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Frederick Mosteller , 'Fifty challenging problems in probability', Q.43:

a bar is broken at random in 2 places. Find the average size of the smallest, middle, and largest pieces

I would like to discuss the solution offered in the book for working out the avergae size of the smallest segment, which says (summarized):

We might as well work with a bar of unit length $1$. let $X$ and $Y$ be the positions of the 2 points dropped. For convenience let us suppose that $X \lt Y$.

If we want to get the distribution of the smallest piece, then either $X$, $Y-X$ or $1-Y$ is smallest. Let us suppose $X$ is smallest, then: $$X \lt Y-X \text{ ie } 2X \lt Y$$ and $$X \lt 1-Y \text{ ie } X+Y \lt 1$$ this corresponds to the triangle shown in the picture, therefore the mean of X over that area will correspond to the X-coordinate of the centroid of that triangle, which as we know from plane geometry is $1 \over 3$ of the way from the base to the opposite vertex, here the point $(\frac{1}{3},\frac {2}{3})$ , therefore $\frac{1}{3} * \frac{1}{3} = \frac{1}{9}$

Therefore the mean of the smallest segment is $\frac{1}{9}$

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My question is: even restricting ourselves to the area $X < Y$, why don't we need to consider the cases where Y-X is smallest, and 1-Y is smallest ?

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    I think the author meant/you should write "is broken into 3 pieces" (title) and "is broken in 2 places" (body) – 2017-01-29

2 Answers 2

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Look at it this way: you have a circle of circumference $1$, and you cut it three times at random points. Then you can always assume that $X$ is the smallest part.

Another approach:

Let $z = \min\{x_1, \dots x_n\}$ were $x_i$ all have the same distribution and the same probability of being the minimum. Then $$\mathbb{E}(z) = \sum_{k=1}^n \mathbb{E}(z|z=x_k)Pr(z=x_k)=\frac{1}{n} n\mathbb{E}(z|z=x_1)=\mathbb{E}(z|z=x_1).$$ So one may as well assume that $z = x_1$.

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    I understand why X, Y-X and 1-Y have the same distribution (by the Principle of Symmetry, aka points-on-a-circle method you describe), but does it follow that computing the average of Z = min (X,Y-X,1-Y) we can just assume X is the smallest in the first place ? – 2017-01-29
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    I updated the answer. – 2017-01-29
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    very clear now , many thanks ! – 2017-01-29
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My question is: even restricting ourselves to the area $X

We do have to consider that. What you are missing is what $X,Y$ represent.

Choose an end of the bar, measure the distance to its nearest break: that is $X$

From the other end of the bar, measure the distance to its nearest break: that is $1-Y$.

The distance between the breaks will be $Y-X$.

$$\overset 0\vert\underbrace{\underline{\;\to\;}}_{X} \overset X\vert \underbrace{\underline{\;\to\;}}_{Y-X}\overset Y\vert \underbrace{\underline{\;\to\;}}_{1-Y}\overset 1\vert$$

Clearly that makes $X

Now, we could arbitrarily have chosen to measure from the other end:

$$\overset 1\vert\underbrace{\underline{\;\gets\;}}_{1-V}\overset V\vert \underbrace{\underline{\;\gets\;}}_{V-U}\overset U\vert \underbrace{\underline{\;\gets\;}}_{U}\overset 0\vert$$

But... this just gives the same result.