In the city of Ikozapolisz, the road network graph matches the graph of the edge network of an icosahedron. Jorgosz's lodging is located at one vertex of the icosahedron, while his favorite theater is at the opposite vertex. After dark, when leaving the theater and heading home, he becomes uncertain at each vertex about which direction to take next. Suppose that at each vertex, the probability $p$ is that he encounters someone who points him in the direction that will take him to his lodging with the fewest edges. Otherwise, he continues in a random direction, with no direction being favored, meaning it is even possible for him to turn back. For what value of $p$ will there be a $50 \%$ probability that he reaches his lodging before returning to the theater?
Solution. From the perspective of the problem, the nodes of the road network graph, excluding the initial and final states, can be of two types: either adjacent to the theater (initial state) or to the accommodation (goal). In the first case, we are talking about a 1-state, and in the second case, a 2-state. From the initial state, Jorgosz approaches the goal with a probability of 1, so the probability of reaching the goal from the initial state and from the 1-state is the same. From the 1-state, one edge leads back, two edges lead to the same state, and two edges lead forward to the 2-state. From the 2-state, two edges lead to the same state, two edges lead back to the 1-state, and one edge leads to the goal. If the probability of reaching the goal from the 1-state is $x$, and from the 2-state is $y$, then the following equations can be written:

\[
\begin{aligned}
x & =\left(\frac{1}{5} \cdot 0+\frac{2}{5} x+\frac{2}{5} y\right)(1-p)+p y \\
y & =p \cdot 1+\left(\frac{1}{5} \cdot 1+\frac{2}{5} x+\frac{2}{5} y\right)(1-p)
\end{aligned}
\]

According to the condition, $x=\frac{1}{2}$, so the second equation becomes:

\[
y=p+\frac{2}{5}(1-p)+(1-p) \frac{2}{5} y, \quad \text { from which } \quad y=\frac{3 p+2}{2 p+3}
\]

Substituting $x$ and $y$ into the first equation:

\[
\begin{aligned}
\frac{1}{2} & =(1-p)\left(\frac{1}{5}+\frac{2}{5} \cdot \frac{3 p+2}{2 p+3}\right)+\frac{3 p^{2}+2 p}{2 p+3} \\
10 p+15 & =(1-p)(4 p+6+4(3 p+2))+30 p^{2}+20 p \\
10 p+15 & =16 p+14-16 p^{2}-14 p+30 p^{2}+20 p \\
0 & =14 p^{2}+12 p-1
\end{aligned}
\]

Of the two solutions, only the positive solution is valid, as the probability is a number between 0 and 1:

\[
p=\frac{-12+\sqrt{12^{2}+4 \cdot 14}}{28}=\frac{-6+5 \sqrt{2}}{14} \approx 0.0765
\]

Therefore, if $p \approx 0.0765$, Jorgosz will reach the goal with a probability of 0.5.