AlmostSurely
Problem 13.8 Easy Free

Piece Marked by a Dot

Several thin glass rods, each 99 inches long with one end marked by a blue dot and the other by a red dot, are dropped onto a hard floor. Each rod that breaks does so into exactly three pieces, with the two break points chosen independently and uniformly at random along its length. Averaging over all such broken rods, what is the expected length of the fragment containing the blue dot?

Work it out, then compare with the answer.

SolutionShow the worked solution

Place the rod on [0,9][0,9] with the blue dot at 00. Let the break points be U,VU,V, i.i.d. Unif(0,9)\mathrm{Unif}(0,9). The fragment containing the blue dot is [0,min⁡(U,V)][0,\min(U,V)], so its length is L=min⁡(U,V)L=\min(U,V).

For 0≤x≤90\le x\le 9, by independence,

P(L>x)=P(U>x) P(V>x)=(1−x9)2.\mathbb{P}(L>x)=\mathbb{P}(U>x)\,\mathbb{P}(V>x)=\Bigl(1-\frac{x}{9}\Bigr)^2 .

Since L≥0L\ge 0,

E[L]=∫09(1−x9)2 dx=9∫01(1−t)2 dt=93=3.\mathbb{E}[L]=\int_0^9 \Bigl(1-\frac{x}{9}\Bigr)^2\,dx=9\int_0^1 (1-t)^2\,dt=\frac{9}{3}=3 .
E[L]=3 inches\boxed{\mathbb{E}[L]=3\text{ inches}}

Alternative. Let X(1)≤X(2)X_{(1)}\le X_{(2)} be the ordered break points. The three spacings (X(1), X(2)−X(1), 9−X(2))(X_{(1)},\,X_{(2)}-X_{(1)},\,9-X_{(2)}) are exchangeable, because they have the same joint law as the three arcs cut from a circle of circumference 99 by three i.i.d. uniform points, read from one of those points. The three pieces therefore have equal means. These means sum to 99, so the blue-dotted piece has mean 33.

Key idea. Uniform break points cut the rod into exchangeable pieces. The particular piece we ask about, here the one carrying the blue dot, does not change its mean.

Spotted an error? Email us