The answer is 0.00389099.

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The angle at the vertex Pi is acute if and only if all other points lie on an open semicircle facing Pi. We first deduce from this that if there are any two acute angles at all, they must occur consecutively. Otherwise, the two arcs that these angles subtend would overlap and cover the whole circle, and the sum of the measures of the two angles would exceed 180◦. So the polygon has either just one acute angle or two consecutive acute angles. In particular, taken in counterclockwise order, there exists exactly one pair of consecutive angles the second of which is acute and the first of which is not. We are left with the computation of the probability that for one of the points Pj , the angle at Pj is not acute, but the following angle is. This can be done using integrals. But there is a clever argument that reduces the geometric probability to a probability with a finite number of outcomes. The idea is to choose randomly( n − 1 )pairs of antipodal points, and then among these to choose the vertices of the polygon. A polygon with one vertex at Pj and the other among these points has the desired property exactly when (n−2)vertices lie on the semicircle to the clockwise side of Pj and one vertex on the opposite semicircle. Moreover, the points on the semicircle should include the counterclockwise-most to guarantee that the angle at Pj is not acute. Hence there are (n − 2 )favorable choices of the total 2^(n−1) choices of points from the antipodal pairs. The probability for obtaining a polygon with the desired property is therefore (n − 2)×2^(−n+1). Integrating over all choices of pairs of antipodal points preserves the ratio. The events j = 1, 2,...,n are independent, so the probability has to be multiplied by n. The answer to the problem is therefore n×(n − 2)×2^(−n+1.)

( Solution is taken solely from that book)