Thanks Pedro ... I did not say you wrote so .. I was just asking if
what I am saying is correct :)

On 16 May 2011 15:51, Pedro Osório <[email protected]> wrote:
> @Brats:
> I didn't write E[n=a+1] = 1 + E(n=a), I wrote for the specific case of a=3
> (which is true for all a>1) that:
> E[n=a] = 1+ sum( p(n=0) * E[n=0] + p(n=1) * E[n=1] + ... + p(n=a) * E[n=a])
> Where the n on the LHS is the number of elements in the wrong position
> before hitting and on the RHS the same after hitting.
> For a=3 the probabilities are trivial to compute, and hence I used it as an
> example (because Eagle insisted that E[n=3] > 3), for bigger a this
> computation is not so easy, but it's still correct.
> I know the notation is strange, but relating to lemma 1, my E[n=a] is simply
> x(A), when A has 'a' elements out of place, and my p(n=b) is simply pb.
>
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