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Consider a permutation $S$ of $n$ numbers between 1 to $n$. We know probability $P(S[i]=j)=a_{ij}$ for $1\leq i,j \leq n$. We want to find $P(S[4]=6 | S[30]=25)$. We can use approximations like $P(S[4]=6)+\frac{P(S[4]=25)}{(n-1)}$ or $\frac{P(S[4]=6)}{(1-P(S[4]=25))}$.

Can we approxiamate better or other than these?

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There is nothing stopping an arbitrary distribution on permutations from having the conditional probability you are interested in be zero. You need to say more about the exact setup you are considering. –  Louis Jan 2 '12 at 9:38
I can understand the problem. Thank you very much for pointing out it. –  user12290 Jan 3 '12 at 5:38

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