Why the expectation of distance to center of disk is $r/3$ and not $r/2$? A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $\frac{r}{3}$, whereas I found $\frac{r}{2}$ as follow
We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$\mathbb E[Y]=\int_0^r (r-x)f_X(x)dx=\frac{1}{r}\int_0^r (r-x)dx=\frac{r}{2}.$$
Maybe there is a subtlety than I don't see ?
 A: $$F_X(x) = \frac{\pi x^2}{\pi r^2}$$
$$f_X(x)=\frac{2x}{r^2}$$
\begin{align}
E[r-X]&=r-E[X] \\
&=r - \frac1{r^2}\int_0^r 2x^2\, dx\\
&= r - \frac1{r^2}\frac{2r^3}3\\
&= \frac{r}{3}
\end{align}
A: You may first construct the probability density as follows:


*

*At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
$$\frac{1}{\pi r^2}\cdot 2\pi \cdot x \cdot dx$$
So, you get 
$$E(Y) = \frac{1}{\pi r^2} \int_0^r (r-x)2\pi \cdot x\; dx = \cdots = \frac{r}{3}$$
A: $$\frac{\int_0^r(r-x)x.dx}{\int_0^r x.dx}$$
This gives $r/3$.
The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2\pi x.dx$ ... but then the $2\pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.
Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.
