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I got an interesting question which I'm clueless about. so I thought maybe someone here could give me a direction. The questions is:

A 1­ meter long stick is broken at a (uniformly) random point to two pieces. The largest piece is then broken randomly again to two pieces, and you keep the largest of the two. what is the probability that this piece is longer than 1⁄2 meter ?

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closed as off-topic by heropup, Pragabhava, iadvd, user91500, Claude Leibovici Sep 13 '16 at 5:36

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  • $\begingroup$ clement - it is not the same question, and not the same answer, sorry I am responding to you here, but I am a guest so I can't respond directly. The answer there is 1/4 and the answer here should be 2-2*ln2. $\endgroup$ – user368344 Sep 12 '16 at 18:37
  • $\begingroup$ Indeed. Retracting my close vote. $\endgroup$ – Clement C. Sep 12 '16 at 18:47
  • $\begingroup$ Daniel, you should be able to comment under your own question. But you seem to have created two accounts, so that's why you could not. Read this for instructions on how to get the accounts merged. $\endgroup$ – Jyrki Lahtonen Sep 12 '16 at 18:53
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Consider the following process:

  • Draw uniformly at random $X$ in $[0,1]$, and set $A\stackrel{\rm def}{=} \max(X,1-X)$. This is the length of the largest piece in the first step.

  • Draw uniformly at random $Y$ in $[0,A]$, and set $B\stackrel{\rm def}{=} \max(Y,1-Y)$. This is the length of the largest piece in the second step.

  • You are interested in $\mathbb{P}\{ B > \frac{1}{2} \} = 1 - \mathbb{P}\{ B \leq \frac{1}{2} \}$.

We will rely on the following (easy to prove) fact:

Fact. If $U\sim\operatorname{Uniform}([0,a])$, then $\max(U,a-U)\sim\operatorname{Uniform}([\frac{a}{2},a])$.

From there, we get that $A\sim\operatorname{Uniform}([\frac{1}{2},1])$. Since, conditioned on $A$, $Y\sim\operatorname{Uniform}([0,A])$, we also get that $B\sim\operatorname{Uniform}([\frac{A}{2},A])$.

From there, we get that $$\begin{align} \mathbb{P}\left\{ B \leq \frac{1}{2} \right\} &= \mathbb{E}[\mathbb{1}_{\{B \leq \frac{1}{2}\}}] = \mathbb{E}[\mathbb{E}[\mathbb{1}_{\{B \leq \frac{1}{2}\}}\mid A]] = \mathbb{E}\left[\frac{\frac{1}{2}-\frac{A}{2}}{A-\frac{A}{2}}\right]\\ &= \mathbb{E}\left[\frac{1-A}{A}\right] = \frac{1}{1-\frac{1}{2}}\int_{1/2}^1 \frac{1-a}{a} da = 2\left(\int_{1/2}^1 \frac{da}{a}-\frac{1}{2}\right) \\ &= 2\ln 2-1 \end{align}$$ so that finally $$\mathbb{P}\left\{ B > \frac{1}{2} \right\} = 2-2\ln 2.$$

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