Since
\begin{align*}
p(p(1)) &= p(1) = 1, \\
p(p(2)) &= p(3) = 2, \\
p(p(3)) &= p(2) = 3,
\end{align*}three of the four solutions to $p(p(x))$ are $x = 1,$ 2, and 3.

Also, the quadratic equation $p(x) = x$ has $x = 1$ as a root.  Let $r$ be the other root.  Then
\[p(p(r)) = p(r) = r,\]so $r$ must be the fourth root we seek.

Since $p(x) - x = 0$ for $x = 1$ and $x = r,$
\[p(x) - x = c(x - 1)(x - r)\]for some constant $c.$  Setting $x = 2$ and $x = 3,$ we get
\begin{align*}
1 &= c(2 - r), \\
-1 &= 2c(3 - r).
\end{align*}Dividing these equations, we get
\[-1 = \frac{2(3 - r)}{2 - r}.\]Solving for $r,$ we find $r = \boxed{\frac{8}{3}}.$
Let $p(x)$ be the second degree polynomial such that $p(1) = 1,$ $p(2) = 3,$ and $p(3) = 2.$  Then $p(p(x)) = x$ has four real solutions.  Find the only such solution which is not an integer.
{'answer': '\\frac{8}{3}', 'subject': 'Intermediate Algebra', 'level': 4, 'unique_id': 'test/intermediate_algebra/2095.json'}