Permutation: how many random permutations needed to have every possible values in every possible places

36 Views Asked by At

I'm doing random permutations of a set (shuffling) and I want to generate the minimum number of permutations to have every values at every places, at least once.

Say I have a set of $n$ distinct elements.

There's $n!$ possible permutations of that set.

In one permutation, there's $\frac{(n - 1)!}{n!} = \frac{1}{n}$ probability to get one value at a given place.

There's $\frac{n - 1}{n}$ probability of not having that value at that place.

So the probability of having one value at a given place in $m$ random permutations is $1 - (\frac{n - 1}{n})^m$

Now what's the probability of having every values at given place in $m$ random permutations ?

And what's the probability of having every values at every place in $m$ random permutations ?