= Polygon probability for independent uniform rods
{title2=$1-1/(n-1)!$}
For $n\ge3$ independent lengths with the <uniform distribution> on $[0,1]$, the <probability> of forming a nondegenerate <polygon> is $1-1/(n-1)!$. The necessary and sufficient condition is that every length is less than the sum of the others. The failure events are disjoint except on sets of probability zero. The sum of $n-1$ lengths has <probability density function> $s^{n-2}/(n-2)!$ on $0<s<1$, by repeated <convolution of probability densities>. A specified rod is too long with probability $\int_0^1(1-s)s^{n-2}/(n-2)!\,ds=1/n!$; summing over the $n$ rods proves the formula. For three and four rods the probabilities are $1/2$ and $5/6$.
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