Usually in mathematics, you have some question you want answered, and you often want to find a constructive answer. You are questioning if something exists, and so you want an example of it (if it does exist), or you know something exists but want to know how many of them exist, and so you would like a way to easily find multiple examples of it. However, sometimes its easier to come up with a nonconstructive answer where you show something exists (maybe even that many of them do), but you don’t produce even one example of it. Still, nonconstructive answers can provide useful insights ^{1}.
In this post I want to show a couple examples of what’s called the probabilistic method. The basic idea is you want to study some property of a set of objects. In order to do this, you consider a random distribution over your set of objects, and investigate the probability that a randomly chosen object has the desired property. In particular, the the probability that the object has the property is nonzero, then some object with that property must exist! In this way, you can prove the existence of certain types of objects without needed to construct an example.
Primes
As a warm up, let’s prove something that hopefully doesn’t seem too farfetched. Before we state, let’s introduce a useful function. Let . That is, counts the number of primes numbers there are lequal^{2} to .
Theorem
There are infinitely many prime numbers.
Fix any postive integer . Consider possible factorizations of all the integers with . Note that each such can be factored as ^{3} where is squarefree^{4}. Since each has a unique factorization of this form, if we can get an upper bound for the number of ways of picking and (such that ), then we will have an upper bound for the value of . This is because each such choice corresponds to a different value for , and so the number of choices is the number of values of which is .
We’ll bound choices for and separately. Since is easier, we’ll start with it. Since , we must have that , and so there are at most possible choices for . For bounding , note that being squarefree means that all of its prime factors are unique. Thus, we can count the number of choices for by counting the number of choices for its prime factors. Each such choice corresponds to some (finite) set of prime numbers. Since we require that , every prime in ’s factorization must be lequal to . Thus, the possible choices for the primes that make up (and hence the possible choices for the values of ) correspond to subsets of where is the largest prime lequal to . By definition, the size of this set is and so there are at most valid choices for . Since has at most possible values and has at most possible values, the factorization has at most possible choices, and hence . A little bit of algebra shows that this implies that . Since the right hand side approaches infinity as , there must be infinitely many primes.
I know what you are thinking. I spent the first bit of this post talking about the probabilistic method and how it’s great and junk, but there was no probability in the proof I just presented. This was intentional. See, many probabilstic proofs don’t actually require probability (although it’s often useful to use it), and can be transformed into equivalent proofs purely using counting arguments. Instead of showing the probability of something is positive, you can count how many of them there are, and so that this is positive directly^{5}. Now that I have said this, I will present an alternate proof of this theorem that makes explicit use of probability.
We will form a distribution on the numbers for some fixed positive integer . This distribution will be formed by noticing that all of these numbers can be factored into the form where is squarefree. Our distribution works by picking an uniformly. For choosing , we once again rely on the fact that there exists a 11 correspondence between choices of and subsets of where is the largest prime lequal to . As such, is chosen by uniformly selecting one of these subsets and taking the product of all its members ^{6}. Every choice of leads to a different number, and every choice of is equally likely, so our distribution is actually just the uniform distribution on !^{7} Since we are interested in the number of primes, let’s investigate how likely drawing from this distribution is the give us a prime number. Note that is prime iff and corresponds to a singleton set. and . Calling these events and , respectively^{8}, it is easily seen that and so, letting be a random variable drawn from this distribution,
Rearranging this equation shows that , and so we once again have that there are infinitely many primes.
And so we see our counting argument has an equivalent probabilistic argument. I’ll admit that we did not gain many insights from using a probabilistic approach here ^{9}, but this was just a warm up and the next proof will rely more on probability.
Graphs
Before getting to the theorem here, I’m gonna do a quick introduction to graph theory.
Definition
A graph is a pair where is a set of vertices and is a set of edges.
Here, we restrict ourselves to simple, undirected graphs. That means that edges go both ways (), no vertices are adjacent to themselves (for all , ), and each edge appears at most once ( is a set and not a multiset).
Definition
A path in a graph from to is an alternating sequences of vertices and edges where every edge is unique, and every vertex is unique. The length of a path is its number of edges.
Below is an example of a simple, undirected graph with a path of length 3 highlighted.
Definition
A graph is connected if for any distinct , there exists a path in from to .
The graph above is connected. The graph below is not connected.
Definition
A graph is edgeconnected for a positive integer if you must remove edges in order to disconnect it. More formally, for any with , the graph is connected. In particular, a graph is 1edgeconnected if and only if it is connected.
Now that we have all these definitions, we are almost ready to state the theorem. First, we say almost all graphs satisfy a given property if the probability that a random graph on vertices satisfies this property approaches 1 as approaches infinity. In other words, if as you consider more and more graphs, you start to become guaranteed than one chosen at random satisfies the property, then almost all graphs satisfy that property. We now just need to formalize what we mean by random graph.
Definition
Fix some number . The Erdős–Rényi random graph on vertices is constructed by choosing each of the possible edges independently where each edge appears with probability .
Note that although we usually talk about a random graph, or the random graph, we actually definied a distribution on graphs, and what we mean by a random graph is some graph drawn from this distribution. Also note that for , we get the uniform distribution on simple, undirected graphs with vertices. Finally, the theorem.
Theorem
For any positive integer , and any choice of with , almost all graphs are edgeconnected.
Fix a choice of and , and consider an Erdős–Rényi random graph with vertices and any . Denote by the set of edges where exactly one of its vertices is in . Note that is not edgeconnected iff for some . Thus, we will investigate . Let , so there are at most edges in . Using the union bound, we have that
Now that we have this, let be the probability that is not edgeconnected. Summing over all possible , we have^{10}
We want to show that almost all graphs are edgeconnected, so we want to show that . As things are right now, does not look particularly nice to work with, so prepare for some bounds. First note that since this expression is symmetric in and , we can sum up to and multiply by 2 to get the same result. Furthermore, for we have that and that ^{11}. Using these facts, we get
Focussing on the sum over , note that when . Thus
Still focussing on this sum, note that its largest term is roughly . Thus, for sufficiently large , this sum is less than and since , the sum is also less than . Thus, for such large ,
Finally, note that and so, for sufficiently large , . For such , we have
As , the right hand side of this inequality approaches 0. Since the right hand side is an upper bound, and for all , this shows that as desired. Hence, almost all graphs are edgeconnected.

And their accompanying proofs are often really nice ↩

Less than or equal to ↩

We allow s=1 ↩

No perfect square (other than 1) divides it ↩

In many probabilistic proofs, you actually show that the probability of an object not having the property you desire is less than 1 (thus showing the probability it does have the property is positive). To translate this into a counting argument, you could count the number of objects without that property and show that this is less than the total number of objects. Thus, some objects with that property must exist. No probability required. ↩

The product of the empty set is 1 ↩

For correctness, we do not allow invalid choices of s,r (ones where rs^2>n). So, techincally speaking, we pick s,r as described with the caveat that we repick whenever our latest drawing is invalid. ↩

A is the event that s=1 ↩

Other than replacing lequality with strict less than. ↩

The 1/2 is so we do not double count. We do not include m=0 and m=n because separating the entire graph from nothing is not disconnecting it. ↩

This does not require m<=n/2 ↩