When I read this question (problem restated below), and the first comment, I was drawn to the great similarities between this problem and the Monty Hall problem (asking for the winning probability if you switch). Both include a choice with partial information about what's hidden behind the doors / in the compartments. In both cases it's very easy for lay-people to assume the probability is $1/2$, when in fact it's $2/3$.
Is there an easy translation between the two problems? What do the goats and car correspond to in the boxes-with-gold-and-silver-bars problem?
I have some ideas, stated in an answer below, but it's far from complete, and I would like to hear other ideas as well. If there is some abstract, literal translation then I would love to hear it.
I have three boxes, each with two compartments.
One has two gold bars
One has two silver bars
One has one gold bar and one silver bar
You choose a box at random, then open a compartment at random.
If that bar is gold, what is the probability that the other bar in the box is also gold?


This isn't a full translation, but it's half-way.
Say, in the metal-bars-problem, that you're after the double-gold box. Restate it like this: you pick one box, and then open a random compartment in one of the other boxes. This turns out to be gold. What is the probability that you get the double gold if you switch to that box?
This immediately looks more like the Monty Hall problem, but instead of opening a full box that isn't double-gold, we're opening half a box and seing one gold bar.
In some sense we're told that the half-opened box is most likely more valuable than the unopened box you didn't pick, which is exactly what mr. Hall tells us when he opens a door with a goat: telling you which of the two other doors is probably more valuable.