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 ?
$$F_X(x) = \frac{\pi x^2}{\pi r^2}$$

$$f_X(x)=\frac{2x}{r^2}$$

\begin{align}
E[r-X]&amp;=r-E[X] \\
&amp;=r - \frac1{r^2}\int_0^r 2x^2\, dx\\
&amp;= r - \frac1{r^2}\frac{2r^3}3\\
&amp;= \frac{r}{3}
\end{align}
$$\frac{\int_0^r(r-x)x.dx}{\int_0^r x.dx}$$

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 &amp; bottom &amp; obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc &amp; 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.