Very difficult combinatorics problem
Assume that there are n items (numbered from 1 to n) in an urn.
We select b items from the urn and record their numbers.
We return the selected b items into the urn and perform another selection.
We do in total m such selections.
At the end of the m selections we check the recorded numbers. For every unit number chosen in the first selection, if it is also chosen in any of the rest m-1 selections we have a loss of 1 unit (the loss is 1 unit irrespectively if the same item has been re-chosen more than 1 times).
Selection 1 | Selection 2| Selection 3
item1 | item1 | item3
item5 | item7 | item1
item8 | item2 | item9
item7 | item3 | item10
In this example item1 is re-chosen in Selection 2 and Selection 3 so we loose 1 unit from it. Also item7 is re-chosen in Selection 2 so we loose another 1 unit from it. In total we have lost 2 units.
We want to compute:
a) the formula that gives the average loss that we will have.
b) the average number of times that an item is re-chosen.
Actually I think the formula must be of the form:
Probability(1 unit re-chosen in >=2 selections)*(-1)+Probability(2 units re-chosen in >=2 selections)*(-2)+...+Probability(b units re-chosen in >=2 selections)*(-b)= ???