What's the probability distribution of an edge length of a polygon, whose other edges are less than l?

Question

A polygon (not necessarily convex), with $m$ edges. $m-1$ of these edges are less than $l$.

What's the probability distribution of the other edge length? When $$m=3$$ I'm not sure. But it seems the answer is $$F(x) = P(M_3 < x) = \frac{\arccos(\frac{m_1^2+m_2^2-x^2}{2m_1m_2})}{\pi}$$

If it's correct. How about $m>3$?

So it's wrong.

Thanks!

EDIT:

$(m+1)$ Points are randomly placed on an infinite plane. But we have priori knowledge, that there has been $m$ edges ( a chain ) shorter than $l$

Answer 1

Not an answer just too long for comments:

In your example, it seems the one vertex is placed on $(0, 0)$, another on $(M_1, 0)$ and the third one is $(M_2\cos\Theta, M_2\sin\Theta)$, where $M_1, M_2 \sim \text{Uniform}(0, l)$, $\Theta \sim \text{Uniform}(0, \pi)$ and they are independent. By Law of Cosine,

$$ M_3^2 = M_1^2 + M_2^2 - 2M_1M_2\cos\Theta$$

Therefore $$\begin{align} \Pr\{M_3 \leq x|M_1 = m_1, M_2 = m_2\} &= \Pr\{M_3^2 \leq x^2|M_1 = m_1, M_2 = m_2\} \\ &= \Pr\{m_1^2 + m_2^2 - 2m_1m_2\cos\Theta\leq x^2\} \\ &= \Pr\left\{\cos\Theta \geq \frac {m_1^2 + m_2^2 - x^2} {2m_1m_2}\right\} \\ &= \Pr\left\{\Theta \leq \arccos\left(\frac {m_1^2 + m_2^2 - x^2} {2m_1m_2}\right)\right\} \\ &= \frac {1} {\pi}\arccos\left(\frac {m_1^2 + m_2^2 - x^2} {2m_1m_2}\right) \end{align}$$

As said if you want to generalize the problem you need to specify more.