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Notes -
Two intuitive ways to formulate this come to mind.
First, law of conditional probability. P(A | B) * P(B) = P(A and B), or alternatively, P(A | B) = P(A and B) / P(B). P(2nd Coin Gold | 1st Coin Gold) = P(Both Coins Gold) / P(1st Coin Gold). P(1st Coin Gold) is 50% or 1/2, the probability of selecting the box with both gold coins (1/3) plus the probability of selecting the mixed coin box times one half (1/3 * 1/2 = 1/6). Probability both coins are gold is 1/3, as it can only occur if you pick the one box of three with two gold coins. 1/3 / (1/2) = 2/3.
Second, just thinking of all the possible outcomes before any coins are picked. Let's arbitrarily designate one coin in each box X, and the other Y. So boxes one to three contain coins Gold X and Gold Y, Gold X and Silver Y, and Silver X and Silver Y, respectively. There are only six possible outcomes for first and second pick, comma separated:
If you picked gold first, you know you can't be in outcomes 3, 5, or 6. Out of outcomes 1, 2, and 4, two of those have gold as the second pick. Thus, 2/3 once again.
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