shravya.nitk
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I have solved this problem partially but I am unable to continue at this point:
- shravya.nitk August 22, 2013case 1:
There are 2 approaches: logical and recursive.
if you pick a number i, to find a number smaller than the one you have is:
Exp(num<i)=1*(i-1)/(n-1)+2*(n-i)(i-1)/(n-1)^2+3*(n-i)^2(i-1)/(n-1)^3.......
which reduces to Exp(num<i)=(n-1)/(n-i)
Also, recursively
E(i)=1*(i-1)/(n-1)+(E(i)+1)*(n-i)/(n-1)
when you solve it, u get
E(i)=(n-1)/(n-i)
Now, this is where i get stuck.
to find final expectation value
E=sum(E(i)) where i=1:n =infinite :/
and if u exclude the case of picking up 1, u get a harmonic sum. so, what do i do next?
case 2:
this doesn't lead to infinite but it poses a logical question: you know that if u pick the number 1, even if we look through the remaining n-1 chits, there is nothing u can find smaller than 1. so it's a failure. meaning u never find it, not... u find it after n-1, so we can't assume it as any number....
so what do i do for that??
Continuing, if i exclude 1 and try to solve logically, cause the recursion formula i couldn't get
if u pick
n -> 1
n-1 -> 1*(n-2)/(n-1)+2*1/n-1
n-2 -> 1*(n-3)/(n-1)+2*2/(n-1)(n-2)
and so on
2 -> 1*1/(n-1)+2*(n-2)*1/(n-1)(n-2)+3*(n-2)(n-3)*1/(n-1)(n-2)(n-3)+.......
how do i simply it?