6
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ 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 $(1/6)\sum_{i=1}^{6}p_i$? $\endgroup$ – mjqxxxx Sep 10 '12 at 14:24
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Oct 6 '12 at 2:52
1
$\begingroup$

Suppose the prime bubble rolls an $n$. Then the cascader bubble rolls will be $\{r_1, r_2, ... r_{n-1},1\}$ with probability $$ \frac{1}{6}\cdot\frac{1}{6r_1}\cdot\frac{1}{6r_1r_2}\cdots\frac{1}{6r_1r_2\cdots r_{n-1}}=\frac{1}{6^n r_1^{n-1} r_2^{n-2} \cdots r_{n-1}} $$ for any $r_1\in[6]=\{1,2,3,4,5,6\}$, $r_2\in[6r_1]$, $r_3\in[6r_1r_2]$, etc., up to $r_{n-1}\in[6r_1r_2\cdots r_{n-2}]$. The total probability is then $$ p_{n}=\frac{1}{6^n}\sum_{r_1=1}^{6}\frac{1}{r_1^{n-1}}\sum_{r_2=1}^{6r_1}\frac{1}{r_2^{n-2}}\cdots\sum_{r_{n-1}=1}^{6r_1r_2\cdots r_{n-2}}\frac{1}{r_{n-1}}. $$ In particular, $$ \begin{eqnarray} p_1&=&\frac{1}{6}=0.166666... \\ p_2&=&\frac{1}{36}\sum_{r_1=1}^{6}\frac{1}{r_1}=\frac{1}{36}H_{6}=\frac{49}{720}=0.06805555... \\ p_3&=&\frac{1}{216}\sum_{r_1=1}^{6}\frac{1}{r_1^2}\sum_{r_2=1}^{6r_1}\frac{1}{r_2} \\ &=&\frac{1}{216}\left(H_{6}+\frac{1}{4}H_{12}+\frac{1}{9}H_{18}+\frac{1}{16}H_{24}+\frac{1}{25}H_{30}+\frac{1}{36}H_{36}\right) \\ &=&\frac{97493779762855253}{5104009215002880000}=0.01910141... \end{eqnarray} $$ Using these first three terms gives the lower bound $(p_1+p_2+p_3)/6 = 0.04230...$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.