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It's well known that the probability of a $n$-permutation being a derangement is about $e^{-1}$ when $n$ is large. But how about if we have $k$ of each number, i.e. we have multipermutations of $M_{k,n}:=\{1\times k, 2\times k, \dots, n\times k\}$. Is the probability of a derangement then $e^{-k}$?

This answer has a generalization for the cycle-lengths. But how do cycles correspond to blocks except in $k=1$ case? It certainly feels that Poisson approximation should work here too, since if there are many, many elements, picking of an element of a permutation is nearly independent of other elements and it has probability $1\over n$ of being mapped to its "own block" and there are $kn$ elements so the mean is $k$.

My try (case $k=2$ calculated).

By rook polynomial theory we know that the probability $p_{k,n}$ of a random permutation of $M_{k,n}$ being a derangement is given by

$$ p_{k,n} = \frac{1}{(kn)!}\sum_{j=0}^{kn}(-1)^jc_j(kn-j)! $$

where $c_j$ are the coefficients of $r_k(x)^n$ where $r_k(x)$ is the rook polynomial

$$ r_k(x) = \sum_{j=0}^k {k \choose j}^2 j! x^j $$

(Note: There's a question of whether we consider the same values in the permutation to be distinct or not, but it doesn't matter as it cancels in the probability: $\frac{1}{(k!)^n} \left({(kn)! \over (k!)^n}\right)^{-1} = \frac{1}{(kn)!}$)

In order to write the sum in a nicer form, let's homogenize the rook polynomial:

$$ r_k(x, t) = \sum_{j=0}^k {k \choose j}^2 j! x^jt^{k-j}. $$

Then because $\int_0^\infty e^{-t}t^m dt = m!$, we get

$$ p_{k,n} = \frac{1}{(kn)!} \int_0^\infty e^{-t}r_{k,n}(-1,t)^n dt. $$

For $k=2$, the rook polynomial is $2x^2 + 4tx + t^2$. Let's start to approximate and manipulate (using Stirling, scale variable by $n$, Taylor approx log at $t=2$,...)

$$ p_{2,n} = \frac{1}{(2n)!} \int_0^\infty e^{-t}(t^2 - 4t + 2)^n dt \\ \approx \frac{e^{2n}}{\sqrt{2\pi n}(2n)^{2n}} \int_0^\infty e^{-t}(t^2 - 4t + 2)^n dt \\ = \frac{1}{\sqrt{2\pi n}} \int_0^\infty e^{2n-t}\left(\frac{1}{4n^2}(t^2 - 4t + 2)\right)^n dt \\ = \frac{1}{2}\sqrt{\frac{n}{\pi}} \int_0^\infty e^{2n-nt}\left(\frac{1}{4}t^2 - \frac{1}{n}t + \frac{1}{2n^2}\right)^n dt \\ \approx \frac{1}{2}\sqrt{\frac{n}{\pi}} \int_1^3 \exp \left(2n-nt + n\log\left(\frac{1}{4}t^2 - \frac{1}{n}t + \frac{1}{2n^2}\right) \right) dt \\ \approx \frac{1}{2}\sqrt{\frac{n}{\pi}} \int_1^3 \exp \left(2n-nt + n\log\left(\frac{1}{4}t^2 - \frac{1}{n}t\right) \right) dt \\ \approx \frac{1}{2}\sqrt{\frac{n}{\pi}} \int_1^3 \exp \left(2n-nt + n \left( \log\left( \frac{n - 2}{n} \right) + \frac{n - 1}{n-2}(t - 2) - \frac{n^2 - 2 n + 2}{4(n-2)^2} (t - 2)^2 \right) \right) dt \\ \approx \frac{1}{2}\sqrt{\frac{n}{\pi}} \int_{-\infty}^{\infty} \exp \left(2n-nt + 2 + n(t - 2) - \frac{n}{4} (t - 2)^2 \right) dt \\ = \frac{1}{2}\sqrt{\frac{n}{\pi}} \int_{-\infty}^{\infty} \exp \left(-2 - \frac{n}{4} (t - 2)^2 \right) dt \\ = e^{-2} $$

There were a lot of guesstimates here, but I think it should hold. By the way I did the calculation without plugging in $x=-1$ at first and it was nice to see it converge to $e^{2x}$.

ploosu2
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1 Answers1

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This answer is still somewhat unrigorous.

From the integral representation it is quite clear that we can approximate the rook polynomial $r_{k,n}(x,t)$ by first two terms: $t^k-k^2t^{k-1}x$. Then we could do the integration approximation (a better change of variable is to scale by $kn$ so the point of interest becomes $t=1$, maybe even translate it to origin...)

But actually, just from the sum representation binomial formula yields

$$ \frac{1}{(kn)!} \sum_{j=0}^n (-1)^j {n \choose j}k^{2j}(kn-j)! \\ = \sum_{j=0}^n \frac{(-k)^j}{j!} \frac{n!k^j(kn-j)!}{(kn)!(n-j)!} $$

This already looks like expansion of $e^{-k}$ and indeed in the limit it is, since the multiplier

$$ \frac{n!k^j(kn-j)!}{(kn)!(n-j)!} = \frac{(kn)(k(n-1))\dots(k(n-j+1))}{(kn)(kn-1)\dots(kn-j+1)} \xrightarrow[{n\to\infty}]{} 1 $$

Maybe we could even keep the whole rook polynomial and use multinomial theorem and then see the non-significant terms vanish.

ploosu2
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