We choose two groups where group A can be of size X where X ranges from 1 to n-1. Group B is the remaining (n-X).
All values of X are equally likely and all group sizes are equally likely
If we choose one item from 1 to n where all choices are equally likely, what is the average size of the group containing the item?
3. The attempt at a solution
Originally I tried the following:
Average size of group A is simply n/2 as it is uniformly distributed and therefore, since the item can be in either group, the size must also be n/2
However, given that the item is more likely to be in the larger of the two groups, we must skew the average to greater than n/2 and indeed, when I run an Excel simulation, that proves correct. Any thoughts on this?
The solution can be obtained form considering the first 1 to n-1 samples which must all be equally likely to be chosen
1/n(n-1) * [sum from 1 to (n-1) (x^2 + (n-x)^2)]
and has been verified. Can be simplified further. There must be another way to do this though? A more elegant method perhaps?
Hmm I think I've got it although maybe in a roundabout way
Originally Posted by Gekko
and although I haven't proven it, based on the numbers this simplifies to .
The way I arrived was a bit complicated, involved using Bayes' theorem, if you want I can explain but maybe someone will post something simpler.