Bertrand's Box puzzle

Basically, the puzzle is:
You have three boxes, each with 2 coins:

Gold/Gold
Gold/Silver
Silver/Silver

Assuming you chose a Gold Coin first, what is the probability that the second coin is also Gold?
The answer appears to be 1/2, but is actually 2/3

How?

After all of the machinations, the problem comes down to this: You have twice as much likelihood to choose Gold/Gold as you do Gold/Silver. (There are two gold vs one gold). And three boxes? Not relevant (if you look at it from the question after the first gold coin is pulled.) You have a 50/50 chance after a 3/4 chance of getting a gold coin.

Or, another way to look at it: out of the three remaining coins that are possible, two of them are Gold.

(Three coins? But only one box of two!)

But not really: the coin you grabbed must be one of three available gold coins (out of 4 possible coins to pick). There are now two available gold coins in that original set (of 3 remaining coins) = 2/3.

No, you don't go to 1/2 just because the box is chosen and there is one other coin left. Your original choice is one of 4 coins, three of which are Gold. Assuming that the first pick is gold (a 3 in 4 chance), there are now three coins that are left from what you could have chosen. Two of those are Gold. (2/3).