This is called Bertrand's box paradox. It's a "paradox" in the sense that the correct answer, P(other_coin_is_gold) = 2/3, is unintuitive to most people at first.
To get a feel for the correct answer, break it down into cases. Let's call the box with 2 gold coins "G", the box with two silver coins "S", and the box with one of each "M" (for "mixed"). Further, imagine each coin in each box being labeled "1" or "2".
Then, one of six things can happen when you pick the first coin out.
Case 1: You have box G, and you pick coin number 1; the coin is gold.
Case 2: You have box G, and you pick coin number 2; the coin is gold.
Case 3: You have box S, and you pick coin number 1; the coin is silver.
Case 4: You have box S, and you pick coin number 2; the coin is silver.
Case 5: You have box M, and you pick coin number 1; the coin is gold.
Case 6: You have box M, and you pick coin number 2; the coin is silver.
You're interested in the conditional probability of the other coin being gold given that the first coin pulled was gold. Since the first coin you picked is gold, you can eliminate cases 3, 4, and 6. So, case 1, 2, or 5 must be true. And in 2/3 of those cases, the other coin is gold.
If you wanted to approach these and other similar problems more formally, you could use Bayes' theorem:
EVENTS
* box_g: you are holding the box that has two gold coins
* first_is_g: the first coin that you drew was gold
We're interested in the conditional probability of us having box G (and hence that that other coin is gold), given that the first coin is gold ("box_g given first_is_g").
BAYES' THEOREM:
P(first_is_g | box_g) * P(box_g)
P(box_g | first_is_g) = --------------------------------
P(first_is_g)
WHAT WE KNOW
* We know that P(first_is_g), the prior probability of the first coin being gold is 1/2.
* We also know that P(box_g), the prior probability of selecting box G is 1/3.
* Finally, we know that P(first_is_g | box_g), the probability of the first coin being gold given we chose box G is 1.
Plugging in all these values yields the correct result of P(box_g | first_is_g) = 2/3
You wrote:
My initial credence (C1) before knowing that I pull a gold coin is
that all scenarios (drawing gold & gold; gold & silver; silver & gold;
silver & silver) have a 1/3 chance of occurring. Since I have pulled a
gold coin, it is certain that I did not pull out of the silver box.
Taking that into consideration, I am thinking that the probability
that the other coin in the box is also gold is 1/2. You already have
the gold coin in your hand, so that leaves you with two options:
either a gold coin or a silver coin in the box, so that gives you 1/2
chance that the other coin is gold.
You are correct that there are two possible events, you can either draw a silver coin or a gold coin. But what you did not take into account is that there are more ways for the second coin to be gold, as can be seen in our analysis of the six cases above.
