How many resources I can get rolling two dice N times Two dice rolling N times.
If the sum of the two dice is equal to 6 we get 1 resource.
If the sum of the two dice is equal to 4 we get 2 resources.
Resources are the same.
How many times (N) should I roll two dice in order to get at least one (case 1), at least two (case 2), at least three (case 3) resource(s) with probability 0.6?
 A: For "at least one."
Find the probability of the complement (in $N$ rolls, you get zero resources) as a function of $N$, and set it equal to $0.4$.

 The probability of rolling a sum of 6 or 4 is $p = \frac{8}{36}$. The probability of getting no resources in $N$ rolls is $(1-p)^N$. Set this equal to $0.4$ and solve for $N$.


For "at least two."
Again, consider the complement (in $N$ rolls, you get zero resources or exactly one resource). You already computed the probability of zero resources. For one resource, it is $Np_6(1-p_6-p_4)^{N-1}$ where $p_6$ and $p_4$ are the probabilities of rolling a sum of$6$ and $4$  respectively. [There are $N$ ways to choose which roll is the $6$, and the other $N-1$ rolls must be neither $6$ nor $4$.] Sum these two probabilities and set equal to $0.4$.

For "at least three" can be solved using further casework on the complement: either zero resources, or one resource, or two resources. In the case of exactly two resources, this can be achieved either by rolling $4$ once, or rolling $6$ twice.
