For n points choose any given point and evaluate the probability that the other n-1 lie within a semicircle going clockwise. This probability is (1/2)^(n-1).
Given that there are n points to start with the overall probability is n/2^(n-1).
This may seem like an abuse of taking the sum of probabilities, but in this situation there are only two cases (only zero or one of the events may be true), which eliminates the problem of joint probabilities.
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