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Breaking a Stick into a Triangle II

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A stick of length $1$ is broken at a point chosen uniformly at random. Then the longer of the two pieces is broken again, at a point chosen uniformly at random along it. What is the probability that the three pieces can form a triangle?

解析

先自己试一试，卡住了再打开提示。

提示 1

As in part I, three pieces adding up to $1$ form a triangle exactly when every piece is shorter than $\tfrac12$ .

提示 2

Let $L$ be the length of the longer piece after the first break. It is uniform on $[\tfrac12, 1]$ . For a fixed $L$ , where must the second break land?

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Step 1: The condition. As in part I, three pieces that add up to $1$ form a triangle exactly when every piece is shorter than $\tfrac12$ .

Step 2: The first break. Break at $U$ , uniform on $[0, 1]$ . The longer piece has length $L = \max(U, 1 - U)$ , which is uniform on $[\tfrac12, 1]$ . The shorter piece, $1 - L$ , is already at most $\tfrac12$ , so it never causes a problem.

Step 3: The second break, for a fixed $L$ . Break the longer piece at a distance $V$ from one end, with $V$ uniform on $[0, L]$ . The two new pieces are $V$ and $L - V$ . Both must be shorter than $\tfrac12$ :

V < \tfrac12 \quad\text{and}\quad L - V < \tfrac12 \iff L - \tfrac12 < V < \tfrac12.

That window has length $\tfrac12 - (L - \tfrac12) = 1 - L$ , out of a total length $L$ , so

P(\text{triangle} \mid L) = \frac{1 - L}{L} = \frac{1}{L} - 1.

Top: the first break leaves a shorter piece 1 − L and a longer piece L. Bottom: on the longer piece, the second break must land between L − ½ and ½, a window of length 1 − L.

Step 4: Average over $L$ . $L$ is uniform on $[\tfrac12, 1]$ , so its density is $2$ there:

P(\text{triangle}) = \int_{1/2}^{1} 2\left(\frac{1}{L} - 1\right) dL = 2\Big[\ln L - L\Big]_{1/2}^{1} = 2\left(-1 - \ln\tfrac12 + \tfrac12\right) = 2\ln 2 - 1.

In the picture below, the curve is the chance of a triangle for each $L$ . The answer is its average height over $L$ from $\tfrac12$ to $1$ : the shaded area divided by the width $\tfrac12$ .

The curve 1/L − 1 falls from 1 at L = ½ to 0 at L = 1; its average height is about 0.386.

Answer: $2\ln 2 - 1 \approx 0.38629$ .

Why it beats part I ( $\tfrac14$ ): breaking the longer piece avoids the worst case, where both breaks land on the same half and leave one piece longer than $\tfrac12$ .

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