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<blockquote data-quote="Andy Nguyen" data-source="post: 6087" data-attributes="member: 1"><p>Correct. Since the sides are from a unit length, we can calculate that any side of the triangle must be strictly less than 1/2. The prob that the length of each stick being less than 1/2 is 50% and this condition must hold for all three of them. Hence I have 1/8</p><p></p><p> </p><p>Not totally convinced they are related or merely coincidence but I thought of the semi-circle as follow: </p><p>Any two points on a circle will be on the same semi-circle (just connect one point to the center, the second point must be on one of the two semi-circles. Pick the one that contains two points). The problem now becomes finding the probability that the third point is on that same semi-circle. There are only 2 semi-circles to begin with so the chance is 50%</p><p></p><p>Welcome any opinions since with 1,2 minutes to come up with an answer, there are pretty good chance that I may have big holes in my reasonings. I hope to learn from my mistakes and not to repeat them again in future occasions.</p></blockquote><p></p>
[QUOTE="Andy Nguyen, post: 6087, member: 1"] Correct. Since the sides are from a unit length, we can calculate that any side of the triangle must be strictly less than 1/2. The prob that the length of each stick being less than 1/2 is 50% and this condition must hold for all three of them. Hence I have 1/8 Not totally convinced they are related or merely coincidence but I thought of the semi-circle as follow: Any two points on a circle will be on the same semi-circle (just connect one point to the center, the second point must be on one of the two semi-circles. Pick the one that contains two points). The problem now becomes finding the probability that the third point is on that same semi-circle. There are only 2 semi-circles to begin with so the chance is 50% Welcome any opinions since with 1,2 minutes to come up with an answer, there are pretty good chance that I may have big holes in my reasonings. I hope to learn from my mistakes and not to repeat them again in future occasions. [/QUOTE]
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