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   Expecto Inversio!
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   Author  Topic: Expecto Inversio!  (Read 507 times)
TenaliRaman
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Expecto Inversio!  
« on: Jul 1st, 2004, 8:47pm »
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In a permutation a1..an of n distinct integers, an inversion is a pair(ai,aj) such tht i<j and ai>aj.
 
If all permutations are equally likely , what is the expected no. of inversions in a randomly chosen permutation of 1..n?
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pedronunezmd
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Re: Expecto Inversio!  
« Reply #1 on: Jul 2nd, 2004, 7:28am »
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Sounds like another way of asking "How many combinations of the n integers taken 2 at a time are there?"
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Barukh
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Re: Expecto Inversio!  
« Reply #2 on: Jul 2nd, 2004, 7:46am »
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What is E(n) - E(n-1)?
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Grimbal
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Re: Expecto Inversio!  
« Reply #3 on: Jul 2nd, 2004, 12:08pm »
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I would just compare the expected number of inversions and the expected number of non-inversions.  (a non-inversion being a pair(ai,aj) such tht i<j and ai<aj).  They should be equal and the sum is the number of pairs.
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towr
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Re: Expecto Inversio!  
« Reply #4 on: Jul 2nd, 2004, 1:57pm »
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::And of course the non-inversion change to inversions and vice versa if you take the mirror image of the permutation, which is itself another permutation.
So on average there must be half of each..
::
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TenaliRaman
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Re: Expecto Inversio!  
« Reply #5 on: Jul 2nd, 2004, 2:22pm »
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gee i knew this was too easy even for easy  Undecided
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towr
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Re: Expecto Inversio!  
« Reply #6 on: Jul 2nd, 2004, 2:39pm »
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I wouldn't say it's too easy for easy. We're just too smart for the easy section, really Tongue
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Grimbal
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Re: Expecto Inversio!  
« Reply #7 on: Jul 2nd, 2004, 2:57pm »
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Expecto Inversio!  sounds like a Harry Potter spell.
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Barukh
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Re: Expecto Inversio!  
« Reply #8 on: Jul 3rd, 2004, 4:31am »
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I like that, Grimbal and towr. Here's another approach: E(n) - E(n-1) = (n-1)/2.
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TenaliRaman
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Re: Expecto Inversio!  
« Reply #9 on: Jul 3rd, 2004, 9:31am »
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Grimbal,
yes i am a harry potter fun obviously excited abt the 6th book.  Grin
 
Barukh,
Nice approach! It never hit me though i think it should have been obvious!
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