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.
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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. |
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