6
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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.

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    Formally: let $X_0,X_1, X_2, X_3, ...$ be a sequence of random variables. $X_{0}$ is equal to $n$ with probability one (where $n$ is the number of sides on the initial cascading dice). Each successive variable $X_{k+1}$ is uniformly distributed on $\{1,2,3,...,\Pi_{i=0}^{k} X_{i}\}$. Define $p_{k}=P[X_{k}=1]$; what is $(1/6)\sum_{i=1}^{6}p_i$?2012-09-10
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    I found ambiguous whether there were $6$ more dice all the time or the number matched the original roll, not the number of sides on the dice. The chance if the prime bubble rolls higher than $2$ that you get a $1$ at the end will be very small, so in practice you can ignore it.2012-10-06

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