If there are N items and you are looking for precisely one of them, with replacement, then you want to work out the sum to infinity of 1/N + 2*(1/N)*(N-1)/N + 3*(1/N)*[(N-1)/N]^2 + 4*(1/N)*[(N-1)/N]^3 ... etc. Write as (1/N)*Sum(i=1,infinity) i [(N-1)/N]^(i-1) This is the same as N* d/dN{Sum(i=1,infinity) [(N-1)/N]^(i)} which is the same as N* d/dN...
11:06 Mon 25th Jul 2022