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.
 A: Here's a more intuitive answer.
You have two random things going on: your choice of coin is random, and the side facing up is random. So in fact you're picking one of 8 coin-sides out of the bag, and each one occurs with equal weighting.
So let's label the sides: G1 and G2, S1 and S2, G3 and S3, and G4 and S4 (where G = gold, S = silver, and they happen to be fused together as I have written them). You choose a random coin-side. It's gold, so, it must be G1, G2, G3 or G4 - all equally probable. The reverses of those coins are, in order, G2, G1, S3 or S4 - all still equally probable, of course. So the probability is 1/2.
This simply corresponds to the fact that, if you pick a random coin-side out of the bag and it's gold, it's twice as likely to be the G1-G2 coin than the G3-S3 coin (because that coin has twice as many gold sides).
A: 50%. GIVEN that the first side you see is gold, what is the chance that you have the double-gold coin?
Assume you do this experiment a hundred times. In 50% of the cases you pull out a coin and see a gold side; the other 50% you see a silver side. In the latter case we have to discard the experiment and only count the cases where we see gold.
There is initially a 25% chance of double-gold, 25% chance double-silver, and 50% chance half-and-half. We discard the 25 cases where you draw the double-silver, and the 25 cases where you draw a half-and-half silver-side up. So of the 50 cases remaining, half are double-gold and half are gold-up silver-down.
Hence given that you have drawn a coin and see gold on top, there is a 50% there is gold on the bottom.  
A: 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 $\Pr(D|S)=\frac{1/4}{1/2}=1/2$. 
A: How many coins in the bucket are the same on both sides? 2
How many coins in the bucket are different on the two sides? 2
So probability that you selected a coin that is the same on both sides ( gold in your example) is 1/2.
A: 
I randomly grab a coin from the bucket and see that the side facing me is gold.

As any good brain-teaser, it distracts you from the fact that, for all purposes, you randomly selected a coin side that happened to be golden (4 total), not a coin with at least one goden side (3 total). There is 2/4 chance of having selected a coin with both sides golden, and 1/4 chance each for the G/S coins - meaning that there's a total 1/2 chance flipping the coin would reveal the other side is golden as well.
This is a variation of Bertrand's box paradox.
