least {{mvar|k}} independent sets, and, hence, has chromatic number at least {{mvar|k}}.

This result gives a hint as to why the computation of the chromatic number of a graph is so difficult: even when there are no local reasons (such as small cycles) for a graph to require many colors the chromatic number can still be arbitrarily large.