What's the probability that the other side of the coin is gold? 4 coins are in a bucket: 1 is gold on both sides, 1 is silver on both sides, and 2 are gold on one side and silver on the other side.
I randomly grab a coin from the bucket and see that the side facing me is gold. What is the probability that the other side of the coin is gold?
I had thought that the probability is $\frac{1}{3}$ because there are 3 coins with at least one side of gold, and only 1 of these 3 coins can be gold on the other side. However, I suspect that the sides might be unique, which derails my previous logic.

If one is careful, one can find an informal but correct argument that gives the right answer. However, it is useful to know how to do a formal conditional probability calculation. One reason is that the intuition can be treacherous. 
Let $S$ be the event that the side we see is gold, and let $D$ be the event we have drawn the double-gold coin. We want $\Pr(D|S)$. By a standard formula, we have
$$\Pr(D|S)\Pr(S)=\Pr(D\cap S).\tag{$1$}$$
We first find $\Pr(S)$. The event $S$ can happen in two ways: (i) We drew the double-gold, and the side we see is gold or (ii) We drew a "mixed" coin, and the side we see is gold. 
To find the probability of (i), note that with probability $1/4$ we draw the double-gold. If this is the case,  then the side we see is gold with probability $1$. So the probability of (i) is $(1/4)(1)$.
To find the probability of (ii), note that with probability $2/4$ we draw a mixed coin. given that we do, the  probability of seeing the gold side is $1/2$. So the probability of (ii) is $(2/4)(1/2)$.
Thus $\Pr(S)=(1/4)(1)+(2/4)(1/2)=1/2$.
Note that $\Pr(D\cap S)$ is just the probability of (i), which is $(1/4)(1)$.
Now from Equation $(1)$ we find that $\Pr(D|S)=\frac{1/4}{1/2}=1/2$. 
