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?

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.
