Kamalukala

Hello, fellow puzzle solvers! I see that many of you have found numerical solutions to the question. Using some hand-wavy mathmatics I came up with the following theoretical approach: The question is "Recall that we have a unit circle where three random points define three arcs. What is the probability that the longest arc contains the point (1,0)?" Assigning three random points on a circle to form three arcs is equivalent to defining two random points on a straight line to form three linear segments. (After placing the first dot on the circle it more or less becomes a 1D line.) The probability can be found by knowing the avarage length of the longest segment: p = L1 / L, where L1 is the average length of the longest segment and L is the total length of the circle. ------------------------------------------- If we randomly place a single dot on the line it will divide the line into two segments; a long segment and a short one. Due to symmetry the segment will (on average) be divided into a (1/4)L part and a (3/4)L part. This can be explained by assuming, without loss of generality (due to the fact that the halfs are equivalent), that the dot will be placed in the left half of the line. On average it would be placed at exactly L/4. This seems to agree with bananova's numerically obtained value for n=2, which is an L1 value of 0.754. Now comes a hand-wavy part: If an additional dot is placed on the line it has (on average) a probability of 1/4 to be placed in the small segment, and, (on average) a probablility of 3/4 to be placed in the large segment. This would (on average) devide that segment into pieces of length: (1/4*1/4)L and (1/4*3/4)L in case of the short segment (note that in this case the overall longest segment is (3/4)L )(3/4*1/4)L and (3/4*3/4)L in case of the long segmentThe average value of L1 is given by: L1 = 1/4*(3/4)*L + 3/4*(3/4)^2*L = 0.6094*L --> p = L1/L = 0.6094 = 60.94% This seems to be in agreement with previously posted numerical values. ---------------------------------------- Although I don't posses the notational skills to make this into a hard proof, I think that the general approach may very well be correct. I'd like to hear how you solved it analytically.