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 I've a broken stick (Posted on 2003-12-07)
I've a straight stick which has been broken into three random-length pieces.
What is the probability that the pieces can be put together to form a triangle?
If not, perhaps this will help: here are several methods to break the stick into the three random length pieces:
1. I select, independently, and at random, two points from the points that range uniformly along the stick, then break the stick at these two points.
2. I select one point, independently, and at random (again uniformly), and break the stick at this point. I then randomly (with even chances) select one of the two sticks and randomly select a point (again uniformly) along that stick, and break it at that point.
3. I select one point, independently, and at random (again uniformly), and break the stick at this point. I then select the larger stick, and randomly select a point (again uniformly) along that stick, and break it at that point.
If this clarifies the problem, please show how this affects your work.

 No Solution Yet Submitted by SilverKnight Rating: 3.6000 (5 votes)

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 shorter solution | Comment 4 of 26 |
For all three pieces to be able to form a triangle, they have to each be less than 1/2 of the full stick's length. Otherwise, the piece greater than 1/2 would be longer than the sum of the other two.

The first break must be somewhere between the middle and a side. The possibilities for the second break to form a triangle are anywhere in the area between the middle and exactly 1/2 away from the first break. Since all cases determine the first break in the same way, it is the second break that makes them different.

Case one: The second break has equal probability anywhere, but the winning possibilities have an area equal to the distance between break 1 and the closest side. The average area then is 1/4.

Case two: The second break has 1/2 chance that it will not form a triangle, because if it splits the shorter side, the bigger side will be greater than 1/2 length. If it splits the other side, the probability is n/(1-n), where n is the distance between break 1 and the closest side. Take the integral from 0 to 1/2, divide by 1/2 and get approximately .38629436 (this was done by a calculator, but someone else may try to do this themselves). Since this only happens 1/2 the time, the probability is approximately .19314718.

Case three: This is the same as case two, except the second break is always on the larger piece, so the probability is approximately .38629436.

I apologize if this solution was not as clear or as short as I had hoped.
 Posted by Tristan on 2003-12-07 23:00:28

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