# Broken stick riddle

There is a famous mathematical riddle called The Broken Stick Problem. Here’s the extension: If a straight stick is accidentally broken into three pieces, the probability of being able to form a triangle with the three pieces is 1/4. If a straight stick is accidentally broken into four pieces, the probability of being able to form a quadrilateral with the four pieces is 1/2. If a straight stick is accidentally broken into seven pieces, What is the probability of being able to form a heptagon (7-sided polygon) with the seven pieces?

0.8906

For a triangle, the length of the largest side must be smaller than the sum of the other two. If it were larger or equal, it would be impossible to have any closed shape with an interior.

Similarly, for a heptagon, the length of the largest side must smaller than the sum of the other six. Thus, this is what we need to calculate: the probability that the largest of the seven pieces is less than the sum of the other six.

The flip-side of this: the probability that at least one segment is at least half the length of the stick. If it is, a heptagon is impossible, but if it isn't, a heptagon is possible.

One way to calculate this probability is the probability that n points will be on the same half of a circle when placed about its perimeter randomly. This probability is given by n/(2^[n−1]). See this post for explanation.

Thus, we just need the complement of this probability, which gives us:

1 - n/(2^[n−1]) =
1 - 7/(2^[7−1]) =
0.8906

This paper discusses the question well.

• I knew I was overthinking it... – AxiomaticSystem Jun 30 '20 at 19:28

The problem reduces to finding the probability that

none of the seven segments has length greater than $$\frac{1}{2}$$.
This is due to a generalization of the triangular inequality, which essentially states that no polygon can have a side greater than half its perimeter - in this case, 1.
Say we have $$n$$ pieces, one of which is has length $$L > \frac{1}{2}$$. (Clearly, we cannot have two such pieces.) We denote the probability that we still have such a piece after the next cut by $$P_n(L)$$.
On our large piece, there are two zones of length $$L-\frac{1}{2}$$ on either side: if the next cut falls within either of those zones, then we will be in the same situation, but with a large length $$x. "Sliding" the cut from one end of either zone to the other gives us a term $$2\int\limits_\frac{1}{2}^L P_{n+1}(x) dx$$ in $$P_n$$.
There are two other possibilities: either the cut misses the large piece entirely, or the cut hits the large piece outside of the above "zones", breaking it into two pieces which are not large. Both happen with probability $$1-L$$, so the former adds another term $$(1-L)P_{n+1}(L)$$ to $$P_n$$. Therefore,
$$P_n(L) = (1-L)P_{n+1}(L) + 2\int\limits_\frac{1}{2}^L P_{n+1}(x) dx$$
Since we can't make any more cuts - and therefore can't break any large pieces - once we have seven pieces, we have that $$P_7(L)=1$$. The remainder of the problem consists of calculating $$P_1(1)$$ with this boundary condition, and then subtracting the result from 1 to get the probability that no large piece remains.