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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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 solution | Comment 1 of 26
Let the stick in each case be 1 unit long.

1. Let x and y be the two random unformly distributed points from 0 to 1. First take the case where x is the smaller. The three pieces have lengths x, y-x and 1-y. In order to form a triangle, the sum of any two piece lengths must be greater than the length of the third piece. In this case

x+y-x>1-y which is the same as y > 1/2
x+1-y>y-x which is the same as y<x+1/2
1-y+y-x>x which is the same as x<1/2

So within the bounds of a unit square from x=0 to x=1 and y=0 to y=1, further limited to the part above y=x, the above three inequalities fall within a triangle of area 1/8, for a probability of (1/8)/(1/2) = 1/4 that the triangle can be formed given that x<y. By symmetry the conditional probability given that y is the smaller, is the same. Thus the probability is 1/4 overall.

Then, let's do case 3 next, because it will provide a conditional probability that will be used in solving case 2:

Let L1 be the length of the shorter piece and L2 be the length of the longer, and x be the point along L2 that the second break will be made. We know already from the fact that the second break is on the larger piece that the total of those two resulting from the second break will be greater than L1, so we need only care that x+L1>L2-x and L1+L2-x>x. These ultimately come out to x>L2-1/2 and x < 1/2. This leaves a length of 1/2 - (L2-1/2) where x is good, out of a total length of L2, for a probability that then, algebraically comes out to (1-L2)/L2, or 1/L2 - 1. We need to integrate this function from 1/2 to 1, these being the uniformly distributed lengths for the longer original piece, and then divide by 1/2 (i.e., multiply by 2). For convenience let's rename L2 w:

∫{1/2 to 1}((w^-1) - 1) dw / (1/2)

= 2[ln w - w]{1/2 to 1}

= 2ln2 - 1 =
.38629436111989....

2. You have a 1/2 probability of choosing the shorter stick, in which case you have 0 conditional probability of making a triangle. There is the remaining 1/2 probability of choosing the larger original piece, in which case the probability found above now holds. Since you have a 1/2 probability of getting there, the overall probability is 1/2 that of case 3, or .193147180559945....

Edited on December 7, 2003, 2:50 pm
 Posted by Charlie on 2003-12-07 14:42:19

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