A straw of length 1 is first broken into two pieces uniformly at random... 
Question:
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:$\mathbb P\{X_N\le x |N=n\}=\frac{\mathbb P\{X_N\le x ,N=n\}}{\mathbb P\{N=n\}}=2^{N}\mathbb P\{X_N\le x\}\mathbb P\{N=n\}=2^{N}\mathbb P\{X_N\le x\}\frac{1}{2^N}=\mathbb P\{X_N\le x\}$
Furthermore
\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
$$f_{X_N|N}(x,n)=\frac{(-1)^{n+1}}{2^{n-1}}.$$
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
$$E[X_N|N=n]=\int_{0}^{1}xf_{X_N|N}(x,n)dx=\frac{(-1)^{n+1}}{2^{n-1}}\int_{0}^{1}xdx=\frac{(-1)^{n+1}}{2^{n}}$$
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.

Notes/Queries
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 :)
 A: 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 $g(1)=c=\frac{1}{3}$.
A: 2nd Attempt:
$\text{Let } Y_{n}\sim Unif([0,1]), \text{ with } n \text{ being the } n^{th} \text{ flip}$
$\text{Also, let } X_N \text{ denote the length after } N \text{ flips, with }\mathbb P\{N=n\}=\frac{1}{2^n}.$
$\text{Notice that } X_n=\prod_{k=1}^{n}Y_k$
$\text{Since all } Y_k \text{ are all i.i.d , we can compute the following:}$
$\mathbb E[X_n]=\mathbb E[\prod_{k=1}^{n}Y_k]=\prod_{k=1}^{n}\mathbb E[Y_k]=\frac{1}{2^n}$
$\text{Now using the Law of Total Expectation, we have}$
$\mathbb E[L]=\sum_{n=1}^{\infty}\mathbb E[X_n]\mathbb P\{N=n\}=\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{1}{2^n}=\sum_{n=1}^{\infty}\frac{1}{2^{2n}}=\sum_{n=1}^{\infty}\frac{1}{4^n}=\frac{1/4}{1-1/4}=\frac{1/4}{3/4}=\frac{1}{3}.$
