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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 $\leq{\frac{3}{4}}$.

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    @Raskolnikov... just a buddy... what do you mean by "why is he teasing you?"2011-03-29

2 Answers 2

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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 $n \to \infty$

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    I had a small error in my calculation for $n$ even, and agree with Shai Covo. Corrected. In essence, you break at the middle with probability $\frac{1}{n-1}$ and get $\frac{1}{2}$ or you don't break in the middle with probability $\frac{n-2}{n-1}$ and get on average $\frac{3}{4}$, again by symmetry.2011-03-29
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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.

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    @Hristo: As I will show later on today, the answer for the finite case is $(3n-1)/(4n)$ for $n$ odd, and $(3n-4)/(4n-4)$ for $n$ even (which can be written as $[3(n-1)-1]/[4(n-1)]$, thus similar to the odd case).2011-03-29