First, categorize the squares:
\[
\begin{aligned}
&\text{Squares with 2 neighbors (corners): } 4.\\
&\text{Squares with 3 neighbors (edge squares, not corners): } 6\times 4 = 24.\\
&\text{Squares with 4 neighbors (interior squares): } 6\times 6 = 36.
\end{aligned}
\]
Next, note the edges (pairs of adjacent squares):
\[
\begin{aligned}
&\text{Between a 2-neighbor square and a 3-neighbor square: } 4\times2 = 8\text{ edges.}\\
&\text{Between two 3-neighbor squares: } 5\times 4 = 20\text{ edges.}\\
&\text{Between a 3-neighbor square and a 4-neighbor square: } 6\times4 = 24\text{ edges.}\\
&\text{Between two 4-neighbor squares: } 5\times 6\times 2 = 60\text{ edges.}
\end{aligned}
\]
When two adjacent squares have degrees \(d_1\) and \(d_2\), the probability that the person on the first square faces the second is \(\tfrac{1}{d_1}\), and simultaneously the person on the second square faces the first with probability \(\tfrac{1}{d_2}\). Hence, for each such edge, the probability of a handshake is \(\tfrac{1}{d_1}\cdot \tfrac{1}{d_2}\).
Summing over all edges:
\[
\begin{aligned}
\mathbb{E}[\text{handshakes}]
&= 8\left(\tfrac{1}{2}\cdot \tfrac{1}{3}\right)
+ 20\left(\tfrac{1}{3}\cdot \tfrac{1}{3}\right)
+ 24\left(\tfrac{1}{3}\cdot \tfrac{1}{4}\right)
+ 60\left(\tfrac{1}{4}\cdot \tfrac{1}{4}\right)\\
&= 8\left(\tfrac{1}{6}\right)
+ 20\left(\tfrac{1}{9}\right)
+ 24\left(\tfrac{1}{12}\right)
+ 60\left(\tfrac{1}{16}\right)\\
&= \tfrac{8}{6} + \tfrac{20}{9} + 2 + \tfrac{60}{16}
\;\approx\; 1.33 + 2.22 + 2 + 3.75
\;\approx\; 9.3.
\end{aligned}
\]
Thus, the expected number of handshakes is about \(\boxed{9.3}\).
