Number of ways to roll a 6-sided die and a 12-sided die (alternating) to sum to n

Question

There are two dice, a six-sided die and a twelve-sided die.

You roll the dice, alternating between the two, starting with the six-sided die. You keep track of the running total and stop when the total reaches $n$.

Show that the number of different ways to reach $n$ is:

$[x^n]$$ 1+x+x^2+\dots+x^{6}\over 1-(x+x^2+\dots+x^{6})(x+x^2+\dots+x^{12})$

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Here is my attempt:

Let $S$ be the set of sets of ways to roll the dice any number of times (set of combinations).

Define the weight function, $w$, to be the sum of the set. $w((c_1,c_2,\dots,c_k))=c_1+c_2+\dots+c_k$

Let $\mathbb{N}_6 =\{1,2,3,\dots,6\}$

Let $\mathbb{N}_{12} =\{1,2,3,\dots,12\}$

Then

$\Phi_{\mathbb{N}_6} = x+x^2+\dots+x^{6} =$$ x-x^7\over 1-x$

$\Phi_{\mathbb{N}_12} = x+x^2+\dots+x^{12} =$$ x-x^{12}\over 1-x$

And

$ \begin{align}S & = \mathbb{N}_6(\mathbb{N}_{12})^k(\mathbb{N}_6)^k\cup(\mathbb{N}_6)^k(\mathbb{N}_{12})^k \\ & = (\mathbb{N}_6)^{k+1}(\mathbb{N}_{12})^k\cup(\mathbb{N}_6)^k(\mathbb{N}_{12})^k \\ \end{align}$

This is because the six sided die can be rolled the same number of times as the twelve sided die or the six sided die can be rolled one more time than the twelve sided die.

Then

$\begin{align} \Phi_{\mathbb S} & =(\Phi_{\mathbb {N}_6})^{k+1}(\Phi_{\mathbb{N}_{12}})^k+(\Phi_{\mathbb {N}_6})^k(\Phi_{\mathbb{N}_{12}})^k \\ & = (x+x^2+\dots+x^{6})^{k+1}(x+x^2+\dots+x^{12})^k+(x+x^2+\dots+x^{6})^k(x+x^2+\dots+x^{12})^k \\ \end{align}$

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I am unsure how to proceed past this point. I tried using the geometric series formula but got stuck.

Answer 1

It's actually very simple:

• $x+x^2+\dots+x^6$ gives the number of ways to roll $n$ on the d6, and similarly $x+x^2+\dots+x^{12}$ for the d12
• Their product $(x+x^2+\dots+x^6)(x+x^2+\dots+x^{12})$ gives the number of ways to roll $n$ as the sum of a d6-d12 pair
• $\frac1{1-(x+x^2+\dots+x^6)(x+x^2+\dots+x^{12})}$ gives the number of ways to roll $n$ on any number of d6-d12 pairs (i.e. an even number of individual die rolls), based on $\frac1{1-x}=1+x+x^2+x^3+\dots$ summing the number of ways to roll $n$ on any non-negative integer number of pairs
• Multiplying this last expression by $(x+x^2+\dots+x^6)$ gives the number of ways to roll $n$ on any number of pairs and one further d6 (an odd number of die rolls)
• Summing the last two expressions yields the desired result: the coefficient of $x^n$ in $\frac{1+x+x^2+\dots+x^6}{1-(x+x^2+\dots+x^6)(x+x^2+\dots+x^{12})}$ is the number of ways to roll $n$ on the sum of alternating d6s and d12s, the d6 being rolled first.