Random Permutation Statistics - Probability That A Random Subset of Lies On The Same Cycle

Probability That A Random Subset of Lies On The Same Cycle

Select a random subset Q of containing m elements and a random permutation, and ask about the probability that all elements of Q lie on the same cycle. This is another average parameter. The function b(k) is equal to, because a cycle of length k contributes subsets of size m, where for k < m. This yields

\frac{\partial}{\partial u} g(z, u) \Bigg|_{u=1} = \frac{1}{1-z} \sum_{k\ge m} {k \choose m} \frac{z^k}{k} = \frac{1}{1-z} \frac{1}{m} \frac{z^m}{(1-z)^m} = \frac{1}{m} \frac{z^m}{(1-z)^{m+1}}.

Averaging out we obtain that the probability of the elements of Q being on the same cycle is

{n \choose m}^{-1} \frac{1}{m} \frac{z^m}{(1-z)^{m+1}} = {n \choose m}^{-1} \frac{1}{m} \frac{1}{(1-z)^{m+1}}

or

\frac{1}{m} {n \choose m}^{-1} {(n-m) \; + \; m \choose m} = \frac{1}{m}.

In particular, the probability that two elements p < q are on the same cycle is 1/2.

Read more about this topic:  Random Permutation Statistics

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