Bertrand's Box Paradox Generalized Original Problem
There are three boxes, each with a different combination of gold and silver bars in them: One has two gold bars, one has a gold and a silver bar, and one has two silver bars. If a bar from a box is chosen at random, if that bar is a gold bar, what is the chance that the other bar in the box is gold?
Although the solution may seem to be $\frac{1}{2}$, it is actually $\frac{2}{3}$. There are three gold bars that could've been chosen:

There are two gold bars that could've been in the same box, and one silver bar, therefore the probability is $\frac{2}{3}$. (This problem is very similar to the Monty Hall problem.)

Generalization
There are $n+1$ boxes, each with $n$ bars that are either gold or silver, so that the first one has $n$ gold bars, the second has $n-1$ gold bars and $1$ silver bar, and so on. If a bar chosen at random is gold, what is the probability that another gold bar is chosen at random from the same box?

Arranging all the boxes side by side, we see an $n$ by $n+1$ array of bars, so there are $n(n+1)$ bars total. Half of them are gold, so there are $\frac{n(n+1)}{2}$ gold bars. The number of gold bars is also equal to $n+(n-1)+...+2+1$, which is also equal to $\frac{n(n+1)}{2}$. This means that the chance of each gold bar being chosen is $\frac{1}{\frac{n(n+1)}{2}}=\frac{2}{n(n+1)}$.
The probability for the gold bar chosen being in the first box is $\left(\frac{2}{n(n+1)}\right)(n)$, because there are $n$ gold bars there, and a $\frac{2}{n(n+1)}$ chance of each. If the gold bar chosen is in the first box, the probability for the other bar chosen later being gold is $\frac{n-1}{n-1}$.
the probability for the gold bar being in the second box is $\left(\frac{2}{n(n+1)}\right)(n-1)$, and the probability for the other bar chosen later being gold is $\frac{n-2}{n-1}$.
Continuing the pattern, we get the sum
$$\left(\frac{2}{n(n+1)}\right)(n)\left(\frac{n-1}{n-1}\right)+\left(\frac{2}{n(n+1)}\right)(n-1)\left(\frac{n-2}{n-1}\right)+\left(\frac{2}{n(n+1)}\right)(n-2)\left(\frac{n-3}{n-1}\right)+...+\left(\frac{2}{n(n+1)}\right)(2)\left(\frac{1}{n-1}\right)+\left(\frac{2}{n(n+1)}\right)(1)\left(\frac{0}{n-1}\right)+\left(\frac{2}{n(n+1)}\right)(0)\left(\frac{-1}{n-1}\right).$$
The last two terms in the sum can be canceled, and we can factor out the $\left(\dfrac{2}{n(n+1)}\right)$ to get
$$\left(\frac{2}{n(n+1)}\right)\left((n)\left(\frac{n-1}{n-1}\right)+(n-1)\left(\frac{n-2}{n-1}\right)+(n-2)\left(\frac{n-3}{n-1}\right)+...+(2)\left(\frac{1}{n-1}\right)\right).$$
Now we can also factor out $\frac{1}{n-1}$ to get
$$\left(\frac{2}{(n-1)n(n+1)}\right)\left((n)(n-1)+(n-1)(n-2)+(n-2)(n-3)+...+(2)(1)\right).$$
Because $(1)(2)+(2)(3)+(3)(4)+\cdots+(n)(n+1)=\dfrac{(n)(n+1)(n+2)}{3}$ (enter link description here), $(n)(n-1)+(n-1)(n-2)+(n-2)(n-3)+...+(2)(1)=\dfrac{(n-1)n(n+1)}{3}$. Therefore the answer becomes
\begin{align*}
\left(\frac{2}{(n-1)n(n+1)}\right)((n)(n-1)+(n-1)(n-2)+(n-2)(n-3)+...+(2)(1)) &= \left(\frac{2}{(n-1)n(n+1)}\right)\left(\frac{(n-1)n(n+1)}{3}\right)
&= \boxed{\frac{2}{3}}.
\end{align*}
This is the same answer as the original problem. Why is this the case?
Further Generalization
How would you solve the problem for $k$ different metals? Also, how many different boxes with $k$ different metals and $n$ bars per box are possible, without order mattering?
Note: drhab below has found the answer to another generalization; can anybody explain why the answer is so neat?
 A: I will set up a further generalization.

