On the whole switch thing, what I’ve never understood is how it accounts for the third of the time I selected the right door in the first place.

That happens 1/3 of the time; in that case, you lose. But that is the only case in which you lose, so if you switch, you now have only a 1/3 chance of losing, which means you have a 2/3 (better than 50-50) chance of winning, so you should switch.

The argument is that if I choose door A out of A/B/C, and then I’m informed C is definitely wrong, I must change my selection to B, right? I get that chances went from 1/3 to 1/2, but doesn’t that apply to both A and B?

This is the incorrect understanding that most people have. Most people think "now I have picked one door out of 2, instead of one out of 3, so the probability of winning is 1/2 instead of 1/3". This would be correct if this was all the information you had, but the way in which the host chose which door to eliminate changes how the probability is distributed among the two doors.

There are three cases:

A. The prize is behind door A. The host eliminates a random door, you switch to the other one, and lose.

B. The prize is behind door B. The host eliminates door C, you switch to B and win.

C. The prize is behind door C. The host eliminates door B, you switch to C and win.

These cases are all equally likely, and if you switch, you win in 2 out of 3 of them. If you hadn't switched, you would have won in only 1 out of 3 of them (case A).

The argument is the same as in the example with 1 million doors, just that 3 is a lot smaller than 1 million, so the advantage you get by switching is a lot smaller (and less intuitively obvious).