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I have a question related to permutations and probability. Suppose we have 6 distinct items that we want to arrange in a specific order. We start with the items in some arbitrary order, and we can make a "move" by swapping the positions of any two items.

Given that we can make up to N (for n in [1,2,3,4,5,6]) moves, what is the probability that we can achieve the correct order of items? Assume that each possible ordering is equally likely.

To clarify, a "move" is defined as swapping the positions of two items. So N=2 means we can swap two such swaps. The "correct" order is a specific predetermined order of the items.

I understand that the total number of possible orderings is 6!, or 720. However, I'm not sure how to calculate the number of orderings that can be reached within N moves, or how to translate this into a probability. Any insights would be appreciated.

Jamal Mantburg
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2 Answers2

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Robert Shore's comment is on point. For a given permutation with cycle lengths $K_1, K_2, \dots, K_M$, where $M$ is the number of cycles, the number of swaps you need to reach the identity permutation is $\sum_{i=1}^M (K_i-1)$, which is the same as $N-M$.

To see this, you can check that any swap either increases or decreases this distance by $1$ (and there is always a way to decrease it, unless you are already at the identity permutation).

For a uniformly random permutation, when $N$ is large, the number of cycles $M$ is roughly Poisson with mean $\log(N)$, with fluctuations on the scale of $\sqrt{\log(N)}$. You can find a ton of different discussions about this quantity, for instance:

So the number of swaps you need for a uniformly random permutation is $N-\log(N)\pm O(\sqrt{\log(N)})$.

From a given permutation, one move that is guaranteed to reduce the distance is to move element $i$ to position $i$ (where $i$ is any element that is currently out of position). So for example, you have the following explicit algorithm to get to the identity in minimum time:

for $i=1$ to $N-1$:

  • if element $i$ is not in position $i$, swap $i$ with the element at position $i$.

You can understand the $\log(N)$ nicely in terms of this algorithm. It turns out that after step $i$, the permutation of items $\{i+1, \dots, N\}$ is uniformly random. In particular, the event $B_i$ that a swap is not required at step $i$ has probability $1/(N-i)$. (In fact, these events $(B_i)_{1\leq 1\leq N-1}$ are independent.) The expected number of steps at which a swap is not required is then $1/N+1/(N-1)+\dots+1/3+1/2$ which grows like $\log(N)$.

James Martin
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As Robert Shore indicates in his comment, if a permutation, $\ p\ $, is a product of $\ c\ $ disjoint cycles of lengths $\ \ell_1,$$\,\ell_2,$$\,\dots,$$\,\ell_c\ $, then it can be expressed as a product of $\ n_p=\sum_\limits{j=1}^c\big(\ell_j-1\big)\ $ transpositions, but not as a product of any smaller number of transpositions. Thus, $\ n_p\ $ is the smallest number of swaps needed to return a sequence of numbers that has been permuted by $\ p\ $ back to its original order. The following table lists the $\ 11\ $ possible cycle structures of the permutations on $\ \{1,2,3,4,5,6\}\ $, the number of permutations with that cycle structure, the probability of a random permutation having that cycle structure and the minimum number of transpositions whose product can be equal to a permutation with that cycle structure. The notation $\ c_1\times\ell_1+c_2\times\ell_2+\dots+c_r\times \ell_r\ $ used in the second column of the table indicates a cycle structure comprising $\ c_i\ $ cycles of length $\ \ell_i\ $ for each $\ i=1,2,\dots,r\ $. \begin{array}{|c|c|c|c|c|} \hline {\ \\\text{index }\ i}&{\hspace{1.5em}\text{cycle}\\{\text{structure}\ s_i}}&{\hspace{1.5em}\text{number of}\\\text{permutations }\ n_i}&{\ \\\text{probability }\ \pi_i}&{\hspace{1.5em}\text{number of}\\\text{transpositions }\ t_i}\\ \hline 1&6\times1&1&\frac{1}{720}&0\\ \hline 2&4\times1+1\times2&{6\choose2}=15&\frac{15}{720}=\frac{1}{48}&1\\ \hline 3&2\times1+2\times2&\frac{1}{2}{6\choose2}{4\choose2}=45&\frac{45}{720}=\frac{1}{16}&2\\ \hline 4&3\times2&\frac{1}{6}{6\choose2}{4\choose2}=15&\frac{15}{720}=\frac{1}{48}&3\\ \hline 5&3\times1+1\times3&2{6\choose3}=40&\frac{40}{720}=\frac{1}{18}&2\\ \hline 6&1\times1+1\times2+1\times3&2{6\choose3}{3\choose2}=120&\frac{120}{720}=\frac{1}{6}&3\\ \hline 7&2\times3&2{6\choose3}=40&\frac{40}{720}=\frac{1}{18}&4\\ \hline 8&2\times1+1\times4&3!{6\choose4}=90&\frac{90}{720}=\frac{1}{8}&3\\ \hline 9&1\times2+1\times4&3!{6\choose4}=90&\frac{90}{720}=\frac{1}{8}&4\\ \hline 10&1\times1+1\times5&4!{6\choose5}=144&\frac{144}{720}=\frac{1}{5}&4\\ \hline 11&1\times6&5!=120&\frac{120}{720}=\frac{1}{6}&5\\ \hline \end{array}

Thus, the probabilities that a random permutation of $6$ objects can be restored with at most $\ n\ $ swaps is given by the following table.

\begin{array}{|c|c|c|c|c|c|c|} \hline n&5&4&3&2&1&0\\ \hline {\text{probability that }\ n\\ \text{swaps will suffice}}&1&\sum_\limits{i=1}^{10}\pi_i=\frac{5}{6}&{\pi_8+\sum_\limits{i=1}^6\pi_i\\\hspace{1em}=\frac{163}{360}}&{\pi_6+\sum_\limits{i=1}^4\pi_i\\\hspace{1em}=\frac{101}{720}}&\pi_1+\pi_2=\frac{1}{45}&\pi_1=\frac{1}{720}\\ \hline \end{array}

lonza leggiera
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