What is the other way of finding probability that coin is gold. The question says :
'Given three identical boxes 1,2,3 each containing two. In box1 both coins are gold, box2 both coins are silver and in box3 there is one gold and one silver. Find the probability that the other coin is also gold.
Now the answer equates the probability to be found out with probability that the coin was from box1 and uses Bayes theorem to solve it. 
But can we find the answer without equating it to the another probability and directly find the probability of obtaining the second coin as gold too?
 A: To gain an appreciation for how easily we can be led astray through flawed reasoning, consider the following "intuitive" approach:
Incorrect:  Among all boxes, there are $3$ gold coins and $3$ silver coins.  Given that a gold coin was obtained in the first draw, this means that either the first or last box was chosen.  In only one of these two cases is the other coin gold, therefore the answer is $1/2$.
Now, before we proceed with the correct line of reasoning, ask yourself why the above reasoning is flawed.
Correct:  Label the coins in each box as follows:  $$\begin{array}{ccc}
\text{Box 1}& \text{Box 2} & \text{Box 3} \\
\hline \{g_1, g_2\} & \{s_1, s_2\} & \{g_3, s_3\} \end{array}$$  Then the act of selecting a box at random and drawing one of the two coins in that box, followed by drawing the second coin in that same box, yields the possible outcomes $$(g_1, g_2), (g_2, g_1), \\
(s_1, s_2), (s_2, s_1), \\
(g_3, s_3), (s_3, g_3).$$
Unlike in our previous use of notation, notice here that the order matters.  Each of the six outcomes is equally likely, and for instance, $(g_1, g_2)$ is distinct from $(g_2, g_1)$ even though both coins are gold, because you have labeled the coins.
Then we can immediately see that $3$ of the outcomes have $g$ in the first position:  $$(g_1, g_2), (g_2, g_1), (g_3, s_3).$$  And we can see that in exactly two of the three outcomes, $g$ is also in the second position.  Therefore, the probability is $2/3$.
