One cut, longer piece at least double

Question

A rope of length 1 is cut at one point chosen uniformly at random along it, giving two pieces. What is the probability that the longer piece is at least twice the length of the shorter piece?

Accepted Answer

Let the cut be at U \sim U(0, 1); the pieces are U and 1 - U. Take U \le 	frac12 (shorter piece U, longer piece 1 - U). The condition is 1 - U \ge 2U \quad\Longrightarrow\quad U \le 	frac13 . That is the interval [0, 	frac13], of length 	frac13. By symmetry the case U 	frac12 gives [	frac23, 1], another 	frac13. P = 	frac13 + 	frac13 = 	frac23 . Takeaway. For "longer \ge k times shorter," the favorable region is the outer 	frac{1}{k+1} at each end, total 	frac{2}{k+1}. With k = 2 that is 	frac23.

One cut, longer piece at least double

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