But neither is a proof of the undecidability of anything satisfying. They assume an interpretation/model of programs that employ randomness (and by extension, any external calls) which completely trivializes the halting problem into a meaningless statement. I think this is my main objection with your examples.
Turing machines are deterministic, so if you want to model things like random choices and externally-obtained data, you model them (without loss of generality) as being provided as input on separate tapes. Indeed, this is exactly what is done in the theory of computation. For this reason, I have a gut reaction against random choices and external data being used to "prove undecidability." Those issues are red herrings and don't speak to undecidability in any meaningful way. From the perspective of undecidability, they should be considered as simply additional, different kinds of input to a program.
So in the first example above, you can confidently say that whenever the random tape contains such-and-such, the program is syntactic, and otherwise it is not. In the second example, you can confidently say that if the random tape contains such-and-such, then the program halts, and otherwise it does not. You can efficiently solve the parseability problem and the halting problem for classes of programs of this simple form, for the way of modeling a program's random choices that doesn't already trivialize the halting problem. So they are not satisfying demonstrations of undecidability.
On the other hand, it is impossible to decide the parseability problem (however you want to define it) for the class of programs of the following form:
BEGIN {
my $x = "... arbitrary perl code ...";
eval "sub foo {}" if eval $x;
}
foo();
And this does not rely on things like external system calls, or random choices, since the halting problem is still quite meaningful when those things are not used.