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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(6): help me! correction | Comment 14 of 26 |
(In reply to re(5): help me! correction by Penny)

I think the reason why Charlie brought up Newton and Leibnitz was because he invoked Calculus in his solution.

Charlie did not (as your previous comment title implies) use a computer program in his solution. Charlie only used the computer to verify his solution to a more complicated aspect of the problem.

An aspect of Chalie's solution is that your approximation to the answer is unnecessary, and almost requires a computer to solve it....

I noticed that you first approximated the solution with a 10 meter stick and fixed the points along 1 meter intervals. And you came up with an answer of 29%. Then, I suppose after looking at others' answers, you decided (and accurately so), that if you approximate with higher gradation you will get closer to the actual answer. So you updated your post to make it 10000 meters long.

Unfortunately, it ends up being a very long "by-hand" calculation... and you wrote:
" ...and (1/9999)*[(3/9999) + (5/9998) + (7/9997) + (9/9996) + ... + (5001/5001)] does appear to be approximately 1/4."

When as Charlie pointed out,
"I don't know which case you were addressing in your original post, but your series does not approximate 1/4. It is about 0.3863, as mentioned, which is close to the actual answer found by calculus (not computer..."

The point of all of this, of course, is that it is possible to get an arbitrarily precise numerical answer without resorting to approximating a smooth probability function.
 Posted by SilverKnight on 2003-12-08 11:20:01

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