Theorem 84] \
A $(2,1,1)$ convolutional code is MDP if and only if it holds for its generator matrix $G(z):=\sum_{i=0}^{\delta}g_iz^i=g_0+g_1z$ with $g_0,g_1\in\mathbb F^2$ that $0\notin\{g_{0,1}, g_{0,2},g_{1,1},g_{1,2}\}$ and $g_{1,1}g_{0,2}-g_{1,2}g_{0,1}\neq 0$.\
Consequently, the probability that a random polynomial matrix $G(z)\in\mathbb F[z]^2$ with $\deg(G(z))=1$ generates a $(2,1,1)$ MDP convolutional code is $\frac{(1-t)^2(1-2t)}{1+t}$.

Form this theorem it follows that the number of $G(z)\in\mathbb F[z]^2$ with $\deg(G(z))=1$ that generate a $(2,1,1)$ MDP convolutional code over $\mathbb F$ is $(|\mathbb F|-1)^3\cdot (|\mathbb F|-2)$. Since two such generator matrices generate the same code if and only if they differ by a factor from $\mathbb F\setminus\{0\}$, the number of $(2,1,1)$ MDP convolutional codes over $\mathbb F$ is $(|\mathbb F|-1)^2\cdot (|\mathbb F|-2)$. In particular, there exists a $(2,1,1)$ MDP convolutional code over $\mathbb F$ if and only if $|\mathbb F|\geq 3$.

Next, we want to investigate reverse and complete MDP convolutional codes with these parameters.

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The dual of a $(2,1,1)$ convolutional code is again a $(2,1,1)$ convolutional code and it is easy to see that one could formulate the criterion for the MDP property in the same way if using the parity-check matrix: If the code has parity-check matrix $H(z)=\sum_{i=0}^{\delta}h_iz^i=h_0+h_1z$ with $h_0,h_1\in\mathbb F^{1\times 2}$, the code is MDP 