The probability that exactly / at-least $k$ numbers are in the correct position

Asked Jul 25 '14 at 11:06

Active Jul 25 '14 at 11:24

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Given a sequence of $[1,\dots,n]$ in random order:

Let $P_k$ be the probability that exactly $k$ numbers are in the correct position
Let $Q_k$ be the probability that at least $k$ numbers are in the correct position

How can we calculate $P_k$ and $Q_k$?

These are the facts that I've managed to establish so far:

Partial answers (for example: $P_0,P_1,Q_1$) or any other insights will also be appreciated...

Thanks

edited Jul 25 '14 at 11:13

asked Jul 25 '14 at 11:06

barak manos

http://mathoverflow.net/questions/9484/number-of-permutations-with-a-specified-number-of-fixed-points – Jack D'Aurizio Jul 25 '14 at 11:16
2

http://en.wikipedia.org/wiki/Rencontres_numbers – Henry Jul 25 '14 at 11:26
As a check, $\sum\limits_{k=0}^{n} P_k =\sum\limits_{k=0}^{n} k P_k = \frac12 \sum\limits_{k=0}^{n} Q_k = n!$ – Henry Jul 25 '14 at 11:28
@Henry: Thank you. So this link gives the answer for $P_k$. But it's a recursive answer, where you need to calculate $P_{n-k}$ first. Does that mean that there is no "straight-forward formula" answering this question? – barak manos Jul 25 '14 at 11:31
I only offered my sums as a check – Henry Jul 25 '14 at 11:33
$\displaystyle P_k = {n \choose k} \times \text{round}\left[\frac{(n-k)!}{e}\right]$, except $P_n=1$ – Henry Jul 25 '14 at 11:37
@Henry $P_n=1$? and $\sum_{k=0}^{n}P_{k}=n!$? $P_{n}=\frac{1}{n!}$ and$\sum_{k=0}^{n}P_{k}=1$. There must be some misunderstanding here. – drhab Jul 25 '14 at 12:07
@Henry: I referred to the Wikipedia link (not to the sum). – barak manos Jul 25 '14 at 12:21
@drhab: Divide my numbers by $n!$ – Henry Jul 25 '14 at 13:23

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