As seen here. Assume that each cascader bubble begins with a d6, as opposed to being determined by the prime bubble along with the number of cascader bubbles; the scene makes it ambiguous.
$\frac{1}{36} = 0.02777...$ provides a simple lower bound for the probability (1 on the prime bubble and the sole cascader bubble). If the prime bubble rolls a 2, the odds are $\frac{1}{6}\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{18}+\frac{1}{24}+\frac{1}{30}+\frac{1}{36}\right) = \frac{49}{720} = 0.0680555...$; multiple that by the 1/6 chance of rolling that 2 and add to the original 1/36, and you get $0.02\bar{7}+0.01134\overline{259} = 0.03912\overline{037}$. After that, it gets way beyond me.