the expectation of a chocolate bar So my buddy claims that if I split a chocolate bar at random into two pieces, then the expected size of the larger piece is $\frac{3}{4}$ of the bar. I can't figure out how he came up with this value...
Can someone explain this? If you can, can you provide some kind of a proof?
p.s. it would be helpful to think of this chocolate bar as a 1D array :)
UPDATE
Imagine the candy bar is a world-famous chocolate bar, the ones that are broken into chunks. However, this special chocolate bar has n chunks. If we broke the chocolate bar randomly along these chunks, what would the expected size of the larger chunk be? My buddy claims it to be $\leq{\frac{3}{4}}$. 
 A: The larger piece is always between 1/2 and 1.  If the break is uniformly distributed along the bar, the larger piece is uniformly distributed between 1/2 and 1.  This gives the expected value of 3/4.
Added when we have break lines:  If there are $n$ pieces, there are $n-1$ break lines, all equally probable.  If $(n-1)$ is even, the size of the largest piece has chance $\frac{2}{n-1}$ of being any step between $\frac{n+1}{2n}$ and $\frac{n-1}{n}$, so the expectation is $\frac{3n-1}{4n}$.  If $(n-1)$ is odd there is $\frac{1}{n-1}$ chance the "larger" piece is $\frac{1}{2}$ and $\frac{2}{n-1}$ for each value between $\frac{n+1}{2n}$ and $\frac{n-1}{n}$, giving $\frac{3n-4}{4n-4}$ as the expected value.  As stated in other answers, this is always below $\frac{3}{4}$, but approaches that as $n \to \infty$
A: Define $Y = \max \{ U,1 - U\} $, where $U$ is a uniform$(0,1)$ random variable.
Then, by the law of total probability (conditioning on $U$),
$$
{\rm E}(Y) = \int_0^1 {{\rm E}(Y|U = u)du}  = \int_0^{0.5} {(1 - u)du}  + \int_{0.5}^1 {udu}  = \frac{3}{4}
$$
EDIT. The discrete case is even more elementary, but involves more calculations. 
If $n$ (the number of chunks) is odd (and greater than $1$), then the ratio is equal to $k/n$, $k=n-1,n-2,\ldots,(n+1)/2$, with probability $2/(n-1)$ (by symmetry). Hence, its expectation is equal to
$$
\sum\limits_{k = (n + 1)/2}^{n - 1} {\frac{k}{n}} \frac{2}{{n - 1}} = \frac{{3n - 1}}{{4n}} < \frac{3}{4}.
$$
If $n$ is even, then the ratio is equal to $k/n$, $k=n-1,n-2,\ldots,n/2+1$, with probability $2/(n-1)$ (again, by symmetry), and to $(n/2)/n = 1/2$ with probability $1/(n-1)$. Hence, its expectation is equal to
$$
\sum\limits_{k = n/2 + 1}^{n - 1} {\frac{k}{n}} \frac{2}{{n - 1}} + \frac{1}{{2(n - 1)}} = \frac{{3n - 4}}{{4n - 4}} < \frac{3}{4}.
$$
It is interesting to note that 
$$
\frac{{3n - 4}}{{4n - 4}} = \frac{{3(n - 1) - 1}}{{4(n - 1)}}.
$$
Thus, the expectations for $n$ and $n-1$ are equal for $n > 2$ even.
