Consider the following game. A straw of length 1 is first broken into two pieces uniformly at random, so that the length of one piece is uniform random variable on $[0,1]$. Only one of the pieces is kept in the game. A fair coin is then tossed, if the result is head the remaining piece of straw is broken again uniformly at random. If the result is tails, the game stops. Let $L$ be the length of the straw at the end of game. Show that $$ \mathbb{E}(L)=\frac{1}{3} . $$

My attempt:

Let $X_N$ denotes length of straw after $N$ flips instead of L. $X_i \sim Unif([0,1])\Rightarrow\mathbb P\{X_1\le x |N=1\}=x$

Note:$\mathbb P\{X_N\le x |N=n\}=\frac{\mathbb P\{X_N\le x ,N=n\}}{\mathbb P\{N=n\}}=2^{N}\mathbb P\{X_N\le x\}\mathbb P\{N=n\}=2^{N}\mathbb P\{X_N\le x\}\frac{1}{2^N}=\mathbb P\{X_N\le x\}$


\begin{align*} \mathbb P\{X_2\le x |N=1\}&=\mathbb P\{N=1\}\mathbb P\{X_{1}\ge x |N=0\}=\frac{1}{2}\mathbb P\{X_{1}\ge x |N=0\} \\ &=\frac{1}{2}(1-\mathbb P\{X_{1}\le x |N=0\}) \\ \Rightarrow \mathbb P\{X_{2}\le x |N=1\}&=\frac{1}{2}-\frac{1}{2}x \\ \Rightarrow \mathbb P\{X_{3}\le x |N=2\}&=\frac{1}{2}-\frac{1}{2}\mathbb P\{X_{2}\le x |N=1\}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^2}x \\ \Rightarrow \mathbb P\{X_{4}\le x |N=3\}&=\frac{1}{2}-\frac{1}{2}\mathbb P\{X_{3}\le x |N=2\}=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^3}x \\ &\vdots\\ \mathbb P\{X_{N}\le x |N=n\}&=\sum_{k=1}^{n-1}\frac{(-1)^{k+1}}{2^k}+\frac{(-1)^{n+1}}{2^{n-1}}x=:F_{X_N|N} \end{align*}

Hence we have


Using the Law of Total Expectation we have

$$\mathbb E[X_N]=\mathbb E[E[X_N|N]]=\sum_{n=1}^{\infty}E[X_N|N]\mathbb P\{N=n|n\}$$

Now computing $E[X_N|N]$, we get the following


Note that $\mathbb P\{N=n|n\}=1$, i.e $E[X_N]=\sum_{n=1}^{\infty}E[X_N|N]=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2^{n}}=-\sum_{n=1}^{\infty}(\frac{-1}{2})^{n}=-\frac{-1/2}{1-(-1/2)}=\frac{1/2}{3/2}=\frac{1}{3}$

Hence we have found that $E[X_N]=\frac{1}{3}$ as required.


This is a quick problem on my random processes course. Even though I have the same answer as stated I feel like I do not have rigorous logic and feel I have accidentally attained the answer. Any help/alternative solutions would be greatly appreciated :)


2 Answers 2


Recursion is powerful! Define $g(x)$ as the expected length of the stick you get by playing your game, starting with a stick of length $x$. You want to compute $g(1)$. Now, let the random variable $X$ denote the stick's length after uniformly breaking a stick of length $1$ once. We have: $$g(1) = \frac{1}{2}E[X] + \frac{1}{2}E[g(X)]$$

But note that $g(x)$ must be proportional to $x$. Assume $g(x)=cx$: $$c = g(1)=\frac{1}{2}E[X]+\frac{1}{2}E[cX] = \frac{c+1}{2}E[X]$$

Since the stick was broken uniformly, $E[X]=\frac{1}{2}$. Hence: $$c = \frac{c+1}{4} \implies c=\frac{1}{3}$$

Hence, the expected length of the final stick is $g(1)=c=\frac{1}{3}$.

  • $\begingroup$ That is clean, I like it. Do you think the way I did it is valid, I know it's long-winded :/ @Haran $\endgroup$ Jan 25 at 13:59
  • 1
    $\begingroup$ There was one thing that I felt was ambiguous when I started reading the proof. You talk about the value of $\mathbb{P}(X_2 \leqslant x)$. Is this probability conditioned on the fact that the first coin toss landed heads? If not, how do you define $X_2$? Do you take it as $X_1$? $\endgroup$
    – Haran
    Jan 25 at 14:07
  • $\begingroup$ Yes correct, Yeah that is very unclear. Hmm I see what you mean. I'll have a think :) @Haran $\endgroup$ Jan 25 at 14:17
  • $\begingroup$ I really like the way you did it. I've just updated my answer, idk if it is more legitimate than it was tbh? Would be great if you could check. Thanks for your help!@Haran $\endgroup$ Jan 25 at 16:43
  • $\begingroup$ I still believe there is some error. You write $\mathbb{P}(X_2 \leqslant x | N=1) = \frac{1}{2}-\frac{1}{2}x$. But this doesn't make sense at $x=0$ and $x=1$. $\endgroup$
    – Haran
    Jan 26 at 7:28

2nd Attempt:

$\text{Let } Y_{n}\sim Unif([0,1]), \text{ with } n \text{ being the } n^{th} \text{ flip}$

$\text{Also, let } X_N \text{ denote the length after } N \text{ flips, with }\mathbb P\{N=n\}=\frac{1}{2^n}.$

$\text{Notice that } X_n=\prod_{k=1}^{n}Y_k$

$\text{Since all } Y_k \text{ are all i.i.d , we can compute the following:}$

$\mathbb E[X_n]=\mathbb E[\prod_{k=1}^{n}Y_k]=\prod_{k=1}^{n}\mathbb E[Y_k]=\frac{1}{2^n}$

$\text{Now using the Law of Total Expectation, we have}$

$\mathbb E[L]=\sum_{n=1}^{\infty}\mathbb E[X_n]\mathbb P\{N=n\}=\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{1}{2^n}=\sum_{n=1}^{\infty}\frac{1}{2^{2n}}=\sum_{n=1}^{\infty}\frac{1}{4^n}=\frac{1/4}{1-1/4}=\frac{1/4}{3/4}=\frac{1}{3}.$

  • $\begingroup$ @Haran I've given it another shot :) $\endgroup$ Jan 26 at 15:24

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