>>13153229>>13153298And to explain:
Before the condition of "the ball you take is gold", you have three random options, each with probability 1/3.
Gold-Gold box
Gold-Silver box
Silver-Silver box
Of those, the Gold-Gold box has probability 1 to pick a gold ball first
The Gold-Silver box has probability 1/2 to pick a gold ball first, and 1/2 to pick a silver ball first
The Silver-Silver box has a probability 1 to pick a silver ball first
The probabilities to pick the second ball are similar and trivial.
Thus, we have four potential outcomes (before the conditional statement):
Gold-Gold box, Gold picked first, Gold picked second: 1/3
Gold-Silver box, Gold picked first, Silver picked second: 1/6
Gold-Silver box, Silver picked first, Gold picked second: 1/6
Silver-Silver box, Silver picked first, Silver picked second: 1/3
Since our condition states that the first ball picked is Gold, we ignore the last two outcomes, and consider only the first two.
The probability of B, given A, is equal to the probability of A and B, divided by the probability of A.
(That is, the probability of B happening if A happens is equal to the probability of B and A both happening (of course), proportional to the probability of A happening at all)
A = the probability of the first ball being gold = 1/3 + 1/6 = 1/2
A and B = the probability of the first and second ball being gold = 1/3
[A and B] / A = (1/3) / (1/2) = 2/3
QED
