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 you can answer this at this point, please do.
If not, perhaps this will help: here are several methods to break the stick into the three random length pieces:
- 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.
- 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.
- 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.
(In reply to solution
As case 3 had the most complicated math, it is the one needing the best verification. The following is a simulation of case 3:
FOR trial = 1 TO 10000000
piece1 = RND(1)
IF piece1 > .5 THEN piece1 = 1 - piece1
piece2 = RND(1) * (1 - piece1)
piece3 = 1 - piece1 - piece2
IF piece1 + piece2 > piece3 AND piece1 + piece3 > piece2 THEN hit = hit + 1
ct = ct + 1
PRINT USING "####### / ######## = #.######;"; hit; ct; hit / ct;
PRINT USING "#.######"; SQR(hit * (1 - hit / ct)) / ct
3862532 / 10000000 = 0.386253;0.000154
That is, 3,862,532 hits out of 10,000,000 trials, for an average of .386253, with std error of the mean of .000154.
Posted by Charlie
on 2003-12-07 17:43:03