# What is the probability of rolling a 1 on the Catenative Doomsday Dice Cascader?

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.

-

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

## 1 Answer

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 $(p_1+p_2+p_3)/6 = 0.04230...$.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9055466651916504, "perplexity_flag": "head"}