How to predict the likelihood of multiple future occurrences based on probability

Some formality and structure should help you tackle the problem you mention in your example.

Take $E$ as the event of getting at least two more fours in the next roll or the one after that, $X$ as the number of fours in the first roll, and $Y$ as the number of fours in the second roll.

Using total law of probability, $$\begin{eqnarray*}\mathbb{P}(E)=\mathbb{P}(E|X=0)\mathbb{P}(X=0)+\mathbb{P}(E|X=1)\mathbb{P}(X=1)+\mathbb{P}(E|X\geq 2)\mathbb{P}(X\geq 2)\end{eqnarray*}$$ Since $\mathbb{P}(E|X\geq 2)=1$ and $X\sim \text{Binomial}(3,1/6)$ the aforementioned reduces to $$\mathbb{P}(E)=\frac{125}{216}\mathbb{P}(E|X=0)+\frac{25}{72}\mathbb{P}(E|X=1)+\frac{2}{27}$$ Now for $x=0,1$ $$\mathbb{P}(E|X=x)=\sum_{y=0}^{3-x}\mathbb{P}(E|X=x,Y=y)\mathbb{P}(Y=y|X=x)$$ Because $\mathbb{P}(E|X=x,Y=y)=1_{\{x+y\geq 2\}}$ and $Y|X\sim \text{Binomial}(3-X,1/6)$ we have $$\mathbb{P}(E|X=0)=\mathbb{P}(Y\geq 2|X=0)=\frac{2}{27}$$ Furthermore, $$\mathbb{P}(E|X=1)=\mathbb{P}(Y\geq 1|X=1)=\frac{11}{36}$$ Putting all this together, $$\mathbb{P}(E)=\frac{5203}{23328}\approx 0.22$$