Breaking Chocolate Bars

What'd the answer be for two pieces? Well, let the cut occur at point $x$ . Clearly every $x \epsilon(0,1)$ is equally likely, and for each $x$ the desired product will be $x(1-x)$ . Then the expected value of the product is just the average value of the function $x(1-x)$ over $(0,1)$ , which is

$\displaystyle\int_{0}^{1}x(1-x)dx=\frac{1}{6}$

So, onto the problem. Let the first cut be at point $x$ . The second cut has probability $x$ to be within the first interval, $(0,x)$ . In this case the expected product of the three pieces will be:

$\frac{1}{6}x^2(1-x)$

because the two pieces within $(0,x)$ have, by the case with the two pieces, an expected product of $\frac{1}{6}$ scaled by $x$ each = $\frac{1}{6}x^2$ , and the third piece has length $(1-x)$ .

If the second cut is in $(x,1)$ , which occurs with probability $1-x$ , analogically the product is expected to be:

$\frac{1}{6}x(1-x)^2$

Combining these results, the expected value of the product of the 3 pieces is:

$\displaystyle\frac{1}{6}\int_{0}^{1}x^3 (1-x)+x (1-x)^3 dx=\frac{1}{60}$

Making the answer $1+60=61$ .

Let be $X$ and $Y$ the two random variables representing the points on the unit bar. If we choose $X$ first and than $Y$ , we'll have that

$\displaystyle 0\leq X \leq 1$

$\displaystyle 0\leq Y \leq x$ .

It's easy to see that the $PDF$ function is $\displaystyle 2$ , since

$\displaystyle \mathbb{P}(\Omega)=\int_{0}^{1} \int_{0}^{x}2 dydx=1$

Let's define $Z=Y(X-Y)(1-X)$ to be the random variable expressing the product of the three bar pieces. Hence,

$\displaystyle \mathbb{E}(Z)=\int_{0}^{1}\int_{0}^{x} 2y(x-y)(1-x) dydx=\frac{1}{60}$

Eventually

$\displaystyle p+q=\boxed{61}$

First, suppose that the chocolate can be cut at $N$ distinct points and that the two points that the chocolate is cut are $\frac{a}{N}$ and $\frac{a+b}{N}$ where $0 ≤ a ≤ a+b ≤ N$ .

Then, the chocolate is cut into three lengths: $\frac{a}{N}$ , $\frac{b}{N}$ , and $\frac{N-a-b}{N}=1-\frac{a}{N}-\frac{b}{N}$ . Their product is $\frac{a}{N}\frac{b}{N}(1-\frac{a}{N}-\frac{b}{N})$ .

We know that $0≤a≤N$ and $0≤b≤N-a$ . Using this, we can sum the products of lengths for all possible pairs $a,b$ : $\displaystyle\sum_{a=0}^{N} \displaystyle\sum_{b=0}^{N-a} \frac{a}{N}\frac{b}{N}(1-\frac{a}{N}-\frac{b}{N})$ . To get the expected value, this must be divided by the total number of values summed: $\frac{N(N+1)}{2}$ .

Find the value as $N$ approaches infinity: $\displaystyle\lim_{N \to \infty} {\frac{2 \displaystyle\sum_{a=0}^{N} \displaystyle\sum_{b=0}^{N-a} \frac{a}{N}\frac{b}{N}(1-\frac{a}{N}-\frac{b}{N})}{N(N+1)}}$ .

Using the rules for limits to infinity, the denominator can be replaced by $N^2$ .

So, we have $\displaystyle\lim_{N \to \infty} {\frac{2 \displaystyle\sum_{a=0}^{N} \displaystyle\sum_{b=0}^{N-a} \frac{a}{N}\frac{b}{N}(1-\frac{a}{N}-\frac{b}{N})}{N^2}}$ .

Substitute $x=\frac{a}{N}$ and $y=\frac{b}{N}$ . Each of the increments are by one, so $dx=dy=\frac{1}{N}$ . Then, switch to integral notation.

Thus, we have $\displaystyle\int_0^1\displaystyle\int_0^{1-x}\ 2xy(1-x-y)\ dy\ dx$ .

Evaluating this integral gives us $\frac{1}{60}$ .

The final answer is $1+60=61$ .