You break a stick of unit length in two. You then subsequently break the biggest of the resulting two sides in two, thus obtaining three pieces.

What is the expected length of the smallest of the three?

(Each breaking of a stick is assumed to be at a random point in that stick, uniformly distributed.)

  • $\begingroup$ Is it possible to answer this? Could it be any value less than $1/3$? $\endgroup$ – Sigur Dec 8 '12 at 14:40
  • $\begingroup$ It certainly is possible - and the result will have to be consistent with a monte carlo simulation. For each particular experiment, it can indeed be any value less than 1/3, but I am asking for the average across a large set of experiments. $\endgroup$ – alexandreC Dec 8 '12 at 14:43
  • $\begingroup$ It depends on the assumptions on where the stick breaks. You said nothing about this so the question cannot be answered. $\endgroup$ – tst Dec 8 '12 at 14:44
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    $\begingroup$ What do you mean by "breaking a stick" in two? I mean what will be the distribution? If you fix it at both ends and bend, it will more likely break in the middle, maybe following a gaussian. Or do you assume it might brake anywhere with the same probability? $\endgroup$ – Simon Markett Dec 8 '12 at 14:46
  • $\begingroup$ Also in the first scenario, the parameters will be different for a smaller stick I guess, so I the first outcome will influence the second one. In short, you need to describe better what you want. $\endgroup$ – Simon Markett Dec 8 '12 at 14:47

If the shorter stick after the first break has length $S$ then $S$ uniformly distributed on $[0,\frac{L}{2}]$.

If $S=s$ then the shorter part of the longer stick will have length $T$ uniformly distributed on $[0,\frac{L-s}{2}]$. You will certainly have $T \lt s$ if $\frac{L}{3} \lt s \le \frac{L}{2}$.

So the expectation of the minimum of $S$ and $T$ is

$$\int_{s=\frac{L}{3}}^\frac{L}{2} \int_{t=0}^\frac{L-s}{2} t \frac{2}{L-s} \frac{2}{L} \,dt\,ds + \int_{s=0}^\frac{L}{3} \int_{t=0}^s t \frac{2}{L-s} \frac{2}{L} \,dt\,ds + \int_{s=0}^\frac{L}{3} \int_{t=s}^\frac{L-s}{2} s \frac{2}{L-s} \frac{2}{L} \,dt\,ds $$ $$= \frac{7 L}{144} +2L\log\left( \frac{3}{2}\right)-\frac{7L}{9} +\frac{5L}{3} -4L\log\left( \frac{3}{2}\right) $$ $$= L\left(\frac{15}{16} -2\log\left( \frac{3}{2}\right) \right)$$ the same as did's answer

  • $\begingroup$ Hi. Aren't you missing a contribution from the possibility that is the shorter part the one that breaks again? $\endgroup$ – drake Jun 16 '15 at 17:00
  • $\begingroup$ @drake - the question says "You then subsequently break the biggest of the resulting two sides in two" $\endgroup$ – Henry Jun 16 '15 at 17:07
  • $\begingroup$ Oh, so sorry, I was trying to solve a different problem and though this one was the same. I didn't read the question carefully. I agree your answer is right. $\endgroup$ – drake Jun 16 '15 at 18:37

When you make the first cut, the bigger piece has length $p$ that is uniformly distributed in $[\frac12, 1]$. (And so the smaller piece has length $1-p$ uniformly distributed in $[0, \frac12]$, but of course not independent of the length $p$ of the bigger piece.)

Similarly, when you cut this piece of length $p$ at a uniformly chosen point, the smaller piece of the two has length $q$ uniformly distributed in $[0, \frac{p}{2}]$.

The smallest of the three pieces is therefore of length $\min(q, 1-p)$. We can calculate its expected value as $$ \int_{1/2}^{1} \int_{0}^{p/2} \min(q, 1-p) \;(\frac{1}{p/2}dq) \;(\frac{1}{1/2}dp) = \int_{1/2}^{1} \frac{4}{p} \int_{0}^{p/2} \min(q, 1-p) \;dq \;dp$$

If $p/2 \le 1-p$ (which happens when $p \le 2/3$), then as $q \le p/2 \le 1-p$, it is always the case that $\min(q, 1-p) = q$, and so the inner integral becomes $$ \int_{0}^{p/2} \min(q, 1-p) \;dq = \int_{0}^{p/2} q \;dq = \frac{p^2}{8}.$$

