Given that there are n points to start with the overall probability is n/2n-1. This may seem like an abuse of taking the sum of probabilities, but in this case only zero or one of the events may be true, which eliminates the problem of joint probabilities.
Occationally somebody challenges my answer to this problem. To verify the results I did ten million trials for various numbers of points on the circle. I assumed the circle to have 32,767 units or degrees. The first point was defined to be at a halfway point, or 16,383. Then a maximum and minimum were taken for the rest of the points. If the difference between the maximum and minimum were less than 16,383 then the trial was considered to be a success. Note that 16,383 is half of 32,767, rounded down. Because of the rounding we should expect a slightly lower number of successes. Here are the results:
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Problem 1 Solution |
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| Number of Points | Actual Successes | Expected Successes |
| 3 | 7,499,013 | 7,500,000 |
| 4 | 4,999,960 | 5,000,000 |
| 5 | 3,123,456 | 3,125,000 |
| 6 | 1,873,092 | 1,875,000 |
| 10 | 194,739 | 195,312.5 |
| 20 | 373 | 381.4697 |
It has been alleged by some that my answer is right but solution is wrong. For a more complete treatment of this problem visit www.mathpages.com/home/kmath327.htm.
This problem has been published in book form in "Contests in Higher Mathematics" by Gabor Szekely, which is a collection of math problems given to the best mathematicians of Poland over the last several decades. The problem (1963 problem 10) is stated as, "Select n points on a circle independently with uniform distribution. Let Pn be the probability that the center of the circle is in the interior of the convex hull of these n points. Calculate the probabilities P3 and P4. I'm told one of the two solutions given for general n uses your method and agrees with my result.
Michael Shackleford, A.S.A.MathProblems.info home