Paths on a dodecahedron Looking at this question, I misread "dodecagon" as "dodecahedron".  I think the latter is a cool problem, so I'm posing it as a question of its own :)

Starting from one vertex of a dodecahedron, an ant wants to reach the opposite vertex of the dodecahedron, moving to adjacent vertices. If $p_n$ is the number of such paths with length $n$, compute $p_1+p_2+\dots+p_{12}$.

 A: Let $A$ be the adjacency matrix. Then $A^k$ gives you the number of paths of length $k$ between corresponding vertices. (This works for any graph.) This can computed very quickly, and you can also get asymptotics and so on, by computing eigenvalues.
For example, there are
171619248
such paths of length 20 (not too far from MJD's guess), and
25768876036573921452762172776956776774411837488
such paths of length 100.
A: It is not necessary to set up the full adjacency matrix of the dodecahedron. Drawing the edge graph with the starting vertex $v_0$ at the center and the opposite vertex at infinity one realizes that due to symmetry there are just six classes $C_i$ $(0\leq i\leq5)$ of vertices. 

It is therefore sufficient to consider the numbers $p_i(n)$, whereby $p_i(n)$ denotes the number of ways to reach a vertex of class $C_i$ in exactly $n$ steps, starting at $v_0$. Looking at my drawing I then obtain the following equations:
$${\bf p}(n+1)=\left[\matrix
{0&3&0&0&0&0\cr
1&0&2&0&0&0\cr
0&1&1&1&0&0\cr
0&0&1&1&1&0\cr
0&0&0&2&0&1\cr
0&0&0&0&3&0\cr}\right]\>{\bf p}(n)\ ,$$
whereby ${\bf p}(n)$ denotes the column vector of the $p_i(n)$.
Since there are so many treatments of the problem already I stop here.
A: I'm still working out a combinatorial method of calculating the answer, and I may not be successful.  In the meantime, here are the results from a computer enumeration of all paths:
$$
\begin{array}{rrr}
  \text{Length} & \text{Simple paths} & \text{All paths} \\
  5 &    6 &     6    \\
  6 &   12 &    12    \\
  7 &    6 &    84    \\
  8 &   12 &   192    \\
  9 &   30 &   882    \\
 10 &   24 &  2\;220    \\
 11 &   42 &  8\;448    \\
 12 &   84 &  22\;704   \\
 13 &   96 &  78\;078   \\
 14 &  132 &  218\;988  \\
 15 &  150 &  710\;892  \\
 16 &   72 &  2\;048\;256 \\
 17 &   48 &  6\;430\;794 \\
 18 &   60 &  18\;837\;516\\
 19 &    6 &  58\;008\;216\\
 \hline
 \text{Total} & 780 & 86\;367\;288
\end{array}
$$
The complete enumeration of paths took around half an hour on my laptop, and the output file is 5.3 GB raw, 0.25 GB compressed.  For theoretical reasons as well as empirical, we can guess that there would be around 180 million paths of length 20, and that computing them would take around 55 minutes.  (After computing the $1\;042\;506$ paths up to length 15, I guessed there would be around 81 times as many paths of length up to 19, or $84\;442\;986$, which is quite close to the correct result.)
