An alternate approach would be to find the probability of the complementary event:
$\textbf{1)}$ The probability of getting no 6's is given by $\big(\frac{5}{6}\big)^3\cdot\big(\frac{5}{6}\big)^3=\big(\frac{5}{6}\big)^6$
$\textbf{2)}$ The probability of getting exactly one 6 is given by
$\hspace{.2 in}\big(\frac{5}{6}\big)^3\cdot3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2+3\big(\frac{1}{6}\big)\big(\frac{5}{6}\big)^2\cdot\big(\frac{5}{6}\big)^2=\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5+\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4$
Therefore the probability of getting at least two 6's is given by
$\hspace{.2 in} 1-\big(\frac{5}{6}\big)^6-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^5-\big(\frac{1}{2}\big)\big(\frac{5}{6}\big)^4=\frac{5203}{23328}\approx.223$
You are both wrong:
You are wrong because the two rolling procedures do not produce the same distributions of probability.
Your DM is wrong, because this has nothing to do with the Monty Hall (three doors) problem.
Tackling these in reverse order: in the Monty Hall problem, the host of the game knows where the goats and prize are. He cannot reveal the prize, so when he reveals a goat, you gain extra information. That is, there is a dependence in between the sequence of events which leads to an outcome (winning or losing the game).
In the dice scenario, the die rolls are independent. There is no difference between rolling three dice and rerolling the lowest, and rolling four dice and throwing away the lowest of the first three. You don't get any extra information after rolling three dice, as the fourth roll is independent of the previous three rolls.
Therefore your DM's explanation is incorrect.
Your error is, perhaps, a bit more subtle. Since the question asks whether the probability distributions are different, and not for the probability distributions themselves, let's consider a much simpler case:
Procedure 1: Flip a coin (a d2, if you will) twice—say tails is $0$ and heads is $1$. Reflip the lowest result. Equivalently, flip three coins, then throw away the lowest result among the first two tosses.
Procedure 2: Flip a coin three times, and throw away the lowest result.
In either case, you flip the coin $3$ times, leading to $2^3=8$ different events, each corresponding to a sequence of flips. For example, possible events are
$$ HHT, \qquad TTT, \qquad\text{and}\qquad HTH. $$
With respect to either procedure, these $8$ events correspond to just three outcomes: you get a sum of $0$, $1$, or $2$ (the total number of heads which "count"). However, the probabilities of these outcomes are not the same. I have chosen this example to be small and tractable, so we can actually show every possible event and outcome, as summarized below:
\begin{array}
.\text{Event} & \text{Proc 1} & \text{Proc 2} \\\hline
TTT & 0 & 0 \\
TTH & 1 & 1 \\
THT & 1 & 1 \\
THH & 2 & 2 \\
HTT & 1 & 1 \\
HTH & 2 & 2 \\
\color{red}{HHT} & \color{red}{1} & \color{red}{2} \\
HHH & 2 & 2
\end{array}
Notice that these two procedures give very similar results, but that Procedure 2 is, on average, slightly better. The difference occurs in the line which I have colored red—under Procedure 1, you get two heads, but then replace one of those heads with tails on your next toss. Under Procedure 2, you get to keep both heads, so you are better off.
The question regarding dice follows a similar pattern. Rolling three dice, rerolling the lowest, then rerolling the lowest again (or, equivalently, rolling five dice, then dropping the lowest two from the first four) is akin to Procedure 1. Rolling five dice and dropping the lowest two is akin to Procedure 2. Procedure 2 is always going to win out—heuristically, this is because you never replace a high roll with a lower roll.
Best Answer
Let $A$ and $B$ be the first two dice rolls. Let $C$ be the re-roll. The final value is $$ X\equiv\max(A,B)+C. $$ By linearity of expectation, $$ \mathbb{E}X=\mathbb{E}\max(A,B)+\mathbb{E}C. $$ Since $A$ and $B$ are IID, $$ \mathbb{P}(\max(A,B)\leq u)=\mathbb{P}(A \leq u, B\leq u)=\mathbb{P}(A\leq u)\mathbb{P}(B\leq u)=\mathbb{P}(A\leq u)^{2}. $$ Note that $\mathbb{P}(A\leq u)=u/6$ for $u=0,1,\ldots,6$. Using a well-known identity for the expectation of a nonnegative random variable, \begin{multline*} \mathbb{E}\left[\max(A,B)\right]=\sum_{u\geq0}\mathbb{P}(\max(A,B)>u)=\sum_{u\geq0}1-\mathbb{P}(\max(A,B)\leq u)\\ =\sum_{u\geq0}1-\mathbb{P}(A\leq u)^{2}=\sum_{u=0}^{6}1-\frac{u^{2}}{36}=7-\frac{1}{36}\sum_{u=1}^{6}u^{2}=\frac{161}{36}. \end{multline*} The final solution is thus $$ \mathbb{E}X=\frac{161}{36}+\frac{7}{2}=\frac{287}{36}=\boxed{7.97\overline{2}}. $$
Here is also some code to verify the answer numerically: