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We have all heard the old problem about forming a triangle from breaking a stick into three pieces, with the breaks randomly distributed. Some variations make the second break contingent on the first in some way. I present a new variation (not original to me).

Problem: Take a stick and break it at a location selected with uniform density along its length. Throw away the left-hand piece and break the right-hand one at a location selected with uniform density along its length. Continue forever. What is the probability that one of the discarded left-hand pieces is more than half as long as the original stick?

I do not have a particularly elegant to solution to this one, and was wondering if one exists. Of course, any answers that solve the problem are welcome.

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Let $f(x)$ be the probability that we eventually cut off at least $x$ from a stick of length $1$, with $1\ge x\ge1/2$. We can either succeed by immediately cutting off at least $x$, with probability $1-x$, or by leaving $t\ge x$ and then cutting off $x$ from a stick of length $t$. Thus we have

$$ f(x)=1-x+\int_x^1f(x/t)\mathrm dt\;. $$

Substituting $u=x/t$ yields

$$f(x)=1-x+x\int_x^1f(u)/u^2\mathrm du\;.\tag1$$

Then differentiating with respect to $x$ yields

$$f'(x)=-1-f(x)/x+\int_x^1f(u)/u^2\mathrm du\;,$$

and differentiating again yields

$$f''(x)=-f'(x)/x+f(x)/x^2-f(x)/x^2=-f'(x)/x\;,$$

so

$$ \frac{f''(x)}{f'(x)}=-\frac1x $$

and thus

$$ \begin{align} \log f'(x)&=-\log x +c\;,\\ f'(x)&=a/x\;,\\ f(x)&=a\log x+b\;. \end{align} $$

Now $f(1)=0$ yields $b=0$, and then substituting into $(1)$ yields $a=-1$, so $f(x)=-\log x$ and

$$f(1/2)=\log2\approx0.693\;.$$

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