I have this problem:
Choose randomly six numbers in $\{1,2,...,56\}$ without replacement. Call $L_0$ to this choosen set.
Fix $N$. For each $n\in\{1,2,...,N\}$, generate a random set $L_n$ choosing six numbers in $\{1,2,...,56\}$ without replacement. Define $X_n$ by $$X_n=\left\{\begin{array}{l}0\mbox{ if }|L_n\cap L_m|\ge 4\mbox{ for some }m\le n-1\\1\mbox{ i.o.c}\end{array}\right.$$
Take $X^{(N)}=\sum_{n=1}^NX_n$. So $X^{(N)}$ is the total of sets that have no more than 3 elements in comune with the previously generated sets.
How do you calcule $E[X^{(N)}]$?
My attemp: for each fixed $n\in\{1,...,N\}$ and $m<n$, I taked $Y_m^{(n)}=|L_n\cap L_m|$. Now, $X_n=1$ iff $Y_m^{(n)}\le 3$ for all $m\in\{0,1,...,n-1\}$,
So $P(X_n=1)=P(Y_0^{(n)}\le3,Y_1^{(n)}\le3,...,Y_{n-1}^{(n)}\le3)$.
But I have reasons (by mean of several simulations in Python and R) to thing that $\{Y_m^{(n)}\}_m$ is not a family of independent random variables or it is not a family of identical distributions (if it would be rviid, then it is easy to conclude that $E[X^{(N)}]\sim\frac{p^{N+1}-1}{p-1}$ where $p=\frac{1}{\binom{56}{6}}\sum_{k=0}^3\binom{6}{k}\binom{50}{6-k}$, which does not agree with the results of the simulations).
What do you think? How would you calculate $E[X^{(N)}]$?