Pando, although your solution gives the right answer somehow, its not completely correct. The problem you solved was if you generate 2 random points, what is the probability that the distance between the two is greater than 0.5. You're ignoring the fact that we're generating a third point which matters. However since you got the correct answer, I think there should be a simple reduction from your solution to the general one... I'll think about it... In the meantime, I'll unlock this thread in case anyone else has some clever insights on this problem.
Hmm yes I see your point...I only thought about the cases in which the points are in abc or cba order (in these two situations, point b does not matters since it is in the middle...ab + bc = ac). Though if other orders were used (bca cab etc)...I think the probability would be higher.
Well it does sorta matter because we're trying to find the sum of the lengths of the two mid-length segments. And the lengths are affected by the position of b.
Ehhh but if b is between a and c...the sum of ab and bc is ac...since b is a point and doesn't have a length... :S The individual lengths are affected but the sum is not...and them sum is all we need. However, everything changes when point b is not between point a and point c.
Let's say we have the following: a = 0.4 b = 0.5 c = 0.7 What you're doing is just taking the distance between a and c which is 0.3 However, the two mid-length segments are 0-a (0.4), and c-1 (0.3)
I totally thought the problem was something different...the sum of the middle two segments...not the sum of the middle two segment lengths. They just happen to have the same answer. D: Anyways as for your real problem...I only spent like 5 minutes thinking about it but...I think this is why (I really did not give it much though so it could be totally wrong XD)... The longest segment must be less than 1/2 (otherwise the sum of the middle two would be less than 1/2) and more than 1/4 (or else it would no longer be possible for it to be the longest segment). That leaves a 1/4 unit interval for the longest segment to vary in length...meaning there's a 25% chance of the sum being more than 1/2 (all the other segments will fall into place...it's 4 pieces that all up to 1, you can arrange however you like). I can explain in more detail if you want...but I don't really feel like doing that now (maybe tonight =P).