A triangle is possible iff no part is $>{1\over2}$. With probability ${1\over2}$ both cuts are on the same side of the midpoint $M={1\over 2}$, M$, in which case no triangle is possible. If the cuts$x$and$y$,$\ x < y$, are on different sides of$M$then with probability${1\over 2}$the point$x+{1\over2}$x$ is to the further left of in its half than $y$, y$is in which the right half. In this case there is no triangle possible either. It follows that only${1\over 4}$of all cuts admit the forming of a triangle. 1 With probability${1\over2}$both cuts are on the same side of the midpoint$M={1\over 2}$, in which case no triangle is possible. If the cuts$x$and$y$,$\ x < y$, are on different sides of$M$then with probability${1\over 2}$the point$x+{1\over2}$is to the left of$y$, in which case there is no triangle possible either. It follows that only${1\over 4}\$ of all cuts admit the forming of a triangle.