probabilty of random points on perimeter containing center

Solution 1:

The answer is in fact always $\frac14$, independent of $n$; you forgot to multiply by $2$ for symmetry and divide by $\tau$ for normalization in the circle result.

Wherever the first point is chosen, the "diameter" on which it lies divides the $n$-gon into two symmetric halves, and the second and third points must be in opposite halves. The line connecting them must also lie above the centre (as seen from the first point), and if the second point is at a distance $x$ from the first point along the perimeter (in units of the length of the perimeter), there's an admissible range of length $\frac12-x$ for the third point. Thus the probability is

$$2\int_0^{1/2}\left(\frac12-x\right)\mathrm dx=2\int_0^{1/2}x\,\mathrm dx=\frac14\;,$$

where the factor of $2$ is for symmetry because the second and third point can be interchanged.

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