Let $[n]$ be the set ${1,2,…,n}$ (we call each element a vertex) and a permutation of $[n]$ be $\pi=[\pi(1),\pi(2),\cdots,\pi(n)]$, i.e., denote $\pi(x)=y$ the fact that in a permutation, position $x$ has vertex $y$. There are $n!$ possible permutations of $[n]$.

For example, permutation of 1,2,3,4 into 3,2,4,1 has a 1-cycle of 2, i.e. $\pi(2) = 2$, and a 3-cycle, $\pi(1) = 3, \pi(3) = 4, \pi(4) = 1$.

A cycle is such that we have a sequence of $\pi(j) = k, \pi(k) = \ell, \cdots, \pi(p) = q, \pi(q) = j$.

expected number of cycles

Assume all permutation is of equal probability. The expected number of cycles in permutation of $[n]$ be $h(n)$. Then obviously $h(1) = 1$ because there is only one permuatation. $h(2) = 1.5$ for the only permuatation are 1,2 and 2,1 which has 2 and 1 cycles respectively.

$h(n) = \sum_{i=1}^n \frac{1}{n}$. Proof as follows. (adapted from ref #1)

Assume we have a permutation of $[n-1]$, $[\pi(1),\pi(2),\cdots,\pi(n-1)]$, we have $n$ ways to generate a permutation of $[n]$ from it: insert $n$ before any $\pi(i)$ or insert $n$ after $\pi(n-1)$. Result of such is either we make $n$ into one of the existing cycle (first $n-1$ ways) or we make $n$ into a 1-cycle (last way). Therefore we have:

And by induction, $h(n) = \sum_{i=1}^n \frac{1}{i}$

Indeed, we can understand it as $h(n) = \sum_{k=1}^n h_i(n)$ where $h_k(n) = \frac{1}{k}$ is the expected number of $k$-cycles in the permutation.

probability of having a n-cycle/1-cycle in a permutation of [n]

$n$-cylce: By counting: $\pi(1)$ can be anything but 1, $\pi(\pi(1))$ can be anything but 1 or $\pi(1)$, and so on. Therefore we have $(n-1)!$ such permutations. Alternatively, there are $(n-1)!$ ways of arranging $n$ symbols in a cycle. Since there are $n!$ possible permutations, the probability of having $n$-cycle is $(n-1)!/n! = 1/n$.

1-cycle: This means $\pi(k)=k$, so for any particular $k$, the probability is $1/n$. But note that, this is the probability of any particular vertex in a 1-cycle, which is different from the probability of a random permutation has any 1-cycle.

The probability that the permutation has only 1-cycle is the probability that the permutation is an identity mapping, $\pi(v)=v$ for all $v$, which is $1/n!$.

probability of a vertex in a k-cycle in a permutation of [n]

If vertex $v_1$ is in a $k$-cycle, it means $\pi(v_1)=v_2, \pi(v_2)=v_3, \cdots, \pi(v_{k-1})=v_k, \pi(v_k)=v_1$. By counting, we have $(n-1)(n-2)\cdots(n-k+1) \times (n-k)! = (n-1)!/(n-k)! \times (n-k)! = (n-1)!$ ways to make a $k$-cycle with $v_1$ in it in a permutation. The term after $\times$ refers to the permutation of remaining $n-k$ vertices not in the cycle. Therefore, given $v_1$, the probability is $(n-1)!/n! = 1/n$.

Alternatively, we can derive the probability by multiplying the probability of correct $\pi(v_i)$ in each step (adapted from ref #2): For $\pi(v_1)$, it can be anything but $v_1$, which has the probability of $\frac{n-1}{n}$. For $\pi(v_2)$, it cant be anything but $v_1,v_2$ but due to the bijective nature of $\pi(\cdot)$ function, $\pi(v_2)=v_2$ has been ruled out by the fact that $\pi(v_1)=v_2$. Therefore the probability is $\frac{n-2}{n-1}$. Similarly, for $\pi(v_i)$ up to $i = k-1$, the probability is $\frac{n-i+1}{n-i+2}$. Finally $\pi(v_k)=v_1$ has probability of $\frac{1}{n-k+1}$. So the overall probability is

probability of a large cycle

There are $\binom{n}{k}$ ways to pick $k$ vertices from $[n]$ and each set of $k$ picked symbols has $(k-1)!$ ways to make a cycle (and the remaining $n-k$ vertices has $(n-k)!$ possible permutations). For $2k > n$, the probability is:

This is also the probability that a permutation has at least one $k$-cycle for $k \le n$.

expected number of k-cycle

In a permutation, there is probability $1/n$ that any vertex is in a $k$-cycle. If we check each of the $n$ vertices whether it is in a $k$-cycle, the expected number of affirmative vertices is $n\times (1/n) = 1$. Since if there is a $k$-cycle, there must be $k$ vertices affirmative to such test, we have the expected number of $k$-cycle as $1/k$.

average length of cycle

A permutation always has $\pi(v_i) = v_j$ for each $v_i$. If we consider that as a graph, there is always $n$ edges. Since we know that the expected number of cycle in a permutation is $\sum_i \frac{1}{i}$, the average cycle length (i.e., counting number of edges, or equivalently the number of vertices), is $n/\sum_i\frac{1}{i} \approx n/\ln n$.

Additional properties

From ref #4:

  • Probability of having $j_k$ number of $k$-cycle, for $k=1,\cdots,n$, in the permutation is $\prod_{k=1}^n \frac{1}{k^{j_k} j_k!}$
  • As $n\to\infty$, the distribution of number of $k$-cycle in permutation converges to Poisson with intensity $1/k$


  1. Math Stackexchange, Name Drawing Puzzle, 2012-07-01
  2. http://www.inference.org.uk/itila/cycles.pdf
  3. Kent E. Morrison, “Random Maps and Permutations”, 1998-04-14
  4. Terence Tao, “The number of cycles in a random permutation”, 2011-11-23