= Solution
At $p=1/2$, explore the directed out-cluster of the origin by a deterministic breadth-first procedure. Maintain the set of reached <graph vertices> and a queue of untested <edges> with one reached and one unreached endpoint. For each such <edge>, reveal its orientation. Add the unreached endpoint if the arrow points from the reached endpoint to it; otherwise discard that <edge>. <edges> whose endpoints are both already reached need no further test.
An untested <edge> has an independent fair orientation even though the choice of the next <edge> depends on earlier discoveries. Its <probability> of pointing toward the unreached endpoint is $1/2$, whichever geometric direction that represents. Thus the discovery procedure has exactly the same transition <probabilities> as the ordinary <bond percolation> cluster exploration at density $1/2$. In particular, for every $m$, the <probability> of reaching at least $m$ <graph vertices> agrees in the two models. The <uniform random orientation out-cluster comparison> implies
$$
\mathbb P(|C^{\to}(0)|=\infty)=\theta(1/2).
$$
A <locally finite graph> has an infinite directed simple ray from the origin exactly when its directed out-cluster is infinite: the exploration predecessor <edges> form a finitely branching rooted <tree>, to which the <König infinity lemma> applies. Conversely, such a ray visits infinitely many reachable <graph vertices>. Hence
$$
\boxed{\psi(1/2)=\theta(1/2)=0.}
$$
Fairness of orientations is essential to this comparison; for general $p$ the success <probability> depends on the geometric direction of exploration. As usual, an infinite directed path means a ray with distinct <graph vertices>. Allowing a finite directed cycle to be traversed repeatedly would define an infinite walk instead and would make the claimed conclusion false.
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