Problem source

Harry the Calculating Horse will do any mathematical problem I set  him, providing the answer is 1, 2, 3 or 4. When I set  him a problem, he   places a  hoof on a large  grid consisting of unit squares and  his   answer is  the number of squares partly covered by his hoof. Harry has circular hoofs, of radius $1/4$ unit.
After many years of collaboration, I suspect that Harry no longer bothers to do the calculations, instead merely placing his hoof on the grid completely at random. I often ask him to  divide 4 by 4, but only  about $1/4$ of his answers are right;  I often ask him to add 2 and 2, but disappointingly only about $\pi/16$ of his answers are right. Is this consistent with my suspicions?
I decide to investigate  further by setting  Harry many problems,  the answers to which are 1, 2, 3, or 4 with equal frequency. If Harry is placing his hoof at random, find the expected value of his answers. The average of Harry's answers turns out to  be 2. Should I get a new horse?

Solution source

Without loss of generality, let's assume that Harry is putting the center of his hoof within one square.


\begin{center}
    \begin{tikzpicture}[scale=5]


        \filldraw[color=red!20] (0.25, .25) rectangle (.75, .75);
        \filldraw[color=green!20] (0.25, 0) rectangle (.75, .25);
        \filldraw[color=green!20] (0.75, .25) rectangle (1, .75);
        \filldraw[color=green!20] (0.25, .75) rectangle (.75, 1);
        \filldraw[color=green!20] (0, 0.25) rectangle (.25, .75);

        \filldraw[color=orange!20] (0,0) -- (.25, 0) arc (0:90:.25) -- cycle;
        \filldraw[color=orange!20] (1,0) -- (.75, 0) arc (180:90:.25) -- cycle;
        \filldraw[color=orange!20] (0,1) -- (.25, 1) arc (0:-90:.25) -- cycle;
        \filldraw[color=orange!20] (1,1) -- ( .75, 1) arc (180:270:.25) -- cycle;

        \filldraw[color=blue!20] (.25,.25) -- (.25, 0) arc (0:90:.25) -- cycle;
        \filldraw[color=blue!20] (.75,.25) -- (.75, 0) arc (180:90:.25) -- cycle;
        \filldraw[color=blue!20] (.25,.75) -- (.25, 1) arc (0:-90:.25) -- cycle;
        \filldraw[color=blue!20] (.75,.75) -- ( .75, 1) arc (180:270:.25) -- cycle;
        
        
        \draw (-.25, -.25) grid (1.25, 1.25);
        
        \draw[dashed] (.25, 0) -- (.25, 1);
        \draw[dashed] (.75, 0) -- (.75, 1);
        \draw[dashed] (0, .25) -- (1, .25);
        \draw[dashed] (0, .75) -- (1, .75);
        \draw[dashed] (.25, 0) arc (0:90:.25);
        \draw[dashed] (.75, 0) arc (180:90:.25);
        \draw[dashed] ( .25, 1) arc (0:-90:.25);
        \draw[dashed] ( .75, 1) arc (180:270:.25);

        
    \end{tikzpicture}
\end{center}

Based on the colour he places his foot in (red $1$, green $2$, blue $3$ and orange $4$) we can see that the probability of him hitting $1$ is $\frac14$ and the probability of him getting $4$ is $\pi 0.25^2 = \frac{\pi}{16}$ just as you expected.

The expected value of randomly placinging his hoof is:

\begin{align*}
\E[A] &= \frac14 \cdot 1 + \frac{4}{8} \cdot 2 + \left ( \frac14 - \frac{\pi}{16}\right) \cdot 3 + \frac{\pi}{16} \cdot 4 \\
&= 2 + \frac{\pi}{16}
\end{align*}

The expected value we should get is $2.5$. That he is \textit{worse} than random means we should probably investigate further. There is probably some bias, which might be solvable (it's hard for the horse to answer $3$ for example), but it may just be we need a new horse.