Back at the start of a tetrahedron

Idea. From A the bug always leaves; from any other vertex it reaches A with probability $\frac{1}{3}$. So the probability of being at A obeys a one-line recurrence.

Step 1: the recurrence. With $p_0=1$, $$p_{n+1}=\frac{1}{3}(1-p_n).$$

Step 2: solve. The fixed point is $p=\frac{1}{4}$. The difference $p_n-\frac{1}{4}$ is multiplied by $-\frac{1}{3}$ at each step, so $$p_n=\frac{1}{4}+\frac{3}{4}{\left(-\frac{1}{3}\right)}^n.$$

Step 3. For $n=7$: $$\begin{array}{rl}{p}_7& =\frac{1}{4}-\frac{3}{4}\cdot \frac{1}{2187}=\frac{1}{4}-\frac{1}{2916}\\ & =\frac{182}{729}\approx \mathrm{0.2497.}\end{array}$$

Check by stepping. $p_1=0$, $p_2=\frac{1}{3}$, $p_3=\frac{2}{9}$, $p_4=\frac{7}{27}$, $p_5=\frac{20}{81}$, $p_6=\frac{61}{243}$, $p_7=\frac{182}{729}$. The values oscillate around $\frac{1}{4}$ and settle there quickly.