First of all, if you're interested in dice probabilities I'd recommend playing around on anydice.com - it's a great tool for calculating these sorts of probabilities. If you run output [lowest 1 of 2d10], you can play around with various probabilities associated with taking the lowest of 2d10.
For a rigorous mathematical answer, though: If you're rolling twice and taking the lower result, you want the probability of both dice being a success. Hence you want to multiply the probability of the first die being successful (70%) with the probability that the second die is successful (also 70%, since the dice are identical and each roll is independent of the others).
Thus, we have $(0.7)^2 = 0.49$, so you have a 49% chance of success.
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
Say you want to compute the probability of getting a $3$.
If the result is $3$, then it is because you got $1$ on every dice, right? There can be no other way to get a $3$ as a result. So,
$$P(\text{result is $3$}) = P(\text{all dice roll $1$}) = \frac16\cdot\frac16\cdot\frac16\cdot\frac16 = \left(\frac16\right)^4$$
(I am assuming you know how to calculate the probability of a specific outcome of a single die);
How does one find $P(18)$? If you get $18$, three die rolled $6$ right? It does not matter what did the fourth die roll, because it will never be higher.
Lets say your dice are coloured, all with different colours. One red, one blue, one black and one white. Let's say the red was the one you ignored. The chances of getting $6$ in all other dice is
$$P(\text{white, black and blue rolled $6$}) = \left(\frac16\right)^3$$
But you chose a specific coloured die to be ignored. In how many ways could you make such a choice? In $4$ ways. So you should multiply that by $4$ right?
That would give $4\cdot\left(\frac16\right)^3 \approx 1.85\%$ which is more than you got. Why? Because you counted one specific event $4$ times: when all $4$ dice roll $6$. So you should subtract $3$ of those events, so as to take that into account only once:
$$4\cdot\left(\frac16\right)^3 - 3\cdot\left(\frac16\right) \approx 1.62\%$$
For other specific outcomes you think similarly. Either by counting every single way of that outcome happening, or by this method of starting by general things and then taking into account what was added or subtracted too many times.