The answer is \(10/512 \simeq 0.02\).
Let us use \(n\) to denote the number of points (in this case \(n = 10\) hods). If all \(n\) points can be covered with a semi-circle arc, then we define the leader of the \(n\) points as the point which is covered first in the clock-wise order. We divide the desirable possibilities into \(n\) disjoint scenarios:

• Point 1 is the leader and all other \(n-1\) points are covered with a half circle arc starting from point 1 in the clock-wise order.

• Point 2 is the leader and all other \(n-1\) points are covered with a half circle arc starting from point 2 in the clock-wise order.

• \(\vdots\)

• Point \(n\) is the leader and all other \(n-1\) points are covered with a half circle arc starting from point \(n\) in the clock-wise order.

Note that all \(n\) scenarios given above are disjoint. Moreover, the probability of each scenario is exactly \(1/2^{n-1}\) as each of the \(n-1\) non-leader points lie on the corresponding semi-circle arc with probability \(1/2\). Thus, the answer to the problem is equal to \(\frac{n}{2^{n-1}} = 10/512 \simeq 0.02\).