10. As shown in Figure 2, let the area of the triangle formed by any three points chosen from the nine intersection points of the "田" shaped grid (consisting of four squares with side length 1) be $\xi$ (when the three points are collinear, $\xi=0$), then the mathematical expectation $E \xi=$ $\qquad$

Figure 2
10. $\frac{5}{6}$.

As shown in Figure 6, the number of ways to choose three points from nine points is $\mathrm{C}_{9}^{3}=84$.

When $\xi=0$, the three points are collinear, with 8 cases.

When $\xi=\frac{1}{2}$, there are 32 cases (12 cases each with $A B, B C, G H, H I$ as sides, totaling 12; 6 cases each with $D E, E F$ as sides, totaling 12; and $\triangle H E A, \triangle H E C, \triangle E B G, \triangle E B I, \triangle G D B, \triangle D A H, \triangle I F B, \triangle H C F$, totaling 32).

When $\xi=1$, there are 32 cases (3 cases each with $G H, H I, G I, A B, B C, A C$ as sides, totaling 18; 6 cases with $D F$ as a side; and $\triangle G D C, \triangle D A I, \triangle I F A, \triangle F C G, \triangle H B F, \triangle H B D, \triangle G A E, \triangle I C E$, totaling 32).

When $\xi=\frac{3}{2}$, there are 4 cases ($\triangle A H F, \triangle G B F, \triangle I D B, \triangle C D H$).

When $\xi=2$, there are 8 cases ($\triangle A G I, \triangle G I C, \triangle I C A, \triangle C A G, \triangle B G I, \triangle D I C, \triangle H A C, \triangle F A G$).
Therefore, the probability distribution of $\xi$ is as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|}
\hline$\xi$ & 0 & $\frac{1}{2}$ & 1 & $\frac{3}{2}$ & 2 \\
\hline$P$ & $\frac{8}{84}$ & $\frac{32}{84}$ & $\frac{32}{84}$ & $\frac{4}{84}$ & $\frac{8}{84}$ \\
\hline
\end{tabular}

Thus, $E \xi=0 \times \frac{8}{84}+\frac{1}{2} \times \frac{32}{84}+1 \times \frac{32}{84}+$
$$
=\frac{5}{6} . \quad \frac{3}{2} \times \frac{4}{84}+2 \times \frac{8}{84}
$$