I know that the following is the standard Randomized Rounding for Set Cover problem:
$\min \sum_{S \in \mathcal{S}} c(S)x(S)$
s.t
$\sum_{S: e \in S} x(S) \geq 1 \ \forall e \in U$
$x(S) \in [0,1] \ S \in \mathcal{S}$
The above linear program cover every element at least one time ($\min \sum_{S: e \in S} x(S) \geq 1 \ \forall e \in U$).
Set $S$ is picked with probability $p(S) = x^*(S)$
$E[ALG] = \sum_{S \in \mathcal{S}} c(S)p(S) = \sum_{S \in \mathcal{S}} c(S)x^*(S) = OPT^{LP} \leq OPT$
Considering $a \in S, \
Pr[a \ is \ covered] = 1 - (1 - p(S_1)) \times \ ... \ \times (1 - p(S_k)) \geq 1 - (1 - \frac{1}{k})^k \geq 1 - \frac{1}{e} $
So each element $a \in U$ is covered with $Prob \geq 1 - \frac{1}{e} $
Picking $d$ log $n$ subcollections $C^\prime = C_1 \cup \ ... \ \cup C_{d \ log \ n}$ with $d$ such that :
$Pr[a \ not \ covered] \leq (\frac{1}{e})^{d \ log \ n} \leq \frac{1}{4n}$
We obtain:
$E[COST(C^\prime)] \leq d \times log \ n \ OPT^{LP}$
$Pr[COST(C^\prime)] \geq 4d \times log \ n \ OPT^{LP} \leq \frac{1}{4}$
$Pr[COST(C^\prime) \ not \ easible] \leq n \times \frac{1}{4n} \leq \frac{1}{4}$
$Pr[COST(C^\prime) \geq 4d \times log \ n \ OPT^{LP} \ AND \ C^\prime \ is \ feasible] \geq \frac{1}{2}$
Expected number of repetitions = 2.
Now my question is: how do thinks change if we consider the following LP?
$\min \sum_{S \in \mathcal{S}} c(S)x(S)$
s.t
$\sum_{S: e \in S} x(S) \geq m \ \forall e \in U$
$x(S) \in [0,1] \ S \in \mathcal{S}$
Considering $m = 1,2,..,n$. Of course for $m = 1$ we re-obtain the above steps but for $m \geq 2$? For example for $m = 3$? Is there a generalized solution $\forall m$?
Hint: the change is in the calculation of $P[\text{a is covered}]$ and its effect on the part of after that. You should consider more cases if $m > 1$ which is different choosing a subset of $S_1, \ldots, S_k$ which are not covered. $$P[\text{a is covered}] = 1 - \left((1-P(S_1))\times \cdots \times (1-P(S_k)) + \binom{k}{1} (p(S_i)\times \text{multiplication of other $1-p(S_j)$}) + \cdots + \binom{k}{m} (p(S_i)\times \cdots \times p(S_{i + m}))\times \text{multiplication of other $1-p(S_j)$}\right) \geq 1- \left((1-\frac{1}{k})^k + k\frac{1}{k}(1-\frac{1}{k-1})^{k-1} + \cdots + \binom{k}{m}(\frac{1}{k})^m(1-\frac{1}{k-m})^{k-m}\right) $$
which $S_i, \ldots, S_{i+k}$ are covered in different cases.
As you can see, simplification of this is not that easy. You can't say it is greater than $1- \frac{m}{e}$ here as the multiplier here is not simple here, in contrast with the first case.