Let it be that there are $k$ different metals and that for every tuple $\langle n_1,\dots,n_k\rangle$ of non-negative integers with $n_1+\cdots+n_k=n$ there is exactly one box containing $n_i$ bars of metal $i$ (among them is metal gold).
Randomly a box is picked out. Then $m$ bars are picked out without replacement and all $m$ bars appear to be gold.
From the same box again a bar is picked out and the question is now: what is the probability that again the bar is of gold?
Let $G$ denote the number of gold bars that are in the selected box, and for $i=1,\dots,m+1$ and let $E_i$ denote the event that the first $i$ bars that are picked out are all gold.
Then to be found is:$$P(E_{m+1}\mid E_m)=P(E_{m+1})/P(E_m)$$
Applying stars and bars we find that total there are $\binom{n+k-1}{k-1}$ boxes and that $\binom{n+k-r-2}{k-2}$ of them contain exactly $r$ gold bars.
Then with total probability we find:
$$\begin{aligned}P\left(E_{i}\right) & =\sum_{r=i}^{n}P\left(E_{i}\mid G=r\right)P\left(G=r\right)\\
 & =\sum_{r=i}^{n}\binom{r}{i}\binom{n-r}{0}\binom{n}{i}^{-1}\binom{n+k-r-2}{k-2}\binom{n+k-1}{k-1}^{-1}\\
 & =\binom{n}{i}^{-1}\binom{n+k-1}{k-1}^{-1}\sum_{r=i}^{n}\binom{r}{i}\binom{n+k-r-2}{k-2}\\
 & =\binom{n}{i}^{-1}\binom{n+k-1}{k-1}^{-1}\binom{n+k-1}{i+k-1}\\
 & =\frac{i!\left(k-1\right)!}{\left(i+k-1\right)!}\\
 & =\binom{i+k-1}{i}^{-1}
\end{aligned}
$$
Then it follows that: $$P(E_{m+1}\mid E_m)=\frac{m+1}{m+k}$$
In your original case we had $m=1$ and $k=2$ leading to probability $\frac23$.

Again the nice equality $P(E_i)=\binom{i+k-1}{i}^{-1}$ strikes me. 
There must(!) be another, more elegant route to this result.
A: Here's an argument by induction.
We prove that for any $n,$ in a set of $n + 1$ boxes filled with $n$ bars per box, $i - 1$ silver bars in the $i$th box and the rest of the bars gold,
the number of gold bars is equal to the number of silver bars;
and that among all possible pairs of bars that can be drawn such that
both bars are drawn from the same box,
the number of pairs with two gold bars, the number with two silver bars,
and the number with one bar of each kind are all equal.
For the base case, $n=1,$ there is $1$ bar of each type (equal numbers of each metal), and the number of possible pairs of any type is zero
(hence all numbers of pairs are equal).
For the inductive case,
assume the statement is true for some integer $n.$
Take the $n+1$ boxes that demonstrate the statement, add a gold bar to each of these boxes, and add a new box with $n+1$ silver bars.
We now have $n + 2$ boxes filled with $n + 1$ bars per box, $i - 1$ silver bars in the $i$th box and the rest of the bars gold.
We added $n+1$ gold bars and $n+1$ silver bars. Since the numbers of gold and silver bars were equal before, they are still equal.
Now consider pairs of bars in which one of the bars is a "new" gold bar
(a bar that was added to a box in this step).
We can form all such pairs by choosing one of the "old" bars from the first
$n+1$ boxes and then choosing the "new" bar from the same box.
If the "old" bar is gold we get a pair of gold bars;
if the "old" bar is silver we get a mixed pair.
Since the "old" bars include equal numbers of gold and silver bars,
the pairs formed in this way include equal numbers of pairs of gold bars
and mixed pairs of bars. Let's say there are $q$ pairs of each type.
Neither kind of pair can be chosen from the new box (which has only silver bars).
The only other pairs of gold bars or mixed pairs that can be chosen are the pairs consisting only of "old" bars,
and there are the same number of each of those pairs of bars.
Let's say there are $r$ pairs of each type.
So the number of pairs of gold bars in the case $n+1$ is $q+r,$
and the number of mixed pairs of bars also is $q+r.$
By symmetry, the number of pairs of silver bars equals the number
of pairs of gold bars.
So the numbers of all three kinds of pairs are equal.
That proves the induction.
Now consider what happens when you choose a box, draw a bar from it,
and then draw another bar. You have then drawn a pair of bars at random,
and each pair of bars out of all possible pairs is equally likely to be chosen.
So you have equal chances to have drawn a pair of gold bars, a pair of silver bars, or one bar of each kind;
and whichever of these events occurred, each bar is equally likely to be the one you drew first.
This is exactly the setup of Bertrand's Box Paradox, and the answer is the same.
