Ladybug walking on a hexagon A ladybug is walking at random on a hexagon. She starts at vertex A. Every minute, she moves to one of the two vertices (chosen at random) adjacent to the one she's currently on. What is the probability that, after 10 minutes, she is back at A?
I thought that at every even minute the ladybug would be at A, C, or E so there is a 1/3 chance. This was wrong and I got the following message: The probability may be close to 1/3, but since there are 2^10 possible paths for the ladybug and 2^10 is not divisible by 3, the probability cannot be exactly 1/3.
 A: The connection with random walks and sums modulo $6$ was already mentioned above, in the comments. 
Consider a sequence of i.i.d Bernoulli random variables $X_i$, where $\mathbb{P}(X_i = \pm 1) = 1/2$. Define $S_0= 0 $ and $S_n = X_1 + ... + X_n$. Thus, the point $A$ on the hexagon is associated with $0$, if the bug moves in the clockwise direction we take $+1$ in $S_n$, otherwise we add $-1$. The sum $S_n $ modulo $6$ gives the location of the bug after $n$ moves, where we enumerate vertices of the hexagon by $0,...,5$ starting from $0\to A$ and moving clockwise.
Hence, for the bug to be at $0$ after $10$ moves we need $S_{10} \mod 6 = 0 $, but the latter is only possible for values $0, 6$ and $-6$ of $S_{10}$, since $-10\leq S_n \leq 10$ for all $n=0,1,...,10$. We conclude that 
$$
\mathbb{P}(S_{10} = 0 \mod 6) = \frac{1}{2^{10}} \left( {{10}\choose{5}} + {{10}\choose{8}} + {{10}\choose{2}} \right),
$$
which gives the probability that the bug is at $A$ after $10$ moves.
A more tedious, but doable approach is to condition on each move, and use symmetries of the hexagon.
A: It is a Markov chain with transition matrix
$$P = \frac{1}{2} \left [ \begin{array}*
0 & 1 & 0 & 0 & 0 & 1 \\
1 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 \\
1 & 0 & 0 & 0 & 1 & 0 \\
\end{array} \right ]$$
The answer is given as the first element of the first row of $P^{10}$. That is
$$\frac{171}{512}$$
By the way you can also diagonilize the matrix $P$ to find a formula for its powers. It is a symmetric circulant matrix and the eigenvalues and -vectors are very easy.
A: The OP made an important observation concerning 'even minutes' with the ladybug on one of the vertices in $\{ A, C,  E \}$.
You can equivalently design the experiment so that you are observing the ladybug walking on an equilateral triangle ('flip two coins' simultaneously to get the next or zero step). With this modelling, and using some elementary probability theory, some algebra, and the figure

you can use recursion to arrive at the answer.
Define 
$$\tag 1 a_{n+1} = \frac{a_n+1}{4} \;\text{ with }\, a_0 = 1$$
ANSWER = $a_5$

Update: Although I got the answer $a_5$ using recursion, it left me feeling a bit uneasy since at each step you'll discover that the recursion is really the 'best binary approximation' of $\frac{1}{3}$ 'from above'.
After seeing Christian Blatter's answer I feel obliged to note that using induction one can also get the closed form solution here:
$$\tag 2 a_{n}={1\over3}\bigl(1+2^{1-2n}\bigr)\qquad(n\geq0)\ $$
Certainly works for $n = 0$.
Assume now that $\text{(2)}$ is true for some fixed $n$. But then
$$ a_{n+1} = \frac{a_n+1}{4} = {1\over3} \bigl( \frac{3a_n+3}{4} \bigr) = {1\over3} \bigl( \frac{(1+2^{1-2n})+3}{4} \bigr) = {1\over3}\bigl(1+2^{1-2(n+1)}\bigr)$$
So we have the result.

You can also use brute force, setting up a google sheets with only a bit of work, mostly dragging/copying the one formula in the C3 cell,
=0.25*B2+0.5*B3+0.25*B4

all over the 'calculation array'.
Here is the formula view of the sheet:

Here are the numbers:

A: After $2n$ moves the bug is either at $A$ with probability $p_{2n}$ or in  $\{C, E\}$ with probability $q_{2n}=1-p_{2n}$. After two more moves it is again at $A$ with probability ${1\over2}p_{2n}$ in the first case and with probability ${1\over4}q_{2n}$ in the second case. It follows that
$$p_{2(n+1)}={1\over2}p_{2n}+{1\over4}(1-p_{2n})={1\over4}p_{2n}+{1\over4}\ .\tag{1}$$
The general solution of $(1)$ is easily seen to be $p_{2n}={1\over3}+{C\over4^n}$ for a suitable $C$, and as $p_0=1$ we definitely obtain
$$p_{2n}={1\over3}\bigl(1+2^{1-2n}\bigr)\qquad(n\geq0)\ .$$
In particular $p_{10}={1\over3}\bigl(1+{1\over512}\bigr)={171\over512}$.
A: Number of ways to end up at vertex $m$ after $n$ moves ($n-k$ clockwise moves and $k$ counterclockwise moves, arranged in $\binom{n}{k}$ ways, placing us $n-2k$ clockwise from the start):
$$
\begin{align}
\sum_{k=0}^n\binom{n}{k}[6\mid n-2k-m]
&=\sum_{k=0}^n\binom{n}{k}\frac16\sum_{j=0}^5e^{2\pi i(n-2k-m)j/6}\tag1\\
&=\frac16\sum_{j=0}^5\left(e^{-2\pi ij/6}+e^{2\pi ij/6}\right)^ne^{-2\pi imj/6}\tag2\\
&=\frac{2^n}6\sum_{j=0}^5\cos^n\left(\frac{2\pi j}6\right)\cos\left(\frac{2\pi mj}6\right)\tag3
\end{align}
$$
Explanation:
$(1)$: $[p\mid n]=\frac1p\sum\limits_{j=0}^{p-1}e^{2\pi inj/p}$
$(2)$: Binomial Theorem
$(3)$: sine is odd, cosine is even
For $n=10$, either calculation gives
$$
\begin{array}{c|c}
m&\text{# ways}\\\hline
0&342\\
1&0\\
2&341\\
3&0\\
4&341\\
5&0
\end{array}
$$
