The chances of any one or more people already excluded in the first month also being excluded in the second month (when there are 104 eligible in the 1st month and 106 eligible in the next month) is:
4 * [1-(100/106)], or ~22.4%
Damn it, I simulated it, and you seem right. I suppose that my model of the situation was wrong because it actually does matter that the first month has already happened, whereas I was trying to eliminate the specificity of this.
Guys I am again on this.
I thought about this problem quite extensively during this weekend on the italian riviera.
I think Quickseller formula is not correct as it does allow repetition, while we must find a solution without repetition (if a user is selected, it cannot be taken out again on the same round).
I think Theymos was then on the right path using binomial coefficients, so I am going to use the same technique.
The right probability of any of the 4 excluded in the first round to be excluded in the second round is equal to 1 - the probability of everyone of such 4 to be selected:
1-C(100,4)/C(106,4)=0.210654248
another less intuitive method give the same exact result:
1-C(102,6)/C(106,6)=0.210654248
(probability of being amongst the 6 excluded from the second extraction chosen by the 100 selected from the first extraction + the 2 new addition).
I am almost sure about this, but please double check me again.
Theymos said he simulated and got a result very similar to Quickseller, this scares me, also because I saw LoyceV and other heavyweight meriting previous solution... shall I go back to school?
EDIT: Forgot to mention, but clearly an hypothesis here: every candidate on the first round is a candidate also for the second round. This simplifies calculations, when we agree on the solution, we'll be able to remove this hypothesis.