Neil Sloane <njasloane@gmail.com> wrote:

Keith, Very interesting!

Thanks.

Sylvester looked at this question, see https://mathworld.wolfram.com/SylvestersFour-PointProblem.html

Sigh. After all that work, I discover that someone did it better the century before last, working with a quill pen by candle-light. At least that lets me check my answers by looking "in the back of the book."

For a square, his theorem is that the prob. of getting a convex quadrilateral is 25/36 , which is exactly the value you found!

The same to the six places I gave, correctly figuring that that's where my luck would run out. I calculated 0.694444085 and posted 0.694444. The correct answer is 0.694444444... with the 4s repeating forever. In the circular case, I calculated 0.7044826905 and posted 0.704483. The correct answer is 0.704479881... so I was off by 3 in the last digit I gave. It should have been 0.704480. Sigh. For five points, I calculated 0.3402739547 and posted 0.340274. The correct answer is 0.3402777777777... with the 7s repeating forever. So I should have posted 0.340278. I was off by 4 in my last digit. For the circular five-point case I calculated 0.3561808623 and posted 0.356181. That page doesn't give the solution. Assuming the solution is in the same form as the circular four-point case, 1 - (i / (j pi^2)), where i and j are integers, (35 and 12 in the 4-point case), if I try 1 - (305 / (48 pi^2)), I get 0.356188312... which differs from my answer by 7 in the sixth digit. Does anyone know if that's the correct formula? But for 1 - (i / (j pi^2)) to be the correct formula for the 3-point case, i would have to be 0. There are no instances of 0,35,305 in OEIS. There are too many instances of 0,35, or even 0,0,35, to look at; is there a way to specify that the sequence must *begin* with 0,35 or 0,0,35? Or to specify that the sequence must be monotonic? Thanks. That page also doesn't address the probabilities of a convex quadrilateral in the 5-point case, for any shape of boundary. At least it agrees that I was right that squares and rectangles would give the same answer, and that (although I didn't say it here) circles and ellipses would, and that equilateral and isosceles triangles would. But apparently *all* triangular boundaries have the same answer? That's not obvious to me.