So, my statistics knowledge is rather poor, so I would welcome a formula explanation to the question: given Nd6 (6-sided dice) what is the probability that the two highest numbers are at least a 4?
Given 2d6, i think it's 6/21, though i shamefully admit reaching it by (possibly faulty) enumeration.
The probability to get at least a four in one die is $1/2$. Therefore, the probability to get three or less in $N$ throws is $$\frac1{2^N}$$ and the probability to get a four or more in exactly one throw is $$\frac N{2^N}$$ The $N$ comes from the fact that the "good throw" can be the first, the second, ..., or the $N$-th.
Thus, the probability to get four or more in two or more dices is $1$ minus the sum of the previous probabilities: $$1-\frac1{2^N}-\frac N{2^N}=\frac{2^N-N-1}{2^N}$$