Let $S = S_{30}$, the set of permutations on $\{1,...,30\}$.

Suppose $A$ is the event: $\{\pi\in S_{30}: \pi(1) = 1\}$ and $B$ is the event $\{\pi\in S_{30}: \pi(2) < \pi(3)\}$.

I am asked to show that these events are independent.

Now $|A| = 29!$ and $|B| = (1+...+29)\times 28!$

Hence $|A|\times|B| = 29!\times 28! \times (1+...+29)$

However $|A\cap B| = (1+...+28)\times 27!$

So $P(A\cap B)< P(A)\times P(B)$.

So these events don't seem to be independent? What am I doing wrong?

  • $\begingroup$ @AdamHughes Thank you, I was thinking of something else. I've written an answer now. $\endgroup$ – DonAntonio Jun 6 '16 at 15:44

By definition: since $\;A\cong S_{29}\;$ :

$$P(A)=\frac{29!}{30!}=\frac1{30}$$ , whereas

$$B:=\{\pi\in S_n\;/\;\pi(2)<\pi(3)\}$$

and we can then count:

$$\begin{cases}\pi(3)=30\implies\pi(2)=1,2,...,29\\{}{}\\\pi(3)=29\implies\pi(2)=1,2,...,28\\\ldots\\\pi(3)=2\implies \pi(2)=1\end{cases}$$

For each option in each line we have $\;(n-2)!\;$ permutations, so all in all we have



Finally: if $\;\pi\in A\cap B\;$ then

$$\begin{cases}\pi(1)=1,\,\pi(2)=2\implies \pi(3)=2,3,....,30\\{}\\\pi(1)=1,\,\pi(2)=3\implies \pi(3)=4,...,30\\\ldots\\{}\\\pi(1)=1,\,\pi(2)=29\implies \pi(3)=30\end{cases}$$

For each option above we have $\;27!\;$ permutations, so in total we have

$$27!\left(1+2+\ldots+28\right)=\frac{29!}2\implies P(A\cap B)=\frac{\frac{29!}2}{30!}=\frac1{60}=P(A)P(B)$$

and the events are independent


I assume that we are picking a permutation $\pi$ at random, with all permutations equally likely.

Let $A$ be the event $\pi(1)=1$ and $B$ the event $\pi(2)\lt \pi(3)$. To show independence, we show that $\Pr(B\mid A)=\Pr(B)$. This is clear, for by symmetry each is $\frac{1}{2}$.

  • $\begingroup$ Thanks for this explanation. May I ask where did I go wrong in my argument? $\endgroup$ – fosho Jun 6 '16 at 15:39
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    $\begingroup$ The calculations seem to be correct, but do not address the issue of independence. $\endgroup$ – André Nicolas Jun 6 '16 at 15:43
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    $\begingroup$ @McFry: At the time I wrote the comment, the OP made no reference to probability. $\endgroup$ – André Nicolas Jun 6 '16 at 16:02

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