Else, for $p/2 > 1-p$, the inner integral can be split as $$\begin{align} \int_{0}^{p/2} \min(q, 1-p) \;dq &= \int_{0}^{1-p} q \;dq + \int_{1-p}^{p/2} (1-p) \;dq \\ &= (1-p)^2/2 \;\;+\;\; p(1-p)/2 - (1-p)^2 \\ &= p(1-p)/2 - (1-p)^2/2 \end{align}$$

So the outer integral is $$\begin{align} &\int_{1/2}^{1} \frac{4}{p} \int_{0}^{p/2} \min(q, 1-p) \;dq \;dp \\ &=\int_{1/2}^{2/3} \frac{4}{p}\frac{p^2}{8} \;dp + \int_{2/3}^{1} \frac{4}{p}(\frac{p(1-p)}{2} - \frac{(1-p)^2}{2})\;dp \\ &=\int_{1/2}^{2/3} p/2 \;dp + \int_{2/3}^{1} \left(2(1-p) - \frac{2(1-p)^2}{p}\right) \; dp \\ &= \frac{7}{144} + \frac{1}{9} - (\log(9/4) - 7/9)\\ &= \frac{15}{16} - \log\left(\frac94\right) \approx 0.12657 \end{align}$$

This looks like a really strange answer, but it is roughly confirmed by the following simulation:

#!/usr/bin/env python
import random
num_samples = 0
sum_samples = 0
for num_samples in xrange(1, 100000000):
    r1 = random.uniform(0, 1)
    p = max(r1, 1 - r1)
    r2 = random.uniform(0, p)
    q = min(r2, p - r2)
    cur_sample = min(q, p - q, 1 - p)
    sum_samples += cur_sample
    average = sum_samples / num_samples
    if num_samples % 100000 == 0:
        print 'Average is %.5f after %d samples' % (average,num_samples)

The lengths of the two pieces of a stick of length $L$ are $\frac12(1\pm U)L$, where $U$ is uniform on $(0,1)$. Hence the lengths of the three pieces are $$ \frac12(1-U),\quad \frac14(1+U)(1+V),\quad\frac14(1+U)(1-V), $$ where $U$ and $V$ are i.i.d. uniform on $(0,1)$. The smallest $S$ of these three lengths is $S=\frac12(1-U)$ on $A=[V\lt a(U)]$ and $S=\frac14(1+U)(1-V)$ on $B=[V\gt a(U)]$, where $a$ is defined by $$ a(u)=\frac{(3u-1)^+}{u+1}. $$ Since $\mathbb E(A\mid U)=a(U)$, $$ \mathbb E(S\mathbf 1_A\mid U)=\frac12(1-U)a(U). $$ Likewise, $$ \mathbb E(S\mathbf 1_B\mid U)=\frac14(1+U)\mathbb E((1-V)\mathbb 1_{V\gt a(U)}\mid U)=\frac14(1+U)\frac12(1-a(U))^2. $$ Summing these yields $$ \mathbb E(S\mid U)=\frac12(1-U)a(U)+\frac18(1+U)(1-a(U))^2, $$ which can be simplified into $$ \mathbb E(S\mid U)=\frac18(1+U)-\frac18\frac{((3U-1)^+)^2}{1+U}. $$ Thus, $\mathbb E(S)=\frac3{16}-\tfrac18t$, with $$ t=\int_{1/3}^1\frac{(3u-1)^2}{1+u}\mathrm du=\int_0^{1}\frac{4x^2}{2+x}\mathrm dx, $$ that is, $$ t=\left[2x^2-8x+16\log(2+x)\right]_{0}^1=-6+16\log\left(\frac32\right). $$ Finally, $$ \mathbb E(S)=\frac{15}{16}-2\log\left(\frac32\right)=0.12657\ldots $$

  • $\begingroup$ good job! anyone for an alternative answer? $\endgroup$ – alexandreC Dec 8 '12 at 16:01
  • $\begingroup$ ?? $ $ $ $ $ $ $ $ $\endgroup$ – Did Dec 8 '12 at 16:14
  • $\begingroup$ Your reasoning was excellent!! result agrees with MC simulation (unsurprisingly) - was just wondering if someone can come up with a different reasoning, that would lead to your correct answer! :) $\endgroup$ – alexandreC Dec 8 '12 at 16:26
  • $\begingroup$ Oh oops, somehow the website didn't show me you had already posted your answer... I need not have spent my time calculating. :-) $\endgroup$ – ShreevatsaR Dec 8 '12 at 17:36
  • $\begingroup$ alexandreC: How is the reasoning in the two other correct answers, different? $\endgroup$ – Did Dec 8 '12 at 18:27

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